Functions Test 45

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Functions Test 45

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Question 1
1 / -0

Directions For Questions

Let $$f :R\rightarrow R$$ is a quadratic function, $$f(x) = ax^{2} + bx + c$$ where $$a, b, c$$ are non zero real numbers and it satisfies certain properties
(i)$$f:\left [ 0,2 \right ]\rightarrow \left [ 0,2 \right ]$$ is bijective
(ii)$$f:\left [ \alpha,2 + \alpha\right ]\rightarrow \left [ 0,2 \right ]$$ is bijective for some non -zero real value of $$\alpha$$.

...view full instructions

$$l= \lim_{x\rightarrow \alpha}\displaystyle \frac{f(x)}{x(x-\alpha)(x-2)}$$ is

A
positive

B
negative

C
$$0$$

D
sign of $$l$$ depends upon $$\alpha\pi$$

Solution

For the quadratic function $$f(x) = ax^2 + bx +c$$, it's given that $$a, b, c$$ are non-zero.

Hence, $$f(0) = c \neq 0$$

It's given that $$f:[0,2] \rightarrow [0,2]$$ is bijective.

Since $$f(0) \neq 0$$, $$f(0) = 2$$ for it to be bijective.

$$ \Rightarrow f(2) = 0$$

So, three cases possible as shown in the figure, which will satisfy the second condition as well.

For cases 1 and 2,

$$ \displaystyle \lim_{x\rightarrow \alpha}\displaystyle \frac{f(x)}{x(x-\alpha)(x-2)}$$

$$ \displaystyle =\lim _{ x\rightarrow \alpha } \frac { 1 }{ x\left( x-2 \right) } \lim _{ x\rightarrow \alpha } \frac { f(x) }{ x-\alpha } \\ \displaystyle =\frac { 1 }{ \alpha \left( \alpha -2 \right) } \lim _{ x\rightarrow \alpha } \frac { f'(x) }{ 1 } =\frac { f'(\alpha ) }{ \alpha \left( \alpha -2 \right) } $$

$$f'(\alpha)$$ is positive for both cases. $$ \alpha( \alpha -2)$$ is also positive for both cases. Hence, the limit value is also positive.

For the case in figure-3,

$$ \displaystyle \lim _{ x\rightarrow \alpha } \frac { f(x) }{ x(x-\alpha )(x-2) } =\lim _{ x\rightarrow \alpha } \frac { 1 }{ x } \lim _{ x\rightarrow \alpha } \frac { f(x) }{ { \left( x-\alpha \right) }^{ 2 } } \\ \displaystyle =\frac { 1 }{ \alpha } \lim _{ x\rightarrow \alpha } \frac { f'(x) }{ 2\left( x-\alpha \right) } =\frac { 1 }{ \alpha } \lim _{ x\rightarrow \alpha } \frac { f''(x) }{ 2 } =\frac { 1 }{ \alpha } \frac { f''\left( \alpha \right) }{ 2 } $$

$$f''(\alpha) > 0$$ since the graph is facing upwards. Also $$ \alpha$$ is positive.

Hence, in this case also , the value of limit is positive. Hence, option A is correct.
Question 2
1 / -0

The domain of the function $$\displaystyle y=\underbrace { \log _{ 10 } \log _{ 10 } ...\log _{ 10 } x }_{ \text{ n times } } $$ is

