Continuity and Differentiability Test

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Continuity and Differentiability Test - 26

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Question 1
1 / -0

If $$x=a\sec^{2}\theta$$, $$y=a\tan^{3}\theta$$ then $$\dfrac {d^{3}y}{dx^{3}}$$

A
$$\dfrac {-3}{8a^{2}}\cot^{3}\theta$$

B
`$$\dfrac {3}{8a^{2}}\cot^{3}\theta$$

C
$$3\sec^{2}\theta \tan \theta$$

D
$$\dfrac {3}{4a^{2}}\cot^{3}\theta$$

Solution

$$\alpha =a\sec^2 \theta , \ y=a\tan 3 \theta ,\ \dfrac {d^3y}{dx^3}=?$$
$$\dfrac {d^3y}{dx^3}=\dfrac {}d^3\theta{d\theta ^3} \dfrac {d^3\theta}{dx^3}$$
$$\dfrac {dy}{d\theta} =\dfrac {d}{d\theta} (a\tan ^3\theta) =a.3\tan^2 \theta \sec^2 \theta$$
$$\dfrac {d^2 y}{d\theta ^2}=a^3 (\sec^2 \theta. \tan \theta .\sec^2 \theta +\tan^2 \theta .2\sec \theta (\sec \theta \tan \theta))$$
$$=3a (2\tan \theta \sec^4\theta +2\tan^3 \theta \sec^2\theta )$$
$$\dfrac {dy}{d\theta^2} =60 [(\tan \theta .4\sec^3\theta \sec \theta +\sec^4 \theta .\sec^2 \theta) +(\tan^3\theta .2\sec \theta \sec \theta \tan \theta +\sec^2 \theta .3\tan^2 \theta \sec^2 \theta)]$$
$$=60 [5\sec^4\theta +\sec^6 \theta +2\tan^4 \theta \sec^2\theta +3\tan^2 \theta \sec^2 \theta]$$
$$=60[7\sec^4\theta \tan^2 \theta +2\tan^4 \theta \sec^2 \theta +\sec^6 \theta]$$
$$\dfrac {dx}{d\theta} \Rightarrow \ \dfrac {d}{d\theta}(a\sec^2 \theta)=a.2\sec \theta \sec \theta \tan \theta$$
$$=2a(\sec^2\theta \tan \theta)$$
$$\dfrac {d^2x}{d\theta} =2a(\tan \theta .2\sec \theta \tan \theta +\sec^2 \theta -\sec^2 \theta )$$
$$=2a((2\tan^2 \theta \sec^2 \theta)+\sec ^4 \theta)$$
$$\dfrac {d^3 x}{d\theta ^3}=2a[(2\tan^2\theta 32\sec \theta \sec \theta \tan \theta+2\sec^2 \theta +2\sec^2\theta 2\tan \theta \sec^2 \theta )+ 4\sec^3\theta \sec \theta \tan \theta]$$
$$=2a [4\tan^3\theta \sec^2\theta +4\sec \theta \sec^4 \theta+4\tan \theta \sec^3\theta]$$
$$=2a.4 [\tan^3\theta \sec^2\theta +2\tan \theta \sec^4 \theta]$$
$$\dfrac {d^3y}{dx^3}\ \Rightarrow \ \dfrac {6a (7\sec^4 \theta \tan^2 \theta +2\tan^4 \theta \sec^2\theta +\sec^6 \theta)}{8a (\tan \theta \sec^2 \theta [\tan^2\theta +2\sec^2 \theta])}$$
$$\Rightarrow \ \dfrac {3}{4} \dfrac {\sec^2\theta [7 \sec^2 \theta \tan ^2 \theta + 2\tan ^4 \theta +\sec^4 \theta]}{\tan \theta \sec^4 \theta (\tan ^2\theta +2 \sec^2 \theta)}$$
$$\Rightarrow \ (Since\ 1+\tan^2\theta =\sin^2\theta)$$
$$\Rightarrow \ \dfrac {3}{4} \left [\dfrac {7(1+\tan^2\theta) (\tan^2 \theta)+2\tan^2 \theta +(1+\tan^2\theta)}{\tan \theta (\tan \theta+2(1+\tan \theta))}\right]$$
on further simplyfying we get
$$\Rightarrow \ 3\sec^2 \theta \tan \theta$$
Question 2
1 / -0

The value of k which makes $$f(x)=\left\{\begin{matrix} \sin\dfrac{1}{x}, x\neq 0\\ k, x=0\end{matrix}\right.$$ continuous at $$x=0$$ is?

