Functions Questions for CAT with Solutions

Question 1:
If $f ( x ) = x ^ { 3 } - \frac { 1 } { x ^ { 3 } } ,$ then $f ( x ) + f \left( \frac { 1 } { x } \right) = ?$

[1] 0

[2] 1

[3] -1

[4] 2

Answer & Solution
Option # 1

We have, $f ( x ) = x ^ { 3 } - \frac { 1 } { x ^ { 3 } } , f \left( \frac { 1 } { x } \right) = \frac { 1 } { x ^ { 3 } } - x ^ { 3 } , f ( x ) + f \left( \frac { 1 } { x } \right) = 0$


Question 2:
a , b , c , d are roots of the equation $x ^ { 4 } - x ^ { 2 } - 1 = 0 .$ If $f ( x ) = x ^ { 6 } - x ^ { 5 } - x ^ { 4 } - x ^ { 2 } - x$ , then $f ( a ) + f ( b ) + f ( c ) + f ( d ) = ?$

[1] 0

[2] 1

[3] 6

[4] -1

Answer & Solution
Option # 4

Let $g ( x ) = x ^ { 4 } - x ^ { 3 } - x ^ { 2 } - 1 = 0$ so we have a, b, c and d as the roots of g(x) so $a ^ { 4 } - a ^ { 3 } - a ^ { 2 } - 1 = 0$ etc.

$f ( x ) = x ^ { 2 } \left\{ x ^ { 4 } - x ^ { 3 } - x ^ { 2 } - 1 - ( 1 / x ) \right\}$, so $f ( a ) = a ^ { 2 } \left( a ^ { 4 } - a ^ { 3 } - a ^ { 2 } - 1 \right) - a = 0 - a = - a$. So f(a) + f(b) + f(c) + f(d)= –(a + b + c + d) = –1.


Question 3:
Let $f \left( x + \frac { 1 } { x } \right) = x ^ { 2 } + 1 / x ^ { 2 } ( x \neq 0 ) ,$ then $f ( x ) = ?$

[1] x2

[2] x2 – 1

[3] x2 – 2

[4] None of these

Answer & Solution
Option # 3

Let $z = x + 1 / x ,$ then

$f ( z ) = f ( x + 1 / x ) = x ^ { 2 } + 1 / x ^ { 2 }$

$= ( x + 1 / x ) ^ { 2 } - 2 = z ^ { 2 } - 2 .$ Hence $f ( x ) = x ^ { 2 } - 2$


Question 4:
If f(x) is a polynomial satisfying f(x) f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) = ?

[1] 63

[2] 65

[3] 17

[4] None of these

Answer & Solution
Option # 2

Any polynomial satisfying the functional equation

f(x).f(1/x) = f(x) + f(1/x) is of the form $1 + x ^ { n }$ or $1 - x ^ { n }$

If $28 = f ( 3 ) = - 3 ^ { n } + 1$ then $3 ^ { n } = - 27$, which is not possible for any n

Hence $28 = f ( 3 ) = 3 ^ { n } + 1 \Rightarrow 3 ^ { n } = 27 \Rightarrow n = 3 .$ Thus $f ( x ) = 4 ^ { 3 } + 1 = 65$


Question 5:
For a real number x,

let f(x) = 1/(1 + x) if x is nonnegative

             = 1 + x, if x is negative

$f ^ { n } ( x ) = f \left( f ^ { n - 1 } ( x ) \right) , n = 2,3 \dots$

What is the value of the product $f ( 2 ) f ^ { 2 } ( 2 ) f ^ { 3 } ( 2 ) f ^ { 4 } ( 2 ) f ^ { 5 } ( 2 ) $?

