Functions Test 57

Result

Functions Test 57

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$f:A\rightarrow B$$ will be an into function if

A
$$f\left(A\right) \subset B$$

B
$$f\left(A\right)=B$$

C
$$B\subset f\left(A\right)$$

D
$$f\left(B\right) \subset A$$

Solution
Question 2
1 / -0

If f (x) = cosx and g (x) = x$$^2$$ then (gof) (x) is ....

A
cos$$^2$$ x

B
cosx$$^2$$

C
both (a)&(b)

D
x$$^2$$ cosx

Solution
Question 3
1 / -0

Suppose that $$g(x)=1+\sqrt { x } and\quad f(g(x))=3+2\sqrt { x } +x\quad then\quad f(x)\quad is$$

A
$$\quad 1+2{ x }^{ 2 }$$

B
$$\quad 2+{ x }^{ 2 }$$

C
$$1+x$$

D
$$2+x$$

Solution
Question 4
1 / -0

Which one of the following is one-one?

A
$$f:R\rightarrow R$$ given by $$f(x) =\left| x-1 \right| $$ for all $$x\in R$$

B
$$g:\left[ -\pi /2,\pi /2 \right] \rightarrow given by$$ $$g(x)=\left| sinx \right| $$

C
$$h:\left[ -\pi /2,\pi /2 \right] \in R$$ given by $$h(x) =sin x $$ for all $$x\in \left[ -\pi /2,\pi /2 \right] $$

D
$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$x\in R$$
Question 5
1 / -0

Let $$f:R\rightarrow R$$ defined by $$f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ -x^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,$$ then

A
f(x) is one-one but not onto

B
f(x) is neither one-one nor onto

C
f(x) is many one but onto

D
f(x) is one-one and onto

Solution

$$\text { solution: } \quad f(x)=\frac{e^{x^{2}}-e^{-x^{2}}}{e^{x^{2}}+e^{-x^{2}}}=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1} \\$$

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

$$\text { Since } f^{\prime}(x) \text { is nither positive, nor negative } \\$$

$$\text { for all } x \text { so } f(x) \text { is not one - one function } \\$$

$$\qquad f(x)=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1}$$

Range of $$f(x)$$ is $$[0,1 )$$

since Range of $$f(x)$$ is not equal to co-domain

of $$f(x)$$ so $$f(x)$$ is not onto function.

Answer: option: (B)
Question 6
1 / -0

The domain of derivative of the function $$f\left( x \right) = \left| {{{\sin }^{ - 1}}\left( {2{x^2} - 1} \right)} \right|$$ is:

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

B
$$\left( { - 1,1} \right) \sim \left\{ {0, \pm \frac{1}{{\sqrt 2 }}} \right\}$$

C
$$\left( { - 1,1} \right) \sim \left\{ 0 \right\}$$

D
$$\left( { - 1,1} \right) \sim \left\{ { \pm \frac{1}{{\sqrt 2 }}} \right\}$$

Solution
Question 7
1 / -0

Let $$f:R\rightarrow R$$, be defined as $$f(x)={ e }^{ x^{ 2 } }+cosx$$ then f is

A
One-one and onto

B
One-one and into

C
Many-one and onto

D
many-one and into

Solution
Question 8
1 / -0

If the mapping $$ f(x)=a x+b, a<0 $$ maps $$ [-1,1] $$ onto $$ [0,2], $$ then for all values of $$ \theta, A=\cos ^{2} \theta+\sin ^{4} \theta $$ is such that

A
$$

f\left(\frac{1}{4}\right) \leq A \leq f(0)

$$

B
$$

f(0) \leq A \leq f(-2)

$$

C
$$

f\left(\frac{1}{3}\right) \leq A \leq f(0)

$$

D
$$f(-1) < A < f(-2)$$

Solution
Question 9
1 / -0

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

A
$$\frac { 1 }{ 4 } (e-7)$$

B
$$\frac { 1 }{ 4 } (e-2)$$

C
$$\frac { 1 }{ 4 } (e-3)$$

D
none of the above

Solution
Question 10
1 / -0

$$f : R \rightarrow R$$ defined by $$f ( x ) = \dfrac { x } { x ^ { 2 } + 1 } , \forall x \in R$$ is

A
one-one

B
onto

C
bijective

D
neither one one nor onto

Solution

$$\begin{aligned} f(x) &=\dfrac{x}{1+x^{2}} \\-f(4)=& \dfrac{4}{1+16}=\dfrac{4}{17} \\ f\left(\dfrac{1}{4}\right) &=\frac{\dfrac{1}{4}}{1+\dfrac{1}{16}}=\dfrac{4}{17} \end{aligned}$$
Hence two different input has Same output $$\cdot$$ so not one-one
$$f(x)=y$$
$$y=\dfrac{x}{1+x^{2}}=1 \quad y+y x^{2}-x=0$$
$$x=\dfrac{1 \pm \sqrt{1-4 y^{2}}}{2 y}$$
$$1-4 y^{2} \geq 0$$
$$(1+2 y)(1-x y) \geq 0$$
$$\dfrac{1}{-3} \leq y<\dfrac{1}{2}$$
Range of $$f(x)$$ is $$\left[\dfrac{-1}{2}, \dfrac{1}{2}\right]$$
Range of $$f(x) \neq$$ codomain $$f(x)$$ Hence f(x) is neither one-one nor onto option D is correct.