# Polynomial Questions for CAT

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## Polynomial Questions for CAT

Download important Polynomial Questions for CAT PDF based on previously asked questions in CAT exam. Practice Polynomial Questions PDF for CAT exam.

Question 1: If $f_{1}(x)=x^{2}+11x+n$ and $f_{2}(x)=x$, then the largest positive integer n for which the equation $f_{1}(x)=f_{2}(x)$ has two distinct real roots is

Question 2: If $f(x^2 – 1) = x^4 – 7x^2 + k_1$ and $f(x^3 – 2) = x^6 – 9x^3 +k_2$ then the value of $(k_2 – k_1)$ is

a) 6

b) 7

c) 8

d) 9

e) None of the above

Question 3: Let $f(x) = ax^2 – b|x|$ , where a and b are constants. Then at x = 0, f(x) is
[CAT 2004]

a) maximized whenever a > 0, b > 0

b) maximized whenever a > 0, b < 0

c) minimized whenever a > 0, b > 0

d) minimized whenever a > 0, b< 0

Question 4: If $f(x)=x^3-4x+p$ , and f(0) and f(1) are of opposite signs, then which of the following is necessarily true[CAT 2004]

a) -1 < p < 2

b) 0 < p < 3

c) -2 < p < 1

d) -3 < p < 0

Question 5: Let $f(x) = ax^2 + bx + c$, where a, b and c are certain constants and $a \neq 0$ ?It is known that $f(5) = – 3f(2)$. and that 3 is a root of $f(x) = 0$.What is the other root of f(x) = 0?[CAT 2008]

a) -7

b) – 4

c) 2

d) 6

e) cannot be determined

Question 6: Let $f(x) = ax^2 + bx + c$, where a, b and c are certain constants and $a \neq 0$ ?It is known that f(5) = – 3f(2). and that 3 is a root of f(x) = 0.What is the value of a + b + c?[CAT 2008]

a) 9

b) 14

c) 13

d) 37

e) cannot be determined

Question 7: Let $f(x) = x^{2}$ and $g(x) = 2^{x}$, for all real x. Then the value of f[f(g(x)) + g(f(x))] at x = 1 is

a) 16

b) 18

c) 36

d) 40

$f_{1}(x)=x^{2}+11x+n$ and $f_{2}(x) = x$
$f_{1}(x)=f_{2}(x)$
=> $x^{2}+11x+n = x$
=> $x^2 + 10x + n = 0$
=> For this equation to have distinct real roots, b$^2$-4ac>0
$10^2 > 4n$
=> n < 100/4
=> n < 25
Thus, largest integral value that n can take is 24.

$f(x^2 – 1) = x^4 – 7x^2 + k_1$

Put $x^2 = 1$ to make it 0

=> $f(0) = (1)^2 – 7(1) + k_1 = k_1 – 6$ ——–(i)

Also, $f(x^3 – 2) = x^6 – 9x^3 +k_2$

Put $x^3 = 2$

=> $f(0) = (2)^2 – 9(2) + k_2 = k_2 – 14$ ———–(ii)

Equating (i) & (ii), we get :

=> $k_1 – 6 = k_2 – 14$

=> $k_2 – k_1 = 14 – 6 = 8$

$f(x) = ax^2 – b|x|$. When $x=0, f(x) = 0$
When $a > 0$ and $b < 0$,
For x > 0, $f(x) = ax^2 – bx$, will be greater than 0 as $ax^2 > 0$ and $bx<0$ as $b$ is negative and $x$ is positive.
For x < 0, $f(x) = ax^2 + bx$ will again be greater than 0 as $ax^2 >0$ and $bx>0$ as both $b$ and $x$ are negative.

Therefore, the function $f(x)$ is positive when $x<0$ and when $x>0$ but becomes 0 when $x=0$.

Therefore, for $a > 0$ and $b < 0$, f(x) will attain its minimum value at $x = 0$.

f(1) = 1-4+p = p-3
f(0) = p
Since they are of opposite signs, p(p-3) < 0
=> 0 < p < 3

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 –> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 –> (2)
From (1) and (2), a – b = 0 => a = b
=> c = -12a
The equation is, therefore, $ax^2 + ax – 12a = 0 => x^2 + x – 12 = 0$
=> -4 is a root of the equation.

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 –> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 –> (2)
From (1) and (2), a – b = 0 => a = b
=> c = -12a
The equation is, therefore, $ax^2 + ax – 12a = 0 => x^2 + x – 12 = 0$
a + b + c = a + a – 12a = -10a.
But the value of a is not given. Therefore, the value cannot be determined.

$f[f(g(1)) + g(f(1))]$
= $f[f(2^1) + g(1^2)]$
= $f[f(2) + g(1)]$
= $f[2^2 + 2^1]$
= $f(6)$
= $6^2 = 36$