Functions Test 14

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Functions Test 14

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle f(x)=\frac{1}{1-x},x\neq 0,1$$ then the graph of the function $$\displaystyle y=f\left \{ f(f(x)) \right \},x> 1,$$ is

A
a circle

B
an ellipse

C
a straight line

D
a pair of straight lines

Solution

$$f(f(x))=\dfrac{1}{1-\dfrac{1}{1-x}}$$

$$=-\dfrac{1-x}{x} =g(x)$$

$$f(g(x))=\dfrac{1}{1+\dfrac{1-x}{x}}$$

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

$$=x$$

$$y=x$$ is an equation of straight line.

Hence, 'C' is correct.
Question 2
1 / -0

If $$f$$ and $$g$$ are two functions such that $$\displaystyle \left ( fg \right )\left ( x \right )=\left ( gf \right )\left ( x \right )$$ for all $$x$$. Then $$f $$ and $$g$$ may be defined as

A
$$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$$

B
$$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$$

C
$$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$$

D
$$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$$ where $$m, n$$ are unequal integers

Solution

A$$)f(x)=$$$$\sqrt{x},g(x)=cosx then (fg)(x) = \sqrt{\cos x} and (gf)(x) = \cos \sqrt{x}$$.They are not equal.

B)$$f(x)=x^{3},g(x)=x+1 then (fg)(x) = (x+1)^{3} and (gf)(x)= x^{3}+1$$.They are not equal.

C)$$f(x)=x1,g(x)=x^{2}+1 then (fg)(x)=x^{2} and (gf)(x)=x^{2}-2x$$. They are not equal.

D)$$f(x)=x^{m}, g(x)=x^{n} then (fg)(x)=x^{n+m} and (gf)(x)=x^{n+m}$$.Hence $$(fg)(x)= (gf)(x)$$
Question 3
1 / -0

If $$\displaystyle f(x)=x^{n},n\in N$$ and $$(gof)(x)=ng(x)$$ then $$g(x)$$ can be

A
$$n\:|x|$$

B
$$3.\sqrt[3]{x}$$

C
$$e^{x}$$

D
$$\log\:|x|$$

Solution

$$f(x)=x^{n}$$

$$g(f(x))=ng(x)$$ ...$$(i)$$

$$\log(f(x))=n\log(x)$$ ...$$(ii)$$

Taking $$\log(|x|)$$ as $$g(x)$$ the above expression is reduced to eq $$i$$.

Hence

$$g(x)=\log(|x|)$$.
Question 4
1 / -0

The domain of the function $$\displaystyle f\left ( x \right )= \sqrt{\sin ^{-1}\left ( 2x \right )+\frac{\pi }{6}}$$ for real $$x$$, is

A
$$\displaystyle \left [ -\frac{1}{4}, \frac{1}{2} \right ]$$

B
$$\displaystyle \left [ -\frac{1}{2}, \frac{1}{2} \right ]$$

C
$$\displaystyle \left ( -\frac{1}{2}, \frac{1}{9} \right )$$

D
$$\displaystyle \left [ -\frac{1}{4}, \frac{1}{4} \right ]$$

Solution

For real values of $$f\left(x\right)$$
$$\displaystyle \sin^{-1}\left(2x\right)+\frac{\pi}{6}\geq0$$
$$\displaystyle \sin^{-1}\left(2x\right)+\sin^{-1}\left(\frac{1}{2}\right)\geq0$$
$$\displaystyle \sin^{-1}\left(2x\right)\geq-\sin^{-1}\left(\frac{1}{2}\right)$$
$$\displaystyle 2x\geq \sin\left(-\sin^{-1}\left(\frac{1}{2}\right)\right)$$
$$\displaystyle 2x\geq-\frac{1}{2}$$
$$\displaystyle x\geq\frac{-1}{4}$$
Also
$$\displaystyle \sin^{-1}\left(x\right)$$ has a domain of $$\left[-1,1\right]$$
Hence for $$\displaystyle \sin^{-1}{2x}$$ it will be $$\left[\dfrac{-1}{2},\dfrac{1}{2}\right]$$
Hence the domain for $$\displaystyle f\left(x\right)$$ will be
$$\displaystyle \left[\frac{-1}{4},\frac{1}{2}\right]$$
Hence, option 'A' is correct.
Question 5
1 / -0

If $$\displaystyle f\left ( x \right )=\left\{\begin{matrix}
x^{2} x \geq 0\\
x x < 0
\end{matrix}\right.$$
then $$\displaystyle (f o f)(x)$$ is given by

A
$$x^{2}$$ for $$x\geq 0$$ and $$x$$ for $$ x< 0$$

B
$$\displaystyle x^{4}$$ for $$\displaystyle x\geq 0$$ and $$x^{2}$$ for $$x< 0$$

C
$$ \displaystyle x^{4}$$ for $$ \displaystyle x\geq 0$$ and $$-x^{2} $$ for $$x < 0$$

D
$$\displaystyle x^{4}$$ for $$ x\geq 0$$ and $$x $$ for $$ x< 0$$

Solution

For $$x\ge0$$,we have $$\displaystyle f \circ f\left( x \right)= {\left( {{x^2}} \right)^2}=\left( {{x^4}} \right)$$
For $$x<0$$,we have $$\displaystyle f \circ f\left( x \right)=x$$
Question 6
1 / -0

