Continuity and Differentiability Test

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Continuity and Differentiability Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

If $$xlog_e(log_ex)-x^2+y^2=4(y>0)$$, then dy/dx at $$x=e$$ is equal to:

A
$$\dfrac{e}{\sqrt{4+e^2}}$$

B
$$\dfrac{1+2e}{2\sqrt{4+e^2}}$$

C
$$\dfrac{2e-1}{2\sqrt{4+e^2}}$$

D
$$\dfrac{1+2e}{\sqrt{4+e^2}}$$

Solution

When $$x=e$$,then $$0-e^2+y^2=4,y=\sqrt{e^2+4}$$
Differentiating with respect to x, We get:
$$x.\dfrac{1}{\ell n x}.\dfrac{1}{x}+\ell n(\ell n x)-2x+2y.\dfrac{dy}{dx}=0$$ at $$x=e$$ we get
$$1-2e+2y\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=\dfrac{2e-1}{2y}$$
$$\Rightarrow\dfrac{dy}{dx}=\dfrac{2e-1}{2\sqrt{4+e^2}}$$ as $$y(e)=\sqrt{4+e^2}$$
Question 2
1 / -0

If $$x=\cos^3\theta $$ and $$y=\sin^3\theta$$, then $$1+\left(\displaystyle\frac{dy}{dx}\right)^2$$ is equal to:

A
$$\tan^2\theta$$

B
$$\cot^2\theta$$

C
$$\sec^2\theta$$

D
$$\text{cosec}^2\theta$$

Solution

$$\Rightarrow \dfrac{dx}{d\theta}=3\cos^2\theta(-\sin\theta)=-3\cos^2\theta\sin\theta$$ ------(1)
and
$$\Rightarrow \dfrac{dy}{d\theta}=3\sin^2\theta\cos\theta$$ -----(2)
Hence, $$1+\left( \dfrac{dy}{dx}\right)^2=1+\left( \dfrac{dy/d\theta}{dx/d\theta}\right)^2=1+\dfrac{\sin^2\theta}{\cos^2\theta}=\sec^2\theta$$
Question 3
1 / -0

For the curve $$x = t^2 - 1, y = t^2 - t$$, tangent is parallel to $$x$$ - axis where,

A
$$t = 0$$

B
$$t=\dfrac{1}{\sqrt{3}}$$

C
$$t=\dfrac{1}{2}$$

D
$$t=-\dfrac{1}{\sqrt{3}}$$

Solution

$$x = t^2 - 1, y = t^2 - t$$
If tangent is parallel to $$x$$ axis, $$\cfrac{dy}{dx} = 0$$
$$\Rightarrow \cfrac{dy}{dt} \times \cfrac{dt}{dx} = 0$$
i.e. $$(2t - 1) \times \cfrac{1}{2t} = 0$$
i.e. $$t = \cfrac{1}{2}$$
Question 4
1 / -0

$$\displaystyle \frac{d}{dx}(\tan ^{-1}x)$$

A
$$ \displaystyle \frac{1}{1+x^{2}}.$$

B
$$\displaystyle \frac{-1}{1+x^{2}}.$$

C
$$\displaystyle \frac{-1}{1-x^{2}}.$$

D
$$\displaystyle \frac{1}{1-x^{2}}.$$

Solution

Let $$\displaystyle y=\tan ^{-1}x\therefore x=\tan y$$
$$\displaystyle y+\delta y=\tan ^{-1}\left ( x+\delta x \right )$$
$$\displaystyle \therefore x+\delta x=\tan\left ( y+\delta y \right )$$
$$\therefore \displaystyle \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}$$
$$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\delta y}{\tan\left ( y+\delta y \right )-\tan y}= \frac{1}{\sec^{2}y}$$
$$\displaystyle =\frac{1}{1+\tan ^{2}y}= \frac{1}{1+x^{2}}.$$
Question 5
1 / -0

