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 $\dfrac{1}{2}$ .

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Relations and Functions Test - 52

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Relations and Functions Test - 52

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Question 1

an equivalence relation

reflexive and transitive but not symmetric

reflexive and transitive but not transitive

symmetric and transitive but nor reflexive

Solution

Question 2

(0,1e)\left(0, \dfrac{1}{e} \right)(0,e1​)

(1e,e)\left(\dfrac{1}{e}, e \right)(e1​,e)

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

None of these

Solution

Question 3

[a, b)

[a, b]

(a, b]

(a, b)

Solution

Question 4

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

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

(3,3),(3,5),(5,3),(5,5)(3,3),(3,5),(5,3),(5,5)(3,3),(3,5),(5,3),(5,5)

(3,3),(3,4),(4,5)(3,3),(3,4),(4,5)(3,3),(3,4),(4,5)

Solution

Question 5

613x23(1+x1−x)136^{\frac{1}{3}}{x^{\frac{2}{3}}}{\left( {\frac{{1 + x}}{{1 - x}}} \right)^{\frac{1}{3}}}631​x32​(1−x1+x​)31​

x13(1−x1+x)13{x^{\frac{1}{3}}}{\left( {\frac{{1 - x}}{{1 + x}}} \right)^{\frac{1}{3}}}x31​(1+x1−x​)31​

x23(1−x1+x)13{x^{\frac{2}{3}}}{\left( {\frac{{1 - x}}{{1 + x}}} \right)^{\frac{1}{3}}}x32​(1+x1−x​)31​

x(1+x1−x)13x{\left( {\frac{{1 + x}}{{1 - x}}} \right)^{\frac{1}{3}}}x(1−x1+x​)31​

Solution

Question 6

444

222

−2-2−2

111

Solution

Question 7

7n2\dfrac{{7n}}{2}27n​

7(n+1)2\dfrac{{7\left( {n + 1} \right)}}{2}27(n+1)​

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

7n(n+1)2\dfrac{{7n\left( {n + 1} \right)}}{2}27n(n+1)​

Solution

Question 8

(0,∞)\left( {0,\infty } \right)(0,∞)

R−(−1,1)R - \left( { - 1,1} \right)R−(−1,1)

(−∞, ∞)\left( { - \infty ,\,\infty } \right)(−∞,∞)

N.O.T

Solution

Question 9

1100\frac{1}{100}1001​

100

does not exist

0

Solution

Question 10

2/72/72/7

1/21/21/2

222

7/27/27/2

Solution

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

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

$\left(0, e \right)$

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

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

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

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

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

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

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

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

$4$

$2$

$-2$

$1$

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

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

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

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

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

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

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

$\frac{1}{100}$

$2/7$

$1/2$

$2$

$7/2$