CAT Algebra Questions

Set A={2,3,5,7,11,13} so |A|=6
Set B={1, 8, 27} so |B|=3

Without any restrictions, each element in A can map to any of the 3 elements in B. Thus, the total number of functions is: $$3^6=729$$

Excluding Functions That Miss One Element in B: If a function does not map to an element in B, there are 2 elements in B left for mapping. The total number of such functions (for each specific element not mapped) is: $$2^6=64$$

Since there are 3 elements in B, the total number of such functions is:3x64=192

Adding Back Functions That Miss Two Elements in B: If a function misses two elements in B, there is only 1 element left for mapping. The total number of such functions is: 1^6=1. 

Since there are $$^3C_2$$ ways to choose which two elements are missed, the total number of such functions is: 3

Using the inclusion-exclusion principle, the number of functions where all elements of B are mapped by at least one element of A is:
729-192+3=540. 

Opening the square on the left-hand side, we get $$a^2+3b^2+2ab\sqrt{\ 3}$$
Comparing the rational part on both side,s we get: $$a^2+3b^2=52$$
And comparing the irrational par,t we get: $$2ab\sqrt{\ 3}=30\sqrt{\ 3}$$

$$ab=15$$, Since we are given that a and b are natural numbers, the possible values of a and b are (1,15), (3, 5), (5, 3), or (15,1)

Putting these values in the first relation we got, we see that 15 squared would exceed the required value and would not be the case. 
We need not check if a=5, b=3 or a=3, b=5 since the answer would be the same. 

(a=5 and b=3 would satisfy it)

a+b = 5+3 = 8

Therefore, Option B is the correct answer. 

We have two equations,

$$4(x^{2}+y^{2}+z^{2}) = a$$ ---(1)

$$4(x - y - z) = 3 + a$$ ---(2)

Substituting the value of a from equation 1 in equation 2, we get,

$$4\left(x\ -\ y\ -\ z\right)\ =\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)$$

$$\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)\ -4\left(x\ -\ y\ -\ z\right)\ =\ 0$$

$$\ 3\ +\ 4x^2\ +\ 4y^2\ +\ 4z^2\ -4x\ \ +\ 4y\ +\ 4z\ =\ 0$$

It can be written as,

$$4x^2\ -4x\ +\ 1+\ 4y^2\ +\ 4y\ +\ 1\ +\ 4z^2\ +\ 4z\ +\ 1\ =\ 0$$

$$\left(2x\ -\ 1\right)^2\ +\ \left(2y\ +\ 1\right)^2\ +\ \left(2z\ +\ 1\right)^2\ \ =0$$

We know that if the sum of the squares of terms is 0, then all the terms must be equal to 0

2x - 1 = 0

x = $$\dfrac{1}{2}$$

2y + 1 = 0

y = $$-\dfrac{1}{2}$$

2z + 1 = 0

z = $$-\dfrac{1}{2}$$

Substituting the values in equation 2, we get,

$$4\left(\dfrac{1}{2}\ -\ \left(-\dfrac{1}{2}\right)\ -\ \left(-\dfrac{1}{2}\right)\right)\ =\ 3\ +\ a$$

$$4\left(\dfrac{3}{2}\right)\ =\ 3\ +\ a$$

$$6\ =\ 3\ +\ a$$

$$\ a\ =\ 3$$

Therefore, the correct answer is option A.

When given more than one equations, stating the fact that there is a common root, 
We need to equate the two equations to get discernible values for $$x$$

Here, we are given three equations with the values of $$m$$, $$n$$

$$x^2+mx+9=x^2+\left(m+n\right)x+35$$
$$mx+9=mx+nx+35$$
$$nx=-26$$

Similarly, we can do it for the other equation as well, 
$$x^2+nx+17=x^2+\left(m+n\right)x+35$$
$$mx=-18$$

Substituting the value of either $$mx$$ or $$nx$$ in the original equations, we get 
$$x^2-18+9=0$$
$$x^2=9$$
$$x=\pm\ 3$$
Since we are given that the root is negative, $$x=-3$$

$$n=-\dfrac{26}{-3}$$
$$m=-\dfrac{18}{-3}$$

$$3n=26$$
$$2m=12$$

$$2m+3n=38$$

The first term of the expression can be rewritten as $$\frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c}$$
Using the property $$\frac{m}{n}\log_ab=\log_ab^{\frac{m}{n}}$$ this can be rewritten as

$$\frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c}$$

And finally using the property $$\frac{\log_ba}{\log_bc}=\log_ca$$, we can rewrite the expression as 

$$\log_c\left(a+b\right)^{\frac{1}{3}}$$

Doing identical operations in the second term, we get the entire left-hand side to be:

$$\log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}}$$
Using property $$\log_ca+\log_cb=\log_c\left(ab\right)$$ we get

$$\log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right]$$
$$\log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}}$$
$$\log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}}$$

This expression is given to be equal to 2/3
Using the definition of log: $$\log_cN=a\ $$ which is $$c^a=N$$
we get:$$c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}}$$

Cubing both sides:
$$c^2=a^2-b^2$$
Finally giving $$a^2=b^2+c^2$$

We have upper limits on b and c as 10, and we want to maximize the value of a squared. 
This can be thought of as a right-angled triangle, and the value of a will be maximum when both b and c are equal to 10, giving $$a^2=200$$, but this would not give an integer value of a
We need to adjust $$a^2$$ to the biggest square less than 200, which is 196
Giving the value of $$a$$ as 14. 

Therefore, 14 is the correct answer.  

Using the arithmetic progression formula for the nth term, where
$$x_n=a+\left(n-1\right)d$$
Substituting the value for n and using that in the equation that is given, 
$$2x_{6}+2x_{9}=x_{11}+x_{13}$$, Then,$$x_{100}$$ equals

We get, $$2\left(a+5d\right)+2\left(a+8d\right)=a+10d+a+12d$$
$$4a+26d=2a+22d$$
$$2a=-4d$$
$$a=-2d$$

We are given, $$x_5=-4$$
$$a+4d=-4$$
Substituting the value for a in terms of d, 
$$2d=-4$$
$$d=-2$$
$$a=4$$

$$x_{100}=a+99d$$
$$x_{100}=4-198=-194$$

Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question. 
f(16000) can be written as $$f\left(2^8\times\ 5^4\right)$$

Now, we can try to find these individual values:

For any prime p: f(p)=1
$$f\left(p^2\right)=f\left(p\right)f\left(p\right)+f\left(p\right)+f\left(p\right)=1+1+1=3$$
$$f\left(p^3\right)=f\left(p^2\right)f\left(p\right)+f\left(p^2\right)+f\left(p\right)=3+3+1=7$$

This way, we can find the function output for any prime number raised to a power. 

We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8
$$f\left(p^4\right)=7+7+1=15$$
$$f\left(p^5\right)=15+15+1=31$$
$$f\left(p^6\right)=31+31+1=63$$
$$f\left(p^7\right)=63+63+1=127$$
$$f\left(p^8\right)=127+127+1=255$$

Using these values in the original expression of $$f\left(2^8\times\ 5^4\right)=f\left(2^8\right)f\left(5^4\right)+f\left(2^8\right)+f\left(5^4\right)$$ we get
$$f\left(2^8\times\ 5^4\right)=\left(255\times\ 15\right)+255+15=4095$$

Therefore, Option A is the correct answer. 

From the sum and product of roots, we get: $$\alpha\ +\beta\ =-\dfrac{\lambda}{3}$$ and $$\alpha\ \beta\ =-\dfrac{1}{3}$$

Simplifying the expression given in the question, we get: $$\dfrac{\alpha^2+\beta^2\ }{\alpha^2\beta^2\ }=15$$
and substituting the denominator's value as 1/9, we get:$$\alpha^2+\beta^2\ =\dfrac{15}{9}$$

We want the expression $$\alpha^3+\beta^3\ $$, so multiplying both sides by $$\alpha+\beta$$, we get:
$$\alpha^3+\beta^3+\alpha\beta\left(a+\beta\ \right)=\dfrac{15}{9}\left(\alpha\ +\beta\ \right)$$
$$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =\dfrac{15}{9}\left(-\dfrac{\lambda}{3}\ \right)$$
$$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =-\dfrac{5\lambda}{9}-\dfrac{\lambda}{9}=-\dfrac{2\lambda\ }{3}\ \ $$

We would still need to find the value of $$\lambda$$
This we can do from the initial relation we had:

$$\alpha^2+\beta^2\ =\dfrac{15}{9}$$
$$\alpha^2+\beta^2\ =\left(\alpha+\beta\right)^2-2\alpha\ \beta\ \ \ =\dfrac{15}{9}$$
$$\dfrac{\lambda^2}{9}+\frac{2}{3}\ \ \ =\dfrac{15}{9}$$
$$\dfrac{\lambda^2}{9}\ =\dfrac{15-6}{9}=\dfrac{9}{9}=1$$
This would finally give us $$\lambda^2=9$$

Using this in our required expression, we get:

$$\left(\alpha^3+\beta^3\right)^2=\left(-\dfrac{2\lambda}{3}\ \ \right)^2=\dfrac{4\times\ 9}{9}=4$$

Therefore, Option B is the correct answer. 

Finding the terms in the sequence, we see that $$t_3=0$$, $$t_4=-\dfrac{1}{3}$$, $$t_5=0$$
We would notice that all the odd terms are 0, and we are also asked the sum of only even terms, so we do not need to consider those

$$t_6=-\dfrac{1}{5}$$
We see that the even terms are in an HP: $$-1,\ -\dfrac{1}{3},\ -\dfrac{1}{5},\ -\dfrac{1}{7},\ ...$$

The sum we are asked is the inverse of these terms, that is: -1, -3, -5, -7, up to 1012 terms

The sum of this AP would be $$\dfrac{\left[-\left(2\times\ 1\right)+\left(1012-1\right)\left(-2\right)\right]}{2}\times\ 1012$$
Which is equal to $$-1012\times\ 1012\ =\ -1024144$$

Therefore, Option A is the correct answer. 

