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
We have, $f ( x ) = x ^ { 3 } - \frac { 1 } { x ^ { 3 } } , f \left( \frac { 1 } { x } \right) = \frac { 1 } { x ^ { 3 } } - x ^ { 3 } , f ( x ) + f \left( \frac { 1 } { x } \right) = 0$
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
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 $f ( x ) = x ^ { 2 } - 2$
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
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 $f ( x ) = 4 ^ { 3 } + 1 = 65$
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
$f ( 2 ) = \frac { 1 } { 3 } , f ^ { 2 } ( 2 ) = \frac { 3 } { 4 } , f ^ { 3 } ( 2 ) = \frac { 4 } { 7 } , f ^ { 4 } ( 2 ) = \frac { 7 } { 11 } , f ^ { 5 } ( 2 ) = \frac { 11 } { 18 }$
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] $- 2 + \sqrt { 3 } \leq x \leq - 2 - \sqrt { 3 }$
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 $- 1 - \sqrt { 3 } \leq x \leq - 1 + \sqrt { 3 }$
$f ( x ) = \sqrt { \left( \frac { ( x + 1 ) ( x - 3 ) } { ( x - 2 ) } \right) }$ 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
$\sqrt { \{ g ( x ) \} }$ is real if $g ( x ) \geq 0$
so we should have
$( x + 1 ) ( x - 3 ) / ( x - 2 ) \geq 0$
Which hold in the domain
$[ - 1,2 ) \cup [ 3 , \infty )$
The minimum value of f(x) = |10 – x| + |x – 2| – |4 – x| is attained at x = ?
[1] 2
[2] 4
[3] 8
[4] 10
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
[ 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
X = 5 (minimum) and X = 29 (maximum). Sum = 34.
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
x must be a multiple of 30. So there are 33 solutions.
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
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 $\frac { 5 } { 16 }$
