Functions Test 16

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Functions Test 16

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Weekly Quiz Competition

Question 1
1 / -0

Given $$\displaystyle f\left ( x \right )=\log \left ( \frac{1+x}{1-x} \right )$$ and $$\displaystyle g\left ( x \right )=\frac{3x+x^{3}}{1+3x^{2}}, fog (x)$$ equals

A
$$-f(x)$$

B
$$3f(x)$$

C
$$\displaystyle \left [ f\left ( x \right ) \right ]^{3}$$

D
none of these

Solution

Given $$f(x)\, =\, \log\, \left( \displaystyle \dfrac{1\, +\, x}{1\, -\,

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

+\, 3x^{2}}$$

$$\therefore fog (x)\, =\, \log \left( \displaystyle \dfrac{1\, +\, \dfrac{3x\, +\,

x^{3}}{1\, +\, 3x^{2}}}{1\, -\,\displaystyle \dfrac{3x\, +\, x^{3}}{1\,

+\, 3x^{2}}} \right)$$

$$=\, \log\, \left(\displaystyle \dfrac{(1\, +\,

x)^{3}}{(1\, -\, x)^{3}} \right)\, =\, 3\, \log\, \left(\displaystyle

\dfrac{1\, +\, x}{1\, -\, x} \right)\, =\, 3\, f(x)$$
Question 2
1 / -0

The domain of definition of the function $$\log \cos x$$ is

A
$$\displaystyle \left \{ x;n\pi -\frac{\pi}2< x< n\pi +\dfrac{\pi}2 \right \}, n \in I.$$

B
$$\displaystyle \left \{ x;2n\pi -\dfrac{\pi}2< x< 2n\pi +\dfrac{\pi}2 \right \}, n \in I.$$

C
$$\displaystyle \left \{ x;2n\pi -\pi < x< 2n\pi +\pi \right \}, n \in I.$$

D
$$\displaystyle \left \{ x;2n\pi -\dfrac{\pi}4< x< 2n\pi +\dfrac{\pi}4 \right \}, n \in I.$$

Solution

For given function to be defined $$\displaystyle \cos x$$ must be +ive and $$\displaystyle \cos x\neq 0,

i.e. x\neq \dfrac{\pi}2$$ or $$\displaystyle -\dfrac{\pi}2.$$

Also we know that $$\displaystyle \cos x$$ is +ive in 1st and 4th quadrants.

$$\displaystyle \therefore $$ Domain $$\displaystyle = \left

\{ x;2n\pi -\dfrac{\pi}2< x< 2n\pi +\dfrac{\pi}2 \right \}, n \epsilon I.$$
Question 3
1 / -0

A function $$R$$ on the set $$N$$ of natural numbers is defined as $$\displaystyle R= \left \{ \left ( 2n,2n+1 \right ):n \epsilon N \right \}$$ The domain of $$R$$ is

A
$$\displaystyle \left \{ 1,2,3,4,5,.... \right \}$$

B
$$\displaystyle \left \{ 2,4,6,8,.... \right \}$$

C
$$\displaystyle \left \{ 1,3,5,7.... \right \}$$

D
all integers

Solution

Given function is $$\displaystyle R= \left \{ \left ( 2n,2n+1 \right ):n \epsilon N \right \}$$

Domain of $$R=2n$$ where $$n\epsilon N$$

$$\Rightarrow$$ Domain $$=\displaystyle \left \{ 2,4,6,8,.... \right \}$$

Hence, option B.
Question 4
1 / -0

If $$\displaystyle y= \sin ^{-1}\frac{x-1}{x+1}+\log \left ( 2-x \right )$$, then its domain is:

A
$$(1, 2)$$

B
$$(-1, 2)$$

C
$$[0, 2)$$

D
none

Solution

$$\displaystyle y= \sin ^{-1}\frac{x-1}{x+1}+\log \left ( 2-x \right )$$
For this function to be defined $$\displaystyle -1\leq \frac{x-1}{x+1}\leq 1 \mbox{and} 2-x > 0$$
$$\Rightarrow \displaystyle \frac{x-1}{x+1} \leq 1, \frac{x-1}{x+1} > -1 \mbox{and} x-2 <0 0$$
$$\Rightarrow \displaystyle \frac{-2}{x+1} \leq 0, \frac{2x}{x+1} > 0 \mbox{and} x<2 $$
$$\Rightarrow \displaystyle x> -1, x\in (-\infty,-1)\cup (0,\infty), \mbox{and} x< 2$$
Thus taking intersection of above range of $$y$$ is $$x \in (0,2)$$
Question 5
1 / -0

