Functions Test 44

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Functions Test 44

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following functions is/are injective map(s) ?

A
$$f(x)=x^2+2, x \in (-\infty,\infty)$$

B
$$f(x)=|x+2|, x \in [-2,\infty)$$

C
$$f(x)=(x-4)(x-5), x \in (-\infty,\infty)$$

D
$$f(x)=\dfrac{4x^2 + 3x -5}{4+3x-5x^2}, x\in(-\infty, \infty)$$

Solution

The function $$f(x)=x^2 + 2, x \in (-\infty, \infty)$$ is not injective as
$$f(1)=f(-1) $$ but $$1 \neq -1.$$

The function $$f(x)=(x-4)(x-5), x \in (-\infty, \infty)$$ is not one-one as $$f(4)=f(5)$$ but $$4 \neq 5.$$

The functin $$f(x)=\dfrac{4x^2 + 3x -5}{4 + 3x - 5x^2}, x \in (-\infty,\infty)$$ is also not injective as $$f(1)=f(-1)$$ but $$1 \neq -1$$.

For the function , $$f(x)=-|x+2|, x\in [-2,\infty)$$

Let $$f(x)=f(y),x,y \in [-2,\infty) \Rightarrow |x+2| = |y+2| $$
$$\Rightarrow x+2 = y+2$$
$$\Rightarrow x=y$$
So , $$f$$ is an injective map.
Question 2
1 / -0

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

A
$$\displaystyle \left(-\frac{\pi}{4} ,\frac{\pi}{4}\right)$$

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

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

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

Solution

Let $$x=\tan(a)$$
Hence
$$\Rightarrow$$$$f(x)=\tan^{-1}\left(\dfrac{2\tan(a)}{1-\tan^2(a)}\right)$$
$$=\tan^{-1}(\tan(2a))$$

$$=2a$$

$$=2\tan^{-1}x$$

Now, since f is one-one and onto,
$$ -1 < x<1$$

$$\Rightarrow$$$$ 2\tan^{-1}(-1) < 2\tan^{-1}x < 2\tan^{-1}(1)$$

$$\displaystyle -\frac{\pi}{2} <f(x) < \frac{\pi}{2}$$

So, the range of $$\displaystyle \tan^{-1}(x)=\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$$.

So, $$B=\displaystyle \left(\frac{-\pi}{2},\frac{\pi}{2}\right)$$.
Question 3
1 / -0

The domain of the function $$f(x)=\log_3 \log_{1/3}(x^2+10x+25)+\dfrac {1}{[x]+5}$$ where [.] denotes the greatest integer function) is

A
$$(-4, -3)$$

B
$$(-6, -5)$$

C
$$(-6, 4)$$

D
None of these

Solution

For given function to be defined,
(i) $$x^2+10x+25>0\Rightarrow (x+5)^2>0\Rightarrow x\neq -5=D_1$$
(ii) $$\log_{1/3}(x^2+10x+25)>0\Rightarrow x^2+10x+25<1$$
$$\Rightarrow x^2+10x+24<0\Rightarrow (x+4)(x+6)<0\Rightarrow x\in(-6,-4)=D_2$$
and
(iii) $$[x]+5\neq 0\Rightarrow x\notin [-5,4)=D_3$$
Hence domain of f is $$D_!\cap D_2\cap D_3 =(-6,-5)$$
Question 4
1 / -0

If $$f:R\rightarrow \left [\dfrac {\pi}{6}, \dfrac {\pi}{2}\right ), f(x)=\sin^{-1}\left (\dfrac {x^2-a}{x^2+1}\right )$$ is a onto function, then set of values of $$a$$ is

A
$$\left \{-\dfrac {1}{2}\right \}$$

B
$$\left [-\dfrac {1}{2}, -1\right )$$

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

D
none of these

Solution

Given, $$f(x)$$ is onto.
$$\therefore \displaystyle \frac {\pi}{6}\leq \sin^{-1}\left (\frac {-x^2-a}{x^2+1}\right ) < \frac {\pi}{2}$$
$$\Rightarrow \displaystyle \frac {1}{2}\leq \frac {x^2-a}{x^2+1} < 1$$
$$\Rightarrow \displaystyle \frac {1}{2}\leq 1-\frac {(a+1)}{x^2+1} < 1, \forall x\epsilon R $$
$$\Rightarrow a+1 > 0$$
$$\Rightarrow a\in (-1, \infty)$$
Hence, (c) is the correct answer.
Question 5
1 / -0

Which one of the following functions is not one-one?

