Functions Test 6

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Functions Test 6

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=x, g(x)=2x^{2}+1$$ and $$h(x)=x+1$$ then $$(hogof)(x)$$ is equal to

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

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

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

D
$$2(x^{2}+1)$$

Solution

Given $$f(x)=x, g(x)=2x^{2}+1$$ and $$h(x)=x+1$$

$$hogof(x)$$

$$=h(g(f(x)))$$
$$=h\left ( g\left ( x \right ) \right )\dots\dots \left[ \because f\left ( x \right )=x \right ]$$
$$=h\left ( 2x^{2}+1 \right )\dots\dots\left [ \because g\left ( x \right )=2x^{2}+1 \right ]$$
$$=2x^{2}+1+1$$
$$=2x^2+2$$
$$=2\left (x ^{2} +1\right )$$

$$\therefore hogof(x)=2(x^2+1)$$
Question 2
1 / -0

The domain of $$f(x) =\displaystyle \frac{1}{\left | x \right |+x}$$ is

A
$$(-\infty ,0)$$

B
$$(0,\infty )$$

C
$$(-\infty ,1)$$

D
$$(-2,-1)$$

Solution

For $$f$$ to be defined
$$\left | x \right |+x\neq 0$$
$$\Rightarrow \left | x \right |\neq -x$$
$$\Rightarrow x>0.$$
$$\therefore x\in \left ( 0,\infty \right )$$
Question 3
1 / -0

The domain of $$f(x)=\sqrt{1-\left | x \right |}$$ is

A
$$[-1,1]$$

B
$$(-1,1)$$

C
$$(0,1)$$

D
$$R$$

Solution

$$1-\left | x \right |\geqslant 0$$
$$\Rightarrow \left | x \right |\leqslant 1$$
$$\Rightarrow -1\leq x \leq 1$$
$$\Rightarrow x\in \left [ -1,1 \right ]$$
Question 4
1 / -0

If $$f(x)=(1-x)^{1/2}$$ and $$g(x)= \ln(x)$$ then the domain of $$(gof)(x)$$ is

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

B
$$(-1,1)$$

C
$$(-\infty ,1]$$

D
$$(-\infty ,1)$$

Solution

Given $$f(x)=(1-x)^{\frac{1}{2}}$$ and $$g(x)=ln(x)$$

$$gof(x)$$
$$=g(f(x))$$
$$=\ln\left ( 1-x \right )^{1/2}$$
$$=\dfrac{1}{2}\ln\left ( 1-x \right )$$
$$\therefore$$ For the composite function to be defined $$1-x>0$$
$$x<1$$
$$\therefore$$ Domain is $$\left ( -\infty ,1 \right )$$
Question 5
1 / -0

The domain of $$f(x) = \displaystyle \frac{1}{\log\left | x \right |}$$ is

A
$$R-\left \{ 0 \right \}$$

B
$$R-\left \{ 0,1 \right \}$$

C
$$R-\left \{ -1,0,1 \right \}$$

D
$$(-\infty ,\infty )$$

Solution

For $$f$$ to be defined,
$$\left | x \right |>0$$
$$\left | x \right |\neq 1$$ as $$\left | \log 1\right |=0$$
$$\therefore x\notin \left \{ -1,0,1 \right \}$$
$$\therefore x\in R-\left \{ -1,0,1 \right \}$$
Question 6
1 / -0

The domain of $$f(x)=\sin^{-1}(3x)$$ is

A
$$\left [ -\dfrac{1}{3},\dfrac{1}{3} \right ]$$

B
$$\left ( -\dfrac{1}{3},\dfrac{1}{3} \right )$$

C
$$\left ( 0,\dfrac{1}{3} \right )$$

D
$$\left ( -\dfrac{1}{3},0 \right )$$

Solution

$$\sin ^{-1}y $$ is defined for $$ y\in \left [ -1,1 \right ]$$

$$\therefore\sin ^{-1}3x $$ is defined for $$3x\in \left [ -1,1 \right ]$$

$$\Rightarrow -1\leq 3x \leq 1$$

$$\Rightarrow \dfrac{-1}{3}\leq x \leq \dfrac{1}{3}$$

$$x\in \left [ \dfrac{-1}{3},\dfrac{1}{3} \right ] $$
Question 7
1 / -0

The domain of $$f(x)= \displaystyle \frac{1}{\sqrt{x^{2}-4}} $$ is

A
$$\left ( -\infty ,4 \right )$$

B
$$(-2,3)$$

C
$$(4,\infty )$$

D
$$(-\infty ,-2)\cup (2,\infty )$$

Solution

Given $$\displaystyle f(x)=\int \dfrac{1}{\sqrt{x^2-4}}$$

for $$f(x)$$ to be defined denominator $$\sqrt{x^2-4}>0$$

$$x^{2}-4> 0$$

$$\left ( x-2 \right )\left ( x+2 \right )> 0$$

$$\Rightarrow x\in \left ( -\infty ,-2 \right ) \cup \left ( 2,\infty \right )$$ as $$(x-2)$$ and $$(x+2)$$ should have same sign always
Question 8
1 / -0