[1] 1/3

[2] 3

[3] 1/18

[4] None of These

Answer & Solution
Option # 3

$f ( 2 ) = \frac { 1 } { 3 } , f ^ { 2 } ( 2 ) = \frac { 3 } { 4 } , f ^ { 3 } ( 2 ) = \frac { 4 } { 7 } , f ^ { 4 } ( 2 ) = \frac { 7 } { 11 } , f ^ { 5 } ( 2 ) = \frac { 11 } { 18 }$


Question 6:
The domain of the function $f ( x ) = \sqrt { ( } 2 - 2 x - x ^ { 2 } )$ is

[1] $- \sqrt { 3 } \leq x \leq \sqrt { 3 }$

[2] $- 1 - \sqrt { 3 } \leq x \leq - 1 + \sqrt { 3 }$

[3] $- 2 \leq x \leq 2$

[4] $- 2 + \sqrt { 3 } \leq x \leq - 2 - \sqrt { 3 }$

Answer & Solution
Option # 2

We must have $2 - 2 x - x ^ { 2 } \geq 0 \Rightarrow x ^ { 2 } + 2 x - 2 \leq 0$

$\Rightarrow ( x + 1 ) ^ { 2 } - 3 \leq 0$ or $- \sqrt { 3 } \leq ( x + 1 ) \leq \sqrt { 3 }$ or $- 1 - \sqrt { 3 } \leq x \leq - 1 + \sqrt { 3 }$


Question 7:
$f ( x ) = \sqrt { \left( \frac { ( x + 1 ) ( x - 3 ) } { ( x - 2 ) } \right) }$ is a real value function in the domain

[1] $( - \infty , - 1 ] \cup [ 3 , \infty )$

[2] $( - \infty , - 1 ] \cup [ 2,3 ]$

[3] $[ - 1,2 ) \cup [ 3 \infty )$

[4] None of these

Answer & Solution
Option # 3

$\sqrt { \{ g ( x ) \} }$ is real if $g ( x ) \geq 0$

so we should have

$( x + 1 ) ( x - 3 ) / ( x - 2 ) \geq 0$

Which hold in the domain

$[ - 1,2 ) \cup [ 3 , \infty )$


Question 8:
The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at

[1] x = 2.3

[2] x = 2.5

[3] x = 2.7

[4] None of these

Answer & Solution
Option # 2

For x = 2.5, we have least value.

Defining different values of x makes the question easier


Question 9:
The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained at x = ?

[1] 2

[2] 4

[3] 8

[4] 10

Answer & Solution
Option # 4

In all such questions, the minimum value will be obtained at one of the critical points. So check the value of f(x) at x = 10, 2 and 4 and see which gives the least value. Here f(10) = 2, f(2) = 6, f(4) = 8 and f(8) = 4. The minimum occurs at x = 10


Question 10:
[ x ] denotes the greatest integer less than or equal to X. If X is a positive integer and $\left[ \frac { \mathbf { X } } { 5 } \right] - \left[ \frac { \mathbf { X } } { 7 } \right] = \mathbf { 1 }$. If the minimum value of X is a and the maximum value is b, then a + b = ?

[1] 40

[2] 33

[3] 35

[4] 34

Answer & Solution
Option # 4

X = 5 (minimum) and X = 29 (maximum). Sum = 34.


Question 11:
If $0 < x < 1000 ,$ and $\left[ \frac { x } { 2 } \right] + \left[ \frac { x } { 3 } \right] + \left[ \frac { x } { 5 } \right] = \frac { 31 x } { 30 }$ where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be

[1] 30

[2] 33

[3] 43

[4] 45

Answer & Solution
Option # 2

x must be a multiple of 30. So there are 33 solutions.


Question 12:
The maximum possible value of $y = \min \left( \frac { 1 } { 2 } - \frac { 3 x ^ { 2 } } { 4 } , \frac { 5 x ^ { 2 } } { 4 } \right)$ for the range $0 < x < 1$ is

[1] 1/3

[2] 1/2

[3] 5/27

[4] 5/16

Answer & Solution
Option # 4

As x changes from 0 to 1, as long as x is less than 1/2, (1/2 – 3x2 /4) is greater than 5x2/4 and is the value of y. As x becomes greater than 1/2, 5x 2 /4 is greater than (1/2 – 3x2 /4), x become equal to $\frac { 1 } { 2 } , \frac { 5 x ^ { 2 } } { 4 }$ and $\left( \frac { 1 } { 2 } - \frac { 3 x ^ { 2 } } { 4 } \right)$ are equal to $\frac { 5 } { 16 }$


Functions Questions for CAT with Solutions
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