Continuity and Differentiability Test

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Continuity and Differentiability Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If $$y=(x^{x})^{x}$$ then $$\dfrac {dy}{dx}=$$

A
$$(x^{x})^{x}(1+2\log x)$$

B
$$(x^{x})^{x}(1-2\log x)$$

C
$$x(x^{x})^{x}(1+2\log x)$$

D
$$x(x^{x})^{x}(1-2\log x)$$

Solution

We have,

$$ y={{\left( {{x}^{x}} \right)}^{x}} $$

$$ y={{x}^{{{x}^{2}}}} $$

On taking $$\log $$ both sides, we get

$$\log y={{x}^{2}}\log x$$ …… (1)

On differentiating w.r.t $$x$$, we get

$$ \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{x}+\log x\left( 2x \right) $$

$$ \dfrac{1}{y}\dfrac{dy}{dx}=x+2x\log x $$

$$ \dfrac{dy}{dx}=y\left( x+2x\log x \right) $$

$$ \dfrac{dy}{dx}={{\left( {{x}^{x}} \right)}^{x}}\left( x+2x\log x \right) $$

$$ \dfrac{dy}{dx}=x{{\left( {{x}^{x}} \right)}^{x}}\left( 1+2\log x \right) $$

Hence, this is the answer.
Question 2
1 / -0

If $$ x=a(\cos\theta+log\ \tan\dfrac{\theta}{2})$$ and $$y=a\sin\theta$$, then$$\dfrac{dy}{dx}$$ is equal to

A
$$\cot\theta$$

B
$$\tan\theta$$

C
$$\sin\theta$$

D
$$\cos\theta$$

Solution

$$x=a(\cos\theta+\log\tan\cfrac{\theta}{2})$$
$$\cfrac{dx}{d\theta}$$$$=a(-\sin\theta+\cfrac{\sec^2 \cfrac{\theta}{2}}{2\tan\cfrac{\theta}{2}})$$
$$=a(-\sin\theta+\cfrac{1}{\sin\theta})$$
$$y=a(sin\theta)$$
$$\cfrac{dy}{d\theta}$$$$=a(cos\theta)$$
$$\cfrac{dy}{dx}=\cfrac{\cfrac{dy}{d\theta}}{\cfrac{dx}{d\theta}}$$
$$=\cfrac{a\sin\theta}{-a(\sin\theta-\cfrac{1}{\sin\theta})}$$
$$=\cfrac{\sin\theta}{\cos\theta}$$
$$=\tan\theta$$
Question 3
1 / -0

If $$f(x)=\dfrac{a^x}{x^a}$$ then $$f'(a)=$$?

A
$$log a-1$$

B
$$log a-a$$

C
$$a log a-a$$

D
$$a log a+a$$

Solution

According to question,

$$ f' (x)= \dfrac{ a^{x} loga x^{a} - x^{a+1} a^{x} }{ x^{2a} } $$

$$ \Rightarrow f' (a)= \dfrac{ a^{a} loga. a^{a} - a^{a+1} a^{a} }{ a^{2a} }$$

Hence,

$$ \Rightarrow f' (a)=loga-a$$
Question 4
1 / -0

If $$y=\tan^{-1}x+\cot^{-1}x+\sec^{-1}x+\csc^{-1}x$$,then $$\dfrac {dy}{dx}$$ is equal to

A
$$-1$$

B
$$\pi$$

C
$$0$$

D
$$1$$

Solution

Consider the given equation.

$$y={{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\sec }^{-1}}x+{{\csc }^{-1}}x$$

On differentiating w.r.t $$x$$, we get

$$ \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{1+{{x}^{2}}}+\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}-\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}$$

$$ \dfrac{dy}{dx}=0$$

Hence, this is the answer.
Question 5
1 / -0

Solve this:-$$\dfrac{d}{{dx}}\left( {\tan^{ - 1}\left( {\sinh \,X} \right)} \right) = $$

A
$$\operatorname{csch} x$$

B
$$\operatorname{sech} x$$

C
$$\sinh x$$

D
$$\cosh x$$

Solution
Question 6
1 / -0

If $$f(x)=\sin^4x+\cos^4x$$. Then $$f$$ is an increasing function in the interval