A
$$[10^{n},+\infty)$$

B
$$(10^{n-1},+\infty)$$

C
$$(10^{n-2},+\infty)$$

D
none of these

Solution

$$\displaystyle y=\underbrace { \log _{ 10 } \log _{ 10 } .........\log _{ 10 } x }_{ \text{ n times } } $$

$$\underbrace { \log _{ 10 } \log _{ 10 } .........\log _{ 10 } x }_{ \text{ n-1 times } }>0$$

$$\underbrace { \log _{ 10 } \log _{ 10 } .........\log _{ 10 } x }_{ \text{ n-2 times } }>10^0$$

$$\underbrace { \log _{ 10 } \log _{ 10 } .........\log _{ 10 } x }_{ \text{ n-3 times } }>10^1$$

$$\underbrace { \log _{ 10 } \log _{ 10 } .........\log _{ 10 } x }_{ \text{ n-4 times } }>10^{10}$$

similarly

$$\displaystyle x>10^{10^{10^{10...(n-2) times}}}$$

Hence, option D.
Question 3
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If $$\displaystyle f \left ( x \right ) = px + q$$ and $$\displaystyle f \left ( f\left ( f\left ( x \right ) \right ) \right ) = 8x + 21$$, where $$p$$ and $$q$$ are real numbers, the $$ p + q$$ equals

A
$$3$$

B
$$5$$

C
$$7$$

D
$$11$$

Solution

$$f(x)=px+q$$
$$f(f(x))=p(px+q)+q$$
$$=p^2x+q(p+1)$$
$$f(f(f(x)))=p(p^2x+q(p+1))+q$$
$$=p^{3}x+q(p(p+1)+1)$$
$$=8x+21$$
By comparing coefficients, we get
$$p^{3}=8$$
$$p=2$$
And $$q(2(3)+1)=21$$
$$7q=21$$
$$q=3$$
Hence
$$f(x)=2x+3$$
Therefore, $$p+q=2+3=5$$
Question 4
1 / -0

Directions For Questions

Let $$\displaystyle f(x) = x^2 - 2x - 1 \: \forall \: x \: \in \: R$$. Let $$\displaystyle f: (-\infty, a] \rightarrow [b, \infty)$$, where $$a$$ is the largest real number for which $$f(x)$$ is bijective.

...view full instructions

The value of $$(a + b)$$ is equal to

A
$$-2$$

B
$$-1$$

C
$$0$$

D
$$1$$

Solution

$$x^{2}-2x-1$$ is an upwards facing parabola. Therefore, it has a point of minimum $$=-\dfrac{b}{2a}=\dfrac12\times$$ Sum of roots
After the point of minimum the curve slopes up and the function no longer remain one-one. Hence the largest interval for it to be injecvtive is upto the point of minimum.
The roots of the quadratic are,
$$\rightarrow x=1+\sqrt{2} $$ and $$x=1-\sqrt{2}$$
So $$a=\dfrac{1+\sqrt{2}+1-\sqrt{2}}{2}$$
$$=1$$
$$f(1)=(1-1)^{2}-2$$
$$=-2$$
$$=b$$
Hence $$a+b$$
$$=1-2$$
$$=-1$$
Question 5
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The domain of the function $$\displaystyle f\left( x \right)=\sin ^{ -1 }{ \left\{ \log _{ 2 }{ \left( \frac { 1 }{ 2 } { x }^{ 2 } \right) } \right\} } $$ is