The domain of $$ f(x) =\dfrac{1}{\sqrt{x^{2}+2x+9}}$$ is

A
$$(1,9)$$

B
$$(1,2)$$

C
$$(-\infty ,\infty )$$

D
$$(0,\infty )$$

Solution

For $$g(x)=x^{2}+2x+9$$, $$a >0$$
Hence for $$g(x) >0$$, $$D <0$$
$$D= b^{2}-4ac=4-4\left ( 9 \right )=-27<0.$$
$$\therefore$$ $$f$$ is defined for all $$x \in (-\infty ,\infty )$$
Question 9
1 / -0

The domain of $$\cosh^{-1} 5x $$ is

A
$$R$$

B
$$[0,\infty )$$

C
$$\displaystyle \left ( \frac{1}{5},\infty \right )$$

D
$$\displaystyle \left [ \frac{1}{5},\infty \right )$$

Solution

$$\cosh ^{-1}x=\log \left [ x+\sqrt{x^{2}-1} \right ]$$
$$\cosh ^{-1}5x$$$$=\dfrac{1}{5}\log \left [ x+\sqrt{x^{2}-1} \right ]$$
Range of $$\log$$ is from $$1$$ to $$\infty $$ as $$ x>0 $$ for $$\cosh^{-1}x$$ to be defined
$$\therefore$$ range is $$=\left [ \dfrac{1}{5},\infty \right ]$$
Question 10
1 / -0

The domain of $$\log_{a} \sin^{-1}x$$ is $$(a>0,a\neq 1)$$

A
$$ 0< x \leq 1$$

B
$$0\leq x\leq 1$$

C
$$0\leq x<1$$

D
$$ 0 < x < 1$$

Solution

$$\sin ^{-1}x>0$$ $$\Rightarrow x\in \left ( 0,1 \right ]$$
$$\therefore$$ domain is $$ (0,1]$$

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Re-Attempt Weekly Quiz Competition

Functions Test 6

Result

Functions Test 6

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SHARING IS CARING

Question 1

$$x^{2}+2$$

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

$$x^{2}+1$$

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

Solution

Question 2

$$(-\infty ,0)$$

$$(0,\infty )$$

$$(-\infty ,1)$$

$$(-2,-1)$$

Solution

Question 3

$$[-1,1]$$

$$(-1,1)$$

$$(0,1)$$

$$R$$

Solution

Question 4

$$(-\infty ,2)$$

$$(-1,1)$$

$$(-\infty ,1]$$

$$(-\infty ,1)$$

Solution

Question 5

$$R-\left \{ 0 \right \}$$

$$R-\left \{ 0,1 \right \}$$

$$R-\left \{ -1,0,1 \right \}$$

$$(-\infty ,\infty )$$

Solution

Question 6

$$\left [ -\dfrac{1}{3},\dfrac{1}{3} \right ]$$

$$\left ( -\dfrac{1}{3},\dfrac{1}{3} \right )$$

$$\left ( 0,\dfrac{1}{3} \right )$$

$$\left ( -\dfrac{1}{3},0 \right )$$

Solution

Question 7

$$\left ( -\infty ,4 \right )$$

$$(-2,3)$$

$$(4,\infty )$$

$$(-\infty ,-2)\cup (2,\infty )$$

Solution

Question 8

$$(1,9)$$

$$(1,2)$$

$$(-\infty ,\infty )$$

$$(0,\infty )$$

Solution

Question 9

$$R$$

$$[0,\infty )$$

$$\displaystyle \left ( \frac{1}{5},\infty \right )$$

$$\displaystyle \left [ \frac{1}{5},\infty \right )$$

Solution

Question 10

$$ 0< x \leq 1$$

$$0\leq x\leq 1$$

$$0\leq x<1$$

$$ 0 < x < 1$$

Solution

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