Continuity and Differentiability Test

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Continuity and Differentiability Test - 39

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Question 1
1 / -0

If a curve is given by $$x=a\cos t+\displaystyle\frac{b}{2}\cos 2t$$ and $$y=\sin t +\displaystyle\frac{b}{2}\sin 2t$$, then the points for which $$\displaystyle\frac{d^2y}{dx^2}=0$$, are given by.

A
$$\sin t=\displaystyle\frac{2a^2+b^2}{3ab}$$

B
$$\cos t=-\left [\displaystyle\frac{a^2+2b^2}{3ab}\right]$$

C
$$\tan t=a/b$$

D
None of the above

Solution

We have,
$$x=a\cos t+\displaystyle\frac{b}{2}\cos 2t$$
and $$y=a\sin t+\displaystyle\frac{b}{2}\sin 2t$$
On differentiating both equations w.r.t. t, we get
$$\displaystyle\frac{dx}{dt}=-a\sin t -b\sin 2t$$
and $$\displaystyle\frac{dy}{dt}=a\cos t +b\cos 2t$$
$$\therefore \displaystyle\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t +b\cos 2t}{-a\sin t-b\sin 2t}$$
$$\Rightarrow \displaystyle\frac{dy}{dx}=-\frac{(a\cos t+b\cos 2t)}{(a\sin t+b\sin 2t)}$$
Now, $$\displaystyle\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\displaystyle\frac{dy}{dx}\right)=\frac{d}{dt}\left(\displaystyle\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$$
$$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\frac{d}{dt}\left(\displaystyle\frac{a\cos t+b\cos 2t}{a\sin t+b\sin 2t}\right)\frac{dt}{dx}$$
$$=\displaystyle \frac{\displaystyle \begin{bmatrix} (a\sin t+b]sin 2t)(-a\sin t-2b\sin 2t) \\ -(a\cos t+b\cos 2t)(a\cos t+2b\cos 2t)\end{bmatrix}}{(a\sin t+b\sin 2t)^2}\times \frac{1}{-a\sin t-b\sin 2t}$$
$$=\displaystyle\frac{\displaystyle\begin{bmatrix} a^2\sin^2t+3ab\sin t\sin 2t+2b^2\sin^22t\\ +a^2\cos^2t+3ab\cos t\cos 2t+2b^2\cos^22t\end{bmatrix}}{(a\sin t+b\sin 2t)^3}$$
$$=\displaystyle\frac{a^2(\sin^2t+\cos^2t)+b^2(\sin^22t+\cos^22t)+3ab\cos(2t-t)}{(a\sin t+b\sin 2t)^3}$$
$$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\left[\displaystyle\frac{a^2+2b^2+3ab\cos t}{(a\sin t+b\sin 2t)^3}\right]$$
Given, $$\displaystyle\frac{d^2y}{dx^2}=0$$
$$\Rightarrow a^2+2b^2+3ab\cos t=0$$
$$\Rightarrow \cos t=-\left(\displaystyle\frac{a^2+2b^2}{3ab}\right)$$
Question 2
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Let f be a function which is continuous and differentiable for all real x. If $$f(2)=-4$$ and $$f'(x)\geq 6$$ for all $$x\epsilon [2, 4]$$, then.

A
$$f(4)<8$$

B
$$f(4)\geq 8$$

C
$$f(4) > 12$$

D
$$f(4) >8$$

Solution

As $$f(2)=-4$$ and $$f'(x) \geq 6 \: \forall x\in [2,4] $$

Applying Lagrange's mean value theorem,

$$\dfrac{f(4)-f(2)}{4-2}=f'(c) \geq 6$$

$$\Rightarrow \dfrac{f(4)-f(2)}{2}\geq 6$$

$$\Rightarrow f(4)-f(2)\geq 12$$

$$\Rightarrow f(4)\geq 12+f(2)$$

$$\Rightarrow f(4)\geq 12-4$$

$$\Rightarrow f(4)\geq 8$$
Question 3
1 / -0

Let $$f(x)$$ be a differentiable function such that $$f(x) = x^{2} + \int_{0}^{x} e^{-t} f(x - t)dt$$, then $$\int_{0}^{1}f(x) dx=$$

