Continuity and Differentiability Test

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Continuity and Differentiability Test - 19

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Question 1
1 / -0

If $$y=1-\cos { \theta } ,x=1-\sin { \theta } $$, then $$\cfrac { dy }{ dx } $$ at $$\theta =\cfrac { \pi }{ 4 } $$ is

A
$$-1$$

B
$$1$$

C
$$\cfrac{1}{2}$$

D
$$\cfrac{1}{\sqrt{2}}$$

Solution

Given, $$ y=1-\cos { \theta } $$
$$\Rightarrow \cfrac { dy }{ d\theta } =\sin { \theta } $$
Also given, $$x=1-\sin { \theta } $$
$$\Rightarrow \cfrac { dx }{ d\theta } =-\cos { \theta } $$
Therefore, $$\cfrac { dy }{ dx } = \cfrac{\sin\theta}{-\cos\theta} =-\tan { \theta } $$ at $$\theta =\cfrac { \pi }{ 4 } $$ will be
$$\therefore { \left[ \cfrac { dy }{ dx } \right] }_{ \theta =\frac { \pi }{ 4 } }=-\tan { \cfrac { \pi }{ 4 } } =-1$$
Question 2
1 / -0

Consider the parametric equation $$x = \cfrac {a(1 - t^{2})}{1 + t^{2}}, y = \cfrac {2at}{1 + t^{2}}$$.
What is $$\cfrac {dy}{dx}$$ equal to?

A
$$\dfrac {y}{x}$$

B
$$-\dfrac {y}{x}$$

C
$$\dfrac {x}{y}$$

D
$$-\dfrac {x}{y}$$

Solution

Given : $$x=\cfrac { a\left( 1-{ t }^{ 2 } \right) }{ 1+{ t }^{ 2 } } \; and\; y=\cfrac { 2at }{ 1+{ t }^{ 2 } } $$

First consider $$x=\cfrac { a\left( 1-{ t }^{ 2 } \right) }{ 1+{ t }^{ 2 } } $$
Diferentiating w.r.t $$'t'$$

$$\cfrac { dx }{ dt } =a\left[ \cfrac { \left( 1+{ t }^{ 2 } \right) \left( -2t \right) -\left( 1{ -t }^{ 2 } \right) \left( 2t \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } \right] $$

$$=-2at\left[ \cfrac { \left( 1+{ t }^{ 2 }+1-{ t }^{ 2 } \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } \right] =\cfrac { -4at }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } \; \rightarrow \; (1)$$

secondly $$y=\cfrac { 2at }{ 1+{ t }^{ 2 } } $$

$$\Rightarrow \cfrac { dy }{ dt } =2a\left[ \cfrac { \left( 1+{ t }^{ 2 }.1 \right) -t\left( 2t \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } \right] \; \rightarrow \; (2)$$

divide $$(2)\div (1)$$

$$\cfrac { \cfrac { dy }{ dt } }{ \cfrac { dx }{ dt } } =\cfrac { 2a\left[ \cfrac { \left( 1+{ t }^{ 2 }.1 \right) -t\left( 2t \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } \right] \; }{ \cfrac { -4at }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } \; } $$

$$=\cfrac { -1\left[ 1+{ t }^{ 2 }-2{ t }^{ 2 } \right] }{ -4t } \\ \cfrac { dy }{ dx } =\cfrac { -1-{ t }^{ 2 } }{ 2t } \; \rightarrow \; (3)$$

Given : $$x=\cfrac { a\left( 1-{ t }^{ 2 } \right) }{ 1+{ t }^{ 2 } } \; and\; y=\cfrac { 2at }{ 1+{ t }^{ 2 } } $$

$$\cfrac { y }{ x } =\cfrac { \cfrac { 2at }{ 1+{ t }^{ 2 } } }{ \cfrac { a\left( 1-{ t }^{ 2 } \right) }{ 1+{ t }^{ 2 } } } $$

$$\cfrac { y }{ x } =\cfrac { 2t }{ 1-{ t }^{ 2 } } \; \rightarrow (4)$$

Therefor substitute equation $$(4)$$ in $$(3)$$

$$\cfrac { dy }{ dx } =\cfrac { -x }{ y } $$

Hence option $$(4)$$ is the correct answer.
Question 3
1 / -0

Consider the function $$f(x)=\left\{\begin{matrix} x^2-5, & x\leq 3\\ \sqrt{x+13}, & x > 3\end{matrix}\right.$$.

