Functions Test 13

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Functions Test 13

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Weekly Quiz Competition

Question 1
1 / -0

The function $$\displaystyle f(x)=\sqrt{e^{ \cos^{-1}(\log_{4}x^{2})}}$$ is real valued. It is defined if

A
$$\displaystyle x\:\in\:\left [ \frac{1}{2},2 \right ]$$

B
$$\displaystyle x\:\in\:\left [-2 ,\frac{-1}{2} \right ]\cup \left [ \frac{1}{2},2 \right ]$$

C
$$\displaystyle x\:\in\:\left [-2 ,\frac{1}{2} \right ]$$

D
none of these

Solution

$$\displaystyle f(x)=\sqrt{e^{ \cos^{-1}(\log_{4}x^{2})}}$$
For $$f(x)$$ to be defined $$x^2> 0 \Rightarrow x \neq 0$$
and $$-1\leq \log_4x^2 \leq 1\Rightarrow \cfrac{1}{4}\leq x^2\leq 4\Rightarrow \displaystyle x\:\in\:\left [-2 ,\frac{-1}{2} \right ]\cup \left [ \frac{1}{2},2 \right ]$$
Hence we get required domain as $$\displaystyle x\:\in\:\left [-2 ,\frac{-1}{2} \right ]\cup \left [ \frac{1}{2},2 \right ]$$
Question 2
1 / -0

The domain of function $$\displaystyle y=\log_{3}(5+4x-x^{2})$$ is

A
$$(0,2]$$

B
$$(-\infty,-1)\cup(5,\infty)$$

C
$$(0,9]$$

D
$$(-1,5)$$

Solution

$$\displaystyle y=\log_{3}(5+4x-x^{2})$$

$$\therefore 5+4x-x^2>0$$

$$x^2-4x-5<0$$

$$(x-5)(x+1)<0$$

$$x<5$$ and $$x>-1$$

$$x\in(-1,5)$$
Question 3
1 / -0

The domain of the function $$\displaystyle f(x)=\sin^{-1}(x+[x]),$$ where $$[\cdot]$$ denotes the greatest integer function is

A
$$[0,1)$$

B
$$[-1,1]$$

C
$$(-1,0)$$

D
none of these

Solution

We know that the range of $$\sin^{-1}$$ is $$\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$$ and domain is $$[-1,1]$$

Now in $$f(x)=\sin(x+[x])$$

The input should be in the domain $$[-1,1]$$

$$x+[x]\geq-1$$

$$\Rightarrow x\geq0$$

and $$x+[x]\leq1$$

$$\Rightarrow x\leq1$$

However, $$[x]$$ is discontinuous at $$x=1$$

Hence domain, of $$f(x)=\sin^{-1}(x+[x])$$ is $$[0,1)$$
Question 4
1 / -0

The domain of the function $$\displaystyle f(x)=\log_{e}(x-[x]),$$ where $$[x]$$ denotes the greatest integer function, is

A
$$R$$

B
$$R-Z$$ where $$Z$$ is the set of all integers

C
$$(0,+\infty)$$

D
None of these

Solution

For the function to be defined the argument of the logarithm must be greater than zero.

$$\therefore x-[x]>0$$

$$x-[x]=\{ x \}$$ where $$\{\cdot\}$$ denotes the fractional part

$$\therefore \{x\}>0$$

This is always true unless $$x$$ is an integer.

Therefore, answer is B
Question 5
1 / -0

Let $$\displaystyle f(x)=\log_{x^{2}}25$$ and $$g(x)=\log_{x}5$$, then $$f(x)=g(x)$$ holds for $$x$$ belonging to

A
$$R$$

B
$$(0,1)\cup(1,+\infty)$$

C
$$\phi$$

D
none of these

Solution

$$f(x)=log_{x^{2}}(25)$$

$$=\dfrac{log(25)}{logx^{2}}$$ ...$$(log(x)\neq 0),x>0$$

$$=\dfrac{2log5}{2logx}$$

$$=log_{x}(5)$$
$$=g(x)$$

Since $$log(x)\neq 0$$
$$x\neq 1$$
Also input of log cannot be negative.
Hence $$x>0$$
Therefore the solution set for which f(x) is equal to g(x) is
$$R^{+}-\{1\}$$

$$=(0,1)\cup(1,\infty)$$.
Question 6
1 / -0

Let $$\displaystyle f:(-1,1)\rightarrow B$$ be a function defined by $$\displaystyle f(x)=\tan^{-1}\dfrac{2x}{1-x^{2}}$$. Then $$f$$ is both one-one and onto function when $$B$$ is in the interval

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

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

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

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

Solution
Question 7
1 / -0

The composite mapping $$fog$$ of the map $$f: R\rightarrow R,f(x)=\sin x$$ and $$g: R\rightarrow R, g(x)=x^2$$ is

