Relations and Functions Test

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Relations and Functions Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f(x)=\cot ^{ -1 }{ \left( \cfrac { 1-{ x }^{ 2 } }{ 2x } \right) } +\cot ^{ -1 }{ \left( \cfrac { 1-3{ x }^{ 2 } }{ 3x-{ x }^{ 3 } } \right) } -\cot ^{ -1 }{ \left( \cfrac { 1-6{ x }^{ 2 }+{ x }^{ 4 } }{ 4x-4{ x }^{ 3 } } \right) } $$, the $$F'(x)$$ equals

A
$$\cfrac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

B
$$\cfrac { -1 }{ 1+{ x }^{ 2 } } $$

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

D
$$\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

Solution

$$F(x)=\tan ^{ -1 }{ \left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) } +\tan ^{ -1 }{ \left( \cfrac { 3x-{ x }^{ 3 } }{ 1-3{ x }^{ 2 } } \right) } -\tan ^{ -1 }{ \left( \cfrac { 4x-4{ x }^{ 3 } }{ 1-6{ x }^{ 2 }+{ x }^{ 4 } } \right) } $$
$$=2\tan ^{ -1 }{ x } +3\tan ^{ -1 }{ x } -4\tan ^{ -1 }{ x }
\\=\tan ^{ -1 }{ x } $$
$$\Rightarrow F'(x)=\cfrac { 1 }{ 1+{ x }^{ 2 } } $$
Question 2
1 / -0

If $$f(x)=\left| x \right| ,x\in R$$, then

A
$$f(x)=\left( f\times f \right) \left( x \right) $$

B
$$f(x)=x$$

C
$$f(x)=\left( f\times f \right) \left( x^2 \right) $$

D
$$f(x)=\left( f\circ f \right) \left( x \right) $$

Solution
Question 3
1 / -0

If $$g(x)={ x }^{ 2 }+x-2$$ and $$\cfrac { 1 }{ 2 } (g\circ f)(x)=2{ x }^{ 2 }-5x+2$$, then $$f(x)$$ is

A
$$2x-3$$

B
$$2x+3$$

C
$$2{ x }^{ 2 }+3x+1$$

D
$$2{ x }^{ 2 }+3x-1$$

Solution

$$\cfrac { 1 }{ 2 } \left( g\circ f \right) (x)=2{ x }^{ 2 }-5x+2$$

$$\Rightarrow \cfrac { 1 }{ 2 } g\left[ f(x) \right] =2{ x }^{ 2 }-5x+2\quad $$

$$\left[ { \left\{ f(x) \right\} }^{ 2 }+\left\{ f(x) \right\} -2 \right] =2\left[ 2{ x }^{ 2 }-5x+2 \right] \quad \left[ \because g(x)={ x }^{ 2 }+x-2 \right] $$

$$\Rightarrow f{ \left( x \right) }^{ 2 }+f(x)-\left( 4{ x }^{ 2 }-10x+6 \right) =0$$ ....represents quadratic equation

We have a formula for solving quadratic equation $$ax^2+bx+c=0$$ is

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$\therefore f(x)=\cfrac { -1\pm \sqrt { 1+4\left( 4{ x }^{ 2 }-10x+6 \right) } }{ 2 } $$

$$=\cfrac { -1\pm(4x-5) }{ 2 } =2x-3$$ or $$2-2x$$
Question 4
1 / -0

If $$(ax^2 + bx + c)y +a'x^2+b'x+c=0$$, then the condition that x may be a rational function of y is

A
$$(ac'-a'c)^2=(ab'-a'b)(bc'-b'c)$$

B
$$(ab'-a'b)^2=(ab'-a' c)(bc'-b'c)$$

C
$$(bc'-b'c)^2=(ab'-a'b)(ac'-a'c)$$

D
None of these

Solution
Question 5
1 / -0

If $$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) } $$ and $$g\left( \cfrac { 5 }{ 4 } \right) =1$$, then $$g\circ f(x)$$ is equal to