Let $$\displaystyle g(x)=1+x-[x]$$ and $$\displaystyle f(x)=\left\{\begin{matrix}{-1}\quad {x< 0} \\ {0} \quad {x=0}\\{1} \quad {x> 0} \end{matrix}\right.$$ Then for all $$\displaystyle x, f\left \{ g\left ( x \right ) \right \}$$ is equal to

A
$$x$$

B
$$1$$

C
$$\displaystyle f(x)$$

D
$$\displaystyle g(x)$$

Solution

Let $$g(x) = 1+x-[x] = 1+\{x\}> 0$$

since $$\{x\}\in [0,1) \forall x\in R$$

Hence $$f\{g(x)\} = 1$$
Question 7
1 / -0

Let $$\displaystyle f(x)=\frac{ax}{x+1}$$, where $$\displaystyle x\neq -1$$. Then for what value of $$\displaystyle a$$ is $$\displaystyle f( f(x))=x$$ always true

A
$$\displaystyle \sqrt{2}$$

B
$$\displaystyle -\sqrt{2}$$

C
$$1$$

D
$$-1$$

Solution

$$\displaystyle f\left( {f\left( x \right)} \right) = \dfrac{{a\dfrac{{ax}}{{x + 1}}}}{{\dfrac{{ax}}{{x + 1}} + 1}} = \dfrac{{\dfrac{{{a^2}x}}{{x + 1}}}}{{\dfrac{{ax + x + 1}}{{x + 1}}}} = \dfrac{{{a^2}x}}{{ax + x + 1}}$$

Since, $$\displaystyle f\left( {f\left( x \right)} \right) =x$$, we have,

$$\displaystyle\dfrac{{{a^2}x}}{{ax + x + 1}}=x$$.

Simplifying the equation we get,

$${a^2}x = \left( {a + 1} \right){x^2} + x$$

$$\therefore \left( {a + 1} \right){x^2} + \left( {1 - {a^2}} \right)x = 0$$

or $$\left( {a + 1} \right)x\left( {x + 1 - a} \right) = 0$$

Hence the only possible value is $$a=-1$$
Question 8
1 / -0

If $$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$$; $$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$$ then $$(fog)y$$ is equal to

A
$$\displaystyle \frac{y}{\sqrt{1-y^2}}$$

B
$$\displaystyle \frac{y}{\sqrt{1+y^2}}$$

C
$$y$$

D
$$2f(x)$$

Solution

Given $$\displaystyle f(y)=\frac{y}{\sqrt{1-y^2}}$$ and $$\displaystyle g(y)=\frac{y}{\sqrt{1+y^2}}$$
$$\therefore \displaystyle (fog)y = f(g(y))=\dfrac{\dfrac{y}{\sqrt{1+y^2}}}{\sqrt{1-\dfrac{y^2}{1+y^2}}}=y$$
Question 9
1 / -0

If $$\displaystyle f(x)= (x-1)+(x+1)$$ and
$$\displaystyle g(x)= f\left \{ f(x) \right \}$$ then $$\displaystyle {g}'(3)$$

A
equals $$1$$

B
equals $$0$$

C
equals $$3$$

D
equals $$4$$

Solution

Simplifying,$$f(x)$$ we get

$$f(x)=2x$$

Hence

$$f(f(x))=2(f(x))$$

$$=2(2x)$$

$$=4x$$

Hence

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

$$=4x$$

Thus

$$g'(x)=4$$

Hence $$g'(3)=4$$ .
Question 10
1 / -0

Let f(x)=tan x, x$$\displaystyle \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$$ and $$\displaystyle g\left (x \right )=\sqrt{1-x^{2}}$$ Determine $$g o f(1)$$.

A
1

B
0

C
-1

D
not defined

Solution

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Re-Attempt Weekly Quiz Competition

Functions Test 14

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Functions Test 14

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SHARING IS CARING

Question 1

a circle

an ellipse

a straight line

a pair of straight lines

Solution

Question 2

$$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$$

$$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$$

$$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$$

$$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$$ where $$m, n$$ are unequal integers

Solution

Question 3

$$n\:|x|$$

$$3.\sqrt[3]{x}$$

$$e^{x}$$

$$\log\:|x|$$

Solution

Question 4

$$\displaystyle \left [ -\frac{1}{4}, \frac{1}{2} \right ]$$

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

$$\displaystyle \left ( -\frac{1}{2}, \frac{1}{9} \right )$$

$$\displaystyle \left [ -\frac{1}{4}, \frac{1}{4} \right ]$$

Solution

Question 5

$$x^{2}$$ for $$x\geq 0$$ and $$x$$ for $$ x< 0$$

$$\displaystyle x^{4}$$ for $$\displaystyle x\geq 0$$ and $$x^{2}$$ for $$x< 0$$

$$ \displaystyle x^{4}$$ for $$ \displaystyle x\geq 0$$ and $$-x^{2} $$ for $$x < 0$$

$$\displaystyle x^{4}$$ for $$ x\geq 0$$ and $$x $$ for $$ x< 0$$

Solution

Question 6

$$x$$

$$1$$

$$\displaystyle f(x)$$

$$\displaystyle g(x)$$

Solution

Question 7

$$\displaystyle \sqrt{2}$$

$$\displaystyle -\sqrt{2}$$

$$1$$

$$-1$$

Solution

Question 8

$$\displaystyle \frac{y}{\sqrt{1-y^2}}$$

$$\displaystyle \frac{y}{\sqrt{1+y^2}}$$

$$y$$

$$2f(x)$$

Solution

Question 9

equals $$1$$

equals $$0$$

equals $$3$$

equals $$4$$

Solution

Question 10

1

0

-1

not defined

Solution

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