A
$$x^2 \sin x$$

B
$$(\sin x)^2$$

C
$$\sin x^2$$

D
$$\dfrac{ \sin x}{x^2}$$

Solution

Composite mapping will be $$f(g(x))$$

$$=f(x^2)$$

$$=\sin(x^2)$$

Hence option $$'C'$$ is the answer.
Question 8
1 / -0

If $$\displaystyle \left [ x+\left [ x \right ] \right ]\leq 2$$ where $$\displaystyle \left [ x \right ]$$ denotes the greatest integer $$\displaystyle \leq x,$$ then $$x$$ lies in the interval,

A
$$\displaystyle \left ( -\infty , 1 \right )$$

B
$$\displaystyle \left ( -\infty , 2 \right )$$

C
$$\displaystyle \left ( -\infty , -2\right )$$

D
$$\displaystyle \left ( -\infty , -1 \right )$$

Solution

$$\displaystyle \left [ x+\left [ x \right ] \right ]\leq 2$$

$$\Rightarrow \displaystyle x+\left [ x \right ] < 3$$

We know $$x = [x]+\{x\}$$, where $$0 \leq \{x\} <1$$

$$\Rightarrow \displaystyle \{x\}+2\left [ x \right ] < 3\Rightarrow 2[x]< 3-\{x\}$$

$$\Rightarrow 2[x]< 3\Rightarrow [x] < 1.5\Rightarrow x < 2$$
Question 9
1 / -0

Let $$ f:R \rightarrow R$$ and $$g:R \rightarrow R$$ be defined by $$f(x)=x^2+2x-3,g(x)=3x-4$$ then $$(gof) (x)=$$

A
$$3x^2+6x-13$$

B
$$3x^2-6x-13$$

C
$$3x^2+6x+13$$

D
$$-3x^2+6x-13$$

Solution

Given that, $$f(x)=x^2+2x-3$$ and $$g(x)=3x-4$$

$$(gof) (x)=g\left(f(x)\right)$$

$$=3(x^2+2x-3)-4$$

$$(gof) (x)=3x^2+6x-13$$

Hence, option A.
Question 10
1 / -0

If $$f(x)=ax+b$$ and $$g(x)=cx+d$$, then $$f(g(x))=g(f(x))$$ implies

A
$$f(a)=g(c)$$

B
$$f(b)=g(b)$$

C
$$f(d)=g(b)$$

D
$$f(c)=g(a)$$

Solution

$$f(g(x))=a(cx+d)+b$$

$$=acx+ad+b$$...(i)

$$g(f(x))=c(ax+b)+d$$

$$=ac(x)+bc+d$$ ...(ii)

$$f(g(x))=g(f(x))$$

From (i) and (ii)

$$ac(x)+bc+d=acx+ad+b$$

$$cb+d=ad+b$$

$$g(b)=f(d)$$

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Re-Attempt Weekly Quiz Competition

Functions Test 13

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Functions Test 13

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SHARING IS CARING

Question 1

$$\displaystyle x\:\in\:\left [ \frac{1}{2},2 \right ]$$

$$\displaystyle x\:\in\:\left [-2 ,\frac{-1}{2} \right ]\cup \left [ \frac{1}{2},2 \right ]$$

$$\displaystyle x\:\in\:\left [-2 ,\frac{1}{2} \right ]$$

none of these

Solution

Question 2

$$(0,2]$$

$$(-\infty,-1)\cup(5,\infty)$$

$$(0,9]$$

$$(-1,5)$$

Solution

Question 3

$$[0,1)$$

$$[-1,1]$$

$$(-1,0)$$

none of these

Solution

Question 4

$$R$$

$$R-Z$$ where $$Z$$ is the set of all integers

$$(0,+\infty)$$

None of these

Solution

Question 5

$$R$$

$$(0,1)\cup(1,+\infty)$$

$$\phi$$

none of these

Solution

Question 6

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

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

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

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

Solution

Question 7

$$x^2 \sin x$$

$$(\sin x)^2$$

$$\sin x^2$$

$$\dfrac{ \sin x}{x^2}$$

Solution

Question 8

$$\displaystyle \left ( -\infty , 1 \right )$$

$$\displaystyle \left ( -\infty , 2 \right )$$

$$\displaystyle \left ( -\infty , -2\right )$$

$$\displaystyle \left ( -\infty , -1 \right )$$

Solution

Question 9

$$3x^2+6x-13$$

$$3x^2-6x-13$$

$$3x^2+6x+13$$

$$-3x^2+6x-13$$

Solution

Question 10

$$f(a)=g(c)$$

$$f(b)=g(b)$$

$$f(d)=g(b)$$

$$f(c)=g(a)$$

Solution

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