Limits and Derivatives Test

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Limits and Derivatives Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

lf $$f(x)=\displaystyle \frac{x}{\sqrt{1-x^{2}}},g(x)=\frac{x}{\sqrt{1+x^{2}}}$$, then $$\displaystyle \frac{d}{dx}(fog (x))=$$

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$2$$

Solution

$$\displaystyle f(g(x)) = \frac{g(x)}{\sqrt{1-(g(x))^2}}$$ provided $$g(x) \neq 1$$

$$\displaystyle \quad = \frac{\dfrac{x}{\sqrt{1+x^2}}}{\sqrt{1-\dfrac{x^2}{1+x^2}}} =x$$

$$\therefore \dfrac{d}{dx}(fog(x)) = \dfrac{d}{dx}(x) =1$$
Question 2
1 / -0

The integer $$n$$ for which $$\displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)(\cos x-e^{x})}{x^{n}}$$ is finite non zero number is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { (\cos { x } -1)\left( \cos { x } -{ e }^{ x } \right) }{ { x }^{ n } } } =\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { \left( \cos { x } -1 \right) }{ { x }^{ 2 } } \right] } \times \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } } $$
$$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } } } =\cfrac { 0 }{ 0 } $$ form

$$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { -\sin { x } -{ e }^{ x } } }{ (n-2){ x }^{ n-3 } } } =\cfrac { 1 }{ 2(n-2) } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { 1 } }{ { x }^{ n-3 } } } $$

$$\therefore$$ for the above limit to be finite $$n-3=0$$
$$\Rightarrow n=3$$
Question 3
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\cos ^{ -1 }{ \left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) } =$$

A
0

B
1

C
2

D
does not exist

Solution

$$\lim_{x\rightarrow 0^{-}} \dfrac{-2}{\dfrac{2x}{(1+x^{2})}}\dfrac{sin^{-1}}{(1+x^{2})}\dfrac{(2x)}{1+x^{2}}=-2$$

$$\lim_{x\rightarrow 0^{+}} \dfrac{2}{(\dfrac{2x}{1+x^{2}})(1+x^{2})} sin^{-1}(\dfrac{2x}{1+x^{2}})=2$$

$$L.H.S\neq R.H.S$$
Question 4
1 / -0

The right-hand limit of the function $$\sec{x}$$ at $$\displaystyle x=-\frac { \pi }{ 2 } $$ is

A
$$-\infty$$

B
$$-1$$

C
$$0$$

D
$$\infty$$

Solution

Function $$\displaystyle f\left( x \right)=\sec { x } $$ and point $$\displaystyle x=-\left( \frac { \pi }{ 2 } \right) $$.
We know that right$$-$$hand limit of the function $$f(x)$$ at $$x=-\left( \frac { \pi }{ 2 } \right) $$ is $$\displaystyle \lim _{ x\rightarrow { \left[ -\left( \pi /2 \right) \right] }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { \left[ -\left( \pi /2 \right) \right] } }{ \sec { x } } .$$
Substituting $$\displaystyle x=h+\left( -\frac { \pi }{ 2 } \right) $$ and $$h\rightarrow 0,$$ we get
$$\displaystyle \lim _{ h\rightarrow 0 }{ \sec { \left[ h+\left( -\frac { \pi }{ 2 } \right) \right] } = } \lim _{ h\rightarrow 0 }{ \sec { \left( \frac { \pi }{ 2 } -h \right) } } =\lim _{ h\rightarrow 0 }{ \csc { h } } =\csc { 0 } =\infty $$
$$\displaystyle \left[ \because \sec { \left( -\theta \right) = } \sec { \theta } \right] $$
Question 5
1 / -0

$$\displaystyle \lim_{x\rightarrow 1}(2-x)^{\displaystyle \tan( \frac{\pi x}{2})}=$$

A
$$e^{\displaystyle \frac{1}{\pi}}$$

B
$$e^{\displaystyle \frac{2}{\pi}}$$

C
$$-e^{\displaystyle \frac{2}{\pi}}$$

D
$$\mathrm{e}$$

Solution

Let $$k=\lim _{ x\rightarrow 1 } (2-x)^{ \tan \left(\dfrac { \pi x }{ 2 } \right) }$$
$$\log { k } =\lim _{ x\rightarrow 1 } tan\left( \dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } $$
$$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } tan\left( \displaystyle\dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } }$$
$$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { \log { \left( 2-x \right) } }{ \cot { \left( \dfrac { \pi x }{ 2 } \right) } } }$$
$$k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { -1 }{ -(2-x)\csc ^{ 2 }{ \left( \dfrac { \pi x }{ 2 } \right) \left( \dfrac { \pi }{ 2 } \right) } } }$$
$$\Rightarrow k={e}^{\displaystyle\dfrac{2}{\pi}}$$
Question 6
1 / -0

If $$f(x)=(ax+b)\cos x + (cx+d)\sin x$$ and $$f^{'}(x)=x \cos x$$, for all values of $$x\in R$$, then $$a,b,c,d$$ are given by

