Relations and Functions Test

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Relations and Functions Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\quad cos(\log { \quad x } )$$, then
$$f({ x }^{ 2 })f({ y }^{ 2 }) -\frac { 1 }{ 2 } \left[ f(\frac { { x }^{ 2 } }{ 2 } )+f(\frac { { x }^{ 2 } }{ { y }^{ 2 } } ) \right] $$ has the value :

A
-2

B
-1

C
1/2

D
None of these

Solution
Question 2
1 / -0

If $$f\left( x+\frac { 1 }{ 2 } \right) +f\left( x-\frac { 1 }{ 2 } \right) =f\left( x \right) ,\quad \forall x\in \quad R$$, then $$f\left( x-3 \right) +f\left( x+3 \right)$$ is equal to

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

B
$$2f\left( x \right)$$

C
$$0$$

D
$$6f\left( x \right)$$

Solution
Question 3
1 / -0

$$f(x) = \frac{{1 - x}}{{1 + x}}$$ for $$x \ne - 1$$ then $$f(x) + \left( {\frac{1}{x}} \right)$$ is $$(for\,\,x \ne 0)$$

A
$$0$$

B
$$1/2$$

C
$$1$$

D
$$-1$$
Question 4
1 / -0

Let $$A$$ be set of first ten natural numbers and $$R$$ be a relation on $$A$$ defined by (x , y) $$\in$$ $$R$$ $$\Rightarrow$$ x + 2y = 10 , then domain of $$R$$ is

A
$$\left \{ 1 , 2 , 3 ,............, 10 \right \}$$

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

C
$$\left \{ 1 , 2 , 3 , 4 \right \}$$

D
$$\left \{ 2 , 4 , 6 , 8 , 10 \right \}$$

Solution

$$\begin{aligned} A &=\{1,2,3, \cdots \cdots, 10\} \\ R &=\{(x, y): x+2 y=10, x, y \in A\} \\ & \therefore y=\dfrac{10-x}{2} \\ R &=\{(2,4),(4,3),(6,2),(8,1)\} \\ \therefore & \text { Domain of } R \text { is }\{2,4,6,8\} \end{aligned}$$
Correct option is (B).
Question 5
1 / -0

If $$f(x)=x^2-5x+6$$, find $$f(A)$$ if $$A=\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$$.

A
$$\begin{bmatrix} 1 & -1 & -3 \\ 1 & 1 & 10 \\ -5 & 4 & -4 \end{bmatrix}$$

B
$$\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ 5 & -4 & 4 \end{bmatrix}$$

C
$$\begin{bmatrix} 1 & -1 & -3 \\ 1 & -1 & 10 \\ -5 & -4 & -4 \end{bmatrix}$$

D
$$\begin{bmatrix} -1 & -1 & 3 \\ -1 & -1 & -10 \\ 5 & 4 & -4 \end{bmatrix}$$

Solution

$$f(x)=x^{2}-5 x+6$$
$$f(A)=A^{2}-5 A+6$$
$$A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$$
$$\begin{aligned} A^{2} &=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right] \\ &=\left[\begin{array}{ccc}4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0\end{array}\right] \\ \end{aligned}$$
$$=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]$$
$$5 A=\left[\begin{array}{lll}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right]$$
On Putting in $$f(A)$$
$$\begin{aligned} f(A) &=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-\left[\begin{array}{ccc}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right]+\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right] \\ &=\left[\begin{array}{ccc}5-10+6 & -1+0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2+0+6\end{array}\right] \\ &=\left[\begin{array}{ccc}1 & -1 & -3 \\-1 & -1 & -10\\ -5 & 4 & -4 \end{array}\right] \end{aligned}$$
Question 6
1 / -0

If $$ f ( x ) = x ^ { 3 } + x ^ { 2 } f ^ { \prime } ( 1 ) + x f ^ { \prime \prime } ( 2 ) + f ^ { \prime \prime \prime } ( 3 ) . $$ Then $$ f ( 2 ) $$ is

A
-2

B
1

C
30

D
7

Solution

$$\textbf{Step 1: Finding the derivative upto three orders.}$$
$$f(x)=x^{3}+x^{2}f^{'}(1)+xf^{''}(2)+xf^{'''}(3)$$
$$\text{Differentiating it with respect to x ,}$$
  $$f^{'}(x)=3x^{2}+2xf^{'}(1)+f^{''}(2)$$ $$\text{...(i)}$$
  $$\text{Differentiating (i) with respect to x ,}$$
  $$f^{''}(x)=6x+2f^{'}(1)$$ $$\text{...(ii)}$$
$$\text{Differentiating (ii) with respect to x ,}$$
  $$f^{'''}(x)=6$$ $$\text{...(iii)}$$
$$\textbf{Step 2: Solving for exact values.}$$
  $$\text{From (i),}$$
  $$f^{'}(1)=3+2f^{'}(1)+f^{''}(2)$$ $$\text{(Replacing x by 1 in (i))}$$
  $$\Rightarrow f^{'}(1)+f^{''}(2)+3=0$$ $$\text{...(iv)}$$
  $$\text{From (ii)}$$,
  $$f{''}(2)=12+2f{'}(1)$$ $$\text{(Replacing x by 2 in (ii))}$$
  $$\text{On substituting it in eq. (iv), we get,}$$
  $$-f^{'}(1) -3=12+2f^{'}(1)$$
  $$\Rightarrow 3f^{'}(1)=-15$$
$$\Rightarrow f^{'}(1)=-5$$    $$\text{...(v)}$$
$$\text{Putting the value of}$$ $$f^{'}(1)$$ $$\text{we get}$$ $$f{''}(2)=12+2f{'}(1),$$ $$\text{we get}$$,
$$f^{''}(2)=12+2 \times (-5)$$
  $$\Rightarrow f^{''}(2)=12-10=2$$    $$\text{...(vi)}$$
$$\textbf{Step 3: Substituting these values in given function.}$$
  $$\therefore$$ $$f(x)=x^{3}-5x^{2}+2x+6$$
$$\text{Putting x=2 in above equation,}$$
  $$\therefore$$ $$f(2)= 8-20+4+6=-2$$
$$\textbf{Hence the value of}$$ $$\mathbf{f(2)= -2}$$.
Question 7
1 / -0

If f(x) =$$\dfrac{1 - x}{1 + x} , then f [f (cos 2\theta)] =$$

A
tan 2$$\theta$$

B
sec 2$$|theta$$

C
cos 2$$\theta$$

D
cot 2$$\theta$$

Solution
Question 8
1 / -0

If f (x) = $$\dfrac{x -|x|}{|x|}, then f (-1) = $$

A
1

B
-2

C
0

D
+2

Solution
Question 9
1 / -0

If $$f\left( x \right) ={ x }^{ n }$$, then the value of $$f(1)-\dfrac { f'\left( 1 \right) }{ 1! } +\dfrac { f''\left( 1 \right) }{ 2! } -\dfrac { f'''\left( 1 \right) }{ 3! } +...+\dfrac { \left( -1 \right) ^{ n }{ f }^{ n }\left( 1 \right) }{ n! } $$ is

A
$${ 2 }^{ n }$$

B
$${ 2 }^{ n-1 }$$

C
0

D
1

Solution
Question 10
1 / -0

Let $$F(x)$$ be the primitive of $$\cfrac{3x+2}{\sqrt{x-9}}$$ with respect $$'x'$$. If $$F(10)=60$$. Then the sum of digits of the value of $$F(13)$$ is

A
6

B
13

C
24

D
26