If $$\displaystyle A= \left \{ x:-1\leq x\leq 1 \right \}=B.$$ Discuss the following function w.r.t one-one onto bijective and write their characteristics.
$$\displaystyle f\left ( x \right )=\frac{x}{2}$$

A
one one, into

B
one one, onto

C
many one, into

D
many one, onto

Solution

$$f(x)=\dfrac{x}{2}$$

$$f(x_{1})=f(x_{2})$$

Implies
$$\dfrac{x_{1}}{2}=\dfrac{x_{2}}{2}$$

Or
$$x_{1}=x_{2}$$
Hence
$$f(x_{1})=f(x_{2})$$ implies $$x_{1}=x_{2}$$
Hence $$f(x)$$ is a one one function.
Now
$$y=\dfrac{x}{2}$$
Or
$$2y=x$$
For $$x\epsilon R$$, $$y\epsilon R$$.
Hence
$$f:R\rightarrow R$$.
Thus there exists atleast one value of x for an element 'y' in the co domain Y.
That is , all the elements in the co domain are mapped by the function.
Hence the function is onto.
Purely linear (straight line ) functions are always one-one and onto.
Question 6
1 / -0

If $$\left[ \log _{ 10 }{ x } \right] =1$$, then $$x$$ lies in which of the following interval
(Here $$[. ]$$ is the greatest integer less than or equal to the number.)

A
$$[10,100)$$

B
$$(10,20)$$

C
$$(100,\infty)$$

D
none of these

Solution

$$\quad \left[ \log _{ 10 }{ x } \right] =1$$

$$\Rightarrow \quad 1\le \log _{ 10 }{ x } <2\quad$$

$$ \Rightarrow \quad 10\le x<100$$

Therefore, $$x\in [10,100)$$

Ans: A
Question 7
1 / -0

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are functions defined by $$f(x)=3x-1; g(x)=\sqrt{x+6}$$, then the value of $$(g\circ f^{-1})(2009)$$ is

A
$$26$$

B
$$29$$

C
$$16$$

D
$$15$$

Solution

Given $$f(x)=3x-1, g(x)=\sqrt{x+6}$$
Let $$f(x)=y$$
$$ \Rightarrow y=3x-1\Rightarrow y+1-3x\Rightarrow x=\cfrac{y+1}{3}$$
Now, $$ f^{-1}(y)=x=\cfrac{y+1}{3}$$
$$\Rightarrow f^{-1}(x)=\cfrac{x+1}{3}$$ and $$g(x)=\sqrt {x+6}$$
Consider, $$(g\circ f^{-1})(2009)=g[{f^{-1}(2009)}]$$
$$ =g\left(\cfrac{2009+1}{3}\right)=g\left(\cfrac{2010}{3}\right)$$
$$=\sqrt{\cfrac{2010}{3}+6}=\sqrt{\cfrac{2028}{3}}=\sqrt{676}=26$$
Hence, option $$A$$ is correct.
Question 8
1 / -0

If $$\displaystyle y= 2^{-x}+\cos ^{-1}\left ( \frac{x}{2}-1 \right )+\log \sqrt{x-\left [ x \right ]},$$then its domain is given by

A
$$\displaystyle \left ( 0, 4 \right )-\left \{ 1 \right \}$$

B
$$\displaystyle \left ( 0, 4 \right )-\left \{ 1, 2 \right \}$$

C
$$\displaystyle \left ( 0, 4 \right )-\left \{ 1, 2, 3 \right \}$$

D
none

Solution

Given function is, $$\displaystyle y= 2^{-x}+\cos ^{-1}\left ( \frac{x}{2}-1 \right )+\log \sqrt{x-\left [ x \right ]}$$

For this function to be defined $$\displaystyle -1\leq \frac{x}{2}-1\leq 1 \,\,{and}\,\, x-[x] > 0$$