A
$$[-2,-1) \cup \left[ 1,2 \right] $$

B
$$(-2,-1]\cup \left[ 1,2 \right] $$

C
$$\left[ -2,-1 \right] \cup \left[ 1,2 \right] $$

D
$$\left( -2,-1 \right) \cup \left( 1,2 \right) $$

Solution

For $$f(x)=\displaystyle \sin ^{ -1 }{ \left\{ \log _{ 2 }{ \left( \frac { 1 }{ 2 } { x }^{ 2 } \right) } \right\} } $$ to be defined, we must have
$$\displaystyle -1\le \log _{ 2 }{ \left( \frac { 1 }{ 2 } { x }^{ 2 } \right) } \le 1\Rightarrow { 2 }^{ -1 }\le \frac { 1 }{ 2 } { x }^{ 2 }\le { 2 }^{ 1 }\Rightarrow 1\le { x }^{ 2 }\le 4$$ ...(1) $$[\because$$ the base $$=2>1]$$
Now, $$1\le { x }^{ 2 }\Rightarrow { x }^{ 2 }-1\ge 0\Rightarrow \left( x-1 \right) \left( x+1 \right) \ge 0\Rightarrow x\le -1$$ or $$x\ge 1$$ ...(2)
Also, $${ x }^{ 2 }\le 4\Rightarrow { x }^{ 2 }-4\le 0\Rightarrow \left( x-2 \right) \left( x+2 \right) \le 0\Rightarrow -2\le x\le 2$$ ...(3)
From (2) and (3), we get the domain of $$f$$
$$=\left( (-\infty ,-1]\cup [1,\infty ) \right) \cap \left[ -2,2 \right] =\left[ -2,-1 \right] \cup \left[ 1,2 \right] $$
Question 6
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$$K(x)$$ is a function such that $$K(f(x))=a+b+c+d$$,
Where,
$$a=\begin{cases}
0 & \text{ if f(x) is even} \\
-1 & \text{ if f(x) is odd} \\
2 & \text{ if f(x) is neither even nor odd}
\end{cases}$$
$$b=\begin{cases}
3 & \text{ if f(x) is periodic} \\
4 & \text{ if f(x) is aperiodic}
\end{cases}$$
$$c=\begin{cases}
5 & \text{ if f(x) is one one} \\
6 & \text{ if f(x) is many one}
\end{cases}$$
$$d=\begin{cases}
7 & \text{ if f(x) is onto} \\
8 & \text{ if f(x) is into}
\end{cases}$$
$$h:R\rightarrow R,h(x)=\left ( \displaystyle \frac{e^{2x}+e^{x}+1}{e^{2x}-e^{x}+1} \right )$$

On the basis of above information, answer the following questions.$$K(\phi(x)) $$

A
$$15$$

B
$$16$$

C
$$17$$

D
$$18$$

Solution

$$\phi(x)\rightarrow \left(\dfrac{-\pi}{2},\dfrac{\pi}{2}\right)$$
Now we know that for the given domain, $$\tan(x)$$ is one-one, periodic, odd and an into function.
Hence $$f(\phi(x))$$
$$=-1+3+5+8$$
$$=16-1$$
$$=15$$
Question 7
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Let $$f:{x, y, z}\rightarrow (a, b, c)$$ be a one-one function. It is known that only one of the following statements is true:
(i) $$f(x)\neq b$$
(ii)$$f(y)=b$$
(iii)$$f(z)\neq a$$

A
$$f=\{(x, a), (y, b), (z, c)\}$$

B
$$f=\{(x, b), (y, a), (z, c)\}$$

C
$$f=\{(x, b), (y, c), (z, c)\}$$

D
$$f=\{(x, b), (y, c), (z, a)\}$$

Solution

When (i) is true, then $$f(x) \neq b, f(y) \neq b , f(z) = a $$

$$\Rightarrow$$ Two ordered pair function is possible $$ f(x) = a, f(y) = c, f(z) = a$$ or $$f(x) = c, f(y) = a, f(z) = a$$

But given $$f$$ is one-one and only one such function is possible. Hence (i) can't be true.

When (ii) is true, then $$f(y) = b, f(z) =a , f(x) = b$$. This is also not possible.

Clearly if (iii) is true then it is satisfying every condition.
Hence ordered pair of $$f$$ is $$\{(x,a), (y,b), (z,c)\}$$
Question 8
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The domain of function $$\displaystyle f(x)=\sqrt{x-\sqrt{1-x^{2}}}$$ is

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

B
$$\displaystyle [-1,1]$$

C
$$\displaystyle \left ( -\infty,-\frac{1}{2} \right ]\cup \left [ \frac{1}{\sqrt{2}},+\infty \right )$$

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

Solution

$$\displaystyle f(x)=\sqrt{x-\sqrt{1-x^{2}}}$$

$$ x-\sqrt{1-x^{2}} \geq 0 $$ and $$ 1 -x^{2}\geq0 $$

$$\Rightarrow x\geq\sqrt{1-x^{2}}$$ and $$ -1\leq x \leq1$$ ... (i)