Consider the following statements.
$$1$$. The function is discontinuous at $$x=3$$.
$$2$$. The function is not differentiable at $$x=0$$.
Which of the statements is$$/$$ are correct?

A
$$1$$ only

B
$$2$$ only

C
Both $$1$$ and $$2$$

D
Neither $$1$$ nor $$2$$

Solution

$$1. \displaystyle \lim _{ x\rightarrow { 3 }^{ - } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow { 3 }^{ - } }{ \left( { x }^{ 2 }-5 \right) } =9-5=4$$

$$\displaystyle \lim _{ x\rightarrow { 3 }^{ + } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow { 3 }^{ + } }{ \sqrt { x+13 } } =\sqrt { 16 } =4$$

LHL$$=$$RHL

$$\Rightarrow \displaystyle \lim _{ x\rightarrow { 3 } }{ f\left( x \right) } =4$$

$$\Rightarrow$$ It is continuous

$$2.f\left( x \right) ={ x }^{ 2 }-5$$

$$f'\left( x \right) =2x$$ for $$x\le 3$$

$$f'\left( x \right) $$ is continuous for $$x\le 3$$

At $$x=0\quad f'\left( x \right) $$ is continuous

$$\Rightarrow f\left( x \right) $$ is differentiable at $$x=0$$
Question 4
1 / -0

If $$f(x)=\sqrt{25-x^2}$$, then what is $$\displaystyle \lim_{ x\rightarrow 1 } \dfrac{f(x)-f(1)}{x-1}$$ equal to?

A
$$\dfrac{1}{5}$$

B
$$\dfrac{1}{24}$$

C
$$\sqrt{24}$$

D
$$-\dfrac{1}{\sqrt{24}}$$

Solution

Consider $$\displaystyle \lim_{x\rightarrow 1}{\cfrac{f(x)-f(1)}{x-1}}$$
We know that, $$\displaystyle \lim_{x\rightarrow a}{\cfrac{f(x)-f(a)}{x-a}}=f'(a)$$
So, we get
$$\displaystyle \lim_{x\rightarrow 1}{\cfrac{f(x)-f(1)}{x-1}}$$
$$={\cfrac{f'(1)}{1}}=\left|{\cfrac{1(-2x)}{2\sqrt{25-x^2}}}\right|_{at\ x=1}$$
$$=-\cfrac{1}{\sqrt {24}}$$
Question 5
1 / -0

If $$x=\sec { \theta } -\cos { \theta } $$ and $$y=\sec ^{ n }{ \theta } -\cos ^{ n }{ \theta } $$, then $${ \left( \dfrac { dy }{ dx } \right) }^{ 2 }$$ is

A
$$\dfrac { { n }^{ 2 }\left( { y }^{ 2 }+4 \right) }{ { x }^{ 2 }+4 } $$

B
$$\dfrac { { n }^{ 2 }\left( { y }^{ 2 }-4 \right) }{ { x }^{ 2 } } $$

C
$$n\dfrac { \left( { y }^{ 2 }-4 \right) }{ { x }^{ 2 }-4 } $$

D
$${ \left( \dfrac { ny }{ x } \right) }^{ 2 }-4$$

Solution

Given, $$x=\sec { \theta } -\cos { \theta } $$ and $$y=\sec ^{ n }{ \theta } -\cos ^{ n }{ \theta } $$

Differentiate given equations with respect to $$\theta $$, we get

$$\dfrac { dy }{ d\theta } =n\sec ^{ n-1 }{ \theta } \cdot \sec { \theta } \cdot \tan { \theta } -n\cdot \cos ^{ n-1 }{ \theta } \left( -\sin { \theta } \right)$$

$$=n\tan { \theta } \left( \sec ^{ n }{ \theta } +\cos ^{ n }{ \theta } \right)$$

$$\dfrac { dx }{ d\theta } =\sec { \theta } \tan { \theta } +\sin { \theta } =\tan { \theta } \left( \sec { \theta } +\cos { \theta } \right) $$

$$\therefore$$, $$ \dfrac { dy }{ dx } =\dfrac { n\tan { \theta } \left( \sec ^{ n }{ \theta } +\cos ^{ n }{ \theta } \right) }{ \tan { \theta } \left( \sec { \theta } +\cos { \theta } \right) } $$

