Limits and Derivatives Test

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Limits and Derivatives Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

If $$y=x^2+sin^{-1}x+log_ex$$, find $$\dfrac {dy}{dx}$$

A
$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

B
$$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

C
$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$$

D
$$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

Solution

$$y=x^2+sin^{-1}+log_ex$$
On differentiating, we get
$$\dfrac {dy}{dx}=\dfrac {d}{dx}(x^2)+\dfrac {d}{dx}sin^{-1}x)+\dfrac {d}{dx}(log_ex)$$
or $$\dfrac {dy}{dx}=2(x)^{2-1}+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {d}{dx}(log_ex)$$
$$\dfrac {dy}{dx}=2x+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {1}{x}$$
Question 2
1 / -0

$$\displaystyle \lim_{x\to0}\left( x^{-3}\sin{3x} + ax^{-2} + b \right)$$ exists and is equal to 0, then

A
$$a = -3$$ and $$b = \dfrac{9}{2}$$

B
$$a = 3$$ and $$b = \dfrac{9}{2}$$

C
$$a = -3$$ and $$b = -\dfrac{9}{2}$$

D
$$a = 3$$ and $$b = -\dfrac{9}{2}$$

Solution

$$\displaystyle\lim_{x\to\infty}\displaystyle\frac{\sin3x}{x^3}+\displaystyle\frac{a}{x^2}+b

= \lim_{x\to0}\displaystyle\frac{\sin3x+ax+bx^3}{x^3}$$

$$=\displaystyle\lim_{x\to0}\displaystyle\frac{3\displaystyle\frac{\sin3x}{3x}+a+bx^2}{x^2}$$

$$\mbox{For existence, }$$

$$(3+a)=0 \mbox{ or } a=-3$$

$$\therefore L=\displaystyle\lim_{x\to0}\displaystyle\frac{\sin3x-3x+bx^3}{x^3}$$

$$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\sin t - t}{t^3}+b = 0 ........ (3x=t ) $$
The left hand side reduces to $$\dfrac{0}{0}$$ form by substituting the limit
Then, using L'Hospital's rule
$$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{\cos t - 1}{3t^2}+b = 0$$
Again, applying L'Hospital's rule
$$\Rightarrow 27\displaystyle\lim_{t\to0}\displaystyle\frac{-\sin t}{6t}+b = 0$$
$$\Rightarrow -\displaystyle\frac{27}{6}+b=0 \mbox{ or } b=\frac{9}{2}$$
Question 3
1 / -0

If $$y=logx^3+3 sin^{-1}x+kx^2$$, then find $$\displaystyle \frac {dy}{dx}$$

A
$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

B
$$3\cdot \displaystyle \frac {1}{x^3}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

C
$$3\cdot \displaystyle \frac {1}{x}-3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

D
$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+2x$$

Solution

Here, $$y=\log x^3+3 \sin^{-1} x+kx^2$$

On differentiating we get

$$\displaystyle \dfrac {dy}{dx}=\dfrac {d}{dx}[\log x^3]+\dfrac {d}{dx}[3 \sin^{-1}x]+\dfrac {d}{dx}[kx^2]$$

$$=3\displaystyle \dfrac {d}{dx}[\log x]+3\dfrac {d}{dx}(\sin^{-1}x)+k\dfrac {d}{dx}(x^2)$$

$$=3\cdot \displaystyle \dfrac {1}{x}+3\cdot \dfrac {1}{\sqrt {1-x^2}}+k(2x)$$
Question 4
1 / -0

The value of $$\displaystyle\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$$ for $$(a>1)$$ is equal to?

A
$$1$$

B
$$0$$

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

D
Does not exist

Solution

$$\mathrm{L} \displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{(x^{-a}\log_a{x})}}{\sec^{-1}{a^x\log_x{a}}}}$$

$$\displaystyle = \lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{\log{x}}{\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log a}{\log x}}}}=\lim_{x\rightarrow\infty}{\frac{\cot^{-1}{\displaystyle \frac{1}{a\log a. x^a}}}{\sec^{-1}{\displaystyle \frac{a^x\log^2 a. x}{1}}}}=\frac{\cot^{-1}0}{\sec^{-1} \infty}=\frac{\pi/2}{\pi/2}=1$$
Note: L-Hospital's rule has been used in step two in both numerator and denominator alone.
Question 5
1 / -0