$$\displaystyle \frac{d(sin^{-1}x)}{dx}$$

A
$$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{2} \right )}}$$

B
$$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{4} \right )}}$$

C
$$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{3} \right )}}$$

D
$$\displaystyle \frac{1}{\sqrt{\left ( 1+x^{2} \right )}}$$

Solution

$$\displaystyle y= \sin ^{-1}x,$$
$$\displaystyle y+\delta y= \sin ^{-1}\left ( x+\delta x \right )$$
$$\displaystyle \therefore \left ( x+\delta x \right) = \sin \left ( y+\delta y \right )$$
$$\displaystyle \frac{dy}{dx}= \lim_{x\rightarrow 0}\frac{1}{\delta x}(\sin ^{-1}\left ( x+\delta x \right )-\sin ^{-1}x)$$
$$\displaystyle = \lim_{x\rightarrow 0}\frac{\delta y}{\sin \left ( y+\delta y \right )-\sin y}= \frac{1}{\cos y}$$
$$\displaystyle

= \frac{1}{\sqrt{\left ( 1-\sin ^{2}y \right )}}= \frac{1}{\sqrt{\left (

1-x^{2} \right )}}$$
Note that when $$\displaystyle :\delta x\rightarrow 0, \delta y$$ also tends to zero.
Question 6
1 / -0

If the tangent to the curve $$x=a \, (\theta + \sin \, \theta), y=a (1+ \cos \,\theta)$$ at $$ \theta=\dfrac{\pi}{3}$$ makes an angle $$\alpha (0 \leq\alpha < \pi)$$ with x-axis, then $$\alpha$$ =

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{2 \pi}{3}$$

C
$$\dfrac{\pi}{6}$$

D
$$\dfrac{5 \pi}{6}$$

Solution

$$x=a\left( \theta +\sin { \theta  } \right) $$

$$y=a\left( 1+\cos { \theta  } \right) $$

If tangent to curve at $$\theta = { \pi }/{ 3 }$$, makes angle $$\alpha$$ with x-axis,

$$\Rightarrow$$ slope of tangent = $$\tan { \alpha  } $$

$$\Rightarrow$$ $$\left. \begin{matrix} \tan { \alpha } ={ { \dfrac { { dy }/{ { d\theta } } }{ { dx }/{ { d\theta } } } } }{ { } } \\ \end{matrix} \right| _{ \theta ={ \pi }/{ 3 } }$$

$$=\dfrac {-a\sin \theta}{ a\left( 1+\cos { \theta  }  \right)  }$$

$$\left. \begin{matrix}\dfrac{-\sin \theta} { 1+\cos { \theta } } \\ \end{matrix}\right|_{ \theta ={ \pi }/{ 3 } }$$

$$=-\dfrac{\sqrt{3}/2} { 1+{ 1 }/{ 2 } }$$

$$\tan { \alpha } \ =-\dfrac{1}{\sqrt { 3 }} $$

Ans. $$\alpha \ =\ { 5{ \pi  } }/{ 6 }$$
Question 7
1 / -0

If $$y$$ is expressed in terms of a variable $$x$$ as $$y = f(x)$$, then $$y$$ is called

A
Explicit function

B
Implicit function

C
Linear function

D
Identity function

Solution

If $$y$$ is expressed in terms of a variable $$x$$ as $$y=f(x)$$, then $$y$$ is called explicit function.
Question 8
1 / -0

If $$f(x)=\begin{cases}
ax &a<1&\\
ax^2+bx+2 &a\ge 1&.
\end{cases}$$
Then the values of $$a$$, $$b$$ for which $$f(x)$$ is differentiable, are

A
$$a=\large{\frac{3}{4}}$$, $$b=\large{\frac{1}{4}}$$

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

C
$$a=\large{\frac{3}{2}}$$, $$b=\large{\frac{1}{4}}$$

D
$$a=\large{\frac{3}{4}}$$, $$b=-2$$

Solution

Given the function $$f(x)$$ is differentiable, then it is differentiable at $$x=1$$ and it will be continuous there at.
Then using continuity condition at $$x=1$$,
$$\lim\limits_{x\to 1^-}f(x)=\lim\limits_{x\to 1+}f(x)$$
or, $$a=a+b+2$$
or, $$b=-2$$.
Now using differentiability at $$x=1$$ we get,
$$\lim\limits_{x\to 1^-}f'(x)=\lim\limits_{x\to 1^+}f'(x)$$
or, $$a=2a+b$$
or, $$a=-b$$
or, $$a=2$$.
$$\therefore a=2,b=-2$$.
so, option (B) is true.
Question 9
1 / -0