$$=\dfrac { n\left( \sec ^{ n }{ \theta } +\cos ^{ n }{ \theta } \right) }{ \left( \sec { \theta } +\cos { \theta } \right) } $$
Square of the given differential would be $$ { \left( \dfrac { dy }{ dx } \right) }^{ 2 }=\dfrac { { n }^{ 2 }{ \left( \sec ^{ n }{ \theta } +\cos ^{ n }{ \theta } \right) }^{ 2 } }{ { \left( \sec { \theta } +\cos { \theta } \right) }^{ 2 } } $$

$$=\dfrac { { n }^{ 2 }\left\{ { \left( \sec ^{ n }{ \theta } -\cos ^{ n }{ \theta } \right) }^{ 2 }+4 \right\} }{ { \left( \sec { \theta } -\cos { \theta } \right) }^{ 2 }+4 } $$

$$=\dfrac { { n }^{ 2 }\left( { y }^{ 2 }+4 \right) }{ \left( { x }^{ 2 }+4 \right) } $$
Question 6
1 / -0

If $$x = a \cos^3 \theta$$ and $$y = a\sin^3 \theta$$, then $$1 + \left( \dfrac{dy}{dx} \right )^2$$ is

A
$$\tan \theta$$

B
$$\tan^2 \theta$$

C
$$1$$

D
$$\sec^2 \theta$$

E
$$\sec \theta$$

Solution

Given, $$x=a\cos^3\theta, y=a\sin ^3\theta$$
$$\dfrac {dx}{d\theta}=3a\cos^2\theta. (-\sin\theta)$$
and $$\dfrac {dy}{d\theta}=3a\sin^2\theta. \cos \theta$$
Therefore, $$\dfrac {dy}{dx}=\dfrac {3a\sin^2\theta. \cos \theta}{-3a\cos^2\theta. \sin \theta}$$ $$=-\tan \theta$$
Question 7
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The function $$f(x) = \left\{\begin{matrix}x^{2}/a, & 0\leq x < 1\\ a, & 1\leq x < \sqrt {2}\\ \dfrac {2b^{2} - 4b}{x^{2}}, & \sqrt {2} \leq x < \infty\end{matrix}\right.$$ is continuous for $$0\leq x < \infty$$, then the most suitable values of $$a$$ and $$b$$ are

A
$$a = 1, b = -1$$

B
$$a = -1, b = 1 + \sqrt {2}$$

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

D
None of these

Solution

Since, $$f$$ is continuous for $$0\leq x < \infty, f$$ is continuous at $$x = 1$$.

$$\displaystyle\lim_{x\rightarrow 1^-}\dfrac{x^2}{a}=a$$

$$\therefore a = \pm 1$$

Since, $$f$$ is continuous at $$x = \sqrt {2}$$.

$$\displaystyle\lim_{x\rightarrow \sqrt{2}}\dfrac{2b^2-4b}{x^2}=a$$

$$\therefore \dfrac {2b^{2} - 4b}{2} = a$$

$$\Rightarrow b^{2} - 2b = a$$

When $$a = 1, b^{2} - 2b = 1$$

$$\Rightarrow b = 1\pm \sqrt {2}$$

When $$a = -1, b^{2} - 2b = -1$$

$$\Rightarrow (b - 1)^{2} = 0 \Rightarrow b = 1$$

Hence, $$a = -1$$ and $$b = 1$$ are most suitable values.
Question 8
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If $$2^x + 2^y = 2^{x + y}$$, then $$\dfrac{dy}{dx}$$ is equal to

A
$$\dfrac{2^x + 2^y}{2^x - 2^y}$$

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

C
$$2^{x - y} \left( \dfrac{2^y - 1}{1 - 2^x} \right )$$

D
$$\dfrac{2^{x + y} - 2^x}{2^y}$$

Solution

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

Differentiating both sides

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

So option $$C$$ is correct.
Question 9
1 / -0

If $$s=\sec ^{ -1 }{ \left( \cfrac { 1 }{ 2{ x }^{ 2 }-1 } \right) } $$ and $$t=\sqrt { 1-{ x }^{ 2 } } $$, then $$\cfrac { ds }{ dt } $$ at $$x=\cfrac { 1 }{ 2 }$$ is