A
$$f:(-1,\infty )\rightarrow R$$ given by $$ f(x)={ x }^{ 2 }+2x\quad $$

B
$$g:(2,\infty )\rightarrow R$$ given by $$g(x)={ e }^{ { x }^{ 3 }-3x+2 }\quad $$

C
$$h:R\rightarrow R$$ given by $$h(x)={ 2 }^{ { x }(x-1) }\quad $$

D
$$\phi :(-\infty ,0)\rightarrow R$$ given by $$\phi (x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } $$

Solution

$$ x(x-1)$$ is not a one -one function in whole $$R$$.
This function has a minima at $$x=\dfrac12$$, therefore repeats value after $$x=\dfrac12$$.
We can also see that $$x(x-1) = 0$$ for $$x=0,1$$ and therefore $$e^{x(x-1)} =1$$ for both $$x=0 \&\ 1$$, which rules it out as a one to one function.
Question 6
1 / -0

Let f: $$X\rightarrow Y$$ be a function defined by $$f(x)=a \sin \left (x+\dfrac {\pi}{4}\right )+b \cos x+c$$. If f is both one-one and onto, then find the sets $$X$$ and $$Y$$

A
$$X\in \left [-\dfrac {\pi}{2}-\alpha, \dfrac {\pi}{2}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha= \tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

B
$$X\in \left [-\dfrac {\pi}{1}-\alpha, \dfrac {\pi}{1}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha=\tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

C
$$X\in \left [-\dfrac {\pi}{4}-\alpha, \dfrac {\pi}{4}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha=\tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

D
$$X\in \left [-\dfrac {\pi}{8}-\alpha, \dfrac {\pi}{8}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha=\tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

Solution

$$f\left( x \right) =a\sin { \left( x+\dfrac { \pi }{ 4 } \right) } +b\cos { x } +c\\ =\dfrac { a }{ \sqrt { 2 } } \sin { x } +\dfrac { a }{ \sqrt { 2 } } \cos { x } +b\cos { x } +c\\ =\dfrac { a }{ \sqrt { 2 } } \sin { x } +\dfrac { a+b\sqrt { 2 } }{ \sqrt { 2 } } \cos { x } +c\\ =r\cos { \alpha } \sin { x } +r\sin { \alpha } \cos { x } +c\quad \left( \because \quad r=\sqrt { { a }^{ 2 }+{ b }^{ 2 }+\sqrt { 2 } ab } \quad and\quad \tan { \alpha } =\frac { a+b\sqrt { 2 } }{ a } \right) \\ =r\sin { \left( x+\alpha \right) } +c$$
Since $$\quad f\left( x \right) \quad$$ is a one-one onto function, $$\sin { \left( x+\alpha \right) }$$ values must not be repetitive.
$$\therefore \quad -\dfrac { \pi }{ 2 } \quad \le x+\alpha \le \dfrac { \pi }{ 2 } \quad$$ and $$\quad c-r\le f\left( x \right) \le c+r$$
Question 7
1 / -0

$$f(x)=x^3+3x^2+4x+b \sin x+c \cos x, \forall x\in R$$ is a one-one function, then the value of $$b^2+c^2$$ is

A
$$\geq 1$$

B
$$\geq 2$$

C
$$\leq 1$$

D
none of these

Solution

Here, $$f(x)=x^3+3x^2+4x+b \sin x+c \cos x$$
$$f'(x)=3x^2+6x+4+b \cos x-c \sin x$$
Now, for $$f(x)$$ to be one-one, the only possibility is
$$f'(x)\geq 0, \forall x\in R$$
ie, $$3x^2+6x+4+b \cos x-c \sin x\geq 0, \forall x\in R$$
ie, $$3x^2+6x+4 \geq c \sin x-b \cos x, \forall x\in R$$
As we are approximating lower bound for the function by a number instead of a function, hence we chose the number.
ie, $$3x^2+6x+4 \geq \sqrt {b^2+c^2}, \forall x\in R$$
ie, $$\sqrt {b^2+c^2} \leq 3(x^2+2x+1)+1,\forall x\in R$$
$$\sqrt {b^2+c^2} \leq 3(x+1)^2+1, \forall x\in R$$
$$\sqrt {b^2+c^2} \leq 1, \forall x \in R$$
$$\Rightarrow b^2+c^2 \leq 1, \forall x \in R$$
Hence, (c) is the correct answer.
Question 8
1 / -0

If $$f(x)=2x+|x|, g(x)=\dfrac {1}{3}(2x-|x|)$$ and $$h(x)=f(g(x))$$, then domain of $$\sin^{-1}\underset {\text {n times}}{\underbrace {(h(h(h(h.....h(x).....))))}}$$ is

A
$$[-1, 1]$$

B
$$\left [-1, -\dfrac {1}{2}\right ]\cup \left [\dfrac {1}{2}, 1\right ]$$

C
$$\left [-1, -\dfrac {1}{2}\right ]$$

D
$$\left [\dfrac {1}{2}, 1\right ]$$

Solution

Since, $$f(x)=\left\{\begin{matrix}2x+x, & x\geq 0\\ 2x-x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}3x, & x\geq 0\\ x, & x < 0\end{matrix}\right.$$
and $$g(x)=\dfrac {1}{3}\left\{\begin{matrix}2x-x, & x\geq 0\\ 2x+x, & x < 0\end{matrix}\right.=\left\{\begin{matrix}\dfrac {x}{3}, & x\geq 0\\ x, & x < 0\end{matrix}\right.$$
$$\therefore f(g(x))=\left\{\begin{matrix}3\left (\dfrac {x}{3} \right ) & x\geq0\\ x & x < 0\end{matrix}\right.$$
$$\Rightarrow f(g(x))=x, \forall x\in R$$
$$\therefore h(x)=x$$
$$\Rightarrow \sin^{-1} (h(h(h....h(x)....)))=\sin^{-1}x$$
$$\therefore$$ Domain of $$\sin^{-1} (h(h(h(h.....h(x)....))))$$ is $$[-1, 1]$$
Hence, A is the correct answer.
Question 9
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The function $$f$$ is one to one and the sum of all the intercepts of the graph is $$5$$. The sum of all the intercept of the graph $$\displaystyle y = f^{-1} \left ( x \right )$$ is

