Relations and Functions Test

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Relations and Functions Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

Let R be a relation on the set of integers given by $$ aRb \leftrightarrow a = 2^k .b $$ for some integer $$k.$$ Then $$R$$ is

A
an equivalence relation

B
reflexive and transitive but not symmetric

C
reflexive and transitive but not transitive

D
symmetric and transitive but nor reflexive

Solution
Question 2
1 / -0

Let $$g(x) = f(\log x) + f(2 - \log x)$$ and $$f''(x) < 0\forall x\in (0, 3)$$. Then find the interval in which $$g(x)$$ increases.

A
$$\left(0, \dfrac{1}{e} \right)$$

B
$$\left(\dfrac{1}{e}, e \right)$$

C
$$\left(0, e \right)$$

D
None of these

Solution

Since $$f^{ " }\left( x \right) <0,$$    $$f'\left( x \right) $$ is decreasing function $$g'\left( x \right) , \dfrac { f'\left( \log { x } \right) }{ x } +\dfrac { f'\left( 2-\log { x } \right) }{ -x } $$
$$\Rightarrow \dfrac{1}{x} \left( f'\left( \log { x } \right) -f'\left( 2-\log { x } \right) \right) $$
     $$g'\left( x \right)>0$$     ( increasing )
$$\Rightarrow \left( f'\left( \log { x } \right) >f'\left( 2-\log { x } \right) \right) $$
Since $$f'\left( x \right) $$ is decreasing
$$\Rightarrow \log x < 2-\log x$$
     $$2\log x<1$$
     $$\log x< \log e$$
     $$x<e$$
     $$x\in \left(0,e\right).$$
Hence, the answer is $$x\in \left(0,e\right).$$
Question 3
1 / -0

The domain of the function $$f(x) = \frac{x^2}{\sqrt{(a - x)(x-b)}}, (b > a)$$ is

A
[a, b)

B
[a, b]

C
(a, b]

D
(a, b)

Solution
Question 4
1 / -0

Let R be a relation from $$ A=\{1,2,3,4\}$$ to $$B=\{1,3,5\}$$such that R=[(a,b):a<b where a\in{A} and b\in{B}].What is $$RoR^{-1} $$ equal to

A
$$(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)$$

B
$$(3,1),(5,1),(3,2)(5,2),(5,3),(5,4)$$

C
$$(3,3),(3,5),(5,3),(5,5)$$

D
$$(3,3),(3,4),(4,5)$$

Solution
Question 5
1 / -0

COMPREHENSION :
If $${\left( {f\left( x \right)} \right)^2} \times f\left( {\frac{{1 - x}}{{1 + x}}} \right) = 6x,x \ne 0,1$$, then $$f(x)$$ is equal to

A
$$6^{\frac{1}{3}}{x^{\frac{2}{3}}}{\left( {\frac{{1 + x}}{{1 - x}}} \right)^{\frac{1}{3}}}$$

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

C
$${x^{\frac{2}{3}}}{\left( {\frac{{1 - x}}{{1 + x}}} \right)^{\frac{1}{3}}}$$

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

Solution
Question 6
1 / -0

Let $$f(-2, 2)\rightarrow(-2, 2)$$ be a continuous function given $$f(x)=f{(x}^{2})$$. Given $$f(0)=\dfrac{1}{2}$$ then the $$4f(\dfrac{1}{2})$$

A
$$4$$

B
$$2$$

C
$$-2$$

D
$$1$$

Solution
Question 7
1 / -0

If $$f:R \to R$$ satisfies $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$$ for all $$x,y \in R$$ and $$f\left( 1 \right) = 7$$, then $$\sum\limits_{r = 1}^n {f\left( r \right)} $$