A
$$8$$

B
$$1$$

C
$$-1$$

D
None

Solution

Then k=?
$$\begin{matrix} \Rightarrow f\left( 0 \right) =k \\ \Rightarrow { f\left( { { 0^{ + } } } \right) }_{ \lim \, \, x\to { 0^{ + } } }={ f\left( { { 0^{ +h } } } \right) }_{ \lim \, \, h\to 0 }={ \sin }_{ \lim \, \, h\to 0 }\dfrac { 1 }{ { (o+h) } } =\sin 0 \\ \Rightarrow { f\left( { { 0^{ - } } } \right) }_{ \lim \, \, x\to { 0^{ - } } }={ f\left( { { 0^{ -h } } } \right) }_{ \lim \, \, h\to 0 }={ \sin }_{ \lim \, \, h\to 0 }\dfrac { 1 }{ { (o-h) } } =\sin 0 \\ \Rightarrow f\left( 0 \right) =f\left( { { 0^{ + } } } \right) =f\left( { { 0^{ - } } } \right) =f\left( 0 \right) =k=0 \\ \end{matrix}$$
the value of $$k$$ is zero
Question 3
1 / -0

If $$x=\cos ec \theta-\sin \theta,y=\cos ec^{n}\theta-\sin^{n}\theta$$ then $$(x^{2}+4)\left(\dfrac {dy} {dx}\right)^{2}-n^{2}y^{2}=$$

A
$$n^{2}$$

B
$$2n^{2}$$

C
$$3n^{2}$$

D
$$4n^{2}$$

Solution

$$ x = cosec \theta - sin\theta $$
$$ x^2+4 = (cosec\theta - sin\theta)^2+4 $$
$$ = (cosec\theta + sin\theta)^2$$
$$ y = cosec^n \theta - sin^n \theta $$
$$ y^2+4 = (cosec^n \theta + sin^n \theta)^2$$
$$ \dfrac{dy}{d\theta} = n. cosec^{n} \theta (-cosec \theta cot \theta)+ nsin^{n-1}\theta. cos\theta.$$
$$ = n cot \theta (cosec^n\theta+sin^n \theta)$$
$$ \dfrac{dx}{d\theta} = cot\theta cosec\theta-cos \theta $$
$$ = -cot\theta = (cosec \theta + sin\theta )$$
$$ \dfrac{dy}{dx} = \dfrac{ncot \theta}{-cot \theta} (\dfrac{cosec^n \theta + sin^n \theta}{ cosec\theta + sin\theta}) $$
$$ (\dfrac{dy}{dx})^{2} = \dfrac{n^2(cosec^n \theta + sin^n \theta )^{2}}{(-1)^2 (cosec \theta + sin \theta)^{2}}$$
$$ = n^2 (\dfrac{y^2+4}{x^2+4 })$$
$$ (x^2+4) (\dfrac{dy}{dx})^2 - x^{2}y^{2} = 4x^{2}$$
option D
Question 4
1 / -0

Let $$f(x) = \underset{0}{\overset{x}{\int}} |2t - 3| dt$$, then $$f$$ is

A
continuous at $$x = \dfrac{3}{2}$$

B
continuous at $$x = 3$$

C
differentiable at $$x = \dfrac{3}{2}$$

D
differentiable at $$x = 0$$
Question 5
1 / -0

Let $$f(x) = \left\{\begin{matrix}x &x < 1 \\ 2 - x & 1 \leq x \leq 2\\ -2 + 3x - x^{2} & x > 2\end{matrix}\right.$$ then $$f(x)$$ is

A
Differentiable at $$x = 0$$

B
Differentiable at $$x = 2$$

C
Differentiable at $$x = 1$$ and $$x = 2$$

D
Not differentiable at $$x = 0$$

Solution
Question 6
1 / -0

Directions For Questions

Let two functions $$f\left( x \right) =\begin{cases} { e }^{ x }+a; & x\ge 0 \\ { e }^{ -x }+b; & x<0 \end{cases}$$ & $$g\left( x \right) =\begin{cases} c\sin { x } ; & x\ge 0 \\ d+\cos { x } ; & x<0 \end{cases}$$ are such that $$f(x)+g(x)$$ & $$f(x).g(x)$$ both are differentiable at $$x=0$$.
On the basis of above information, answer the following questions :

...view full instructions

Range of function $$f(x)-g(x)$$ in [-$$\pi, \pi$$] is-

A
$$[0, e^{\pi}+1]$$

B
$$[0, e^{\pi}-1]$$

C
$$e^{\pi}-1, e^{\pi}+1]$$

D
$$[e^{-\pi}, e^{\pi}]$$
Question 7
1 / -0

Directions For Questions

Let two functions $$f\left( x \right) =\begin{cases} { e }^{ x }+a; & x\ge 0 \\ { e }^{ -x }+b; & x<0 \end{cases}$$ & $$g\left( x \right) =\begin{cases} c\sin { x } ; & x\ge 0 \\ d+\cos { x } ; & x<0 \end{cases}$$ are such that $$f(x)+g(x)$$ & $$f(x).g(x)$$ both are differentiable at $$x=0$$.
On the basis of above information, answer the following quaestions :

...view full instructions

Which of the following is NOT CORRECT-

A
$$b=d$$

B
$$a=d$$

C
$$a\neq b$$

D
$$c\neq d$$
Question 8
1 / -0

$$\sqrt{1+\left(\dfrac{d^2y}{dx^2}\right)^3}=\left(2+\dfrac{dy}{dx}\right)^{1/3}$$
Find it's order and degree.