We are told that, for any natural number $$n_{1}$$ let $$a_{n}$$ be the largest integer not exceeding $$\sqrt{n}$$

So for n=1, the largest integer not exceeding $$\sqrt{1}$$ will be 1
For n=2, the largest integer not exceeding $$\sqrt{2}$$ will be 1
For n=3, the largest integer not exceeding $$\sqrt{3}$$ will be 1
For n=4, the largest integer not exceeding $$\sqrt{4}$$ will be 2

We see a pattern here regarding the squares of the numbers, 
Listing down all the perfect squares, 
1, 4, 9, 16, 25, 36, 49, 64, ...
We see that the difference between 4 and 1 is 3 and there were three natural numbers in the given pattern with the value as 1, 
So we can write for the rest of the numbers as well, 
3 numbers will have value 1, giving a total value of 3
5 numbers will have value 2, giving a total value of 10
7 numbers will have value 3, giving a total value of 21
9 numbers will have value 4, giving a total value of 36
11 numbers will have value 5, giving a total value of 55
13 numbers will have value 6, giving a total value of 78
Now, only the values of $$a_{49},\ a_{50}$$ will have the value of 7, total value of 14. 

Adding these values, we get the total sum as 217, which is the answer. 

The expression on the right-hand side will have two critical points: 2 and -2

For any value of x greater than equal to 2, the equation changes to x+2-(x-2) = 4

the value of min{x,2} would be 2, so we would want max{x,2} to be 4+2 = 6

Therefore, x=6 works. 

For any value of x less than equal to -2, the equation changes to -(x+2)+(x-2) = -4
the value of max{x,2} would be 2, so we would want min{x,2} to be 6 again; this cannot be the case.
In general, subtracting the smaller (min) of the two values from the bigger (max) can not lead to a negative number. The max it can lead to is a 0

When x lies between -2 and 2,  the equation becomes 2x
The maximum function will give back 2, and the minimum function will give back x, with the right-hand side giving 2x
Solving this we would get 2-x = 2x which is x=2/3

Therefore, there are two real values of x for which the given equation holds. 

Hence, 2 is the correct answer.

The graph of the function on the right hand side can be visualised as:

Hence, 2 is the correct answer. 

Taking $$10^x=a$$
we get $$a+\frac{4}{a}=\frac{81}{2}$$
This would give the quadratic equation: $$2a^2-81a+8=0$$

We want to find the sum of possible values of x, let the value of x be x1 and x2

these would correspond to log a1, and log a2

The sum of log a1 + log a2 would be log (a1 x a2)

From the quadratic equation we got above, we can see that the product of the possible values of a would-be 8/2 = 4

Threfore, the sum of values of x would be log (4) which would be $$2\ \log_{10}2$$

Therefore, Option A is the correct answer. 

There are two critical points for the inequality to consider: x=-5 and x=3/2

Region I: x is greater than 3/2
In this scenario, both the terms would be positive; cross-multiplying, we get the relation $$2x-3\le x+5$$
Giving the boundary $$x\le8$$, hence giving us the valid range as $$\frac{3}{2}<x\le8$$

Region II: $$-5<x<\frac{3}{2}$$
In this case, the right-hand side will be a negative value, and hence, the sign would change when multiplying, giving the inequality
$$2x-3\ge x+5$$
Which will give x>8, which is out of bounds for this region 

Another way is to put a value in the region to check for the validity of the inequality; by putting x=0, we could see that the inequality does not hold in this region 

Region III: x less than -5
In this scenario, both the terms are negative, essentially giving us the same boundary as region 1; we take the lower bounds, giving us that x has to be less than 5

Therefore, for the given inequality to hold true x<-5 or $$\frac{3}{2}<x\le8$$

Hence, Option A is the correct answer. 

Taking a log for each of the expressions, we get the following:

$$\log_34=a,\ \log_45=b,\ \log_56=c,\ \log_67=d,\ \log_78=e,\ \log_89=f$$

The expression $$abcef$$ would then be: $$\log_34\times\ \log_45\times\ \log_56\times\ \log_67\times\ \log_78\times\ \log_89$$

Next, we can use this property of log: $$\frac{\log_ba}{\log_bc}=\log_ca$$
Using this, we get:

$$\frac{\log\ 4}{\log\ 3}\times\ \frac{\log\ 5}{\log\ 4}\times\ \frac{\log\ 6}{\log\ 5}\times\ \frac{\log\ 7}{\log\ 6}\times\ \frac{\log\ 8}{\log\ 7}\times\ \frac{\log\ 9}{\log\ 8}$$

All the terms will cancel out except: $$\frac{\log\ 9}{\log\ 3}=\log_39=2$$

Therefore, 2 is the correct answer. 

We have two inequalities,

$$y\ \ge\ x\ +\ 4$$

$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$

The second inequality can be written as two separate inequalities,

$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)$$ and $$\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$

The first inequality can be written as,

$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ +4\ \ge\ 0$$

$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -4\ge\ 0$$

$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \ge\ 4$$

The second inequality can be written as,

$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ \ \le\ \ 0$$

$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -8\ \le\ \ 0$$

$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \le\ \ 8$$

Representing all three inequalities in the graph, we get the graph as shown above, and we must calculate the intersection of all three inequalities,

We can see that the line passes through the centre of both circles (-2, 2), and the area obtained from the second inequality is the area between the two circles. So, the area of intersection of all three graphs is half of the area between both the circles as the line divides the circle in half, and we must only consider the area above the line as per the given inequality. We know that the area of the bigger circle is $$\sqrt{\ 8}$$ and the area of the smaller circle is $$\sqrt{\ 4}$$ from the equations of the circles as we know that equation of circle as $$\left(x\ -\ a\right)^2\ +\ \left(y\ -\ b\right)^2\ =\ \left(radius\right)^2$$ where (a,b) is the centre of the circle.

The area of intersection 

= $$\dfrac{1}{2}$$ (Area of bigger circle - Area of smaller circle)

= $$\dfrac{1}{2}\left(\pi\ \left(\sqrt{\ 8}\right)^2\ -\pi\ \left(\sqrt{\ 4}\right)^2\right)$$

= $$\dfrac{1}{2}\left(8\pi\ \ -4\pi\right)$$

= $$\dfrac{1}{2}\left(4\pi\ \right)$$

= $$2\pi\ $$

Therefore, the correct answer is option A.

Using the logarithmic property that $$\log_{a^p}b=\frac{1}{p}\log_ab$$

$$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$$
Can be written as 
$$4\log_{10}x+2\log_{10}x+\frac{8}{3}\log_{10}x=13$$
$$\frac{26}{3}\log_{10}x=13$$
$$\log_{10}x=1.5$$
$$x=10^{1.5}$$
$$x=\sqrt{1000}$$
$$\left[\sqrt{1000}\right]=31$$
Where [.] is Greatest Integer Function since that is what is asked in the question, 
31 is the greatest integer that does not exceed x. 

The moduli will give out only non-negative outputs, and since we are to consider only integer values of x and y, this drastically reduces the possible cases. 

We can get 2 from either 2+0 or 1+1

We get a 2+0 form when either the first term or the second term is 0
The second term is 0; this is when x=y, in this case,|2x|=2, where x can be 1 or -1; therefore, the two cases are (1,1) and (-1,-1)
The first term is 0; this is the case when x = -y, in this case, |x- (-x)|=2, giving x=1 or -1 yet again, here the two cases are (1,-1) and (-1,1)

The other way we can get 2 is through 1+1

This is possible when one of the terms is 0; if y=0, |x|+|x|=2, where x can be 1 or -1, giving two cases (1,0) and (-1,0)
Similarly,y for x=0, we get two cases, (0,1) and (0,-1)

Therefore, there are 8 pairs of (x,y) that satisfy the given equation. 

We are given, $$f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$$
Substituting $$\frac{1}{x}\ for\ x$$

$$f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{3}{x}$$
Multiplying the second equation by 2 we will have 

$$2f\left(\dfrac{1}{x}\right)+4f\left(x\right)=\dfrac{6}{x}$$

Subtracting the first equation from the second equation we have, 

$$3f\left(x\right)=\frac{6}{x}-3x$$

$$f\left(x\right)=\frac{2}{x}-x$$
We want the sum of values when this function equals 3, 

$$\frac{2}{x}-x=3$$
$$x^2+3x-2=0$$
Since the discriminant is greater than zero, both values of x will be real, and we can directly take the sum of values of $$x$$ to be, 
$$-\frac{3}{1}$$

Answer is -3. 

In such questions, we should be trying to complete the squares. 
We see a $$xy$$ term; we need to accommodate that in a square that has both x and y terms. 

Since there is only one other term with x, we also need to have it entirely in the square. 
$$\left(2x-y\right)^{^2}=4x^2+y^2-4xy$$
Using this in the given equation, we are left with $$\left(2x-y\right)^{^2}+3y^2+3-6y$$

This can be written as $$\left(2x-y\right)^{^2}+3\left(y^2+1-2y\right)$$
$$\left(2x-y\right)^{^2}+3\left(y-1\right)^2=0$$

Since both the squares add up to 0, this is only possible when the squares themselves are 0
This would give us y=1 from the second term, and using that, we get x= 1/2 from the first term. 

Therefore the value of 4x+5y will be 2+5 = 7

Hence, 7 is the correct answer. 

Arranging the equation, we know that there are no solutions when the lines are parallel:
for that the condition had to $$\frac{p}{3}=-\frac{4}{k}\ne\ \frac{2}{a}$$

Checking through options: 
Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant. 

Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel. 

Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want. 

Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0

Therefore, B is a condition that is necessary for the lines to be parallel and have no solution. 

Therefore, Option B is the correct answer. 

Squaring on both sides, we get:

$$x+6\sqrt{\ 2}+x-6\sqrt{\ 2}-2\left(x^2-72\right)^{\frac{1}{2}}=8$$
$$x-\left(x^2-72\right)^{\frac{1}{2}}=4$$

Bringing x to the other side, we get:
$$-\left(x^2-72\right)^{\frac{1}{2}}=4-x$$
Squaring on both sides again, we get:

$$x^2-72=16+x^2-8x$$
$$8x=88$$
$$x=11$$

Therefore, 11 is the correct answer. 

We can consider the quadrants of a graph:

First quadrant: Both x and y are positive
This would change the equation to 2x+y=15 and x=20, giving a negative value of y; hence, this is not the case. 

Second quadrant: x is negative, but y is positive
This would change the equations to y=15 and x=20, giving a positive value of x, which hence can not be the case.  

Third quadrant: Both x and y are negative
This would change the equation to y=15 and x-2y=20; this gives a positive value of y and hence can not be the case. 

Fourth quadrant: x is positive, but y is negative
This would change the equations to 2x+y=15 and x-2y=20; this gives the value of x as 10 and y as -5, which would lie in the fourth quadrant. 

The value of x-y would be 10-(-5)=15

Therefore, Option B is the correct answer. 

$$(a + b \sqrt{n})$$ is the positive square root of $$(29 - 12\sqrt{5})$$

So $$29-12\sqrt{5}=\left(a+b\sqrt{n}\right)^2$$
$$29-12\sqrt{5}=a^2+b^2n+2ab\sqrt{n}$$

$$a^2+b^2n=29$$ and
$$ab\sqrt{n}=-6\sqrt{5}$$
$$a^2b^2n=180$$
$$b^2n=\frac{180}{a^2}$$
Substituting this in the above equation, 
$$a^2+\frac{180}{a^2}=29$$
$$a^4-29a^2+180=0$$
$$a^2=\frac{\left(29\pm\sqrt{29^2-4\left(180\right)}\right)}{2}$$
$$a^2=\frac{\left(29+\sqrt{841-720}\right)}{2}$$
$$a^2=9\ or\ 20$$

That means, one of $$a^2\ or\ b^2n$$ is 9 and 20. 

We also have, $$ab\sqrt{n}=-6\sqrt{5}$$ that means one of a or b should be negative
And also the fact that this is a positive square root,
And we need to maximise the value of a, b and n. 

We can have a=-3, b=1 and n=20. 
This satisfies all the above equations, and the value of a+b+n=18. 

Opening the brackets, we get the series as: $$\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^4-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2+\left(\dfrac{1}{5}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6+...$$

These are two infinite GPs when rearranged:
$$\left(\dfrac{1}{5}\right)^2+\left(\dfrac{1}{5}\right)^4+\left(\dfrac{1}{5}\right)^6+...-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2-...$$

The sum of the first series would be $$\dfrac{\dfrac{1}{25}}{1-\dfrac{1}{25}}=\dfrac{1}{24}$$

The sum of the second series would be $$\frac{\dfrac{1}{35}}{1-\dfrac{1}{35}}=\dfrac{1}{34}$$

The answer to the given series would then be $$\dfrac{1}{24}-\dfrac{1}{34}=\dfrac{10}{816}=\dfrac{5}{408}$$

Therefore, Option B is the correct answer. 

It is given that $$x^8 + \left(\frac{1}{x}\right)^8 = 47$$, which can be written as:

=> $$\left(x^4\right)^{^2}+\left(\ \frac{\ 1}{x^4}\right)^{^2}=47$$

=> $$\left(x^4+\frac{\ 1}{x^4}\right)^{^2}-2\cdot x^4\cdot\frac{1}{x^4}=47$$

=> $$\left(x^4+\frac{\ 1}{x^4}\right)^{^2}=49$$

=> $$x^4+\frac{\ 1}{x^4}=7$$

Similarly, $$x^4+\frac{\ 1}{x^4}=7$$ can be expressed as:

=> $$\left(x^2\right)^{^2}+\left(\frac{\ 1}{x^2}\right)^{^2}=7$$

=> $$\left(x^2+\frac{\ 1}{x^2}\right)^{^2}-2\cdot x^2\cdot\frac{1}{x^2}=7$$

=> $$\left(x^2+\frac{\ 1}{x^2}\right)^{^2}=9$$

=> $$x^2+\frac{\ 1}{x^2}=3$$

By the same logic, we get $$x+\frac{1}{x}=\sqrt{\ 5}$$

Now, $$x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^{^3}-3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)$$

=> $$x^3+\frac{1}{x^3}=\left(\sqrt{\ 5}\right)^{^3}-3\sqrt{\ 5}=2\sqrt{\ 5}$$

By the same logic, we can say that 

=> $$x^9+\frac{1}{x^9}=\left(x^3+\frac{1}{x^3}\right)^{^3}-3\cdot x^3\cdot\frac{1}{x^3}\left(x^3+\frac{1}{x^3}\right)$$

=> $$x^9+\frac{1}{x^9}=\left(2\sqrt{\ 5}\right)^{^3}-3\left(2\sqrt{\ 5}\right)$$

=> $$x^9+\frac{1}{x^9}=40\sqrt{\ 5}-6\sqrt{\ 5}=34\sqrt{\ 5}$$

The correct option is D

It is given that $$\frac{x}{y}<\ \frac{\ x+3}{y-3}$$, which can be written as $$\frac{x}{y}-\frac{\ x+3}{y-3}<0$$

=> $$\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$

From this inequality, we can say that, when $$y<0=>y(y-3)>0$$. Now to satisfy the given equation $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$, 

$$(x+y)$$ must be greater than zero Hence, $$x>0$$ and $$\left|x\right|\ >\left|y\right|$$

Therefore, the magnitude of $$x$$ is greater than the magnitude of $$y$$. 

Hence, $$x>y$$, and $$\left|x\right|\ >\ \left|y\right|$$ =>$$-x\ <\ y$$ (Since the magnitude of $$x$$ is greater than the magnitude of $$y$$.)

The correct option is B.

Given, $$\log_{x}(x^2 + 12) = 4$$

=> $$x^2+12=x^4$$

=> $$x^4-x^2-12=0$$

=> $$x^4-4x^2+3x^2-12=0$$

=> $$x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$$

=> $$\left(x^2-4\right)\left(x^2+3\right)=0$$ => since, x is a positive real number (given) => x = 2.

Now, Given $$3 \log_{y} x = 1$$

=> $$\log_yx=\frac{1}{3}$$

=> $$x=y^{\frac{1}{3}}$$

=> $$y=x^3$$ => y = 8.

=> x + y = 2 + 8 = 10.

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between $$8^m$$ and $$8^n$$, (where m, and n are positive integers, and m < n)

Hence, $$2^{3m}\ <\ \text{41 numbers }<\ 2^{3n}$$

Since, m is a positive integer, the least value of m is 1. Therefore, $$2^{3m\ }=2^3$$, hence, the 41 numbers between them are $$2^4,\ 2^5,\ 2^6,...,2^{44}$$.

Then the lowest possible value of $$8^n$$ is $$2^{45}$$. Hence, the smallest value of n is $$2^{45}\ =\ 8^n\ =>\ 2^{3n\ }=2^{45\ }=>\ n\ =\ 15$$

Hence, the smallest value of m+n is (15+1) = 16

The correct option is D

It is given that for some real numbers a and b, the system of equations $$x + y = 4$$ and $$(a+5)x+(b^2-15)y=8b$$ has infinitely many solutions for x and y.

Hence, we can say that 

=> $$\ \frac{\ a+5}{1}=\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}$$

This equation can be used to find the value of a, and b.

Firstly, we will determine the value of b. 

=> $$\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}\ =>\ b^2-2b-15=0$$

=> $$\left(b-5\right)\left(b+3\right)=0$$

Hence, the values of b are 5, and -3, respectively. 

The value of a can be expressed in terms of b, which is $$a+5=b^2-15\ =>\ a\ =b^2-20$$

When $$b=5,a=5^2-20=5$$

When $$b=-3,a=3^2-20=-11$$

The maximum value of $$ab=(-3)\cdot(-11)=33$$

The correct option is A

Given, $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=6+3\sqrt{\ 2}$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=\sqrt{\ 36}+\sqrt{\ 18}$$

Comparing the L.H.S. and R.H.S.

=> $$5x+9=36\ $$ => $$5x=27$$ => $$x=\dfrac{27}{5}$$ (can be verified using the second term as well).

=> $$\sqrt{10x+9}$$ = $$\sqrt{\left(10\times\dfrac{27}{5}\right)+9}$$ = $$\sqrt{\ 63}=3\sqrt{\ 7}$$

It is given that $$\frac{1}{2}, \frac{\log_3(2^x - 9)}{\log_3 4}$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ are in an arithmetic progression.

$$\frac{\log_3(2^x-9)}{\log_34}$$ can be written as $$\log_4\left(2^x-9\right)$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ can be written as $$\log_4\left(2^x+\frac{17}{2}\right)$$ 

Hence, $$2\log_4\left(2^x-9\right)=\frac{1}{2}+\log_4\left(2^x+\frac{17}{2}\right)$$

$$\frac{1}{2}$$ can be written as $$\log_42$$.

