Functions Test 17

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Functions Test 17

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Weekly Quiz Competition

Question 1
1 / -0

If $$g(x)=1+\sqrt { x } $$ and $$f(g(x))=3+2\sqrt { x } +x$$, then $$f(x)=$$

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

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

C
$$1+x$$

D
$$2+x$$

Solution

We have $$g\left( x \right)=1+\sqrt { x } $$ and $$f(g(x))=3+2\sqrt { x } +x$$ ...(1)

Also, $$f\left( g\left( x \right) \right) =f\left( 1+\sqrt { x } \right) $$ ...(2)

From (1) and (2), we get

$$f\left( 1+\sqrt { x } \right) =3+2\sqrt { x } +x$$

Let $$1+\sqrt { x } =y\Rightarrow x={ \left( y-1 \right) }^{ 2 }$$

$$\therefore f\left( y \right) =3+2\left( y-1 \right) +{ \left( y-1 \right) }^{ 2 }\\ =3+3y-2+{ y }^{ 2 }-2y+1=2+{ y }^{ 2 }\\ \therefore f\left( x \right) =2+{ x }^{ 2 }$$
Question 2
1 / -0

Let $$f:R\rightarrow R$$ be a function defined by $$f(x)=\left\{ \left|\cos x\right|\right\},$$ where $$\left\{x\right\}$$ represents fractional part of $$x$$. Let $$ S$$ be the set containing all real values $$x$$ lying in the interval $$[0,2\pi]$$ for which $$f(x)\neq \left|\cos x\right|.$$ Then, number of elements in the set $$ S$$ is

A
$$0$$

B
$$1$$

C
$$3$$

D
infinite

Solution

$$f(x)\neq \left | \cos x \right |$$ is true only when

Function inside the fractional part must be an integer.

$$\cos x$$ has a maximum integer value $$=1$$

$$\left | \cos x \right |=1$$

As the interval is $$[0,2\pi]$$

$$\Rightarrow x=0, \:\pi , \:2\pi $$

So, the number of elements $$=3$$
Question 3
1 / -0

If $$f(x)$$ is defined in $$[-3,3]$$ then find the domain of definition of $$f(|2x+3|)$$

A
$$[-3,0]$$

B
$$[-2,0]$$

C
$$[-1,0]$$

D
$$[0,3]$$

Solution

For the required domain, $$-3 \leq 2x+3 \leq 3$$

Case 1. $$2x+3 \geq -3 \Rightarrow 2x+3 \geq -3 \Rightarrow x \geq -3$$

Case 2. $$2x+3 \leq 3 \Rightarrow 2x+3 \leq 3 \Rightarrow x \leq 0$$

Hence required domain is $$[-3, 0]$$
Question 4
1 / -0

The domain of the function $$ f(x)=\cos^{-1}(\sec (\cos^{-1}x))+\sin ^{-1}(\ \text {cosec} (\sin ^{-1}x))$$is

A
$$x\in R$$

B
$$x=1,-1$$

C
$$-1\leq x\leq 1$$

D
$$x\in \phi$$

Solution

$$\displaystyle f(x)=\cos^{-1}(\sec (\cos^{-1}x))+\sin^{-1}(cosec (\sin^{-1}x))$$

$$\Rightarrow -1\leq \sec (\cos^{-1}x)\leq 1$$

And $$-1\leq \text{cosec} (\sin^{-1}x)\leq 1$$

$$\Rightarrow \sec (\cos^{-1}x)=\pm 1$$

$$\Rightarrow \displaystyle \cos^{-1}x=0, \pi $$ and $$\displaystyle \sin^{-1}x=\frac{\pi }{2}, -\frac{\pi }{2}$$

$$\Rightarrow x=\pm 1$$ and $$x=\pm 1$$

$$\Rightarrow $$ Domain is $$x=\pm 1$$.
Question 5
1 / -0

If $$f_{0}(x)\, =\, \dfrac{x}{(x\, +\, 1)}$$ and $$f_{n\, +\, 1}\, =\, f_{0}\circ f_{n}(x)$$ for $$n = 0, 1, 2,\cdots$$ then $$f_{n}(x)$$ is

A
$$\displaystyle \frac{x}{(n\, +\, 1) x\, +\, 1}$$

B
$$f_{0}(x)$$

C
$$\displaystyle \frac{nx}{nx\, +\, 1}$$

D
$$\displaystyle \frac{x}{nx\, +\, 1}$$

Solution

Given $$f_{0}\left(x\right)=\dfrac{x}{ \left(x\, +\, 1\right)}\cdots\cdots\left(1\right)$$