The value of
$$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 \sin^2 x + \sin x-1}{2 \sin^2 x - 3 \sin x + 1} $$

A
$$3$$

B
$$-3$$

C
$$6$$

D
$$0$$

Solution

We have
$$\displaystyle \lim_{x \rightarrow \pi/6} \frac{2 sin^2 x + sin x -1}{2 sin^2 x-3 sin x + 1}$$

$$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{(2 sin x - 1)(sin x + 1)}{(2 sin x - 1)(sin x - 1)}$$

$$\displaystyle = \lim_{x \rightarrow \pi/6} \frac{sin x + 1}{sin x - 1} = - 3$$
Question 6
1 / -0

If $$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) + \sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$, then $$\displaystyle\frac{dy}{dx}$$ equals

A
$$1$$

B
$$0$$

C
$$\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}$$

D
$$\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}$$

Solution

$$y = sec^{-1}\left(\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}\right) +

\sin^{-1}\left(\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}\right)$$
$$y=\cos ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) +\sin ^{ -1 }{ \left(\displaystyle \frac { \sqrt { x } -1 }{ \sqrt { x } +1 } \right) } } =\displaystyle \frac { \pi }{ 2 } $$ ............. $$\because \sin ^{ -1 }{ \theta +\cos ^{ -1 }{ \theta } =\displaystyle \frac { \pi }{ 2 } } $$
$$\therefore \displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left(\displaystyle \frac { \pi }{ 2 } \right) =0$$
Question 7
1 / -0

If $$f'(x)=\sin x+\sin 4x\cdot \cos x$$, then $$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$ is

A
$$4x\left \{\cos(2x^2)-sin 8x^2\cdot \sin 2x^2\right \}$$

B
$$4x\left \{\cos(2x^2)+\sin 8x^2\cdot \sin 2x^2\right \}$$

C
$$\left \{\cos (2x^2)-\sin 8x\cdot \sin 2x^2\right \}$$

D
none of the above

Solution

$$f'(x)=\sin x+\sin 4x\cdot \cos x$$
$$f'\left (2x^2+\displaystyle \frac {\pi}{2}\right )$$=$$\displaystyle \frac { d }{ dx } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) \left[ \sin { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) +\sin { 4 } \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) .\cos { \left( 2{ x }^{ 2 }+\displaystyle \frac { \pi }{ 2 } \right) } } \right] $$
$$=4x\left[ \cos { 2{ x }^{ 2 }-\sin { 8 } { x }^{ 2 }.\sin { 2{ x }^{ 2 } } } \right] $$
Question 8
1 / -0

If $$y=|\cos x|+|\sin x|$$, then $$\displaystyle \dfrac {dy}{dx}$$ at $$x=\dfrac {2\pi}{3}$$ is

A
$$\displaystyle \dfrac {1}{2}(\sqrt 3+1)$$

B
$$2(\sqrt 3-1)$$

C
$$\displaystyle \dfrac {1}{2}(\sqrt 3-1)$$

D
none of these

Solution

$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { d }{ dx } \left( |cosx|+|sinx| \right) $$
$$=-\sin { x } +\cos { x } $$ at $$x=\displaystyle \frac { 2\pi }{ 3 }$$
$$=\left| -\sin { \displaystyle \frac { 2\pi }{ 3 } } \right| +\left| \cos { \displaystyle \frac { 2\pi }{ 3 } } \right| $$
$$=\displaystyle \frac { \sqrt { 3 } }{ 2 } -\displaystyle \frac { 1 }{ 2 } $$
$$=\displaystyle \frac { 1 }{ 2 } \left( \sqrt { 3 } -1 \right) $$
Question 9
1 / -0

Directions For Questions

If f: $$R\rightarrow R$$ and $$f(x)=g(x)+h(x)$$ where $$g(x)$$ is a polynomial and $$h(x)$$ is a continuous and differentiable bounded function on both sides, then $$f(x)$$ is one-one, we need to differentiate $$f(x)$$. If $$f'(x)$$ changes sign in domain of $$f$$, then $$f $$ is many-one else one-one.

