SRCC GBO 2023 Question Paper

We want the function to be odd i.e, f(-x) = - f(x)
Now we know that sin(-x) = - sin (x) as sine is an odd function 
But cos (-x) = cos (x) as cosine is an even function

So if we put the input as -x in our defined function, we get, 
= $$\sum_{j=1}^k a_{j} \cos(-jx) + \sin (-x) + 2 \sin (-3x) + 4 \sin (-8x)$$
= $$\sum_{j=1}^k a_{j} \cos(jx) - \sin x - 2 \sin 3x - 4 \sin 8x$$
I we want both of these terms to be equal, we would have to remove the cosine term 
Option B does just that, by putting the coefficient of the cosine function as zero for all values of j. And the range given in option B is the same as the range of j in the given function.
This would give us the value of function to be  $$\sin (x) + 2 \sin (3x) + 4 \sin (8x)$$, which is an odd function. 
So option B is the correct answer.

The question asks us the integer values of x for which, the inequality
$$\left|\left|x+3\left|-\right|x-3\right|\right|<6$$ is satisfied. 
We can also interpret the given inequality as, 
$$-6<\left|x+3\left|-\right|x-3\right|<6$$

Now we see that there are two key points to take note of, where the modulus shows it's effects. Those are at x=-3 and x=3
We need to check the value of the middle term for the three cases of x<-3, -3<x<3 and 3<x

For x< -3, 
We see that the first and second modulus terms will have negative values inside them, and will thus reverse the sign. 
We can take a value as an example to see the result. Let's take x= -10
$$\left|-10+3\right|-\left|-10-3\right|$$
$$\left|-7\right|-\left|-13\right|$$
$$7-13\ =\ -6$$
For any negative value of x < -3, the answer will be -6. 
This does not satisfy our condition as we need the resulting values bigger than -6.

For x> 3,
We see that both the terms in the modulus will be positive and hence the modulus will not change any signs.
$$\left(x+3\right)-\left(x-3\right)$$
$$x+3-x+3\ =\ 6$$
For any value of x> 3, the answer would be 6. 
This again does not satisfy our condition and thus is not valid. 

For -3< x< 3, 
For this range, the first modulus term will have positive value inside while the second modulus modulus will have a negative value inside. 
$$\left(x+3\right)-\left(-\left(x-3\right)\right)$$
$$\left(x+3\right)-\left(-x+3\right)$$
$$\left(x+3\right)+x-3$$
$$2x$$
And since the values of x lies between -3 and 3, the values of 2x lies between -6 and 6. 
And thus this range satisfies the inequality.

The integers that lie in this range are -2, -1, 0, 1 and 2.
So a total of 5 integers satisfy this inequality.

Since we know that b is a cube, we should try to express b in terms of its prime factors to get a clear idea on how to proceed. 
$$b=8^8=\left(2^3\right)^8=2^{3\times\ 8}=2^{24}$$
Now, $$b=a^3=2^{24}$$
$$a=\sqrt[\ 3]{2^{24}}=2^{\frac{24}{3}}=2^8$$
Which is equal to 256

Therefore, Option A is the correct answer. 

Let P and Q's income be 16x and 9x respectively, and P and Q's expenditure be 8y and 5y respectively, where x and y are constants. 

P's savings = 16x - 8y
Q's savings = 9x - 5y

We are given that P's savings are 220% more or 320% of Q's savings. Using this we get, 
$$16x-8y=\frac{32}{10}\left(9y-5x\right)$$
$$160x-80y=288x-160y$$
$$80y=128x$$
$$\frac{y}{x}=\frac{8}{5}$$

we are given that R's earning are 25% less than P's earnings, using this we get R's earnings to be $$\frac{3}{4}\times\ 16x=12x$$
We are also given that R's expenses are 20% more than Q, using this we get R's expenditure to be $$\frac{6}{5}\times\ 5y=6y$$

This gives R's savings to be 12x - 6y, using the relation between y and x, we can convert this in terms of x, 
$$12x-6\left(\frac{8}{5}x\right)=12x-\frac{48}{5}x=\frac{12}{5}x$$

The ratio of P's income to R's savings would be $$\frac{16x\times\ 5}{12x}$$
=$$\frac{20}{3}$$

Therefore, Option B is the correct answer. 

Option A: 
$$4a^{2} — 3ab + b^{2} \geq 9$$
= $$4a^{2} — 4ab + b^{2} + ab \geq 9$$
= $$\left(2a\ -\ b\right)^2+ab\ \ge\ 9$$

From this we cannot necessarily say that $$ab\ \ge\ 9$$, because we don't know the value of $$\left(2a-b\right)^2$$

If we look at option D which is very similar, 
It says that we have $$ab\ \ge\ 9$$ and we have to prove $$4a^{2} — 3ab + b^{2} \geq 9$$. 
From the inequality we derived above, $$\left(2a\ -\ b\right)^2+ab\ \ge\ 9$$, 
from the option we know that$$ab\ge\ 9$$ , therefore, this term would absolutely be $$\ge\ 9$$ since the minimum value $$\left(2a-b\right)^2$$ can have is zero and even that satisfies the condition. 

