Functions Test 56

Result

Functions Test 56

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Weekly Quiz Competition

Question 1
1 / -0

Consider the function $$f\left( x \right) ={ e }^{ x }$$ and $$g\left( x \right)=\sin ^{ -1 }{ x } $$, then which of the following is/are necessarily true.

A
Domain of $$gof =$$ Domain of $$f$$

B
Range of $$gof\ \subset$$ Range of $$g$$

C
Domain of $$gof$$ is $$\left( -\infty ,0 \right) $$

D
Range of $$gof$$ is $$\left( -\dfrac{\pi}{2} ,0 \right) $$

Solution

$$\text { solution: } \quad f(x)=e^{x} \quad g(x)=\sin ^{-1}(x) \\$$

$$\text { gof }=\sin ^{-1}\left(e^{x}\right) \\$$

$$\therefore-1 \leqslant e^{x} \leqslant 1 \\$$

$$\text { so we need } x \in(-\infty \text { o] } \\$$

$$\text { domain of gof is }(-\infty, 0] \\$$

$$\text { Range of gof is same as range of } g \\$$

$$\text { Ansuer. : option (B) }$$
Question 2
1 / -0

If $$f(x)=\frac { \alpha x }{ x+1 } $$, where $$x\neq -1$$ and (fof) (x) = x, then $$\alpha =$$

A
$$\sqrt { 2 } $$

B
$$-\sqrt { 2 } $$

C
$$1$$

D
$$-1$$

Solution
Question 3
1 / -0

Domain of function $$y=\sqrt{\log_{0.6}\left(\dfrac{x-1}{x+5}\right)}\cdot\dfrac{1}{(x^2-49)}$$ is?

A
$$(1, 7)\cup (7, \infty)$$

B
$$R-(1, 7)$$

C
$$R-\{7\}$$

D
$$(-7, 7)$$

Solution

$$ \begin{array}{l} \text { Solution - Given that, } \\ \qquad \begin{array}{c} y=\sqrt{\log _{0.6}\left(\frac{x-1}{x+5}\right)} \cdot \frac{1}{\left(x^{2}-49\right)} \\ x^{2}-49 \neq 0 \\ (x-7)(x+7) \neq 0 \\ x \neq 7,-7 \end{array} \\ \text { Now, } \\ \qquad \log _{0.6}\left(\frac{x-1}{x+5}\right) \geqslant 0 \end{array} $$

$\begin{aligned} \Rightarrow & \frac{x-1}{x+5} \leqslant(0,6)^{0} \\ \Rightarrow \frac{x-1}{x+5} \leqslant 1 & \Rightarrow \frac{x-1}{x+5}-1 \leqslant 0 \\ \Rightarrow \frac{x-1-x-5}{x+5} \leqslant 0 & \\ \Rightarrow+\frac{6}{x+5} \geqslant 0 & \Rightarrow x+5>0 \\ \text { Also, for log to be defined, } \frac{x-1}{x+5}>0 \end{aligned}$ $$ \begin{array}{l} x \in(1,7) \cup(7, \infty) \\ \text { Hence, }(A) \text { is the correct option. } \end{array} $$
Question 4
1 / -0

Let N be the set of natural numbers and two functions f and g be defined as
and g(n)=$$n-{ \left( -1 \right) }^{ n }$$ then fog is:

A
one-one but not onto

B
onto but not one-one

C
neither one-one nor into

D
both one-one and onto

Solution
Question 5
1 / -0

let $$f:R\rightarrow R$$ be a function defined by $$f(x)=\frac { { x }^{ 2 }-3x+4 }{ { x }^{ 2 }+3x+4 }$$ then f is

A
one-one but not onto

B
onto but not one

C
onto as well as one-one

D
neither onto nor one-one

Solution
Question 6
1 / -0

$$f (x) = x^4 - 10x^3 + 35x^2 - 50x + c$$ is a constant. the number of real roots of . f (x) = 0 and
f'' (x) = 0 are respectively

A
1 , 0

B
3, 2

C
1 , 2

D
3 , 0

Solution

Given, $$f(x)=x^{4}-10 x^{3}+35 x^{2}-50 x+c$$
Then $$, f^{\prime}(x)=\dfrac{d}{d x}\left(-10 x^{3}+x^{4}+35 x^{2}-50 x+c\right)$$
$$\Rightarrow f^{\prime}(x)=4 x^{3}-30 x^{2}+70 x-50=0$$
$$\therefore 2 x^{3}-15 x^{2}+35 x-25=0$$
First, let us factorise $$f(x)=0$$
$$\Rightarrow x^{4}-10 x^{3}+35 x^{2}-50 x+c=0$$
$$\Rightarrow(x-1)\left(x^{3}-9 x^{2}+26 x-c\right)=0$$
$$\Rightarrow(x-1)\left((x-2)\left(x^{2}-7 x+\dfrac{c}{2}\right)\right)=0$$
$$\Rightarrow f(x)=(x-1)(x-2)(x-4)(x-3)$$
Roots of $$f^{\prime}(x)=0$$
$$\Rightarrow 2 x^{3}-15 x^{2}+35 x-25=0$$
$$\Rightarrow f^{\prime}(x)$$ has 3 roots which are
$$x=\dfrac{5}{2}, \dfrac{5+\sqrt{5}}{2}, \dfrac{-\sqrt{5}+5}{2}$$
Hence $$(A)$$ is the correct option.
Question 7
1 / -0

The function $$f:R\rightarrow R$$ defined by $$f\left( x \right) =\frac { { e }^{ \left| x \right| }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } $$ is

A
One-One and onto

B
One-one but not onto

C
Not one-one but onto

D
Neither one-one nor onto

Solution
Question 8
1 / -0

Let f(x+y)=f f(x) f(y) and f(x) =1+x g(x) G(x), where $$\underset { x\rightarrow 0 }{ lim } g\left( x \right) =a and \underset { x\rightarrow 0 }{ lim } G\left( x \right) =b,$$ then f' (x) is equal to

A
1+ab

B
ab

C
f(x)

D
ab f(x)

Solution
Question 9
1 / -0

$$f:N\rightarrow N\quad where\quad f\left( x \right) =x-{ (-1) }^{ x }$$, then 'f' is

A
one-one and into

B
many- one and into

C
one-one and onto

D
many-one and onto

Solution
Question 10
1 / -0

Let (X) be a function satisfying f' (X) = f (X) with f (0) = 1 and g (X) be a function that satisfies f (X) + g (x) = $${ x }^{ 2 },$$ Then the value of the integral $$\int _{ 0 }^{ 1 }{ f } (x)\quad g\quad (x)\quad dx,\quad is$$

A
$$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 5 }{ 2 } $$

B
$$e+\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } $$

C
$$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } $$

D
$$e+\frac { { e }^{ 2 } }{ 2 } +\frac { 5 }{ 2 } $$

Solution