Quadratic Equations for IIFT PDF
Download important IIFT Quadratic Equations Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Quadratic Equations questions and answers for IIFT exam.
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Question 1: If $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$ , then what is the value of $U+3V$ ?
a) $0$
b) $\dfrac{1}{2}$
c) $\dfrac{-1}{4}$
d) $\dfrac{1}{4}$
Question 2: Consider a function $f(x) = x^4 + x^3 + x^2 + x + 1$, where x is a positive integer greater than 1. What will be the remainder if $f(x^5)$ is divided by f(x)?
a) 5
b) 6
c) 7
d) a monomial in x
e) a polynomial in x
Question 3: If $x+1=x^{2}$ and $x>0$, then $2x^{4}$ is
a) $6+4\sqrt{5}$
b) $3+3\sqrt{5}$
c) $5+3\sqrt{5}$
d) $7+3\sqrt{5}$
Question 4: If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ then the closest approximate values of a and b are
a) 25, 43
b) 52, 43
c) 52, 67
d) None of the above
Question 5: If the roots $x_1$ and $x_2$ are the roots of the quadratic equation $x^2 -2x+c=0$ also satisfy the equation $7x_2 – 4x_1 = 47$, then which of the following is true?
a) c = -15
b) $x_1 = -5$ and $x_2=3$
c) $x_1 = 4.5$ and $x_2 = -2.5$
d) None of these
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Question 6: Davji Shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per samosa up to 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paise per samosa. What should be the maximum size of the box that would maximise the revenue?
a) 240
b) 300
c) 400
d) None of these
Question 7: Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?
a) (6, 1)
b) (-3, -4)
c) (4, 3)
d) (-4, -3)
Question 8: Let p and q be the roots of the quadratic equation $x^2 – (\alpha – 2) x – \alpha -1= 0$ . What is the minimum possible value of $p^2 + q^2$?
a) 0
b) 3
c) 4
d) 5
Question 9: If the roots of the equation $x^3 – ax^2 + bx – c = 0$ are three consecutive integers, then what is the smallest possible value of b?[CAT 2008]
a) $\frac{-1}{\sqrt 3}$
b) $-1$
c) $0$
d) $1$
e) $\frac{1}{\sqrt 3}$
Question 10: A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?
a) -119
b) -159
c) -110
d) -180
e) -105
Answers & Solutions:
1) Answer (C)
Given that $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$
$\Rightarrow$ $U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$ = −$4V(U+V)$
$\Rightarrow$ $U^{2}+(U^2-4UV-2U+4V^2+4V+1)$ = −$4V(U+V)$
$\Rightarrow$ $2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$
$\Rightarrow$ $2U^2-2U+8V^2+4V+1=0$
$\Rightarrow$ $2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$
$\Rightarrow$ $2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$
Sum of two square terms is zero i.e. individual square term is equal to zero.
$U-\dfrac{1}{2}$ = 0 and $V+\dfrac{1}{4}$ = 0
U = $\dfrac{1}{2}$ and V = $-\dfrac{1}{4}$
Therefore, $U+3V$ = $\dfrac{1}{2}$+$\dfrac{-1*3}{4}$ = $\dfrac{-1}{4}$. Hence, option C is the correct answer.
2) Answer (A)
$x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$
$x^5-1=(x-1)f(x)$
$x^5 = 1+(x-1)f(x)$
$f(x^5) = (1+(x-1)f(x))^4+(1+(x-1)f(x))^3+(1+(x-1)f(x))^2+(1+(x-1)f(x))+1$
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.
3) Answer (D)
We know that $x^2 – x – 1=0$
Therefore $x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$
Therefore, $2x^4 = 6x+4$
We know that $x>0$ therefore, we can calculate the value of $x$ to be $\frac{1+\sqrt{5}}{2}$
Hence, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$
4) Answer (C)
If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$
Then x = -5 and x = 2 will give $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ = 0
Substituting x = -5 we get,
$3(-5)^{4}+2(-5)^{3}-a(-5)^{2}+b(-5)-a+b-4 = 0$
Solving we get,
$26a+4b = 1621$…….(i)
Substituting x = 2 we get,
$3(2)^{4}+2(2)^{3}-a(2)^{2}+b(2)-a+b-4 =0$
=> $5a-3b = 60$……..(ii)
Solving i and ii we get
a and b $\approx 52, 67$
Hence, option C is the correct answer.
5) Answer (A)
$x_1 + x_2 = 2$
and $7x_2 – 4x_1 = 47$
So $x_1 = -3$ and $x_2 = 5$
And $c = x_1 \times x_2 = -15$
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6) Answer (B)
Let the optimum number of samosas be 200+20n
So, price of each samosa = (2-0.1*n)
Total price of all samosas = (2-0.1*n)*(200+20n) = $400 – 20n + 40n – 2n^2$ = $400 + 20n – 2n^2$
This quadratic equation attains a maximum at n = -20/2*(-2) = 5
So, the number of samosas to get the maximum revenue = 200 + 20*5 = 300
7) Answer (A)
We know that quadratic equation can be written as $x^2$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $x^2$-(7)*x+(12)=0 where the constant term is wrong.
Keshab got the roots as (3, 2) so the equation is $x^2$-(5)*x+(6)=0 where the coefficient of x is wrong .
So the correct equation is $x^2$-(7)*x+(6)=0. The roots of above equations are (6,1).
8) Answer (D)
Let $\alpha $ be equal to k.
=> f(x) = $x^2-(k-2)x-(k+1) = 0$
p and q are the roots
=> p+q = k-2 and pq = -1-k
We know that $(p+q)^2 = p^2 + q^2 + 2pq$
=> $ (k-2)^2 = p^2 + q^2 + 2(-1-k)$
=> $p^2 + q^2 = k^2 + 4 – 4k + 2 + 2k$
=> $p^2 + q^2 = k^2 – 2k + 6$
This is in the form of a quadratic equation.
The coefficient of $k^2$ is positive. Therefore this equation has a minimum value.
We know that the minimum value occurs at x = $\frac{-b}{2a}$
Here a = 1, b = -2 and c = 6
=> Minimum value occurs at k = $\frac{2}{2}$ = 1
If we substitute k = 1 in $k^2-2k+6$, we get 1 -2 + 6 = 5.
Hence 5 is the minimum value that $p^2+q^2$ can attain.
9) Answer (B)
b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
Therefore, $b = (n-1)n + n(n+1) + (n+1)(n-1) = n^2 – n + n^2 + n + n^2 – 1$
$b = 3n^2 – 1$. The smallest value is -1.
10) Answer (B)
Let the function be $ax^2 + bx + c$.
We know that x=0 value is 1 so c=1.
So equation is $ax^2 + bx + 1$.
Now max value is 3 at x = 1.
So after substituting we get a + b = 2.
If f(x) attains a maximum at ‘a’ then the differential of f(x) at x=a, that is, f'(a)=0.
So in this question f'(1)=0
=> 2*(1)*a+b = 0
=> 2a+b = 0.
Solving the equations we get a=-2 and b=4.
$ -2x^2 + 4x + 1$ is the equation and on substituting x=10, we get -159.
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