A
$$0$$

B
$$1$$

C
$$\sin { { 1 }^{ o } } $$

D
None of these

Solution

Given, $$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) } $$
$$=\sin ^{ 2 }{ x } +\left( \sin { x } \cos { \cfrac { \pi }{ 3 } } +\cos { x } \sin { \cfrac { \pi }{ 3 } } \right) +\cos { x } \left( \cos { \cfrac { \pi }{ 3 } } -\sin { x } \sin { \cfrac { \pi }{ 3 } } \right) $$
$$=\sin ^{ 2 }{ x } +\cfrac { \sin ^{ 2 }{ x } }{ 4 } +\cfrac { 3\cos ^{ 2 }{ x } }{ 4 } +\cfrac { 2\sqrt { 3 } }{ 2.2 } \sin { x } \cos { x } +\cfrac { \cos ^{ 2 }{ x } }{ 2 } -\cos { x } \sin { x } \cfrac { \sqrt { 3 } }{ 2 } $$
$$=\cfrac { 5 }{ 4 } \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) =\cfrac { 5 }{ 4 } \quad $$
Therefore, $$ g\circ f(x)=g(f(x))=g\left( \cfrac { 5 }{ 4 } \right) =1\quad \quad $$
Question 6
1 / -0

If $$f(x)=ax+b $$ and $$g(x)=cx+d$$, then $$f\left( g(x) \right) =g\left( f(x) \right) \Leftrightarrow$$

A
$$f(a)=g(c)$$

B
$$f(b)=g(b)$$

C
$$f(d)=g(b)$$

D
$$f(x)=g(a)$$

Solution

We have $$f(x)=ax+b,g(x)=cx+d$$
Therefore, $$ f\left\{ g(x) \right\} =g\left\{ f(x) \right\} \Leftrightarrow f(cx+d)=g(ax+b)$$
$$\Leftrightarrow a(cx+d)+b=c(ax+b)+d$$
$$\Leftrightarrow ad+b=cb+d\Leftrightarrow f(d)=g(b)$$
Question 7
1 / -0

The inverse of the function $$f(x) = \log (x^{2} + 3x + 1), x\epsilon [1, 3]$$, assuming it to be an onto function, is

A
$$\dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$$

B
$$\dfrac {-3 \pm \sqrt {5 + 4e^{x}}}{2}$$

C
$$\dfrac {-3 - \sqrt {5 + 4e^{x}}}{2}$$

D
None of the above

Solution

Given, $$f(x) = \log (x^{2} + 3x + 1)$$
Therefore, $$ f'(x) = \dfrac {2x + 3}{(x^{2} + 3x + 1)} > 0\forall x\epsilon [1, 3]$$
which is a strictly increasing function. Thus, $$f(x)$$ is injective, given that $$f(x)$$ is onto. Hence, the given function $$f(x)$$ is invertible.
Now, $$f\left \{f^{-1}(x)\right \} = x$$
$$\Rightarrow \log \left \{f^{-1}(x)\right \}^{2} + 3\left \{f^{-1}(x) + 1 \right \} = x$$
$$\Rightarrow \left \{f^{-1}(x)\right \}^{2} + 3\left \{f^{-1}(x)\right \} + 1 - e^{x} = 0$$
Thus $$ f^{-1}(x) = \dfrac {-3\pm \sqrt {9 - 4.1 (1 - e^{x})}}{2}$$
$$= \dfrac {-3\pm \sqrt {5 + 4e^{x}}}{2}$$
$$= f^{-1}(x) = \dfrac {-3\pm \sqrt {5 + 4e^{x}}}{2} [\because f^{-1}(x) \epsilon [1, 3]]$$
Hence, $$f^{-1}(x) = \dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$$
Question 8
1 / -0

If $$\ast$$ is defined by $$a\ast b = a - b^{2}$$ and $$\oplus$$ is defined by $$a\oplus b = a^{2} + b$$, where $$a$$ and $$b$$ are integers, then $$(3\oplus 4)\ast 5$$ is equal to

A
$$164$$

B
$$38$$

C
$$-12$$

D
$$-28$$

E
$$144$$

Solution

Given, $$a\ast b = a - b^{2} $$ .... (i)

and $$a\oplus b = a^{2} + b$$ .... (ii)

where, $$a$$ and $$b$$ are integers

Then, $$(3\oplus 4) \ast 5 = \left \{(3)^{2} + 4\right \}\ast 5$$

$$= (9 + 4) \ast 5 = 13\ast 5$$

$$= 13 - (5)^{2} = 13 - 25 = -12$$
Question 9
1 / -0

If $$f\left( x \right) $$ and $$g\left( x \right) $$ are two functions with $$g\left( x \right) =x-\dfrac { 1 }{ x } $$ and $$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } $$, then $$f^{ ' }\left( x \right) $$ is equal to