...view full instructions

$$f:R\rightarrow R$$ and $$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$$, then $$f(x)$$ is

A
one-one ito

B
many-one onto

C
one-one onto

D
many-one into

Solution

$$\displaystyle f(x)=\frac {x(x^4+1)(x+1)+x^4+2}{x^2+x+1}$$
$$\displaystyle \quad \quad =\frac {x(x^4+1)(x+1)+(x^4+1)+1}{x^2+x+1}$$
$$\displaystyle \quad \quad =\frac {(x^4+1)(x^2+x+1)+1}{x^2+x+1}$$
$$\displaystyle \quad \quad =(x^4+1)+\frac {1}{x^2+x+1}$$
$$\displaystyle f'(x) =4x^3-\frac {2x+1}{(x^2+x+1)^2}=$$ not always positive or negative
Thus, $$f$$ is many one.
Also range and co-domain of $$f$$ are not same,
Hence is many-one into function
Question 10
1 / -0

Which one of the following statement is true?

A
if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

B
if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exist

C
if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

D
if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exists
Solution
The statements can be proved false by giving counter examples

(1) If $$\displaystyle \lim_{x\to c}f(x).g(x)$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists
Let $$g(x) = 1/x $$ and $$f(x) = 2x$$, with $$c = 0$$,
then $$\displaystyle \lim_{x\to c}f(x).g(x) =2 $$
$$\displaystyle \lim_{x\to c}f(x) = 0 $$ ,but $$\displaystyle \lim_{x\to c}g(x)$$ does not exist
(2) Same example as (1)
(4) Let $$g(x) =\dfrac{ x+ 1}{x} $$ and $$f(x) = \dfrac{x-1}{x} $$, with $$c = 0$$,
then $$\displaystyle \lim_{x\to c}[f(x)+g(x)] =2$$
but $$\displaystyle \lim_{x\to c}f(x) \ and \ g(x)$$ do not exist

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 18

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Limits and Derivatives Test - 18

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SHARING IS CARING

Question 1

$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

$$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$$

$$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

Solution

Question 2

$$a = -3$$ and $$b = \dfrac{9}{2}$$

$$a = 3$$ and $$b = \dfrac{9}{2}$$

$$a = -3$$ and $$b = -\dfrac{9}{2}$$

$$a = 3$$ and $$b = -\dfrac{9}{2}$$

Solution

Question 3

$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

$$3\cdot \displaystyle \frac {1}{x^3}+3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

$$3\cdot \displaystyle \frac {1}{x}-3\cdot \frac {1}{\sqrt {1-x^2}}+k(2x)$$

$$3\cdot \displaystyle \frac {1}{x}+3\cdot \frac {1}{\sqrt {1-x^2}}+2x$$

Solution

Question 4

$$1$$

$$0$$

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

Does not exist

Solution

Question 5

$$3$$

$$-3$$

$$6$$

$$0$$

Solution

Question 6

$$1$$

$$0$$

$$\displaystyle\frac{\sqrt x + 1}{\sqrt x - 1}$$

$$\displaystyle\frac{\sqrt x - 1}{\sqrt x + 1}$$

Solution

Question 7

$$4x\left \{\cos(2x^2)-sin 8x^2\cdot \sin 2x^2\right \}$$

$$4x\left \{\cos(2x^2)+\sin 8x^2\cdot \sin 2x^2\right \}$$

$$\left \{\cos (2x^2)-\sin 8x\cdot \sin 2x^2\right \}$$

none of the above

Solution

Question 8

$$\displaystyle \dfrac {1}{2}(\sqrt 3+1)$$

$$2(\sqrt 3-1)$$

$$\displaystyle \dfrac {1}{2}(\sqrt 3-1)$$

none of these

Solution

Question 9

one-one ito

many-one onto

one-one onto

many-one into

Solution

Question 10

if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

if $$\displaystyle \lim_{x\to c}f(x).g(x)$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exist

if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ and $$\displaystyle \lim_{x\to c}f(x)$$ exist, then $$\displaystyle \lim_{x\to c}g(x)$$ exists

if $$\displaystyle \lim_{x\to c}(f(x)+g(x))$$ exists, then $$\displaystyle \lim_{x\to c}f(x)$$ and $$\displaystyle \lim_{x\to c}g(x)$$ exists

Solution

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