Option B is simple calculations, where upon squaring the original statement, we get $$9a^{2} — 12ab + 4b^{2} > 0$$ which is opposite to what are are supposed to check. 

Option C is different, Upon taking a common from the given equation, we get:
$$a\left(9a-12b+4b^2\right)=0$$
Since it's given that a and b are non-zero, the second term must be zero.
Solving this for b we would reach an irrational equation where we cannot solve for b anymore and not reach the value of b= 2a/3
That is, we cannot reach the conclusion that b=2a/3 from the initial equation. 

Therefore, Option D is the correct answer. 

Thinking about this through a number line would make the solution a lot simpler. 

We can eliminate option B and C because we are talking about the distance between two numbers, that would be an absolute value and thus never less than zero. 
Option A comes close but miss out on the case of zero. The distance between zero and double of zero, which would be zero again is also zero. 
That is why Option D would be the correct answer. 
It covers all the cases as the distance between a number and double of that number, for negative numbers, positive numbers and zero. 

After seeing that the initial quadratic equation cannot be easily factorised, we should be looking at the sum and product of the roots, 
Sum of the roots: $$x_1+x_2=2$$
Product of roots: $$x_1x_2=\frac{5}{2}$$

Now for the new equation we need, we can again calculate the sum and product of roots. 
Sum of roots: $$x_1\ +x_2\ +\frac{1}{x1}+\frac{1}{x2}$$
                     =$$\left(x_1+x_2\right)+\left(\frac{x_1+x_2}{x_1x_2}\right)$$
                     =$$\left(x_1+x_2\right)\left(1+\frac{1}{x_1x_2}\right)$$  
Putting in the values, we get: $$2\left(1+\frac{2}{5}\right)=\frac{14}{5}$$

Product of roots: $$\left(x_1+\frac{1}{x_1}\right)\left(x_2+\frac{1}{x_2}\right)$$
                          =$$x_1x_2+\frac{x_1}{x_2}+\frac{x_2}{x_1}+\frac{1}{x_1x_2}$$
                          =$$x_1x_2+\frac{1}{x_1x_2}+\frac{\left(x_1^2+x_2^2\right)}{x_1x_2}$$
                          =$$x_1x_2+\frac{1}{x_1x_2}+\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}$$
Putting in the values, we get $$\frac{5}{2}+\frac{2}{5}+\frac{2\left(\left(2\right)^2-\left(2\times\ \frac{5}{2}\right)\right)}{5}$$
                          =$$\frac{5}{2}+\frac{2}{5}-\frac{2}{5}$$
                          =$$\frac{5}{2}$$

Checking the options for these conditions of sum and product of roots, we see that only option A satisfies all of our criterion.
Therefore, Option A is the correct answer. 

We can rewrite the series as 
$$2+0.12+1+0.0012+0.000012+\ ...$$ and so on
=$$2+1+0.12+0.0012+0.000012+0.00000012+.....$$
=$$3+\frac{12}{100}+\frac{12}{100^2}+\frac{12}{100^3}+\frac{12}{100^4}+.....$$
=$$3+12\left(\frac{1}{100}+\frac{1}{100^2}+\frac{1}{100^3}+.....\right)$$
Using the formula for the sum of an infinite series we get, 
$$3+\frac{12\left(\frac{1}{100}\right)}{\left(1-\frac{1}{100}\right)}$$
=$$3+\frac{12}{99}$$
$$\frac{309}{99}=\frac{103}{33}$$

And hence, Option A is the correct answer. 

$$\sqrt{\ 3-2\sqrt{\ 2}}$$ is the same as $$\sqrt{\ 2}-1$$
So we can rewrite the equation as, $$\sqrt{\ 10+4\sqrt{\ \left(\sqrt{\ 2}-1\right)^2}}=\sqrt{\ 10+4\left(\sqrt{\ 2}-1\right)}$$
$$\sqrt{\ 6+4\sqrt{\ 2}}=a+\sqrt{\ 2}b$$
Squaring on both side, we get
$$6+4\sqrt{\ 2}=a^2+2b^2+2\sqrt{\ 2}ab$$
Equating the rational and irrational parts on both side, we get
$$a^2+2b^2=6$$  and
$$2ab\ =\ 4$$

Trying different values, we can reach that a=2 and b=1

Putting these values in the given formula we get, 
$$\sqrt{\ a^3+b^3+4a^2b}$$
$$\sqrt{\ 8+1+16}\ $$ 
=$$\sqrt{\ 25}=5$$

Therefore, option B is the correct answer. 

In still water, the boat can cover a distance of 10 km with a speed of 5 km/hr in $$\frac{10}{5}=2$$ hours.

In two hours, the boat travels 8 km upstream. 
Therefore, the speed of boat upstream is $$\frac{8}{2}=4$$km/hr.

We know the speed of the boat is 5 km/hr and thus we can find the speed of the stream.

In upstream, effective speed = boat's speed - stream's speed
4 = 5 - stream's speed
Stream's speed = 1 km/hr

Hence, Option C is the correct answer.