Also given $$f_{n\, +\, 1}\, =\, f_{0}\, o\, f_{n}$$ for $$(n = 0, 1, 2,\cdots)\ldots\left(2\right)$$

Put $$n=0$$ in eqn (2)

$$f_1=f_{0} o f_0$$

$$f_1\left(x\right)=f_{0}[f_{0}\left(x\right)]$$

$$\Rightarrow f_{1}\left(x\right)=f_{0}\left(\dfrac{x}{x+1}\right)$$

$$=\dfrac{\dfrac{x}{x+1}}{\dfrac{x}{x+1}+1}$$

$$=\dfrac{x}{2x+1}$$

$$\Rightarrow f_1\left(x\right)=\dfrac{x}{2x+1}$$

Put $$n=1$$ in eqn $$(2)$$

$$f_2=f_{0} o f_1$$

$$f_2\left(x\right)=f_{0}[f_{1}\left(x\right)]$$

$$\Rightarrow f_{2}\left(x\right)=f_{0}\left(\dfrac{x}{2x+1}\right)$$

$$=\dfrac{\dfrac{x}{2x+1}}{\dfrac{x}{2x+1}+1}$$

$$=\dfrac{x}{3x+1}$$

$$\Rightarrow f_2\left(x\right)=\dfrac{x}{3x+1}$$

Hence, $$f_n\left(x\right)=\dfrac{x}{\left(n+1\right)x+1}$$
Question 6
1 / -0

If the function $$f : R \rightarrow$$ A given by $$f(x)\, =\, \displaystyle \frac{x^{2}}{x^{2}\, +\, 1}$$ is a surjection, then A is

A
$$R$$

B
$$[0, 1]$$

C
$$(0, 1]$$

D
$$[0, 1)$$

Solution

As 'f' is surjective,
Range of f = co-domain of 'f'
$$\implies$$ A = range of 'f'
Since, $$f(x)=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y(x^2+1)=x^2$$
$$\implies$$ $$(y-1)x^2+y=0$$
$$\implies$$ $$x^2=\dfrac{-y}{y-1}$$
$$\implies$$ $$x=\sqrt{\dfrac{y}{1-y}}$$
$$\implies$$ $$\dfrac{y}{1-y} \geq 0$$
$$\implies$$ $$y \epsilon [0,1)$$
$$\implies$$ $$A=[0,1)$$
Question 7
1 / -0

If $$\displaystyle \log_{1/3} \frac{3x -1}{x + 2}$$ is less than unity then $$x $$ must lie in the interval

A
$$\left(-\infty, -2\right) \cup \left(\dfrac58, \infty\right)$$

B
$$\left(-2, \dfrac58\right)$$

C
$$\left(-\infty, -2\right) \cup \left(\dfrac13, \dfrac58\right)$$

D
$$\left(-2, \dfrac13\right)$$

Solution

$$\displaystyle\log _{ \frac { 1 }{ 3 } }{ \left( \frac { 3x-1 }{ x+2 } \right) } =\log _{ 3 }{ \left( \frac { x+2 }{ 3x-1 } \right) } <1$$

$$\displaystyle\Rightarrow \frac { x+2 }{ 3x-1 } >0$$ and $$\displaystyle\frac { x+2 }{ 3x-1 } <3$$

$$\displaystyle \frac { x+2 }{ 3x-1 } >0$$

$$\Rightarrow x\in (-\infty,-2)\cup (1/3,\infty)$$ -----(1)

$$\displaystyle\frac { x+2 }{ 3x-1 } <3$$

$$\Rightarrow \displaystyle\frac{-8x+5}{3x-1}<0$$

$$\Rightarrow \displaystyle\frac{8x-5}{3x-1}>0$$

$$\therefore x\in (-\infty,1/3)\cup (5/8,\infty)$$ ------(2)

From (1) and (2), we get

$$x\in(-\infty, -2) \cup (5/8, \infty)$$

Hence, option A.
Question 8
1 / -0

If $$f(x)=\begin{cases} 2x+3\quad \quad x\le 1 \\ a^{ 2 }x+1\quad x>1 \end{cases}$$, then the values of $$a$$ for which $$f(x)$$ is injective.