Therefore, 

=> $$2\log_4\left(2^x-9\right)=\log_42+\log_4\left(2^x+\frac{17}{2}\right)$$

=> $$\log_4\left(2^x-9\right)^{^2}=\log_42\left(2^x+\frac{17}{2}\right)$$

=> $$\left(2^x-9\right)^{^2}=2\left(2^x+\frac{17}{2}\right)$$

=> $$2^{2x}-18\cdot2^x+81=2\cdot2^x+17$$

=> $$2^{2x}-20\cdot2^x+64=0$$

=> $$2^{2x}-16\cdot2^x-4\cdot2^x+64=0$$

=> $$2^x\left(2^x-16\right)-4\left(2^x-16\right)=0$$

=> $$\left(2^x-4\right)\left(2^x-16\right)=0$$

The values of $$2^x$$ can't be 4 (log will be undefined), which implies The value of $$2^x$$ is 16.

Therefore, the common difference is $$\log_4\left(2^x-9\right)-\log_42$$

=> $$\log_47-\log_42\ =\ \log_4\left(\frac{7}{2}\right)$$

The correct option is D

Given, $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$

=> $$x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$$

=> $$\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$$

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

=> x - 2y - 1 = 0 => x - 2y = 1

It is given that $$2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$$, which can be written as:

=>$$\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$$

=> $$\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$$

=> $$2^{2x^2}-2^{x+15}=0$$ (Since $$\left(a-b\right)^2\ =\ 0\ =>\ a-b\ =0$$)

=> $$2x^2\ =\ x+15$$

=> $$2x^2-x-15=0$$ 

=> $$2x^2-6x+5x-15=0$$

=> $$2x\left(x-3\right)+5\left(x-3\right)=0$$

=> $$\left(2x+5\right)\left(x-3\right)\ =\ 0$$

Hence, the possible values of x are $$-\frac{5}{2}$$, and $$3$$, respectively.

Therefore, the sum of the possible values is $$\left(3-\frac{5}{2}\right)=\frac{1}{2}$$

The correct option is D

It is given that $$5^{n-1} < 3^{n + 1}$$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5.

Now, we need to find the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$

For, n =1, the least integer value of m is 3.

For, n = 2, the least integer value of m is 3

For, n = 3, the least integer value of m is 4.

For, n = 4, the least integer value of m is 4.

For, n= 5, the least integer value of m is 5.

Hence, the least integer value of m such that for all the values of n, the equation holds is 5.

Let us consider 3 cases:

1) x = 0, This is a solution, as both L.H.S and R.H.S will be equal (0) when x = 0. (1 solution)

2) x > 0

=> $$2x\left(x^2+1\right)=5x^2$$

=> $$2\left(x^2+1\right)=5x$$

=> $$2x^2-5x+2=0$$ => $$2x^2-4x-x+2=0$$

=> $$2x\left(x-2\right)-1\left(x-2\right)=0$$

=> $$\left(x-2\right)\left(2x-1\right)=0$$ => x = 2 or 1/2 => (1 integer solution)

3) x < 0

=> $$-2x\left(x^2+1\right)=5x^2$$

=> $$2x^2+5x+2=0$$

=> $$2x^2+4x+x+2=0$$

=> $$2x\left(x+2\right)+1\left(x+2\right)=0$$

=> $$\left(x+2\right)\left(2x+1\right)=0$$ => x = -2 or -1/2 => (1 integer solution)

So, the total number of integer solutions are 0, 2, -2 => 3.

It is given that  $$\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$$, which can be written as:

=> $$2\log_3x+\log_{0.008}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{8}{1000}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{1}{125}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{5^{-3}}\left(5\right)^2\ =\ \frac{16}{3}$$

=> $$2\log_3x-\frac{2}{3}=\ \frac{16}{3}$$

=> $$2\log_3x=\frac{16}{3}+\frac{2}{3}$$

=> $$2\log_3x=6$$

=> $$\log_3x^2=6\ =>\ x^2\ =\ 3^6$$

Hence, $$\log_3\left(3\cdot x^2\right)\ =\ \log_3\left(3\cdot3^6\right)\ =\log_33^7\ =\ 7$$

Given that -2 is a root of the given cubic equation.

=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coefficient of $$x^2$$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.

=> The quadratic obtained by dividing the cubic = $$x^2+\left(2r-1\right)x+1=0$$, Since, this equation has 2 real roots => Discriminant should be greater than 0

=> $$\left(2r-1\right)^2>4$$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.

=> Minimum possible non-negative integer value of r is 2.

It is given that $$x^2 + bx + c = 0$$ has two real roots. Let the roots of the equation be $$\alpha\ ,\beta\ $$. ($$\alpha\ \ >\ \beta\ $$)

Then, we can say that $$\frac{1}{\alpha\ }-\frac{1}{\beta\ }=\frac{1}{3}$$ .... Eq(1)

Similarly, $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta^2\ }=\frac{5}{9}$$ .... Eq (2)

Eq(2) can be written as $$\left(\frac{1}{\alpha\ }-\frac{1}{\beta\ }\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\left(\frac{1}{3}\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\frac{2}{\alpha\ \cdot\beta\ }=\frac{4}{9}=>\ \frac{1}{\alpha\ \cdot\beta\ }=\frac{2}{9}$$

=> $$\alpha\ \cdot\beta\ =\frac{9}{2}$$

We know that the product of the roots is equal to c, which implies$$c=\frac{9}{2}$$

We also know that the sum of the roots is equal to -b. 

=> $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta\ ^2}=\left(\frac{1}{\alpha\ }+\frac{1}{\beta\ }\right)^{^2}-\frac{2}{\alpha\ \beta\ }=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}-\frac{4}{9}=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}=\left(1\right)^2$$

=> $$\alpha\ +\beta\ \ =\ \pm\ \alpha\ \beta\ $$

Hence, the maximum value of b is $$\frac{9}{2}$$.

Hence, the maximum value of (b+c) is 9

Given a and b are the distinct roots of the equation $$2x^{2} - 6x + k = 0$$

=> a + b = -(-6/2) = 3 (Sum of the roots)

=> ab = k/2 (Product of the roots)

Now, (a+b) and ab are the roots of the quadratic equation $$x^{2} + px + p = 0$$

=> a + b + ab = -p => 3 + k/2 = -p ---(1)

=> (a + b)(ab) = p => 3(k/2) = p ---(2)

$$3+\dfrac{k}{2}=-\dfrac{3k}{2}$$ => 2k = -3 => k = $$-\dfrac{3}{2}$$

p = $$\dfrac{3k}{2}=\dfrac{3}{2}\left(-\dfrac{3}{2}\right)=-\dfrac{9}{4}$$

=> 8(k-p) = $$8\left(-\frac{3}{2}+\frac{9}{4}\right)=-12+18=6$$

It is given that $$\left(x-1\right)^2+2kx+11=0$$ has no real roots. (Where k is the largest integer)

$$\left(x-1\right)^2+2kx+11=0$$, which can be written as:

=> $$x^2-2x+1+2kx+11=0$$

=> $$x^2+2\left(k-1\right)x+12=0$$

We know that for no real roots, D < 0 => b^2 -4ac < 0

Hence, $$\left\{2\left(k-1\right)\right\}^2-4\cdot1\cdot12\ <0$$

=> $$4\left(k-1\right)^2<48$$

=> $$\left(k-1\right)^2<12$$

Since k is an integer, it implies (k-1) is also an integer. 

Therefore, from the above inequality, we can say that the largest possible value of (k-1) = 3 => The largest possible value of k is 4.

Now we need to calculate the least possible value of $$\frac{k}{4y}+9y$$. 

$$\frac{k}{4y}+9y$$ can be written as $$\frac{4}{4y}+9y\ =\ \frac{1}{y}+9y$$

The least possible value of $$9y\ +\frac{1}{y}$$ can be calculated using A.M-G.M inequality.

Using A.M-G.M inequality, we get:

$$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9y\times\ \frac{1}{y}}$$

=> $$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9}$$

=> $$\ \frac{\ 9y+\frac{1}{y}}{2}\ge3$$

=> $$\ \ 9y+\frac{1}{y}\ge6$$

Hence, the least possible value is 6

It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.

Hence, the population in 2021 is $$100000\left(\ \frac{\ 100-y}{100}\right)$$.

The population in 2022 is $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$$

It is also given that the population in 2022 was greater than the population in 2020 and the difference between x and y is 10.

Hence, 

$$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$$, and (x-y) = 10

=> $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$$

=> $$\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$$

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $$\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$$

=> $$10000-y^2+1000-10y>\ 10000$$

=> $$y^2+10y<1000$$

=> $$y^2+10y+25<1025$$

=> $$\left(y+5\right)^2=1024\ <\ 1025$$

=> $$\left(y+5\right)^2=32^2$$

=> $$y=27$$

Hence, the population in 2021 is 100000*(100-27) = 73000

The correct option is C

Given that the average marks of 4 girls and 6 boys is 24.

Let us assume 'b' is the marks scored by a boy and 'g' is the marks scored by a girl.

=> 4g + 6b = 10*24 = 240 ---(1)

Given that, $$b\le g\le2b$$

We need to find the distinct possible values of 2g + 6b = 2g + 240 - 4g = 240 - 2g.

From (1)

when b = g => 10g = 240 => g = 24

when b = g/2 => 7g = 240 => g = 240/7

=> 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7

=> 171.42 to 192 => Integer values of 172 to 192 => 21 values.

It is given that if a certain amount of money is divided equally among n persons, each one receives Rs 352. Hence, the total amount of money is (352*n) = 352n ... Eq(1)

It is also known that if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receives less than or equal to Rs 330

Hence, the maximum amount of money with them = 506*2 + (n-2)*330 = 1012+330n-660 = 352+330n

Now, $$352+330n\ \ge\ 352n\ $$

=> $$22n\ \le\ 352$$ 

=> $$n\ \le\ 16$$

Hence, the maximum value is 16

Let us assume the initial stock of all the fruits is S.

Let us take we have 'b' and 'a' mangoes initially.

