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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