XAT Quadratic Equations Questions

First, we have to find the roots of the cubical equation.

One root has to be figured out using the trial and error method.

Let's try 1.

$$P(x) = 2x^3 - 11x^2 + 17x - 6$$

$$P(1) = 2*1^3 - 11*1^2 + 17 - 6=1$$

So, 1 is not a root.

Let's try 2,

$$P(2) = 2*2^3 - 11*2^2 + 17*2 - 6=0$$

Hence, two is one of the roots of this equation.

Now, using the synthetic division method cubical equation can be written as (x-2)($$2x^2-7x+3$$)=0

The roots of $$2x^2-7x+3$$ are 1/2, 3.

The area of circles with radii 3, 2, 1/2 are $$9\pi $$, $$4\pi$$, $$\frac{\pi}{4}$$.

Their ratio od areas is 9:4:$$\frac{1}{4}$$

$$\Rightarrow$$ 36:16:1 is the ratio.

Initially considering the number of members = a

The number of ladoos each member is required to make as per the original plan = b.

Given : a*b = 2400.

Given that 10 members were absent and each member had to make an additional 12 ladoos :

(a-10)*(b+12) = 2400.

ab - 10b+12a-120 = 2400.

Since a*b = 2400.

Hence 12a-10b = 120.

Substituting b = 2400/a.   

$$12a\ -\ 10\cdot\left(\dfrac{2400}{a}\right)\ =\ 120$$

= $$\ 12a^2-24000-\ 120a\ =\ 0$$

The roots are a = 50, a = -40.

Hence a = 50, b = 48.

The number of people who took part in making ladoos = a - 10 = 40

Considering the two numbers to a, b :

We were given that :

$$a^3+b^3\ =\ 128$$

$$\frac{1}{a^3}+\ \frac{1}{b^3}=\ 2$$

$$\frac{\left(a^3+b^3\right)}{a^3\cdot b^3}=\ 2\ =\ \frac{128}{k}$$

k = 64.

Hence $$a^3\cdot b^3\ =\ 64$$

a*b = 4 and $$a^2\cdot b^2\ =\ 16$$

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

This function is definitely greater than 0

$$let\ y=\frac{\left(x^2+1\right)}{\left(x+1\right)^2}$$

=> $$x^2\left(y-1\right)+2yx+\left(y-1\right)=0$$ which is quadratic in x

Disctiminant should be greater than 0

$$4y^2-4\left(y-1\right)^2\ge0$$

=> y>=1/2

When x =1, f(x)/g(x) = 1/3

Hence either the value should be greater than 1/2 or should be equal to 1/3

Note: For this question, discrepancy is found in question/answer. Full Marks is being awarded to all candidates.

We know that,

$$x^{3} - 1 = (x - 1)(x^{2} + x + 1)$$

Since, $$x^2 + x + 1 = 0$$

$$\therefore $$ $$x^{3} - 1$$ = 0 

=> $$x^{3}$$ = 1 

Now, x=1 is a solution but its not the only solution, as there are 2 more solutions which are not real numbers. Hence we cant simply substitute x=1 and obtain the value as 2. however we can substitute $$x^3=1$$ because cubinig the other 2 complex solutions will also give us 1. so we use $$x^3=1$$ and simply the expression . 

Now, $$x^{2018} + x^{2019}$$ 

= $$(x^{3})^{672} * x^{2}$$ + $$(x^{3})^{673}$$

= $$1^{672} * x^{2}$$ + $$1^{673}$$

= $$x^{2}$$ + 1

= -x

Hence, option C.

It has been given that A+B+C = 41.
Let the common root be B. 
All the roots are positive integers. 
The product of the roots of one of the equations is 35. 
35 can be obtained only in 2 ways - either as 5*7 or 35*1. 
A+B+C = 41.
If A and B are 5 and 7 in any order, then C = 41 - 5 - 7 = 29.
If A and B are 35 and 1 in any order, then C = 41 - 35 - 1 = 5. 
As we can see, in either case, 5 is one of the 3 roots. 
Therefore, option C is the right answer. 

Expression :  $$ax^{3} + bx^{2 }+ cx + d$$

When it intersects x-axis at x = 1, => Point = (1,0)

=> $$a(1)^3 + b(1)^2 + c(1) + d = 0$$

=> $$a + b + c + d = 0$$ --------Eqn(I)

Similarly at (-1,0)

=> $$a(-1)^3 + b(-1)^2 + c(-1) + d = 0$$

=> $$-a + b -c + d = 0$$ => $$(a + c) = (b + d)$$

Substituting it in eqn(I), we get : 

=> $$2 (b + d) = 0$$ => $$b + d = 0$$ ---------Eqn(II)

When it intersects y-axis at  = 2, => Point = (0,2)

=> $$a(0)^3 + b(0)^2 + c(0) + d = 2$$

=> $$d = 2$$

Substituting it in Eqn(II), => $$b = -2$$

$$x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$$
$$x^5-1=(x-1)f(x)$$
$$x^5 = 1+(x-1)f(x)$$
$$f(x^5) = (1+(x-1)f(x))^4+(1+(x-1)f(x))^3+(1+(x-1)f(x))^2+(1+(x-1)f(x))+1$$
On dividing by $$f(x)$$, every term will leave $$1$$ as the remainder. Therefore, the remainder when $$f(x^5)$$ is divided by $$f(x)$$ is $$1+1+1+1+1=5$$.
Therefore, option A is the right answer.