A
$$2, 3$$

B
$$2, 9$$

C
$$2, 6$$

D
$$2, 2$$

Solution

Consider the given equation.

$$\sqrt{1+{{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}}={{\left( 2+\dfrac{dy}{dx} \right)}^{1/3}}$$

On squaring both sides, we get

$$1+{{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}={{\left( 2+\dfrac{dy}{dx} \right)}^{2/3}}$$

Here, Order $$=2$$ and Degree $$=3$$

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

If $$x$$ $$sin$$ $$y$$$$=3 sin y$$ $$+$$ $$4 cos y$$, then $$\frac { dy }{ dx } =$$

A
$$\frac { { -sin }^{ 2 }y }{ 4 } $$

B
$$\frac { { sin }^{ 2 }y }{ 4 } $$

C
$$\frac { { -cos }^{ 2 }y }{ 4 } $$

D
$$\frac { { cos }^{ 2 }y }{ 4 } $$

Solution

$$x\sin{y}=3\sin{y}+4\cos{y}$$
$$\Rightarrow\,\left(x-3\right)\sin{y}=4\cos{y}$$
$$\Rightarrow\,x-3=4\dfrac{\cos{y}}{\sin{y}}$$
$$\Rightarrow\,x-3=4\cot{y}$$
Differentiating both sides w.r.t $$x$$
$$\Rightarrow\,1=4\left(-{\csc}^{2}{y}\right)\dfrac{dy}{dx}$$
$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{-1}{4{\csc}^{2}{y}}$$
$$\therefore\,\dfrac{dy}{dx}=\dfrac{-{\sin}^{2}{y}}{4}$$
Question 10
1 / -0

Directions For Questions

Let two functions $$f\left( x \right) =\begin{cases} { e }^{ x }+a; & x\ge 0 \\ { e }^{ -x }+b; & x<0 \end{cases}$$ & $$g\left( x \right) =\begin{cases} c\sin { x } ; & x\ge 0 \\ d+\cos { x } ; & x<0 \end{cases}$$ are such that $$f(x)+g(x)$$ & $$f(x).g(x)$$ both are differentiable at $$x=0$$.
On the basis of above information, answer the following quaestions :

...view full instructions

Function $$f(x)-g(x)$$ is-

A
continuous & differentiable at $$x=0$$

B
continuous but not differentiable at $$x=0$$

C
discontinuous at $$x=0$$

D
non differentiable because it is discontinuous at $$x=0$$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 26

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Continuity and Differentiability Test - 26

Score

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Rank

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SHARING IS CARING

Question 1

$$\dfrac {-3}{8a^{2}}\cot^{3}\theta$$

`$$\dfrac {3}{8a^{2}}\cot^{3}\theta$$

$$3\sec^{2}\theta \tan \theta$$

$$\dfrac {3}{4a^{2}}\cot^{3}\theta$$

Solution

Question 2

$$8$$

$$1$$

$$-1$$

None

Solution

Question 3

$$n^{2}$$

$$2n^{2}$$

$$3n^{2}$$

$$4n^{2}$$

Solution

Question 4

continuous at $$x = \dfrac{3}{2}$$

continuous at $$x = 3$$

differentiable at $$x = \dfrac{3}{2}$$

differentiable at $$x = 0$$

Question 5

Differentiable at $$x = 0$$

Differentiable at $$x = 2$$

Differentiable at $$x = 1$$ and $$x = 2$$

Not differentiable at $$x = 0$$

Solution

Question 6

$$[0, e^{\pi}+1]$$

$$[0, e^{\pi}-1]$$

$$e^{\pi}-1, e^{\pi}+1]$$

$$[e^{-\pi}, e^{\pi}]$$

Question 7

$$b=d$$

$$a=d$$

$$a\neq b$$

$$c\neq d$$

Question 8

$$2, 3$$

$$2, 9$$

$$2, 6$$

$$2, 2$$

Solution

Question 9

$$\frac { { -sin }^{ 2 }y }{ 4 } $$

$$\frac { { sin }^{ 2 }y }{ 4 } $$

$$\frac { { -cos }^{ 2 }y }{ 4 } $$

$$\frac { { cos }^{ 2 }y }{ 4 } $$

Solution

Question 10

continuous & differentiable at $$x=0$$

continuous but not differentiable at $$x=0$$

discontinuous at $$x=0$$

non differentiable because it is discontinuous at $$x=0$$

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