A
$$1$$

B
$$2$$

C
$$-2$$

D
$$4$$

E
$$-4$$

Solution

Given, $$s=\sec ^{ -1 }{ \left( \cfrac { 1 }{ 2{ x }^{ 2 }-1 } \right) } $$

$$\Rightarrow s=\cos ^{ -1 }{ (2{ x }^{ 2 }-1) } $$

$$\left[ \because \cos ^{ -1 }{ x } =\sec ^{ -1 }{ \cfrac { 1 }{ x } } \right] $$

$$\Rightarrow s=2\cos ^{ -1 }{ x } $$

$$\left[ \because 2\cos ^{ -1 }{ x } =\cos ^{ -1 }{ (2{ x }^{ 2 }-1) } \right] $$

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

$$\cfrac { ds }{ dx } =2.\cfrac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } =\cfrac { -2 }{ \sqrt { 1-{ x }^{ 2 } } } $$

also $$\quad t=\sqrt { 1-{ x }^{ 2 } } $$

On differentiating again

$$\cfrac { dt }{ dx } =\cfrac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } } } (-2x)=\cfrac { -x }{ \sqrt { 1-{ x }^{ 2 } } } $$

Now, $$\cfrac { ds }{ dx } =\cfrac { ds }{ dx } \times \cfrac { dx }{ dt } $$

$$=\cfrac { -2 }{ \sqrt { 1-{ x }^{ 2 } } } \times \cfrac { \sqrt { 1-{ x }^{ 2 } } }{ -x } =\cfrac { 2 }{ x } \quad $$

$$\therefore { \left( \cfrac { ds }{ dt } \right) }_{ x=\cfrac { 1 }{ 2 } }=\cfrac { 2 }{ 1/2 } =4$$
Question 10
1 / -0

If $$x = \sin t$$ and $$y = \tan t$$, then $$\dfrac{dy}{dx}$$ is equal to

A
$$\cos^3 t$$

B
$$\dfrac{1}{\cos^3 t}$$

C
$$\dfrac{1}{\cos^2 t}$$

D
$$\sin^2 t$$

E
$$\dfrac{1}{\sin^2 t}$$

Solution

Given, $$x = \sin t$$ and $$y = \tan t$$
On differentiating both sides w.r.t. $$t$$ respectively, we get
$$\dfrac{dx}{dt} = \cos t$$ and $$\dfrac{dy}{dt} = \sec^2 t$$
Therefore, $$ \dfrac{dy/dt}{dx/dt} = \dfrac{\sec^2 t}{\cos t} = \dfrac{1}{\cos^3 t}$$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 19

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Continuity and Differentiability Test - 19

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SHARING IS CARING

Question 1

$$-1$$

$$1$$

$$\cfrac{1}{2}$$

$$\cfrac{1}{\sqrt{2}}$$

Solution

Question 2

$$\dfrac {y}{x}$$

$$-\dfrac {y}{x}$$

$$\dfrac {x}{y}$$

$$-\dfrac {x}{y}$$

Solution

Question 3

$$1$$ only

$$2$$ only

Both $$1$$ and $$2$$

Neither $$1$$ nor $$2$$

Solution

Question 4

$$\dfrac{1}{5}$$

$$\dfrac{1}{24}$$

$$\sqrt{24}$$

$$-\dfrac{1}{\sqrt{24}}$$

Solution

Question 5

$$\dfrac { { n }^{ 2 }\left( { y }^{ 2 }+4 \right) }{ { x }^{ 2 }+4 } $$

$$\dfrac { { n }^{ 2 }\left( { y }^{ 2 }-4 \right) }{ { x }^{ 2 } } $$

$$n\dfrac { \left( { y }^{ 2 }-4 \right) }{ { x }^{ 2 }-4 } $$

$${ \left( \dfrac { ny }{ x } \right) }^{ 2 }-4$$

Solution

Question 6

$$\tan \theta$$

$$\tan^2 \theta$$

$$1$$

$$\sec^2 \theta$$

$$\sec \theta$$

Solution

Question 7

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

$$a = -1, b = 1 + \sqrt {2}$$

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

None of these

Solution

Question 8

$$\dfrac{2^x + 2^y}{2^x - 2^y}$$

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

$$2^{x - y} \left( \dfrac{2^y - 1}{1 - 2^x} \right )$$

$$\dfrac{2^{x + y} - 2^x}{2^y}$$

Solution

Question 9

$$1$$

$$2$$

$$-2$$

$$4$$

$$-4$$

Solution

Question 10

$$\cos^3 t$$

$$\dfrac{1}{\cos^3 t}$$

$$\dfrac{1}{\cos^2 t}$$

$$\sin^2 t$$

$$\dfrac{1}{\sin^2 t}$$

Solution

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