Functions Test 35

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Functions Test 35

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Weekly Quiz Competition

Question 1
1 / -0

Difference between the greatest and the least values of the function
$$f(x) = x(ln x - 2)$$ on $$[1, e^{2}]$$ is

A
$$2$$

B
$$e$$

C
$$e^{2}$$

D
$$1$$

Solution
Question 2
1 / -0

The function $$f :\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\rightarrow \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ defined by $$f(x)=\sin^{-1}(3x-4x^{3})$$ is

A
both one-one and onto

B
onto but not one-one

C
one-one but not onto

D
neither one-one nor onto

Solution

$$f:[-\cfrac{1}{2},\cfrac{1}{2}]\rightarrow[-\cfrac{\pi}{2},\cfrac{\pi}{2}]\\f(x)=\sin^{-1}(3x-4x^3)$$
Range of $$\sin^{-1}(3x-4x^3)$$ is $$[-\cfrac{\pi}{2},\cfrac{\pi}{2}]$$
And $$-\cfrac{\pi}{2}\le\sin^{-1}(3x-4x^3)\le\cfrac{\pi}{2}$$
$$\Rightarrow \sin(\cfrac{-\pi}{2})\le3x-4x^3\le\sin\cfrac{\pi}{2}$$
$$\Rightarrow-1\le3x-4x^3\le1$$
$$3x-4x^3\le-1$$
$$\Rightarrow 4x^3-3x-1\le0$$
$$\Rightarrow x\le\cfrac{1}{2}$$
and$$3x-4x^3\le1\Rightarrow 4x^3-3x+1\le0\\ \Rightarrow x\le-\cfrac{1}{2}\\ \therefore x\epsilon [-\cfrac{1}{2},\cfrac{1}{2}]\\f{x}=\sin^{-1}(3x-4x^3)\\ \Rightarrow f^1(x)=\cfrac{1}{\sqrt{1-(3x-4x^3)^2}}\times(3-12x^2)\\ \Rightarrow f^1(x)=3-12x^2\\ \quad=-(12x^2-3)\\x^2\le0\Rightarrow 12x^2\le0\Rightarrow 12x^2-3\le-3\\-3(12x^2-3)\le3\\ \therefore f^1(x)\le0$$
$$\therefore f^1(x)$$ is a decreasing function.
Hence $$f(x)$$ is one-one and range of function $$=$$ its co-domain.
Hence it is both one-one and onto.
Question 3
1 / -0

The value of the function f(x)=$$\dfrac { { x }^{ 2 }-3x+2 }{ { x }^{ 2 }+x-6 } $$ lies in the interval:

A
$$\left( \infty ,\infty \right) -\left\{ \dfrac { 1 }{ 5 } ,1 \right\} $$

B
$$\left( \infty ,\infty \right) $$

C
$$\left( \infty ,\infty \right) -1\left\{ 1 \right\} $$

D
None of these

Solution
Question 4
1 / -0

The domain of $$f(x)=\log(||x-2|-2|-1)$$ is$$

A
$$R-(1,3)$$

B
$$(-\infty,-1)\bigcup(1,3)\bigcup(5,\infty)$$

C
$$(5,\infty)$$

D
$$none of these$$

Solution

Given,
$$f(x)=\log (|| x-2|-2|-1)$$

Now,
$$\mid p(-2|-2|>1$$
$$|x-2|-2>\mid \quad 0 r-(|x-2|-2)>1$$
$$\Rightarrow \quad|x-2|>3 \quad$$ or $$\quad|x-2|-2<-1$$
$$\Rightarrow \quad x-2\rangle 3-(i)$$ or $$|x-2|<-1+2$$
$$-(x-2)>3 \longrightarrow$$ (ii) $$|x-2|<1 - (iii)$$.

From (i) and (ii)
$$\Rightarrow \quad x>5 ; x-2<-3$$
$$\Rightarrow \quad x>5, \quad x<-3+2$$
$$\Rightarrow \quad x>5, \quad x<-1$$
$$x \in[-\infty,-1) \cup(5, \infty)$$ - (iv)

From (iii)
$$\begin{array}{ll}x-2<1 . & \text { or }-x-2<1 \\ x<3 & ; \quad x-2 >-1 \\ x<3 & , x>1\end{array}$$
$$x \in[1,3)$$ - $$(v)$$
From (iv) and (v)
option $$B:(-\infty,-1) \cup(1,3) \cup(5, \infty)$$
Question 5
1 / -0

Let g be the inverse function of differentiable function f and $$G\left( x \right) =\frac { 1 }{ g\left( x \right) } if\quad f\left( 4=2 \right) $$ and $$f'\left( 4 \right) =\frac { 1 }{ 16 } $$, then the value of $${ \left( G'\left( 2 \right) \right) }^{ 2 }$$ equals to:

A
1

B
4

C
16

D
64

Solution
Question 6
1 / -0

If $$f:( - 1,1) \to B$$ , is a function defined by $$f(x) = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$$, then find $$B$$ when $$f(x)$$ is both one-one and onto function.