A
$$-3$$

B
$$1$$

C
$$0$$

D
none of these

Solution

A function is called injective (one-one) if it is monontonic.
Clearly for $$x<1$$ f is increasing and it's maximum values is $$2(1)+3=5$$
Hence for f to be monotonic $$(x>1)$$, $$f(1) \geq 5$$
$$\Rightarrow a^2(1)+1\geq 5 \Rightarrow a^2\geq 2 \Rightarrow a\in R-(-2,2)$$
Question 9
1 / -0

The domain of the function $$f\left( x \right)=_{ }^{ 24-x }{ { C }_{ 3x-1 } }+_{ }^{ 40-6x }{ { C }_{ 8x-10 } }$$ is

A
$$\{2,3\}$$

B
$$\{1,2,3\}$$

C
$$\{1,2,3,4\}$$

D
None of these

Solution

$$_{ }^{ 24-x }{ { C }_{ 3x-1 } }$$ is defined if,

$$24-x>0,3x-1\ge 0$$ and $$24-x \ge 3x-1$$

$$\displaystyle \Rightarrow x<24,x\ge \frac { 1 }{ 3 } $$ and $$\displaystyle x\le \frac { 25 }{ 4 } \Rightarrow \frac { 1 }{ 3 } \le x\le \frac { 25 }{ 4 } $$ ...(1)

$$^{ 40-6x }{ { C }_{ 8x-10 } }$$ is defined if

$$40-6x>0,8x-10\ge 0$$ and $$40-6x\ge 8x-10$$

$$\displaystyle x<\frac { 20 }{ 3 } ,x\ge \frac { 5 }{ 4 } $$ and $$\displaystyle x\le \frac { 25 }{ 7 } \Rightarrow \frac { 5 }{ 4 } \le x\le \frac { 25 }{ 7 } $$ ...(2)

From (1) and (2), we get $$\displaystyle \frac { 5 }{ 4 } \le x\le \frac { 25 }{ 7 } $$

But, $$24-x\in N$$

$$\therefore x$$ must be an integer,

$$\therefore x=2,3.$$

Hence domain of $$f(x)={2,3}$$
Question 10
1 / -0

The domain of $$f(x)\, =\,\sqrt{\displaystyle \frac{1\, -\,|\, x\, |}{2\, -\, |x|}}$$ is

A
$$(-\, \infty,\, \infty)\, -\, [-\, 2\,, 2]$$

B
$$(-\, \infty,\, \infty)\, -\, [-\, 1\,, 1]$$

C
$$[-\, 1,\, 1]\, \cup\, (-\, \infty,\, -2)\, \cup\, (2\,, \infty)$$

D
none of these

Solution

For given function to be defined, $$\displaystyle \frac{1-|x|}{2-|x|} \geq 0$$
$$\Rightarrow \displaystyle \frac{|x|-1}{|x|-2} \geq 0$$
$$\Rightarrow |x| \in (-\infty, 1] \cup (2, \infty)\Rightarrow |x| \in [0, 1] \cup (2, \infty)$$, Since $$|x| \geq 0$$
$$\Rightarrow x \in [-\, 1,\, 1]\, \cup\, (-\, \infty,\, -2)\, \cup\, (2\,, \infty)$$

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Re-Attempt Weekly Quiz Competition

Functions Test 17

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Functions Test 17

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SHARING IS CARING

Question 1

$$1+2{ x }^{ 2 }$$

$$2+{ x }^{ 2 }$$

$$1+x$$

$$2+x$$

Solution

Question 2

$$0$$

$$1$$

$$3$$

infinite

Solution

Question 3

$$[-3,0]$$

$$[-2,0]$$

$$[-1,0]$$

$$[0,3]$$

Solution

Question 4

$$x\in R$$

$$x=1,-1$$

$$-1\leq x\leq 1$$

$$x\in \phi$$

Solution

Question 5

$$\displaystyle \frac{x}{(n\, +\, 1) x\, +\, 1}$$

$$f_{0}(x)$$

$$\displaystyle \frac{nx}{nx\, +\, 1}$$

$$\displaystyle \frac{x}{nx\, +\, 1}$$

Solution

Question 6

$$R$$

$$[0, 1]$$

$$(0, 1]$$

$$[0, 1)$$

Solution

Question 7

$$\left(-\infty, -2\right) \cup \left(\dfrac58, \infty\right)$$

$$\left(-2, \dfrac58\right)$$

$$\left(-\infty, -2\right) \cup \left(\dfrac13, \dfrac58\right)$$

$$\left(-2, \dfrac13\right)$$

Solution

Question 8

$$-3$$

$$1$$

$$0$$

none of these

Solution

Question 9

$$\{2,3\}$$

$$\{1,2,3\}$$

$$\{1,2,3,4\}$$

None of these

Solution

Question 10

$$(-\, \infty,\, \infty)\, -\, [-\, 2\,, 2]$$

$$(-\, \infty,\, \infty)\, -\, [-\, 1\,, 1]$$

$$[-\, 1,\, 1]\, \cup\, (-\, \infty,\, -2)\, \cup\, (2\,, \infty)$$

none of these

Solution

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