Tag: functions and graphs

Questions Related to functions and graphs

Let $f$ be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false :
$f (x) = 1, f (y) \sqrt 1, f (z) \sqrt 2$. The value of $f^{-1} (1)$ is

  1. x

  2. y

  3. z

  4. none of these


Correct Option: B
Explanation:

$f(x)=1,\quad f(y)\neq 1,\quad f(z)\neq 1$

Case 1:
$f(x)=1\ f(z)=2\ f(y)=1$
$\therefore f $ is not injective
Case 2: $f(y)\neq 1,\quad f(z)=2,\quad f(x)=1$
Case 3:
$f(z)\neq 2\quad \quad \quad f(z)=3\ f(x)\neq 1\quad \quad \quad f(x)=2\ f(y)=1\quad \quad \quad f(y)=1\ f(x)=2,f(y)=1,f(z)=3\ f^{ -1 }\left( 2 \right) =x,f^{ -1 }\left( 1 \right) =y,f^{ -1 }\left( 3 \right) =z$ 

$c \to c\,\,is\,defined\,as\,f\left( x \right) = \frac{{ax + b}}{{cx + d}}\,\,bd \ne 0$.then f is a constant function when

  1. a=c

  2. b=d

  3. ad=bc

  4. ab=cd


Correct Option: B

$f:c \to c$ is defined as $f(x) = \dfrac{{ax + b}}{{cx + d}},bd \ne 0$ then $f$ is a constant function when,

  1. a=c

  2. b=d

  3. ad=bc

  4. ab=cd


Correct Option: C
Explanation:

f($x$)=$\frac{ax+b}{cx+d}$ is a constant function,

 then lets say it equal to same constant m. 
$m(cx+d)=ax+b$ 
$a=mc $
$b=md $
$\frac{a}{c}$ =$\frac{b}{d}=m$
$\frac{a}{b}$ =$\frac{c}{d}$
 $ad=bc$
C is correct.

If $f(n+1)=f(n)$ for all $n\in N, f(7)=5$  then  $f(35)=$

  1. $25$

  2. $49$

  3. $35$

  4. $5$


Correct Option: D
Explanation:

$f(n+1)=f(n)$
$\Rightarrow f\left ( n \right )=f\left ( n-1 \right )=f\left ( n-2 \right )=...........=f\left ( 1 \right )$
$\therefore f\left ( 7 \right )=5\Rightarrow f\left ( 35 \right )=5$

Let $f(x)$ is a cubic polynomial with real coefficients, $x\ \in R$ such that $f"(3)=0,\ f'(5)=0$  
If $f(3)=1$ and $f(5)=-3$, then $f(1)$ is equal to

  1. $2$

  2. $3$

  3. $5$

  4. $6$


Correct Option: A

The complete set of values of $x$ for which the function $f(x)=2\tan^{-1}x+\sin^{-1} \dfrac{2x}{1+x^{2}}$ behaves like a constant function with positive output is equal to

  1. $x \in [-1,1]$

  2. $[1,\infty)$

  3. $(-\infty,1]$

  4. $(-\infty, -1] \cup [1,\infty)$


Correct Option: A

Let f be a polynomial function such that $f(3x)=f'(x).f"(x)$, for all $x\epsilon R$. Then :

  1. $f(2)+f'(2)=28$

  2. $f"(2)-f'(2)=0$

  3. $f"(2)-f(2)=4$

  4. $f(2)-f'(2)+f"(2)=10$


Correct Option: A

If  $f \left( \dfrac { x + y } { 2 } \right) = \dfrac { f ( x ) + f ( y ) } { 2 }$  for all  $x , y \in R$  and  $f ^ { \prime } ( o ) = - 1 , f ( o ) = 1$  then  $f(2)=$

  1. $\dfrac { 1 } { 2 }$

  2. $1$

  3. $-1$

  4. $\dfrac { -1 } { 2 }$


Correct Option: A
Explanation:

let $f(x)=ax+b$

$f(0)=1\implies b=1$
$f'(0)=-1 \implies a=-1$
$\implies f(x)=1-x$
$\implies f(2)=-1$

let $f(x)$ be a polynomial of degree $4$ having extreme values at $x=2$.if $\underset { x\rightarrow 0 }{ lim } \left( \frac { f\left( x \right)  }{ { x }^{ 2 } } +1 \right) =3$ then $f(1)$

  1. $\frac { 1 }{ 2 } $

  2. $\frac { 3 }{ 2 } $

  3. $\frac { 5 }{ 2 }$

  4. $\frac { 9 }{ 2 } $


Correct Option: A

If $\alpha$ and $\beta$ are the polynomial  $f(x)=x^2-5x+k$ such that $\alpha-\beta=1$, then value of k is 

  1. $8$

  2. $6$

  3. $\dfrac{13}{2}$

  4. $4$


Correct Option: A