A
$$5$$

B
$$\dfrac15$$

C
$$\dfrac25$$

D
$$-5$$

Solution

Since the function $$f$$ is one-one there exist only one $$x$$-intercept and only one $$y$$-intercept for the function.
Let $$a$$ be the $$y$$-intercept of the function $$f$$. Hence $$f\left( 0 \right) = a$$, but then we have $${f^{ - 1}}\left( a \right) = 0$$.
Therefore $$a$$ is an $$x$$-intercept of the function $${f^{ - 1}}$$.
Similarly the $$x$$-intercept of the function$$f$$ is $$y$$-intercept of the function $${f^{ - 1}}$$, say $$b$$.
Hence the sum of the intercepts of the function $$f$$ is same as the sum of the intercepts of the function $${f^{ - 1}}$$ which is equal to 5.
Question 10
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Let $$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right. $$ where $$[\cdot]$$ denotes the greatest integer function. Then $$\displaystyle f\left \{f(-2.3) \right\}$$ is equal to

A
$$4$$

B
$$2$$

C
$$-3$$

D
$$3$$

Solution

$$f\left( x \right) =\left\{ \begin{matrix} 1+|x|,\; x<-1 \\ \left[ x \right] ,\; x\ge -1 \end{matrix} \right. $$
.

$$f\left( -2.3 \right) =1+\left| -2.3 \right| =3.3\dots ( \because x < -1$$)
.

hence, $$f\left( f\left( -2.3 \right) \right) =f\left( 3.3 \right) =\left[ 3.3 \right] =3$$

$$f\left( f\left( -2.3 \right) \right)=3$$

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Functions Test 44

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Functions Test 44

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Question 1

$$f(x)=x^2+2, x \in (-\infty,\infty)$$

$$f(x)=|x+2|, x \in [-2,\infty)$$

$$f(x)=(x-4)(x-5), x \in (-\infty,\infty)$$

$$f(x)=\dfrac{4x^2 + 3x -5}{4+3x-5x^2}, x\in(-\infty, \infty)$$

Solution

Question 2

$$\displaystyle \left(-\frac{\pi}{4} ,\frac{\pi}{4}\right)$$

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

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

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

Solution

Question 3

$$(-4, -3)$$

$$(-6, -5)$$

$$(-6, 4)$$

None of these

Solution

Question 4

$$\left \{-\dfrac {1}{2}\right \}$$

$$\left [-\dfrac {1}{2}, -1\right )$$

$$(-1, \infty)$$

none of these

Solution

Question 5

$$f:(-1,\infty )\rightarrow R$$ given by $$ f(x)={ x }^{ 2 }+2x\quad $$

$$g:(2,\infty )\rightarrow R$$ given by $$g(x)={ e }^{ { x }^{ 3 }-3x+2 }\quad $$

$$h:R\rightarrow R$$ given by $$h(x)={ 2 }^{ { x }(x-1) }\quad $$

$$\phi :(-\infty ,0)\rightarrow R$$ given by $$\phi (x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } $$

Solution

Question 6

$$X\in \left [-\dfrac {\pi}{2}-\alpha, \dfrac {\pi}{2}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha= \tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

$$X\in \left [-\dfrac {\pi}{1}-\alpha, \dfrac {\pi}{1}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha=\tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

$$X\in \left [-\dfrac {\pi}{4}-\alpha, \dfrac {\pi}{4}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha=\tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

$$X\in \left [-\dfrac {\pi}{8}-\alpha, \dfrac {\pi}{8}-\alpha \right ]$$ and $$Y\in [c-r, c+r]$$ where $$\alpha=\tan^{-1}\left (\dfrac {a+b\sqrt 2}{a}\right )$$ and $$r=\sqrt {a^2+\sqrt 2ab+b^2}$$

Solution

Question 7

$$\geq 1$$

$$\geq 2$$

$$\leq 1$$

none of these

Solution

Question 8

$$[-1, 1]$$

$$\left [-1, -\dfrac {1}{2}\right ]\cup \left [\dfrac {1}{2}, 1\right ]$$

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

$$\left [\dfrac {1}{2}, 1\right ]$$

Solution

Question 9

$$5$$

$$\dfrac15$$

$$\dfrac25$$

$$-5$$

Solution

Question 10

$$4$$

$$2$$

$$-3$$

$$3$$

Solution

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