Relations and Functions Test

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Relations and Functions Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

if $$f\left( x \right) = 3x + 2$$ , $$g\left( x \right) = {x^2} + 1$$,then the values of $$\left( {f_og} \right)\left( {{x^2} - 1} \right)$$

A
$$3{x^4} - 6{x^2} + 8$$

B
$$3{x^4} + 3x + 4$$

C
$$6{x^4} + 3{x^2} - 2$$

D
$$6{x^4} + 3{x^2} + 2$$

Solution

$$f\left(x\right)=3x+2$$

$$g\left(x\right)={x}^{2}+1$$
$$f.g\left(x\right)=f\left({x}^{2}+1\right)$$
$$f.g\left(x\right)=3\left({x}^{2}+1\right)+2=3{x}^{2}+5$$

$$f.g\left(x\right)=3{x}^{2}+5$$
$$f.g\left({x}^{2}-1\right)=3{\left({x}^{2}-1\right)}^{2}+5$$

$$=3\left({x}^{4}-2{x}^{2}+1\right)+5$$

$$=3{x}^{4}-6{x}^{2}+3+5$$

$$=3{x}^{4}-6{x}^{2}+8$$
Question 2
1 / -0

Let A = {1,2,3,4,5} and B={1,2,3,4,5}. If $$f:A\rightarrow B$$ is an one-one function and $$f(x)=x$$ holds only for one value of $$x\epsilon \{ 1,2,3,4,5\} ,$$ then the number of such possible function is

A
$$120$$

B
$$36$$

C
$$45$$

D
$$44$$

Solution
Question 3
1 / -0

The function $$f :\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\rightarrow \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ defined by $$f(x)=\sin^{-1}(3x-4x^{3})$$ is

A
both one-one and onto

B
onto but not one-one

C
one-one but not onto

D
neither one-one nor onto

Solution

$$f:[-\cfrac{1}{2},\cfrac{1}{2}]\rightarrow[-\cfrac{\pi}{2},\cfrac{\pi}{2}]\\f(x)=\sin^{-1}(3x-4x^3)$$
Range of $$\sin^{-1}(3x-4x^3)$$ is $$[-\cfrac{\pi}{2},\cfrac{\pi}{2}]$$
And $$-\cfrac{\pi}{2}\le\sin^{-1}(3x-4x^3)\le\cfrac{\pi}{2}$$
$$\Rightarrow \sin(\cfrac{-\pi}{2})\le3x-4x^3\le\sin\cfrac{\pi}{2}$$
$$\Rightarrow-1\le3x-4x^3\le1$$
$$3x-4x^3\le-1$$
$$\Rightarrow 4x^3-3x-1\le0$$
$$\Rightarrow x\le\cfrac{1}{2}$$
and$$3x-4x^3\le1\Rightarrow 4x^3-3x+1\le0\\ \Rightarrow x\le-\cfrac{1}{2}\\ \therefore x\epsilon [-\cfrac{1}{2},\cfrac{1}{2}]\\f{x}=\sin^{-1}(3x-4x^3)\\ \Rightarrow f^1(x)=\cfrac{1}{\sqrt{1-(3x-4x^3)^2}}\times(3-12x^2)\\ \Rightarrow f^1(x)=3-12x^2\\ \quad=-(12x^2-3)\\x^2\le0\Rightarrow 12x^2\le0\Rightarrow 12x^2-3\le-3\\-3(12x^2-3)\le3\\ \therefore f^1(x)\le0$$
$$\therefore f^1(x)$$ is a decreasing function.
Hence $$f(x)$$ is one-one and range of function $$=$$ its co-domain.
Hence it is both one-one and onto.
Question 4
1 / -0

If $$f\left( x \right) = \frac{{x - 1}}{{x + 1}}$$, then $$f^{-1}\left( x \right)$$ is

A
$$\frac{{f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$

B
$$\frac{{3f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$

C
$$\frac{{f\left( x \right) + 3}}{{f\left( x \right) + 1}}$$

D
$$\frac{{f\left( x \right) + 3}}{{3f\left( x \right) + 1}}$$

Solution
Question 5
1 / -0

Let g be the inverse function of differentiable function f and $$G\left( x \right) =\frac { 1 }{ g\left( x \right) } if\quad f\left( 4=2 \right) $$ and $$f'\left( 4 \right) =\frac { 1 }{ 16 } $$, then the value of $${ \left( G'\left( 2 \right) \right) }^{ 2 }$$ equals to:

A
1

B
4

C
16

D
64

Solution
Question 6
1 / -0

If $$f:( - 1,1) \to B$$ , is a function defined by $$f(x) = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$$, then find $$B$$ when $$f(x)$$ is both one-one and onto function.