A
$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$

B
$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

C
$$\left( {0,\frac{\pi }{2}} \right)$$

D
$$\left[ {0,\frac{\pi }{2}} \right)$$

Solution

For $$x \epsilon (-1,1)$$, we have
$$f(x)=\tan^{-1}\left[\dfrac {2x}{1-x^2}\right]$$
Substituting $$x=\tan\theta$$ in above equation.
Therefore, $$f(\tan \theta)=\tan^{-1}\left[\dfrac {2\tan \theta}{1-\tan^2\theta}\right]$$
$$=\tan^{-1}\tan (2\theta)=2\theta$$
$$=2\tan ^{-1}x$$
Thus $$-\dfrac {\pi}{2}<\tan ^{-1}\left[\dfrac {2x}{1-x^2}\right]<\dfrac {\pi}{2}$$
Thus option B is correct.
Question 7
1 / -0

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 $$f(2)=$$

A
$$\dfrac { 1 } { 2 }$$

B
$$1$$

C
$$-1$$

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

Solution

let $$f(x)=ax+b$$
$$f(0)=1\implies b=1$$
$$f'(0)=-1 \implies a=-1$$
$$\implies f(x)=1-x$$
$$\implies f(2)=-1$$
Question 8
1 / -0

If $$f(x)=\dfrac {4^{x}}{4^{x}+2}$$, then the value of $$f(x)+f(1-x)$$ is

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$can't\ be\ determined$$

Solution

Given $$f{\left( x \right)} = \cfrac{{4}^{x}}{{4}^{x} + 2}$$

$$\therefore f{\left( 1 - x \right)} = \cfrac{{4}^{1-x}}{{4}^{1-x} + 2}$$

$$\Rightarrow f{\left( 1 - x \right)} = \cfrac{\left( \cfrac{4}{{4}^{x}} \right)}{\left( \cfrac{4}{{4}^{x}} \right) + 2}$$

$$\Rightarrow f{\left( 1 - x \right)} = \cfrac{4}{4 + {4}^{x} \cdot 2} = \cfrac{2}{2 + {4}^{x}}$$

Therefore,

$$f{\left( x \right)} + f{\left( 1 - x \right)} = \cfrac{{4}^{x}}{2 + {4}^{x}} + \cfrac{2}{2 + {4}^{x}}$$

$$\Rightarrow f{\left( x \right)} + f{\left( 1 - x \right)} = \cfrac{{4}^{x} + 2}{2 + {4}^{x}} = 1$$
Question 9
1 / -0

Let $$f(x)$$ be a function whose domain is $$[-5,7]$$. Let $$g(x) =|2x+5|$$, then domain of $$(fog) (x)$$ is

A
$$[-4, 1]$$

B
$$[-5, 1]$$

C
$$[-6, 1]$$

D
$$(-6, 1)$$

Solution
Question 10
1 / -0

If $$f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)\ \forall x\ \epsilon \ R$$, then $$f(x)$$ is

A
one-one and onto

B
one-one and into

C
many-one and onto

D
non-invertible

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Re-Attempt Weekly Quiz Competition

Functions Test 35

Result

Functions Test 35

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2$$

$$e$$

$$e^{2}$$

$$1$$

Solution

Question 2

both one-one and onto

onto but not one-one

one-one but not onto

neither one-one nor onto

Solution

Question 3

$$\left( \infty ,\infty \right) -\left\{ \dfrac { 1 }{ 5 } ,1 \right\} $$

$$\left( \infty ,\infty \right) $$

$$\left( \infty ,\infty \right) -1\left\{ 1 \right\} $$

None of these

Solution

Question 4

$$R-(1,3)$$

$$(-\infty,-1)\bigcup(1,3)\bigcup(5,\infty)$$

$$(5,\infty)$$

$$none of these$$

Solution

Question 5

1

4

16

64

Solution

Question 6

$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$

$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

$$\left( {0,\frac{\pi }{2}} \right)$$

$$\left[ {0,\frac{\pi }{2}} \right)$$

Solution

Question 7

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

$$1$$

$$-1$$

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

Solution

Question 8

$$0$$

$$-1$$

$$1$$

$$can't\ be\ determined$$

Solution

Question 9

$$[-4, 1]$$

$$[-5, 1]$$

$$[-6, 1]$$

$$(-6, 1)$$

Solution

Question 10

one-one and onto

one-one and into

many-one and onto

non-invertible

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