Tag: introduction to sets

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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

The set values of $x$ for which function $f(x)=x\ln {x}-x+1$

  1. $\left( 1,\infty \right) $
  2. $\left( \cfrac { 1 }{ e } ,\infty \right) $
  3. $[e,\infty )$
  4. $\left( 0,1 \right) \cup \left( 1,\infty \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$f(x)=x \ln x -x +1$

we use the formula

$\log _{a}f(x)\Rightarrow f(x)>0$

$\Rightarrow x>0$

$\therefore x>0\Rightarrow (1,\infty )$
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

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. $(-\infty, -1] \cup [1,\infty)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given function is f(x) = 2 * tan^-1(x) + sin^-1(2x / (1 + x^2)). For x >= 1, the second term simplifies to pi - 2 * tan^-1(x), making the total function equal to pi, which is a constant function. For x <= -1, the second term simplifies to -pi - 2 * tan^-1(x), which also gives a constant value. The question specifies a positive output, and since pi is positive, the interval where this holds is [1, infinity).

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 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. $f(2)-f'(2)+f"(2)=10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since f(x) is a polynomial and f(3x) = f'(x) * f''(x), comparing degrees shows that if f(x) has degree n, the left side has degree n while the right side has degree 2n - 3, meaning n = 3. Thus f(x) is a cubic polynomial of the form ax^3 + bx^2 + cx + d. Solving the functional equation coefficients leads to specific relations, satisfying f''(2) - f'(2) = 0.

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).