A
$$\left [ \dfrac{5\pi}{8},\dfrac{3\pi}{4} \right ]$$

B
$$\left [ \dfrac{\pi}{2},\dfrac{5\pi}{8} \right ]$$

C
$$\left [ \dfrac{\pi}{4},\dfrac{\pi}{2} \right ]$$

D
$$\left [ 0,\dfrac{\pi}{4} \right ]$$

Solution

$$f{\left( x \right)} = \sin^{4}{x} + \cos^{4}{x}$$
Diferentiating $$f{\left( x \right)}$$ w.r.t. $$x$$, we have
$$f'{\left( x \right)} = 4 \sin^{3}{x} \cos{x} + 4 \cos^{3}{x} \left( -\sin{x} \right)$$
$$f'{\left( x \right)} = 4 \sin^{3}{x} \cos{x} - 4 \cos^{3}{x} \sin{x}$$
$$f'{\left( x \right)} = 4 \sin{x} \cos{x} \left( \sin^{2}{x} - \cos^{2}{x} \right)$$
$$f'{\left( x \right)} = 2 \sin{2x} \left( - \cos{2x} \right)$$
$$f'{\left( x \right)} = - \sin{4x}$$
For $$f{\left( x \right)}$$ to be an increasing function,
$$f'{\left( x \right)} > 0$$
$$\Rightarrow - \sin{4x} > 0$$
$$\Rightarrow \sin{4x} < 0$$
$$\Rightarrow \pi < 4x < 2 \pi$$
$$\Rightarrow \cfrac{\pi}{4} < x < \cfrac{\pi}{2}$$
Hence $$f{\left( x \right)}$$ will be increasing for all $$x \in \left[ \cfrac{\pi}{4}, \cfrac{\pi}{2} \right]$$.
Question 7
1 / -0

Let $$f(x)$$ be a function defined on$$(-a,a)$$ with $$a > 0$$. Amuse that $$f(x)$$ is continuous at $$x=0$$ and $$\underset{x\to 0}{\lim}\dfrac {f(x)-f(kx)}{x}=\alpha$$, where $$k \in (0,1)$$ then compute $$f'(0^{+})$$ and $$f'(0^{-})$$, and comment upon the differentiablity of $$f$$ at $$x=0$$? Denote $$\alpha$$.

A
$$\underset{x\to o}{\lim}\dfrac{f(x) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

B
$$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{kx} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

C
$$\underset{x\to 0}{\lim}\dfrac{f(x) - f(kx)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

D
$$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

Solution

$$\underset{x \to 0^-}{\lim} f(x) = \underset{x\to 0}{\lim} + f(x)$$
$$\underset{x \to 0^-}{\lim} f'(x) = \underset{x\to 0}{\lim} + f'(x)$$
$$f'(0^=) = f'(0^-)$$
$$\underset{x\to 0}{\lim}\dfrac {f(x)-f(kx)}{x}=\alpha$$ ...(1)
$$f'(0^+)= \underset{x\to 0}{\lim} \dfrac{f(x) - f(0)}{x}$$
$$f' (0^-) = \underset{x\to 0}{\lim} \dfrac{f(x) - f(0)}{-x}$$
From eqn (1)
$$\underset{x\to 0}{\lim} \dfrac{f(x)+f(0) - f(o) - f(kx)}{x}$$
$$\underset{x\to o}{\lim}\dfrac{f(x) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$
Question 8
1 / -0

Solve:
$$ \dfrac{\log_{xy}xy}{\log_{xy} \, xyz} \, + \, \dfrac{\log_{yz}yz}{\log_{yz} \, xyz} \, + \, \dfrac{\log_{zx}zx}{\log_{zx} \, xyz} $$

A
$$4$$

B
$$0$$

C
$$2$$

D
$$1$$

Solution

We have,
$$\Rightarrow \dfrac{\log_{xy}xy}{\log_{xy} \, xyz} \, + \, \dfrac{\log_{yz}yz}{\log_{yz} \, xyz} \, + \, \dfrac{\log_{zx}zx}{\log_{zx} \, xyz} $$

We know that
$$\dfrac{\log_a m}{\log_a n}=\log_n m$$

Therefore,
$$\Rightarrow \log_{xyz}xy\, + \, \log_{xyz}yz\, + \,\log_{xyz}zx$$
$$\Rightarrow \log_{xyz}(x^2y^2z^2)$$
$$\Rightarrow \log_{xyz}(xyz)^2$$
$$\Rightarrow 2\log_{xyz}(xyz)$$
$$\Rightarrow 2$$

Hence, this is the answer.
Question 9
1 / -0

If $$x=a\left(\cos t+\log \tan\dfrac{t}{2}\right), y=a\sin t$$, then evaluate $$\dfrac{d^2y}{dx^2}$$ at $$t=\dfrac{\pi}{3}$$.