A
$$\dfrac{{7n}}{2}$$

B
$$\dfrac{{7\left( {n + 1} \right)}}{2}$$

C
$$7n\left( {n + 1} \right)$$

D
$$\dfrac{{7n\left( {n + 1} \right)}}{2}$$

Solution

$$\because f(1)=7$$

$$\therefore f(2)=f(1+1)=f(1)+f(1)=7+7=14$$

Similarly $$f(3)=f(1+2)=f(1)+f(2)=7+14=21$$

$$f(4)=f(1+3)=f(1)+f(3)=7+21=28$$ and so on

$$\therefore f(n)=7+(n-1)7=7n$$

Now, $$\displaystyle \sum_{ r=1 }^{ n }{ f\left( r \right) } =f(1)+f(2)+f(3)+....+f(n)$$

$$=7+14+21+....+7n=7(1+2+3+....+n)$$

$$=\dfrac{7n(n+1)}{2}$$
Question 8
1 / -0

COMPREHENSION :
If $${\left( {f\left( x \right)} \right)^2} \times f\left( {\frac{{1 - x}}{{1 + x}}} \right) = 6x,x \ne 0,1$$, The domain of $$f(x)$$ is

A
$$\left( {0,\infty } \right)$$

B
$$R - \left( { - 1,1} \right)$$

C
$$\left( { - \infty ,\,\infty } \right)$$

D
N.O.T

Solution
Question 9
1 / -0

If $$f(x)=1+x+\frac{x^2}{2} +...+\frac{x^{100}}{100}$$, then f'(1) is equal to

A
$$\frac{1}{100}$$

B
100

C
does not exist

D
0

Solution
Question 10
1 / -0

Let $$3f(x)-2f\left(\dfrac {1}{x}\right)=x$$, then $$f'(2)$$ is equal to

A
$$2/7$$

B
$$1/2$$

C
$$2$$

D
$$7/2$$

Solution

Differentiate $$3f\left( x \right) - 2f\left( {\dfrac{1}{x}} \right) = x$$ with respect to $$x$$,

$$3f'\left( x \right) - 2f'\left( {\dfrac{1}{x}} \right)\left( { - \dfrac{1}{{{x^2}}}} \right) = 1$$

$$3f'\left( x \right) + \dfrac{2}{{{x^2}}}f'\left( {\dfrac{1}{x}} \right) = 1$$ (1)

Put $$x = 2$$ in equation (1),

$$3f'\left( 2 \right) + \dfrac{2}{{{2^2}}}f'\left( {\dfrac{1}{2}} \right) = 1$$

$$3f'\left( 2 \right) + \dfrac{1}{2}f'\left( {\dfrac{1}{2}} \right) = 1$$ (2)

Put $$x = \dfrac{1}{2}$$ in equation (1),

$$3f'\left( {\dfrac{1}{2}} \right) + \dfrac{2}{{{{\left( {\dfrac{1}{2}} \right)}^2}}}f'\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right) = 1$$

$$3f'\left( {\dfrac{1}{2}} \right) + \dfrac{2}{{\dfrac{1}{4}}}f'\left( 2 \right) = 1$$

$$3f'\left( {\dfrac{1}{2}} \right) + 8f'\left( 2 \right) = 1$$

$$3f'\left( {\dfrac{1}{2}} \right) = 1 - 8f'\left( 2 \right)$$

$$f'\left( {\dfrac{1}{2}} \right) = \dfrac{1}{3} - \dfrac{8}{3}f'\left( 2 \right)$$

Substitute above value in equation (2),

$$3f'\left( 2 \right) + \dfrac{1}{2}\left( {\dfrac{1}{3} - \dfrac{8}{3}f'\left( 2 \right)} \right) = 1$$

$$3f'\left( 2 \right) + \dfrac{1}{6} - \dfrac{4}{3}f'\left( 2 \right) = 1$$

$$3f'\left( 2 \right) - \dfrac{4}{3}f'\left( 2 \right) = 1 - \dfrac{1}{6}$$

$$\dfrac{{9f'\left( 2 \right) - 4f'\left( 2 \right)}}{3} = \dfrac{{6 - 1}}{6}$$

$$5f'\left( 2 \right) = \dfrac{5}{2}$$

$$f'\left( 2 \right) = \dfrac{1}{2}$$

Therefore, the value of $$f'\left( 2 \right)$$ is $$\dfrac{1}{2}$$.