Now, $$ x^2\geq(\sqrt{1-x^{2}})^2$$, for positive $$x$$

$$x\leq-\dfrac{1}{\sqrt{2}} $$ or $$ x\geq\dfrac{1}{\sqrt{2}}$$ ... (ii)

From (i) and (ii), we get

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

Hence, option 'D' is correct.
Question 9
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The domain of the function $$\displaystyle f(x)=\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}$$ is

A
$$\displaystyle \left \{ x|x< 1 \right ) $$

B
$$\displaystyle \left \{ x|x> -1 \right \}$$

C
$$[0,1]$$

D
$$[-1,1]$$

Solution

For real range of $$f(x)$$

$$1-x^{2}\geq 0$$

$$x^2\leq 1$$

$$|x|\leq 1$$

$$-1 \leq x \leq 1$$

$$x\in [-1,1]$$

Hence the domain for the above function is $$[-1,1]$$
Question 10
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Let $$\displaystyle f(x)=\sin^{2}\dfrac{x}{2}+\cos^{2}\frac{x}{2}$$ and $$g(x)=\sec^{2}x-\tan^{2}x.$$ The two functions are equal over the set

A
$$\Phi$$

B
$$R$$

C
$$\displaystyle R-\left \{ x|x=(2n+1)\frac{\pi}{2} ,n\:\in Z\right \}$$

D
None of these

Solution

We know,
$$\displaystyle f(x)=sin^{2}\frac{x}{2}+cos^{2}\frac{x}{2}=1 $$ for all values of $$x$$ and $$\:g(x)=sec^{2}x-tan^{2}x=1 $$ except $$ x=(2n+1)\dfrac{\pi}{2} where n \epsilon Z$$

Therefore, two functions are equal in the interval:

$$\displaystyle R-\left \{ x|x=(2n+1)\frac{\pi}{2} ,n\:\epsilon Z\right \}$$

Hence, option 'C' is correct.

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Re-Attempt Weekly Quiz Competition

Functions Test 45

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Functions Test 45

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SHARING IS CARING

Question 1

positive

negative

$$0$$

sign of $$l$$ depends upon $$\alpha\pi$$

Solution

Question 2

$$[10^{n},+\infty)$$

$$(10^{n-1},+\infty)$$

$$(10^{n-2},+\infty)$$

none of these

Solution

Question 3

$$3$$

$$5$$

$$7$$

$$11$$

Solution

Question 4

$$-2$$

$$-1$$

$$0$$

$$1$$

Solution

Question 5

$$[-2,-1) \cup \left[ 1,2 \right] $$

$$(-2,-1]\cup \left[ 1,2 \right] $$

$$\left[ -2,-1 \right] \cup \left[ 1,2 \right] $$

$$\left( -2,-1 \right) \cup \left( 1,2 \right) $$

Solution

Question 6

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 7

$$f=\{(x, a), (y, b), (z, c)\}$$

$$f=\{(x, b), (y, a), (z, c)\}$$

$$f=\{(x, b), (y, c), (z, c)\}$$

$$f=\{(x, b), (y, c), (z, a)\}$$

Solution

Question 8

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

$$\displaystyle [-1,1]$$

$$\displaystyle \left ( -\infty,-\frac{1}{2} \right ]\cup \left [ \frac{1}{\sqrt{2}},+\infty \right )$$

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

Solution

Question 9

$$\displaystyle \left \{ x|x< 1 \right ) $$

$$\displaystyle \left \{ x|x> -1 \right \}$$

$$[0,1]$$

$$[-1,1]$$

Solution

Question 10

$$\Phi$$

$$R$$

$$\displaystyle R-\left \{ x|x=(2n+1)\frac{\pi}{2} ,n\:\in Z\right \}$$

None of these

Solution

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