A
$$\cfrac{4\sqrt3}{a}$$

B
$$\cfrac{\sqrt3}{a}$$

C
$$\cfrac{8\sqrt3}{a}$$

D
$$\cfrac{8\sqrt3}{5a}$$

Solution

$$x = a(cost+logtan\cfrac{t}{2})$$
$$1 = a(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}}) \cfrac{dt}{dx}$$
$$\cfrac{dt}{dx} = \cfrac{1}{a(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}})}$$
$$\cfrac{dy}{dt} = acost$$

$$\cfrac{dy}{dx} = \cfrac{cost}{(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}})}$$

$$\cfrac{dy}{dx} = \cfrac{cost}{(-sint+\cfrac{1}{2sin\cfrac{t}{2}cos\cfrac{t}{2}})}$$
$$\cfrac{dy}{dx} = \cfrac{cost}{(-sint+\cfrac{1}{sint})}$$
$$\cfrac{dy}{dx} = \cfrac{costsint}{1-sin^2t}$$
$$\cfrac{dy}{dx} = tant$$
$$\cfrac{d^2y}{dx^2} = sec^2t \cfrac{dt}{dx}$$
At $$t = \cfrac{\pi}{3}$$
$$\cfrac{d^2y}{dx^2} = \cfrac{4}{a(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}})} = \cfrac{8\sqrt3}{a}$$
Question 10
1 / -0

The derivative of $$\sin^{-1}\left (\dfrac {2x}{1 + x^{2}}\right )$$ with respect to $$\tan^{-1} \left (\dfrac {2x}{1 - x^{2}}\right )$$is

A
$$0$$

B
$$1$$

C
$$\dfrac {1}{1 - x^{2}}$$

D
$$\dfrac {1}{1 + x^{2}}$$

Solution

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Continuity and Differentiability Test - 25

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Continuity and Differentiability Test - 25

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SHARING IS CARING

Question 1

$$(x^{x})^{x}(1+2\log x)$$

$$(x^{x})^{x}(1-2\log x)$$

$$x(x^{x})^{x}(1+2\log x)$$

$$x(x^{x})^{x}(1-2\log x)$$

Solution

Question 2

$$\cot\theta$$

$$\tan\theta$$

$$\sin\theta$$

$$\cos\theta$$

Solution

Question 3

$$log a-1$$

$$log a-a$$

$$a log a-a$$

$$a log a+a$$

Solution

Question 4

$$-1$$

$$\pi$$

$$0$$

$$1$$

Solution

Question 5

$$\operatorname{csch} x$$

$$\operatorname{sech} x$$

$$\sinh x$$

$$\cosh x$$

Solution

Question 6

$$\left [ \dfrac{5\pi}{8},\dfrac{3\pi}{4} \right ]$$

$$\left [ \dfrac{\pi}{2},\dfrac{5\pi}{8} \right ]$$

$$\left [ \dfrac{\pi}{4},\dfrac{\pi}{2} \right ]$$

$$\left [ 0,\dfrac{\pi}{4} \right ]$$

Solution

Question 7

$$\underset{x\to o}{\lim}\dfrac{f(x) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

$$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{kx} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

$$\underset{x\to 0}{\lim}\dfrac{f(x) - f(kx)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

$$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$$

Solution

Question 8

$$4$$

$$0$$

$$2$$

$$1$$

Solution

Question 9

$$\cfrac{4\sqrt3}{a}$$

$$\cfrac{\sqrt3}{a}$$

$$\cfrac{8\sqrt3}{a}$$

$$\cfrac{8\sqrt3}{5a}$$

Solution

Question 10

$$0$$

$$1$$

$$\dfrac {1}{1 - x^{2}}$$

$$\dfrac {1}{1 + x^{2}}$$

Solution

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