Stock of Mangoes = 40% of S = 2S/5

The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

= 2S/10 + 96 + 4a/10 = S/2 (Given)

=> S/5 + 96 + 2a/5 = S/2

=> S = $$\dfrac{\left(4a+960\right)}{3}$$

=> $$\dfrac{4a}{3}+320$$

'a' has to be a multiple of 3 for the above term to be an integer.

But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

=> The smallest value of 'a' satisfying both conditions is 15.

=> $$\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$$ = 340

From the inequality and nature of x, and y, we get the given diagram:

We need to find the area of the quadrilateral ABDE = area of rectangle ABCD + area of triangle CDE

=> Area of ABCD = (7-3)*5 = 20 units, and the area of triangle CDE = (1/2)*10*5 = 25 units.

Hence, the area of the quadrilateral ABDE = (20+25) = 45 units.

Given that $$\dfrac{1}{\sqrt{y}+\sqrt{z}}$$ is the arithmetic mean of $$\dfrac{1}{\sqrt{x}+\sqrt{z}}$$ and $$\dfrac{1}{\sqrt{x}+\sqrt{y}}$$

=> $$\dfrac{2}{\sqrt{\ y}+\sqrt{\ z}}=\dfrac{1}{\sqrt{\ x}+\sqrt{\ z}}+\dfrac{1}{\sqrt{\ x}+\sqrt{\ y}}$$

=> $$2\left(\sqrt{\ x}+\sqrt{\ z}\right)\left(\sqrt{\ x}+\sqrt{\ y}\right)=\left(\sqrt{\ y}+\sqrt{\ z}\right)\left(\sqrt{\ x}+\sqrt{\ y}+\sqrt{\ x}+\sqrt{\ z}\right)$$

=> $$2\left(x+\sqrt{\ xy}+\sqrt{\ xz}+\sqrt{\ yz}\right)=2\sqrt{xy}+y+\sqrt{\ yz}+2\sqrt{\ xz}+\sqrt{\ yz}+z$$

=> $$2x=y+z$$

=> y, x, z are in A.P. as x is the arithmetic mean of y and z.

Let the first term of both series be $$a_1$$, and $$b_1$$, respectively, and the common difference be $$d_1$$, and $$d_2$$, respectively.

It is given that $$a_5=b_9$$, which implies $$a_1+4d_1\ =b_1+8d_2$$

=> $$a_1-b_1\ =8d_2-4d_1$$  ..... Eq(1)

Similarly, it is known that a_{19}=b_{19}, which implies $$a_1+18d_1=b_1+18d_2$$

=> $$a_1-b_1=18d_2-18d_1$$  ...... Eq(2)

Equating (1) and (2), we get:

=> $$18d_2-18d_1=8d_2-4d_1$$

=> $$10d_2=14d_1$$

=> $$5d_2=7d_1$$

Since, $$d_1,\ d_2$$ are the prime numbers, which implies $$d_1=5,\ d_2=7$$.

It is also known that $$b_{2}=0$$, which implies $$b_1+d_2=0\ =>\ b_1=-d_2\ =\ -7$$

Putting the value of $$b_1,d_1,\ \text{and, }d_2$$ in Eq(1), we get:

$$a_1=8d_2-4d_1+b_1=56-20-7\ =\ 29$$

Hence, $$a_{11}=a_1+10d_1\ =\ 29+10\cdot5\ =\ 29+50\ =\ 79$$

The correct option is B

The given sequence can be written as:

$$1\left(1+\ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)+\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+...\right)+\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+...\right)+..$$

We know that the sum of an infinite G.P. is $$\dfrac{a}{1-r}$$, where a is the first term and r is the common ratio.

=> The first term = $$\frac{1}{1-\frac{1}{4}}=\dfrac{4}{3}$$

=> The second term = $$\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{9}$$

=> The third term = $$\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{108}$$

Observing these three terms, we see that they are in G.P. with a common ratio of $$\dfrac{1}{12}$$

=> Sum of this infinite G.P. = $$\dfrac{\left(\dfrac{4}{3}\right)}{1-\left(\dfrac{1}{12}\right)}=\dfrac{16}{11}$$

Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)

Now, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2-2pq=9$$ => $$\left(p-q\right)^2=9$$ => $$p-q=\pm\ 3$$

Also, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2=2pq+9=2\left(18\right)+9=45$$

Now, $$p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$$ = $$\left(p-q\right)\left(45+18\right)$$ = $$\left(p-q\right)\left(63\right)$$

=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.

=> The difference = 189 - (-189) = 378.

The first series goes as follows:

54, 62, 70, 78, 86, 94, 102,....

The second series goes as follows:

102, 106, 110,...

The first common term is 102 (first term of the common terms) and the common difference between them will be LCM(4,8) = 8

=> The required sequence is 102, 110, 118,..... (last term should be less than 468 (100th term of second series))

=> 102 + (n-1)(8) $$\le$$ 498

=> n is less than or equal to 50.5  =>  n = 50

Using the summation of A.P. formula:

Required sum = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)=\dfrac{50}{2}\left(2\times\ 102+49\times\ 8\right)=14900$$

Given on day-1, there are 2 organisms.

On day-2, there are 2*2 + 3 = 7 and on day-3, there are 2*7 + 3 = 17..

Let us try to form a pattern:

2 = 2 + 0 (n = 1)

7 = 4 + 3 (n = 2)

17 = 8 + 9 [8 + 3*3] (n = 3)

37 = 16 + 21[16 + 3*7] n = 4

T(n) = $$2^n+3\left(2^{n-1}-1\right)$$

We know that $$2^{20}=2^{10}\times\ 2^{10}=1024\times\ 1024$$, which is more than 1 million.

Let us check for n = 19

$$2^{19}+3\left(2^{18}-1\right)=2^{19}\ +\ 3\cdot2^{18}-3=2\cdot2^{19}+2^{18}-3=2^{20}+2^{18}-3$$, which is more than 1 million.

Let us check for n = 18

=> $$2^{18}+3\left(2^{17}-1\right)=2^{18}\ +\ 3\cdot2^{17}-3=2\cdot2^{18}+2^{17}-3=2^{19}+2^{17}-3$$ which is not more than a million.

=> n = 19.

It is given that $$a_{n}=13+6(n-1)$$, which can be written as $$a_n=13+6n-6\ =\ 7+6n$$

Similarly, $$b_{n}=15+7(n-1)$$, which can be written as $$b_n=15+7n-7\ =\ 8+7n$$

The common differences are 6, and 7, respectively, The common difference of terms that exists in both series is l.c.m (6, 7) = 42

The first common term of the first two series is 43 (by inspection)

Hence, we need to find the mth term, which is less than 1000, and the largest three-digit integer, and exists in both series.

$$t_m=a+\left(m-1\right)d\ <\ 1000$$

=> $$43+\left(m-1\right)42\ <\ 1000$$

=> $$\left(m-1\right)42\ <\ 957$$

=> m-1 < 22.8 => m < 23.8 => m = 23

Hence, the 23rd term is $$43+22\times\ 42\ =\ 967$$

Given that f(3x + 2y, 2x - 5y) = 19x.

Let us assume the function f(a,b) is a linear combination of a and b.

=> f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x

=> 3m + 2n = 19 and 2m - 5n = 0

Solving we get m = 5 and n = 2

=> f(a,b) = 5a+2b

=> f(x,2x) = 5x + 2(2x) = 9x = 27 => x = 3.

Let $$\frac{x}{y}\ be\ t$$

Therefore, $$c=16t\ +\ \frac{49}{t}$$

Applying AM>= GM

$$\frac{\left(16t\ +\ \frac{49}{t}\right)}{2}\ge\ \left(16t\times\frac{49}{t}\right)^{\frac{1}{2}}$$

$$16t\ +\ \frac{49}{t}\ge56$$

When t is positive then c is greater than equal to 56.

When t is negative then c is less than equal to -56.

Therefore $$c\ \in\ \left(-\infty,\ -56\right]\ ∪\ \left[56,\infty\ \right]$$

As -50 is not in the range of c so it is the answer

$$2x^{2}+kx+5=0$$ has no real roots so D<0

$$k^2-40\ <0$$

$$\left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)<0$$

$$k\in\left(-\sqrt{40},\sqrt{40}\right)$$

$$x^{2}+(k-5)x+1=0$$ has two distinct real roots so D>0

$$\left(k-5\right)^2-4>0$$

$$k^2-10k+21>0$$

$$\left(k-3\right)\left(k-7\right)>0$$

$$k\in\left(-\infty\ ,3\right)∪\left(7,\infty\ \right)$$

Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

In 9 total 9 integer values of k are possible.

$$a_1+a_2+.....+a_N=300N$$

$$6a_1+a_2+.....+a_N=400N$$

$$5a_1=100N$$

$$a_1=20N$$

As the given sequence of numbers is non-decreasing sequence, N can take values from 2 to 15.

N is not equal to 1, if N = 1, then average of N numbers is 300 wouldn't satisfy.

Therefore, N can take values from 2 to 15, i.e. 14 values.

It is given,

$$S_n=2n^2+n$$

$$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$$

$$S_{n-1}=2n^2-3n+1$$

$$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$$

$$T_n=4n-1$$

The terms are 3, 7, 11, 15, 19, 23, 27,......

27 is the first term in the series divisible by 9.

27 is the 7th term.

Therefore, the least possible value of n is 7.

The answer is option C.

$$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{125}{343}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$$

3x-y = -6

$$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$

Therefor, y=6x as the bases are different so the power should be zero for the results to be equal.