A
$$3{ x }^{ 2 }+3$$

B
$${ x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } } $$

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

D
$$3{ x }^{ 2 }+\dfrac { 3 }{ { x }^{ 4 } } $$

Solution

$$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } $$
Writing $${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } $$ using $${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right) $$, we have
$${ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } -3x\cdot \dfrac { 1 }{ x } \left( x-\dfrac { 1 }{ x } \right) $$
$$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right) $$
We have,
$$f\left( g\left( x \right) \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right) $$
As $$g\left( x \right) =x-\dfrac { 1 }{ x } $$, this yields
$$f\left( x-\dfrac { 1 }{ x } \right) ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right) $$
On putting $$x-\dfrac { 1 }{ x } =t$$, we get
$$f\left( t \right) ={ t }^{ 3 }+3t$$
Thus, $$f\left( x \right) ={ x }^{ 3 }+3x$$
and $$f^{ ' }\left( x \right) =3{ x }^{ 2 }+3$$
Question 10
1 / -0

If $$f:R\rightarrow R$$, $$g:R\rightarrow R$$ are defined by$$ f(x)=5x-3$$, $$g(x)=x^{2}+3$$, then $$(gof^{-1})(3)$$=

A
$$\dfrac{25}{3}$$

B
$$ \dfrac{111}{25}$$

C
$$\dfrac{9}{25} $$

D
$$\dfrac{25}{111} $$

Solution

$$f:R\rightarrow R$$
$$ g:R\longrightarrow R$$
$$ f\left( x \right) =5x-3$$
$$g\left( x \right) ={ x }^{ 2 }+3$$
$$\left(gof^{-1}\right)\left(3\right)$$
$$f\left(x\right)={5}{x}-{3}$$
Let $$f\left(x\right)=y$$ $$\Rightarrow{x}=f^{-1}\left(y\right)$$
$$f\left(x\right)=y={5}x-3$$
$$\Rightarrow x=\dfrac{y+3}{5}$$
$$\Rightarrow f^{-1}\left(y\right)=\dfrac{y+3}{5}$$
$$ f^{-1}\left(x\right)=\dfrac{x+3}{5}$$ and $$g\left(x\right)=x^{2}+{3}$$
$$\left(gof^{-1}\right)\left(3\right)=g\left[f^{-1}\left(3\right)\right]=g\left(\dfrac{3+3}{5}\right)=g\left(\dfrac{6}{5}\right)$$
$$=\left[\left(\dfrac{6}{5}\right)^{2}+3\right]=\dfrac{36}{25}+{3}=\dfrac{111}{25}$$
Hence, $$\dfrac{111}{25}$$ is the correct answer.

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 61

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Relations and Functions Test - 61

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SHARING IS CARING

Question 1

$$\cfrac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

$$\cfrac { -1 }{ 1+{ x }^{ 2 } } $$

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

$$\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$

Solution

Question 2

$$f(x)=\left( f\times f \right) \left( x \right) $$

$$f(x)=x$$

$$f(x)=\left( f\times f \right) \left( x^2 \right) $$

$$f(x)=\left( f\circ f \right) \left( x \right) $$

Solution

Question 3

$$2x-3$$

$$2x+3$$

$$2{ x }^{ 2 }+3x+1$$

$$2{ x }^{ 2 }+3x-1$$

Solution

Question 4

$$(ac'-a'c)^2=(ab'-a'b)(bc'-b'c)$$

$$(ab'-a'b)^2=(ab'-a' c)(bc'-b'c)$$

$$(bc'-b'c)^2=(ab'-a'b)(ac'-a'c)$$

None of these

Solution

Question 5

$$0$$

$$1$$

$$\sin { { 1 }^{ o } } $$

None of these

Solution

Question 6

$$f(a)=g(c)$$

$$f(b)=g(b)$$

$$f(d)=g(b)$$

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

Solution

Question 7

$$\dfrac {-3 + \sqrt {5 + 4e^{x}}}{2}$$

$$\dfrac {-3 \pm \sqrt {5 + 4e^{x}}}{2}$$

$$\dfrac {-3 - \sqrt {5 + 4e^{x}}}{2}$$

None of the above

Solution

Question 8

$$164$$

$$38$$

$$-12$$

$$-28$$

$$144$$

Solution

Question 9

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

$${ x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } } $$

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

$$3{ x }^{ 2 }+\dfrac { 3 }{ { x }^{ 4 } } $$

Solution

Question 10

$$\dfrac{25}{3}$$

$$ \dfrac{111}{25}$$

$$\dfrac{9}{25} $$

$$\dfrac{25}{111} $$

Solution

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