Let $$f(x)=x^2e^x$$ then $$f''(0)=$$

A
$$1$$

B
$$0$$

C
$$2$$

D
$$7$$

Solution

Given,
$$f(x)=x^2e^x$$
Now differentiating both sides with respect to $$x$$ we get,
$$f'(x)=x^2e^x+2xe^x$$
Again differentiating both sides with respect to $$x$$ we get,
$$f''(x)=x^2e^x+4xe^x+2e^x$$.
Now $$f''(0)=2$$.
Question 10
1 / -0

If $$f(x)=\dfrac{1}{2}\left [ \left | \sin x \right |+\sin x \right ],\ 0 < x \leq 2\pi$$, then $$f$$ is

A
Increasing in $$\left ( \dfrac{\pi}{2},\dfrac{3\pi}{2} \right )$$

B
Decreasing in $$\left ( 0,\dfrac{\pi}{2} \right )$$ and increasing in $$\left ( \dfrac{\pi}{2},\pi \right )$$

C
Increasing in $$\left ( 0,\dfrac{\pi}{2} \right )$$ and decreasing in $$\left ( \dfrac{\pi}{2},\pi \right )$$

D
Increasing in $$\left ( 0,\dfrac{\pi}{4} \right )$$ and decreasing in $$\left ( \dfrac{\pi}{4},\pi \right )$$

Solution

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 12

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Continuity and Differentiability Test - 12

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Question 1

$$\dfrac{e}{\sqrt{4+e^2}}$$

$$\dfrac{1+2e}{2\sqrt{4+e^2}}$$

$$\dfrac{2e-1}{2\sqrt{4+e^2}}$$

$$\dfrac{1+2e}{\sqrt{4+e^2}}$$

Solution

Question 2

$$\tan^2\theta$$

$$\cot^2\theta$$

$$\sec^2\theta$$

$$\text{cosec}^2\theta$$

Solution

Question 3

$$t = 0$$

$$t=\dfrac{1}{\sqrt{3}}$$

$$t=\dfrac{1}{2}$$

$$t=-\dfrac{1}{\sqrt{3}}$$

Solution

Question 4

$$ \displaystyle \frac{1}{1+x^{2}}.$$

$$\displaystyle \frac{-1}{1+x^{2}}.$$

$$\displaystyle \frac{-1}{1-x^{2}}.$$

$$\displaystyle \frac{1}{1-x^{2}}.$$

Solution

Question 5

$$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{2} \right )}}$$

$$\displaystyle \frac{1}{\sqrt{\left ( 1-x^{4} \right )}}$$

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

$$\displaystyle \frac{1}{\sqrt{\left ( 1+x^{2} \right )}}$$

Solution

Question 6

$$\dfrac{\pi}{3}$$

$$\dfrac{2 \pi}{3}$$

$$\dfrac{\pi}{6}$$

$$\dfrac{5 \pi}{6}$$

Solution

Question 7

Explicit function

Implicit function

Linear function

Identity function

Solution

Question 8

$$a=\large{\frac{3}{4}}$$, $$b=\large{\frac{1}{4}}$$

$$a=2$$, $$b=-2$$

$$a=\large{\frac{3}{2}}$$, $$b=\large{\frac{1}{4}}$$

$$a=\large{\frac{3}{4}}$$, $$b=-2$$

Solution

Question 9

$$1$$

$$0$$

$$2$$

$$7$$

Solution

Question 10

Increasing in $$\left ( \dfrac{\pi}{2},\dfrac{3\pi}{2} \right )$$

Decreasing in $$\left ( 0,\dfrac{\pi}{2} \right )$$ and increasing in $$\left ( \dfrac{\pi}{2},\pi \right )$$

Increasing in $$\left ( 0,\dfrac{\pi}{2} \right )$$ and decreasing in $$\left ( \dfrac{\pi}{2},\pi \right )$$

Increasing in $$\left ( 0,\dfrac{\pi}{4} \right )$$ and decreasing in $$\left ( \dfrac{\pi}{4},\pi \right )$$

Solution

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