$$\displaystyle 0 \leq \frac{x}{2}\leq 2 \,\, {and} \,\, x-[x] > 0 $$

this is always true except at integral values of $$x$$

$$\Rightarrow x \in \left(0,4\right)-\{1,2,3\}$$
Question 9
1 / -0

Let $$f(x)=x^{2}-2x$$ and $$g(x)=f(f(x)-1)+f(5-f(x)),$$ then

A
$$g(x)<0,\forall x\in R$$

B
$$g(x)<0$$ for some $$x\in R$$

C
$$g(x)\leq 0$$ for some $$x\in R$$

D
$$g(x)\geq 0,\forall x\in R$$

Solution

Given, $$ f(x) = x^2 - 2x $$
Now, $$ f(f(x) - 1) \\ = f(x^2 - 2x - 1 ) \\ = (x^2 - 2x - 1 )^2 - 2 (x^2 - 2x - 1 ) $$
And, $$ f(5 - f(x) ) \\ = f(5 - x^2 + 2x ) \\ = (5 - x^2 + 2x )^2 - 2 (5 - x^2 + 2x ) $$
Hence, $$ g(x) = f(f(x) - 1) + f (5 - f(x)) \\ \therefore g(x) = (x^2 - 2x - 1)^2 - 2 (x^2 - 2x -1 ) + (5 - x^2 + 2x )^2 - 2(5 - x^2 + 2x ) \\ \Rightarrow g(x) = (x^2 - 2x - 1)^2 + (5- x^2 + 2x)^2 - 2(x^2 - 2x - 1 + 5 - x^2 + 2x) \\ \therefore g(x) = (x^2 + 2x - 1 )^2 + ( 5 - 2x ^2 + 2 x^2)^2 - 8$$
Since the first two terms are in square,
$$\therefore $$ it can not be negative and if x = 9 then we also get positive value.
$$\therefore g(x) \geq 0 \, \, \forall x \in R $$
$$ \therefore $$ option D is correct
Question 10
1 / -0

The domain of the function $$f(x)\, =\, \displaystyle \frac{1}{\log_{10}(1\, -\, x)}\, +\, \sqrt{x\, +\, 2}$$, is

A
$$[-\, 2, \, 0)\, \cup\, (0,\, 1) $$

B
$$[-\, 2, \, 0)\, \cup\, (0,\, 1 ]$$

C
$$(-\, 2, \, 0)\, \cup\, (0,\, 1 ]$$

D
$$(-\, 2, \, 0)\, \cup\, [0,\, 1 ]$$

Solution

Given,
$$f(x)\, =\, \displaystyle \frac{1}{\log_{10}(1\, -\, x)}\, +\, \sqrt{x\, +\, 2}$$

$$f(x)$$ will be defined if, $$1-x> 0,\neq 1$$ and $$x+2\geq 0$$

Or $$x< 1, \neq 0$$ and $$x\geq -2$$

Combining these conditions we get, $$x\in [-2,0)\cup (0,1)$$

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Functions Test 16

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Functions Test 16

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Question 1

$$-f(x)$$

$$3f(x)$$

$$\displaystyle \left [ f\left ( x \right ) \right ]^{3}$$

none of these

Solution

Question 2

$$\displaystyle \left \{ x;n\pi -\frac{\pi}2< x< n\pi +\dfrac{\pi}2 \right \}, n \in I.$$

$$\displaystyle \left \{ x;2n\pi -\dfrac{\pi}2< x< 2n\pi +\dfrac{\pi}2 \right \}, n \in I.$$

$$\displaystyle \left \{ x;2n\pi -\pi < x< 2n\pi +\pi \right \}, n \in I.$$

$$\displaystyle \left \{ x;2n\pi -\dfrac{\pi}4< x< 2n\pi +\dfrac{\pi}4 \right \}, n \in I.$$

Solution

Question 3

$$\displaystyle \left \{ 1,2,3,4,5,.... \right \}$$

$$\displaystyle \left \{ 2,4,6,8,.... \right \}$$

$$\displaystyle \left \{ 1,3,5,7.... \right \}$$

all integers

Solution

Question 4

$$(1, 2)$$

$$(-1, 2)$$

$$[0, 2)$$

none

Solution

Question 5

one one, into

one one, onto

many one, into

many one, onto

Solution

Question 6

$$[10,100)$$

$$(10,20)$$

$$(100,\infty)$$

none of these

Solution

Question 7

$$26$$

$$29$$

$$16$$

$$15$$

Solution

Question 8

$$\displaystyle \left ( 0, 4 \right )-\left \{ 1 \right \}$$

$$\displaystyle \left ( 0, 4 \right )-\left \{ 1, 2 \right \}$$

$$\displaystyle \left ( 0, 4 \right )-\left \{ 1, 2, 3 \right \}$$

none

Solution

Question 9

$$g(x)<0,\forall x\in R$$

$$g(x)<0$$ for some $$x\in R$$

$$g(x)\leq 0$$ for some $$x\in R$$

$$g(x)\geq 0,\forall x\in R$$

Solution

Question 10

$$[-\, 2, \, 0)\, \cup\, (0,\, 1) $$

$$[-\, 2, \, 0)\, \cup\, (0,\, 1 ]$$

$$(-\, 2, \, 0)\, \cup\, (0,\, 1 ]$$

$$(-\, 2, \, 0)\, \cup\, [0,\, 1 ]$$

Solution

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