A
$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$

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

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

D
$$\left[ {0,\frac{\pi }{2}} \right)$$

Solution

For $$x \epsilon (-1,1)$$, we have
$$f(x)=\tan^{-1}\left[\dfrac {2x}{1-x^2}\right]$$
Substituting $$x=\tan\theta$$ in above equation.
Therefore, $$f(\tan \theta)=\tan^{-1}\left[\dfrac {2\tan \theta}{1-\tan^2\theta}\right]$$
$$=\tan^{-1}\tan (2\theta)=2\theta$$
$$=2\tan ^{-1}x$$
Thus $$-\dfrac {\pi}{2}<\tan ^{-1}\left[\dfrac {2x}{1-x^2}\right]<\dfrac {\pi}{2}$$
Thus option B is correct.
Question 7
1 / -0

Difference between the greatest and the least values of the function
$$f(x) = x(ln x - 2)$$ on $$[1, e^{2}]$$ is

A
$$2$$

B
$$e$$

C
$$e^{2}$$

D
$$1$$

Solution
Question 8
1 / -0

If $$f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)\ \forall x\ \epsilon \ R$$, then $$f(x)$$ is

A
one-one and onto

B
one-one and into

C
many-one and onto

D
non-invertible
Question 9
1 / -0

Let S be a non-empty set and P(S) be the power set of set S. Find the identity element for the union $$(\cup)$$ as a binary operation on $$P(S)$$.

A
$$\phi$$

B
$$1$$

C
$$0$$

D
None of these

Solution

We observe that
$$A \cup \phi = A = \phi \cup A$$ for every subset A of set S.
$$A \cup \phi = A =\phi \cup A$$ for all $$A \in P(S)$$
$$\phi$$ is the identity element for union $$(\cup)$$ on $$P(S)$$.
Question 10
1 / -0

If $$\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}$$ is not invertible, then $$k=$$

A
$$2$$

B
$$\dfrac{1}{2}$$

C
$$1$$

D
$$3$$

Solution

We have,
$$\begin{bmatrix} \sin { \left( \dfrac { \pi }{ 2 } \right) } & \cos { \left( \dfrac { \pi }{ 3 } \right) } \\ 2\tan { \left( \dfrac { \pi }{ 4 } \right) } & 2k \end{bmatrix}$$

Since, $$|A| = 0$$ if it is not invertible,
$$1(2k) - ((\cos 60) ( 2 \tan 45 )) = 0$$
$$2k - 1 = 0$$
$$k = \dfrac{1}{2}$$

Hence, this is the answer.

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 67

Result

Relations and Functions Test - 67

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SHARING IS CARING

Question 1

$$3{x^4} - 6{x^2} + 8$$

$$3{x^4} + 3x + 4$$

$$6{x^4} + 3{x^2} - 2$$

$$6{x^4} + 3{x^2} + 2$$

Solution

Question 2

$$120$$

$$36$$

$$45$$

$$44$$

Solution

Question 3

both one-one and onto

onto but not one-one

one-one but not onto

neither one-one nor onto

Solution

Question 4

$$\frac{{f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$

$$\frac{{3f\left( x \right) + 1}}{{f\left( x \right) + 3}}$$

$$\frac{{f\left( x \right) + 3}}{{f\left( x \right) + 1}}$$

$$\frac{{f\left( x \right) + 3}}{{3f\left( x \right) + 1}}$$

Solution

Question 5

1

4

16

64

Solution

Question 6

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

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

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

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

Solution

Question 7

$$2$$

$$e$$

$$e^{2}$$

$$1$$

Solution

Question 8

one-one and onto

one-one and into

many-one and onto

non-invertible

Question 9

$$\phi$$

$$1$$

$$0$$

None of these

Solution

Question 10

$$2$$

$$\dfrac{1}{2}$$

$$1$$

$$3$$

Solution

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