Differentiation Test 31

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Differentiation Test 31

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Weekly Quiz Competition

Question 1
1 / -0

Sum of the three digit numbers (no digit being zero) having the property that all digits are perfect squares, is

A
$$3108$$

B
$$6216$$

C
$$13986$$

D
$$None\ of\ these$$

Solution

Perfect square digits from $$1 $$ to $$9 $$ are - $$1,4,9$$

The 3 digit nos. having the property that all digits are perfect squares are,

$$149, 194, 491, 419, 914, 941$$

The sum is,

$$149+194+491+419+914+941=3108$$
Question 2
1 / -0

The number of ways in which $$6$$ rings can be worn on the four fingers of one hand is

A
$$4^{6}$$

B
$$^{6}C_{4}$$

C
$$6^{4}$$

D
None of these

Solution

Each ring can be worn on $$4$$ fingers, means each have $$4$$ possibilities.
$$\therefore$$ Total ways$$=4\times4\times4\times4\times4\times4=4^{6}$$
Hence, $$(A)$$
Question 3
1 / -0

Let $$\int _{ 0 }^{ x }{ \left( \cfrac { bt\cos { 4t } -a\sin { 4t } }{ { t }^{ 2 } } \right) } dt=\cfrac { a\sin { 4x } }{ x } $$ then $$a$$ and $$b$$ are given by

A
$$a=\cfrac { 1 }{ 4 } ,b=1$$

B
$$a=2,b=2$$

C
$$a=-1,b=4$$

D
$$a=2,b=4$$

Solution

$$\displaystyle \int_{0}^{x}{\dfrac{bt\cos 4t-a\sin 4t}{t^{2}}}dt=\dfrac{a\sin 4x}{x}$$
$$\text{Differentiating w.r.t. x}$$
$$ \dfrac{bx\cos 4x-a\sin 4x}{x^{2}}=\dfrac{4ax\cos 4x-a\sin 4x}{x^{2}}$$
$$b=4a \Rightarrow a=\dfrac{1}{4},b=1$$
Question 4
1 / -0

Directions for questions 1 to 3: Find the related word/ letters/numbers from given alternatives.
12:72::8:?

A
36

B
32

C
38

D
40

Solution

12 $$\times \, \frac{12}{2}$$ = 72 similarly, 8 $$\times \, \frac{8}{2}$$ = 32
Question 5
1 / -0

Select the missing number from the given alternatives.

A
110

B
115

C
121

D
54

Solution

48 $$\div$$ 2 = 24;
24 $$\times$$ 3 = 72: 72 $$\div$$ 2 = 36;
36 $$\div$$ 3 = 108; 108 $$\div$$ 2 = 54.
Question 6
1 / -0

Choose the correct answer alternatives given.
Select the missing number from the given alternatives.

A
18

B
21

C
5

D
17

Solution

80 = 8 $$\times$$ (8 + 2)
143 = 11 $$\times$$ (11 + 2)
323 = ? $$\times$$ (? + 2)
$$\therefore$$ ? = 17
Question 7
1 / -0

How many different $$4$$-person committees can be chosen form the $$100$$ members of the Senate ?

A
$$25$$

B
$$400$$

C
$$3,921,225$$

D
$$94,109,400$$

Solution

Total number of members $$=100$$.
Members has to be selected $$=4$$.
$$\therefore$$ different committees $$=^{100}C_{4}=\large{\frac{100!}{4!\times96!}}$$ $$=3,921,225$$.
Question 8
1 / -0

Arrange these numbers in ascending order.
$$756, 567, 657, 676$$

A
$$657, 567, 756, 676$$

B
$$567, 657, 676, 756$$

C
$$756, 676, 657, 567$$

D
$$676, 756, 567, 657$$

Solution

$$\Rightarrow$$ Numbers are said to be in ascending order when they are arranged from the smallest to the largest number.
$$\Rightarrow$$ The numbers which we have to arrange in ascending order are $$756,\,567,\,657$$ and $$676$$
$$\Rightarrow$$ $$567<657<676<756$$
$$\therefore$$ Ascending order $$=567,\,657,\,676,\,756$$
Question 9
1 / -0