3x-y=-6

or, 3x - 6x = -6

or x= 2

y= 6x = 12

x+y = 14

Case 1: When $$x^2-3x-10=0$$ and $$x^2-10\ne\ 0$$

$$x^2-3x-10=0\ $$or, $$(x-5)(x+2) = 0$$

or, x= 5 or -2

Case 2: $$x^2-10=1$$

$$x^2-11=0$$

No integer solutions

Case 3: $$x^2-10=-1\ and\ x^2-3x-10\ is\ even$$

$$x^2-9=0$$

or, (x+3)(x-3)=0

or, x= -3 and 3

for x= -3 and +3 $$x^2-3x-10$$ is even

In total 4 values of x satisfy the equations

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

a, b, c, m and n are integers so if one root is $$3+2\sqrt{2}$$ then the other root is $$3-2\sqrt{2}$$

Sum of roots = 6 = -b/a  or b= -6a

Product of roots = 1 = c/a  or c=a

a, b, c, m and n are integers so if one root is $$4+2\sqrt{3}$$ then the other root is $$4-2\sqrt{3}$$

Sum of roots = 8 = -m/a or m = -8a

product of roots = 4 = n/a or n = 4a

$$(\frac{b}{m}+\frac{c-2b}{n})$$

= $$\frac{6a}{8a}+\frac{\left(a+12a\right)}{4a}=\frac{3}{4}+\frac{13}{4}=\frac{16}{4}=4$$

$$f(x) \geq 0$$for all real numbers $$x$$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) = $$a\left(x-2\right)^2$$

f(4)=6

or, 6 = $$a\left(x-2\right)^2$$

a= 3/2

$$f\left(-2\right)=\ -\dfrac{3}{2}\left(-4\right)^2=24$$

Let the roots of the given equation $$5x^{3} + cx^{2} - 10x + 9 = 0$$ be r, -r and p

r - r + p = $$-\frac{c}{5}$$

p = $$-\frac{c}{5}$$ ...... (1)

$$-r^2-pr+pr=-2$$

$$r^2=2$$ ...... (2)

$$-r^2p=-\frac{9}{5}$$

$$p=\frac{9}{10}$$ ...... (3)

Substituting p in (1), we get

$$\frac{9}{10}=-\frac{c}{5}$$

$$-\frac{9}{2}=c$$

The answer is option A.

$$b^2 < 4ac$$ means that the discriminant is less than 0. Therefore, f(x)>0 for all x if the coefficient of $$x^2$$ is positive, and f(x)<0 for all x if the coefficient of $$x^2$$ is negative.

We are given that f(m)<0 and m is an integer.

So the set containing values of m will either be empty if the coefficient of $$x^2$$ is positive, or it will be a set of all integers if the coefficient of $$x^2$$ is negative.

Given, 

$$f\left(x\right)+f\left(x-1\right)=1$$ ...... (1)

$$f\left(x^2-x\right)=5$$ ......  (2)

$$g\left(x\right)=x^2$$

Substituting x = 1 in (1) and (2), we get

f(0) = 5

f(1) + f(0) = 1

f(1) = 1 - 5 = -4

f(2) + f(1) = 1

f(2) = 1 + 4 = 5

f(n) = 5 if n is even and f(n) = -4 if n is odd

f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 ...... (1)

3x - y + z = 97 ...... (2)

(2)-(1) -> x - y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 - z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.

a(a + b + 1) = 14 ...... (1)

b(a + b + 1) = 28 ...... (2)

$$\frac{a}{b}=\frac{1}{2}$$

b = 2a

Substituting in (1), we get

a(3a + 1) = 14

$$3a^2+a-14=0$$

$$3a^2-6a+7a-14=0$$

$$3a\left(a-2\right)+7\left(a-2\right)=0$$

Given, a and b are natural numbers.

Therefore, a = 2 and b = 4

2a + b = 2(2) + 4 = 8

The answer is option A.

Given, the number of particles on day 1 = 100

On day 2, one out of every 2 articles produces another particle.

The number of particles on day 2 will be $$\frac{100}{2}$$, i.e. 50 particles

On day 3, one out of every 3 articles produces another particle.

The number of particles on day 3 will be $$\frac{100+50}{3}$$, i.e. 50 particles

On day 4, one out of every 4 articles produces another particle.

The number of particles on day 4 will be $$\frac{100+50+50}{4}$$, i.e. 50 particles

Series will be 100, 50, 50, 50,....

It is given,

100 + (m-1)50 = 1000

m = 19

The answer is option A.

General term = 38 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4

Each term is in the form of 17k + 4

Least 3-digit number in the form of 17k + 4 is at k = 6, i.e. 106

Highest 3-digit number in the form of 17k + 4 is at k = 58, i.e. 990

106, 123, 140,..........., 990

990 = 106 + 17(n-1)

n = 53

Sum = $$\frac{53}{2}\left(106+990\right)=53\times548$$

Average = $$53\times\frac{548}{53}=548$$

When x< r

f(x) = r

f(x) = f(f(x))

r = f(r)

r= 2r-r

r=r

When x>=r

f(x) = 2x-r

f(x) = f(f(x))

2x-r = f(2x-r)

2x-r = 2(2x-r) - r

2x-r = 4x-3r

or, x=r

Therefore x<= r

In the question, it is given that the equation $$\mid x + a \mid + \mid x - 1 \mid = 2$$ has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.

Case 1:

x + a < 0, x - 1 $$\ge\ $$ 0

- a - x + x - 1 = 2

a = -3

Case 2:

x + a $$\ge\ $$ 0 and x - 1 < 0

x + a - x + 1 = 2

a = 1

Largest value of a is 1.

The answer is option C.

Sum of n terms in an A.P = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)$$

$$A_n=\dfrac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$$

$$\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \dfrac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \dfrac{\ n\left(n+1\right)}{2}$$

Substituting n = 25, we get

$$\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\dfrac{1}{25}\left(\ \dfrac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \dfrac{\ 25\left(25+1\right)}{2}\right)$$

$$\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\ 26\left(17\right)+13$$ = 455

The answer is option A. 

Let $$\frac{x^2-6x+10}{3-x}=p$$

$$x^2-6x+10=3p-px$$

$$x^2-\left(6-p\right)x+10-3p=0$$

Since the equation will have real roots,

$$\left(6-p\right)^2-4\times\left(10-3p\right)\ge0$$

$$p^2-12p+12p+36-40\ge0$$

$$p^2\ge4$$

$$p\ge2\ ,\ p\le-2$$

Now, when $$p=-2$$, $$x = 4$$.  Since it is given that $$x<3$$, thus this value will be discarded.

Now, $$\frac{1}{2}$$ and $$-\frac{1}{2}$$ do not come in the mentioned range.

when $$p=2$$, $$x = 2$$

Thus, the minimum possible value of p will be 2.

Thus, the correct option is B.

Alternate explanation:

Since $$x<3$$,

$$3-x>0$$

Let $$3-x=y$$. So, $$y>0$$.

Now, $$\frac{x^2-6x+10}{3-x}=\frac{x^2-6x+9+1}{3-x}$$

=> $$\frac{\left(3-x\right)^2+1}{3-x}$$

Since $$3-x=y$$, the equation will transform to $$\frac{y^2+1}{y}$$ or $$y+\frac{1}{y}$$

The minimum value of the expression $$y+\frac{1}{y}$$ for $$y>0$$ will at $$y=1$$

i.e., Minimum value = $$1+1=2$$

Thus, the correct option is B.

Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.

We need to find the maximum value of z.

x + y + z = 100 ...... (1)

100x + 250y + 500z = 15250

2x + 5y + 10z = 305 ...... (2)

2x + 2y + 2z = 200 ....... (1)

(2) - (1), we get

3y + 8z = 105

At z = 12, x = 3

Therefore, maximum value z can take is 12.

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

Let $$\ \log_2n=y$$

$$\ \ \dfrac{\ 4-y}{3-\dfrac{y}{2}}<0$$

$$\ \ \left(4-y\right)\left(3-\dfrac{y}{2}\right)<0$$

$$\ \ \left(4-y\right)\left(6-y\right)<0$$

$$\ \ \left(y-4\right)\left(y-6\right)<0$$

$$4 < y < 6$$

$$4<\log_2n<6$$

$$2^4<n<2^6$$

$$16<n<64$$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47

It is given,

$$\Sigma_{n=1}^N\ \left[\dfrac{1}{5}+\dfrac{n}{25}\right]=25$$

$$\Sigma_{n=1}^N\ \left[\dfrac{5+n}{25}\right]=25$$

For n = 1 to n = 19, value of function is zero.

For n = 20 to n = 44, value of function will be 1.

44 = 20 + n - 1

n = 25 which is equal to given value.

This implies N = 44

Let the number of large shirts be l and the number of small shirts be s.

Let the price of a small shirt be x and that of a large shirt be x + 50.

Now, s + l = 64

l (x+50) = 5000

sx = 1800

Adding them, we get,

lx + sx + 50l = 6800

64x +50l = 6800

Substituting l = (6800 - 64x) / 50, in the original equation, we get

$$\frac{\left(6800-64x\right)}{50}\left(x+50\right)=5000$$

(6800 - 64x)(x + 50) = 250000

$$6800x+340000-64x^2-3200x = 250000$$

$$64x^2-3600x-90000=0$$

Solving, we get, x= $$\frac{225\pm\ 375}{8}=\frac{600}{8}or-\frac{150}{8}$$

SO, x = 75

x + 50 = 125

Answer = 75 + 125 = 200.

Alternate approach: By options.

Hint: Each option gives the sum of the costs of one large and one small shirt. We know that large = small + 50

Hence, small + small + 50 = option.

SMALL = (Option - 50)/2

LARGE = Small + 50 = (Option + 50)/2

Option A and Option D gives us decimal values for SMALL and LARGE, hence we will consider them later. 

Lets start with Option B.

