Tag: graphs of the form y=ax^2+bx+c

Questions Related to graphs of the form y=ax^2+bx+c

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

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

Reveal answer Fill a bubble to check yourself
B Correct answer
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:
$f(z)\neq 2\quad \quad \quad f(z)=3\ f(x)\neq 1\quad \quad \quad f(x)=2\ f(y)=1\quad \quad \quad f(y)=1\ f(x)=2,f(y)=1,f(z)=3\ f^{ -1 }\left( 2 \right) =x,f^{ -1 }\left( 1 \right) =y,f^{ -1 }\left( 3 \right) =z$ 

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

$f:c \to c$ 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

Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

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

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

let $f(x)=ax+b$

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

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

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 } $

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given f(x) is a quartic polynomial with extreme values at x=2, f'(2)=0. The limit condition lim(x->0) (f(x)/x^2 + 1) = 3 implies f(0)=0 and f'(0)=0, and f''(0)/2 = 2, so f''(0)=4. Using these conditions, one can determine the coefficients of f(x).