If $$8f(x) + 6f\left( {{1 \over x}} \right) = x + 5$$ and $$y = {x^2}f(x)$$ ,then $${{dy} \over {dx}}$$ at x=-1 is equal to

A
0

B
$${1 \over {14}}$$

C
$$ - {1 \over {14}}$$

D
None of these

Solution

$$8{f(x)}+6{f\bigg(\dfrac{1}{x}\bigg)}=x+5\implies f(-1)=\dfrac{2}{7}$$
Differentiating on both sides
$$8{f'(x)}-\dfrac{6}{x^{2}}f'\bigg(\dfrac{1}{x}\bigg)=1$$
Put $$x=-1\implies f'(-1)=\dfrac{1}{2}$$
$$y=x^{2}f(x)$$
$$\dfrac{d{y}}{d{x}}=2{x}f(x)+x^{2}f'(x)$$
at $$x=-1,\dfrac{d{y}}{d{x}}=2(-1)\times \dfrac{2}{7}+(-1)^{2}\dfrac{1}{2}=-\dfrac{1}{14}$$
Question 10
1 / -0

If $$f:\mathrm{R}\to\mathrm{R}$$ is a differentiable function such that $$f'(x)\gt 2f(x)\forall x\in \mathrm{R}$$ and $$f(0)=1$$, then

A
$$f(x)$$ is increasing in $$(0,\infty)$$

B
$$f(x)$$ is decreasing in $$(0,\infty)$$

C
$$f(x)\gt e^{2x} $$ in $$(0,\infty)$$

D
$$f(x)\lt e^{2x} $$ in $$(0,\infty)$$

Solution

Given that,
$$f'\left( x \right) >2f\left( x \right) ,f\left( 0 \right) =1$$
$$\int { \cfrac { df\left( x \right) }{ f\left( x \right) } } >\int { 2dx } $$
$$\ln { f\left( x \right) } >2x+c$$
$$f\left( x \right) >k{ e }^{ 2x },\quad k>0$$
$$f\left( 0 \right) =1$$
$$1>k\quad \therefore 0<k<1$$
$$\therefore f\left( x \right) >k{ e }^{ 2x }>0$$
$$f'\left( x \right) >2f\left( x \right) >0$$
$$\therefore f\left( x \right) $$ is increasing function (since $$f'\left( x \right) >0$$) in $$\left( 0,\infty \right) $$

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Re-Attempt Weekly Quiz Competition

Differentiation Test 31

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Differentiation Test 31

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SHARING IS CARING

Question 1

$$3108$$

$$6216$$

$$13986$$

$$None\ of\ these$$

Solution

Question 2

$$4^{6}$$

$$^{6}C_{4}$$

$$6^{4}$$

None of these

Solution

Question 3

$$a=\cfrac { 1 }{ 4 } ,b=1$$

$$a=2,b=2$$

$$a=-1,b=4$$

$$a=2,b=4$$

Solution

Question 4

36

32

38

40

Solution

Question 5

110

115

121

54

Solution

Question 6

18

21

5

17

Solution

Question 7

$$25$$

$$400$$

$$3,921,225$$

$$94,109,400$$

Solution

Question 8

$$657, 567, 756, 676$$

$$567, 657, 676, 756$$

$$756, 676, 657, 567$$

$$676, 756, 567, 657$$

Solution

Question 9

0

$${1 \over {14}}$$

$$ - {1 \over {14}}$$

None of these

Solution

Question 10

$$f(x)$$ is increasing in $$(0,\infty)$$

$$f(x)$$ is decreasing in $$(0,\infty)$$

$$f(x)\gt e^{2x} $$ in $$(0,\infty)$$

$$f(x)\lt e^{2x} $$ in $$(0,\infty)$$

Solution

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