Polynomial Questions for CAT

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Polynomial Questions for CAT
Polynomial Questions for CAT

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.

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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

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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

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Answers & Solutions:

1) Answer: 24

$f_{1}(x)=x^{2}+11x+n$ 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.

2) Answer (C)

$f(x^2 – 1) = x^4 – 7x^2 + k_1$

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 :

=> $k_1 – 6 = k_2 – 14$

=> $k_2 – k_1 = 14 – 6 = 8$

3) Answer (D)

$f(x) = ax^2 – b|x|$. 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 $x = 0$.

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4) Answer (B)

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

5) Answer (B)

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.

6) Answer (E)

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.

7) Answer (C)

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

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