A
$$a = b = c = d$$

B
$$0, 1, -1, 0$$

C
$$1, 0, -1, 0$$

D
$$0, 1, 1, 0$$

Solution

Given, $$f(x) = (ax+b)\cos x + (cx+d)\sin x$$
Thus $$f'(x) = (ax+b)(-\sin x) + a\cos x + c\sin x + (cx+d)\cos x$$
$$= (-ax-b+c)\sin x + (a+cx+d)\cos x$$
Also given, $$f'(x) = x\cos x$$
Comparing the coefficients, we get
$$a =0, b=1, c =1, d=0$$
Question 7
1 / -0

If $$\displaystyle f(x) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} $$ then $$\mathop {\lim }\limits_{x \to \infty } f(x)$$ is

A
O

B
$$\infty$$

C
1

D
None of these

Solution

$$\displaystyle f(x) =\sqrt{\frac{x-sin x}{x+ cos^2x}} =\sqrt{\frac{1- \frac{sin x}{x}}{1+ \frac{cos^2x}{x}}}$$
$$\lim_{x \rightarrow \infty}f(x)=1$$ $$\displaystyle \lim_{x \rightarrow \infty} \frac{sin x}{x}=0$$
$$\displaystyle \lim_{x \rightarrow \infty} \frac{cos^2 x}{x}=0$$
Question 8
1 / -0

$$\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x}$$ is

A
$$\ln (n!)$$

B
$$n$$

C
$$n!$$

D
$$0$$

Solution

Consider that $$y=\displaystyle\lim_{x\to\infty}(1^x+2^x+3^x+.........+n^x)^{1/x}$$
take natural logarithm on both sides, we have
$$\ln y=\displaystyle\lim_{x\to\infty}\ln\left[(1^x+2^x+3^x+.........+n^x)^{1/x}\right]$$
$$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{\ln (1^x+2^x+3^x+.........+n^x)}{x}$$
RHS has $$\dfrac{\infty}{\infty}$$ form. Hence, use L'hospital's rule
$$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{1^x\ln 1+2^x\ln 2+3^x\ln 3+.........+n^x\ln n }{1^x+2^x+3^x+.........+n^x}$$
On approximating this form, we have
$$\ln y=\displaystyle\lim_{x\to\infty}\dfrac{n^x\ln n }{n^x}$$
$$\ln y=\ln n$$
$$\therefore y=n$$
Hence, this is required answer.
Question 9
1 / -0

Assertion (A): $$\mathrm{f}(\mathrm{x})=\sin(\pi[x])$$ is differentiable every where $$[\ ]$$ is greatest integer function

Reason (R): lf $$\mathrm{x}=\mathrm{n}\pi\Rightarrow $$ $$\sin x$$ $$=0\ \forall \ \mathrm{n}\in \mathrm{Z}$$ then

A
Both (A) and (R) are true and R is correct explanation for A

B
Both (A) and (R) are true and R is not correct explanation for A

C
(A) is true (R) is false

D
(A) is false (R) is true

Solution

$$ [ x ]$$is an integer$$\ \forall \ x\epsilon R$$
Hence, sin$$(\pi [ x ])=0 \ \forall\ x\epsilon \ R$$
Hence, $$f(x) =0 \ \forall \ x\epsilon\ R$$
Hence, $$f(x)$$ is differentiable everywhere.

The Assertion is correct and the Reason provides the correct explanation for it.
Question 10
1 / -0

$$\displaystyle \lim_{x\to1}{\displaystyle \frac{1-x^2}{\sin 2\pi x}}$$ is equal to

A
$$\displaystyle \frac{1}{2\pi}$$

B
$$\displaystyle \frac{-1}{\pi}$$

C
$$\displaystyle \frac{-2}{\pi}$$

D
None of these

Solution

$$\displaystyle\lim_{x\to1}\frac{1-x^2}{\sin2\pi
x}= -\displaystyle \lim_{x\to1}\frac{2\pi(1-x)(1+x)}{-2\pi\sin(2\pi-2\pi x)} $$
$$=\displaystyle \lim_{x \to 1}\left( \frac{(2\pi-2\pi x}{\sin(2\pi-2\pi x)}\right).\frac{1+x}{-2\pi}=-\frac{1}{\pi}$$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 16

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Limits and Derivatives Test - 16

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SHARING IS CARING

Question 1

$$1$$

$$0$$

$$-1$$

$$2$$

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

0

1

2

does not exist

Solution

Question 4

$$-\infty$$

$$-1$$

$$0$$

$$\infty$$

Solution

Question 5

$$e^{\displaystyle \frac{1}{\pi}}$$

$$e^{\displaystyle \frac{2}{\pi}}$$

$$-e^{\displaystyle \frac{2}{\pi}}$$

$$\mathrm{e}$$

Solution

Question 6

$$a = b = c = d$$

$$0, 1, -1, 0$$

$$1, 0, -1, 0$$

$$0, 1, 1, 0$$

Solution

Question 7

O

$$\infty$$

1

None of these

Solution

Question 8

$$\ln (n!)$$

$$n$$

$$n!$$

$$0$$

Solution

Question 9

Both (A) and (R) are true and R is correct explanation for A

Both (A) and (R) are true and R is not correct explanation for A

(A) is true (R) is false

(A) is false (R) is true

Solution

Question 10

$$\displaystyle \frac{1}{2\pi}$$

$$\displaystyle \frac{-1}{\pi}$$

$$\displaystyle \frac{-2}{\pi}$$

None of these

Solution

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