Large = 150 + 50 / 2 = 100

Small = 150 - 50 / 2 = 50

Now, total shirts = 5000/100 + 1800/50 = 50 + 36 = 86 (X - This is wrong)

Option C - 

Large = 200 + 50 / 2 = 125

Small = 200 - 50 / 2 = 75

Total shirts = 5000/125 + 1800/75 = 40 + 24 = 64 ( This is the right answer)

Given x, y, z are three terms in an arithmetic progression.

Considering x = a, y = a+d, z = a+2*d.

Using the given equation x*y*z = 5*(x+y+z)

 a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)

=a*(a+d)*(a+2*d)  = 5*(3*a+3*d) = 15*(a+d).

= a*(a+2*d) = 15.

Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers.

The common difference is positive and greater than 2.

Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1)

Hence the only possible case satisfying the condition is :

a = 1, a+2*d = 15.

x = 1, z = 15.

z-x = 14.

Let us break this up into 2 inequations [ Let us assume x as m and y as n ]

| 1 + mn | < | m + n |

| m + n | < 5

Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.

Let us try out with the first quadrant and extend the results to the other quadrants.

We will also consider the +X and +Y axes along with the quadrant.

So, the first inequality becomes,

1 + mn < m + n

1 + mn - m - n < 0

1 - m + mn - n < 0

(1-m) + n(m-1) < 0

(1-m)(1-n) < 0

(m - 1)(n - 1) < 0

Let us try to plot the graph.

If we consider only mn < 0, then we get 

But, we have (m - 1)(n - 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.

So, we have,

But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.

So, if we look for only integer values, we get

(0,2), (0,3),.......

(0,-2), (0, -3),......

(2,0), (3,0), ......

(-2,0), (-3,0), .......

Now, let us consider the other inequation as well, in which |x + y| < 5

Since one of the values is always zero, the modulus of the other value is less than or equal to 4.

Hence, we get 

(0,2), (0,3), (0,4)

(0,-2), (0, -3), (0, -4)

(2,0), (3,0), (4,0)

(-2,0), (-3,0), (-4,0)

Hence, a total of 12 values.

We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .

Given $$x_{n+1}\ =\ x_n\ +\ n\ -1$$ and x1 = -1.

Considering 

x1   = -1.      (1)

x2   = x1+1-1 = x1 + 0    (2)

x3   = x2 + 2 - 1  =x2 + 1     (3)

x4   = x3 + 3 - 1 = x3 + 2        (4)

x100 = x99 + 98      (100)

Adding the LHS and RHS for the hundred equations we have:

(x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)

Subtracting this we have :

(x1+...........x100) - (x1+............. x 99) = $$\frac{\left(98\cdot99\right)}{2}$$ - 1.

x100 = 4851 - 1 = 4850

Alternatively

$$x_1=-1$$

$$x_2=x_1+1-1=x_1=-1$$

$$x_3=x_2+2-1=x_2+1=-1+1=0$$

$$x_4=x_3+3-1=x_3+2=0+2=2$$

$$x_5=x_4+4-1=x_4+3=2+3=5$$

......

If we observe the series, it is a series that has a difference between the consecutive terms in an AP.

Such series are represented as $$t\left(n\right)=a+bn+cn^2$$

We need to find t(100).

t(1) = -1

a + b + c = -1

t(2) = -1

a + 2b + 4c = -1

t(3) = 0

a + 3b + 9c = 0

Solving we get,

b + 3c = 0

b + 5c = 1

c = 0.5

b = -1.5

a = 0

Now, 

$$t\left(100\right)=\left(-1.5\right)100+\left(0.5\right)100^2=-150+5000=4850$$

Given a, b, c are rational numbers.

Hence a, b, c are three numbers that can be written in the form of p/q.

Hence if one both the root is 2+$$\sqrt{\ 3}$$ and considering the other root to be x.

The sum of the roots and the product of the two roots must be rational numbers.

For this to happen the other root must be the conjugate of $$2+\sqrt{\ 3}$$ so the sum and the product of the roots are rational numbers which are represented by: $$-\frac{b}{a},\ \frac{c}{a}$$

Since the quadratic equation has negative b, it will cancel the negative sign of the sum of the roots. Hence, the sum of the roots and the products of the roots will be represented by  $$-\frac{b}{a},\ \frac{c}{a}$$

Hence the sum of the roots is 2+$$\sqrt{\ 3}+2-\sqrt{\ 3}$$ = 4.

The product of the roots is $$\left(2+\sqrt{\ 3}\right)\cdot\left(2-\sqrt{\ 3}\right)\ =\ 1$$

b/a = 4, c/a = 1.
b = 4*a, c= a.
Since b = $$c^3$$
4*a = $$a^3$$
$$a^2=\ 4.$$
a = 2 or -2.
|a| = 2

Let the cost of an apple, an orange and a mango be a, o, and m respectively.

Then it is given that:

2a+4o+6m = a+4o+8m

or a = 2m.

Also, a+4o+8m = 8o + 7m

10m-7m = 4o

3m = 4o.

We can now express the cost of a basket in terms of mangoes only:

2a+4o+6m = 4m+3m+6m = 13m.

We have :
$$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$$
we get $$3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$$
we get $$\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$$
we get $$4+\log_4\left(x-1\right)\ =\ 3$$
$$\log_4\left(x-1\right)\ =\ -1$$
x-1 = 4^-1 
x = $$\frac{1}{4}+1=\frac{5}{4}$$
4x = 5

We have $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.

Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.

Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 40$$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference. Thus we have the condition that n>40.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 60$$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $$\ge80$$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64......79. Thus, a total of 19 numbers.

Alternatively

as per the given condition : $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )

1) For n < 20.

|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n 

considering the inequality part :$$\left|n-100\right|<\ \left|n\ -20\right|$$

100 -n < 20 -n,

No value of n satisfies this condition.

2) For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 - n and 100 - n < n - 20

For 100 -n < n - 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3) For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 ................79.

A total of 19 values

4) For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n - 100.

No value of n in the given range satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

$$f(x)=\frac{x^{2}+2x+4}{2x^{2}+4x+9}$$

If we closely observe the coefficients of the terms in the numerator and denominator, we see that the coefficients of the $$x^2$$ and x in the numerators are in ratios 1:2. This gives us a hint that we might need to adjust the numerator to decrease the number of variables.

$$f(x)=\frac{x^2+2x+4}{2x^2+4x+9}=\frac{x^2+2x+4.5-0.5}{2x^2+4x+9}$$

= $$\frac{x^2+2x+4.5}{2x^2+4x+9}-\frac{0.5}{2x^2+4x+9}$$

= $$\frac{1}{2}-\frac{0.5}{2x^2+4x+9}$$

Now, we only have terms of x in the denominator.

The maximum value of the expression is achieved when the quadratic expression $$2x^2+4x+9$$ achieves its highest value, that is infinity.

In that case, the second term becomes zero and the expression becomes 1/2. However, at infinity, there is always an open bracket ')'.

To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.

The minimum value is obtained when 4x + 4 = 0 [d/dx = 0]

x=-1.

The expression comes as 7. 

The entire expression becomes 3/7.

Hence, $$[\frac{3}{7},\frac{1}{2})$$

Now as per the given series :
we get $$x_1=1+2\ =3$$
Now $$x_1-x_2=\ 8$$
so$$x_2=-5$$
Now $$x_1-x_2+x_3\ =\ 15$$
so $$x_3\ =7$$
so we get $$x_n\ =\left(-1\right)^{n+1}\left(2n+1\right)$$
so $$x_{49}\ =\ 99$$ and $$x_{50}\ =-101$$
Therefore $$x_{49\ }+x_{50}\ =-2$$

Let the number of Covid patients in Hospitals A and B be x and x+21, respectively. Then, it has been given that:

$$\dfrac{200}{x}-\dfrac{152}{x+21}=3$$

$$\dfrac{\left(200x+4200-152x\right)}{x\left(x+21\right)}=3$$

$$\dfrac{\left(48x+4200\right)}{x\left(x+21\right)}=3$$

$$16x+1400=x\left(x+21\right)$$

$$x^2+5x-1400=0$$

(x+40)(x-35)=0

Hence, x=35.

Case 1 : $$x\ge\frac{40}{3}$$
we get 3x-20 +3x-40 =20 
6x=80
x=$$\frac{80}{6}$$=$$\frac{40}{3}$$=13.33
Case 2 :$$\frac{20}{3}\le\ x<\frac{40}{3}\ $$
we get 3x-20+40-3x =20
we get 20=20
So we get x$$\in\ \left[\frac{20}{3},\frac{40}{3}\right]$$
Case 3 $$x<\frac{20}{3}$$
we get 20-3x+40-3x =20
40=6x
x=$$\frac{20}{3}$$
but this is not possible
so we get from case 1,2 and 3
$$\frac{20}{3}\le\ x\le\frac{40}{3}$$
Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.

$$f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18}$$<0

$$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$$

We have four inflection points -5, -2, 3, and 9.

For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.

When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.

We are left with the range (-5,-2) and (3,9) where the expression will be negative.

$$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$

We can re-write the equation as: $$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$$

$$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$$

$$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$$

$$5=-5\log_{10}\sqrt{1-x}$$

$$\sqrt{1-x}=\frac{1}{10}$$

Squaring both sides: $$\left(\sqrt{1-x}\right)^2=\frac{1}{100}$$

$$\therefore\ $$ $$x=1-\frac{1}{100}=\frac{99}{100}$$

Hence, $$100\ x\ =100\times\ \frac{99}{100}=99$$

Now we have :
$$f(g(x))-3x$$
so we get f(x+3)-3x
= $$\left(x+3\right)^2-7\left(x+3\right)-3x$$
=$$x^2-4x-12$$
Now minimum value of expression = $$-\frac{D}{4a}$$ $$ \frac{\left(4ac-b^2\right)}{4a}$$
We get - (16+48)/4
= -16

$$x_0=1$$

$$x_1=2$$

$$x_2=\frac{\left(1+x_1\right)}{x_0}=\frac{\left(1+2\right)}{1}=3$$

$$x_3=\frac{\left(1+x_2\right)}{x_1}=\frac{\left(1+3\right)}{2}=2$$

$$x_4=\frac{\left(1+x_3\right)}{x_2}=\frac{\left(1+2\right)}{3}=1$$

$$x_5=\frac{\left(1+x_4\right)}{x_3}=\frac{\left(1+1\right)}{2}=1$$

$$x_6=\frac{\left(1+x_5\right)}{x_4}=\frac{\left(1+1\right)}{1}=2$$

Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5n+1 are 2... and so on, where n=0,1,2,3,....

2021 is of the form 5n+1. Hence, its value will be 2.

We have :
$$x^2-xy-x\ =22\ \ \ \ \ \ \left(1\right)$$ 
And $$y^2-xy+y\ =34\ \ \ \ \ \ \left(2\right)$$         
Adding (1) and (2)
we get $$x^2-2xy+y^2-x+y\ =56$$
we get $$\left(x-y\right)^2-\ \left(x-y\right)\ =56$$
Let (x-y) =t
we get $$t^2-t=56$$
$$t^2-t-56=0$$
(t-8)(t+7) =0
so t=8
so x-y =8

$$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$$

$$(\sqrt[7]{10})^{1+2+...+n}>999$$

$$10^{\frac{1+2+...+n}{7}}>999$$

For minimum value of n,

$$\frac{1+2+...+n}{7}=3$$

1 + 2 + ... + n = 21

We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.

The first number in each group: 1, 2, 5, 10, 17.....

Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.

Let the quadratic equation of the general term be $$ax^2+bx+c$$

1st term = a+b+c=1

2nd term = 4a+2b+c=2

3rd term = 9a+3b+c=5

Solving the equations, we get a=1, b=-2, c=2.

Hence, the first term in the 15th group will be $$\left(15\right)^2-2\left(15\right)+2=197$$

We can see that the number of terms in each group is 1, 3, 5, 7.... and so on. These are of the form 2n-1. Hence, the number of terms in 15th group will be 29. Hence, the last term in the 15th group will be 197+29-1 = 225.

Sum of terms in group 15= $$\frac{29}{2}\left(197+225\right)\ =\ 6119$$

Alternatively,

The final term in each group is the square of the group number.

In the first group 1, second group 4, ............

The final element of the 14th group is $$\left(14\right)^2=\ 196$$, similarly for the 15th group this is : $$\left(15\right)^2=\ 225$$

Each group contains all the consecutive elements in this range.

Hence the 15th group the elements are:

(197, 198, ................................225).

This is an Arithmetic Progression with a common difference of 1 and the number of element 29.

Hence the sum is given by :  $$\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$$

$$\frac{29}{2}\cdot\left(197+225\right)$$

6119.

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start with x >= 0, y >= 0,

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us test x >= 0, y < 0,

3x - y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 - 2 = 0.

The quadratic equation of the form $$\mid x^2 - 4x - 13 \mid = r$$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part : $$\left|x^2-4\cdot x-13\right|$$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function $$\left|x^2-4\cdot x-13\right|$$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC'DE. 

There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

The point C which is the value of the function at x = 2, = $$2^2-8-13$$

= -17, the reflection about the x-axis is 17.

Alternatively,

$$\mid x^2 - 4x - 13 \mid = r$$ .

This can represented in two parts :

$$x^2-4x-13\ =\ r\ if\ r\ is\ positive.$$

$$x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$$

Considering the first case : $$x^2-4x-13\ =r$$

The quadraticequation becomes : $$x^2-4x-13-r\ =\ 0$$

The discriminant for this function is : $$b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$$x^2-4x-13+r\ =\ 0$$ the function inside the modulus is negaitve

The discriminant is $$16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$$

In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

Hence $$\ 68-4r\ =\ 0$$

r = 17, for r = 17 we can have exactly 3 roots.

$$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$$

Given, $$5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$$

=> $$Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$$

=>$$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$$

since Y is a negative number, Y=$$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$$

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.'. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

$$f\left(x\right)=\ x^2+ax+b$$ 

$$f\left(x+1\right)=x^2+2x+1+ax+a+b$$

$$f\left(x-1\right)=x^2-2x+1+ax-a+b$$

$$ g(x)=f(x+1)-f(x-1)= 4x+2a$$

Now $$g(20) = 72$$ from this we get $$a = -4$$ ; $$f\left(x\right)=x^2-4x\ +b$$

For this expression to be greater than zero it has to be a perfect square which is possible for $$b\ge\ 4$$

Hence the smallest value of 'b' is 4.

To have real roots the discriminant should be greater than or equal to 0.

So, $$m^2-8n\ge0\ \&\ 4n^2-4m\ge0$$

=> $$m^2\ge8n\ \&\ n^2\ge m$$

Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.

.'. m+n=6.

The graphs of $$2^{x}+2^{-x} and 2-(x-2)^{2}$$ never intersect. So, number of solutions=0.

Alternate method:

We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

$$(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$$

if $$(x^{2}-13x+42)$$=0 or $$(x^{2}-7x+11)$$=1 or $$(x^{2}-7x+11)$$=-1 and $$(x^{2}-13x+42)$$ is even number

For x=6,7 the value $$(x^{2}-13x+42)$$=0

$$(x^{2}-7x+11)$$=1 for x=5,2.

$$(x^{2}-7x+11)$$=-1 for x=3,4 and for X=3 or 4, $$(x^{2}-13x+42)$$ is even number.

.'. {2,3,4,5,6,7} is the solution set of x.

.'. x can take six values. 

$$x_1=-1$$

$$x_1=x_2+2$$ => $$x_2=x_1-2$$ = -3

Similarly, 

$$x_3=x_1-5=-6$$

$$x_4=-10$$

.

.

The series is -1, -3, -6, -10, -15......

When the differences are in AP, then the nth term is $$-\frac{n\left(n+1\right)}{2}$$

$$x_{100}=-\frac{100\left(100+1\right)}{2}=-5050$$

$$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$$...........(1)

$$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$$.............(2)

and finally $$\log_a2=3$$

Substituting this in (1) we get $$\log_a5+\log_a3=A-3$$

Now we have two equations in two variables (1) and (2) . On solving we get

$$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$$

Let tom's age = x

=> Dick=3x

=>harry = 6x

Given, 

3x+1 = (x+3x+6x)/3 

=> x= 3

Hence, Harry's age = 18 years

The area of the region contained by the lines $$\mid x\mid-y\leq1,y\geq0$$ and $$y\leq1$$ is the white region.

Total area = Area of rectangle + 2 * Area of triangle = $$2+\left(\frac{1}{2}\times\ 2\times\ 1\right)\ =3$$

Hence, 3 is the correct answer.

$$x_0=max(x_1,x_2,....,x_{12})$$

$$x_0$$ will be minimum if x1,x2...x12 are close to each other

100/12=8.33

.'. max$$(x_1,x_2,....,x_{12})$$ will be minimum if $$(x_1,x_2,....,x_{12})$$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.'. Option B is correct.

On expanding the expression we get $$1-\log_ab+1-\log_ba$$

$$or\ 2-\left(\log_ab+\frac{1}{\log_ab}\right)$$

Now applying the property of AM>=GM, we get that  $$\frac{\left(\log_ab+\frac{1}{\log_ab}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ab}\right)\ge2$$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.

Let  $$14^a=36^b=84^c$$ = k

=> a = $$\log_{14}k$$ , b = $$\log_{36}k$$, c=$$\log_{84}k$$

$$6b(\frac{1}{c}-\frac{1}{a})$$ = $$6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$$ = 3

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that

x(2)+x(3)+x(4)+........x(10)= 47*9=423...................(1)

Similarly x(1)+x(2)+x(3)+x(4)................+x(9)= 42*9=378...............(2)

Subtracting both the equations we get x(10)-x(1)=45

Now, the sum of the 10 observations from equation (1) is 423+x(1)

Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5

Hence difference in average will be 46.5-42.5=4 which is the correct answer

$$x=2^{12\left(7+4\sqrt{\ 3}\right)}$$.

$$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$$

$$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$$

$$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$$ = $$2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$$ =$$2^6$$.

Hence C is correct answer.

Let $$a=x+\frac{1}{x}$$
So, the given equation is $$a^2-3a+2=0$$
So, $$a$$ can be either 2 or 1.

If $$a=1$$, $$x+\frac{1}{x}=1$$ and it has no real roots. 
If $$a=2$$, $$x+\frac{1}{x}=2$$ and it has exactly one real root which is $$x=1$$

So, the total number of distinct real roots of the given equation is 1

$$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$$

$$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$$

$$\left(x^2-5x+7\right)^{x+1}=1$$

There can be a solution when $$\left(x^2-5x+7\right)=1$$ or $$x^2-5x\ +6=0$$

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.

Two linear equations ax+by= c and dx+ ey = f have a unique solution if $$\frac{a}{d}\ne\ \frac{b}{e}$$

Therefore, $$\frac{k}{4}\ne\ \frac{1}{k}$$ => $$k^2\ne\ 4$$

=> k $$\ne\ $$ |2|

We have 2x + 5y = 99 or $$x=\frac{\left(99-5y\right)}{2}$$

Now $$x\ge\ y\ \ge\ -20$$ ; So $$\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$$

So $$-20\le y\le14$$. Now for this range of "y", we have to find all the integral values of "x". As the coefficient of "x" is 2,

then (99 - 5y) must be even, which will happen when "y" is odd. However, there are only 17 odd values of "y" be -20 and 14.

Hence the number of possible values is 17.