A
$$\dfrac {1}{3}$$

B
$$\dfrac {1}{4}$$

C
$$\dfrac {7}{12}$$

D
$$\dfrac {5}{12}$$

Solution
Question 4
1 / -0

If $$\sin ^{ -1 }{ \left( \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\log { a } $$, then $$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } $$ equals

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

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

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

D
$$0$$

Solution

We have $$\sin ^{ -1 }{ \left( \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\log { a } $$
$$\Rightarrow \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } =\sin { \left( \log { a } \right) } $$
$$\Rightarrow \dfrac { 1-\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } =\sin { \left( \log { a } \right) } $$
(on putting $$y=x\tan { \theta } $$)
$$\Rightarrow \cos { 2\theta } =\sin { \left( \log { a } \right) } $$
$$\Rightarrow 2\theta =\cos ^{ -1 }{ \left( \sin { \log { a } } \right) }$$
$$\Rightarrow \theta =\dfrac { 1 }{ 2 } \cos ^{ -1 }{ \left\{ \sin { \left( \log { a } \right) } \right\} } $$
$$\tan ^{ -1 }{ \left( \dfrac { y }{ x } \right) } =\dfrac { 1 }{ 2 } \cos ^{ -1 }{ \left\{ \sin { \log { a } } \right\} } $$
$$\Rightarrow \dfrac { 1 }{ 1+\dfrac { { y }^{ 2 } }{ { x }^{ 2 } } } \cdot \dfrac { x\dfrac { dy }{ dx } -y }{ { x }^{ 2 } } =0$$
$$\Rightarrow x\dfrac { dy }{ dx } -y=0$$
$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { y }{ x } $$ ...(i)
$$\Rightarrow \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\dfrac { -y }{ { x }^{ 2 } } +\dfrac { 1 }{ x } \cdot \dfrac { dy }{ dx } =\dfrac { -y }{ { x }^{ 2 } } +\dfrac { 1 }{ x } \left( \dfrac { y }{ x } \right) $$ ....{from (i)}
$$\Rightarrow \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\dfrac { -y }{ { x }^{ 2 } } +\dfrac { y }{ { x }^{ 2 } } =0$$
Question 5
1 / -0

If $$y=\tan ^{ -1 }{ \left( \cfrac { 4x }{ 1+5{ x }^{ 2 } } \right) } +\tan ^{ -1 }{ \left( \cfrac { 2+3x }{ 2-3x } \right) } $$, then $$\cfrac { dy }{ dx } $$ is

A
$$\cfrac { 6 }{ 1+4{ x }^{ 2 } } $$

B
$$\cfrac { 3 }{ 1+4{ x }^{ 2 } } $$

C
$$\cfrac { 5 }{ 1+{ 25x }^{ 2 } } $$

D
$$\dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} - \dfrac{1.5}{(1+2.25x^{2})}$$

Solution

Use properties of inverse trigonometric functions-

y= $$\tan ^{ -1 }{ 5x}+\tan^{-1}{(-x)}$$ +$$\tan^{-1}{1} +\tan^{-1}{(-1.5x)}$$

$$\Rightarrow$$ y= $$\tan ^{ -1 }{ 5x}-\tan^{-1}{x}$$ +$$\tan^{-1}{1} -\tan^{-1}{1.5x}$$

Differentiate with respect to x,

$$\therefore $$ $$\dfrac{d(tan^{-1}x)}{dx}$$ = $$\dfrac{1}{(1+x^{2})}$$

So,

$$\therefore$$ $$\dfrac{dy}{dx} = \dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} + 0 - \dfrac{1.5}{(1+2.25x^{2})}$$

$$\therefore$$ $$\dfrac{dy}{dx} = \dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} - \dfrac{1.5}{(1+2.25x^{2})}$$
Question 6
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Let $$f\left( x \right)$$ and $$g\left( x \right)$$ be defined and differentiable for $$x\ge { x }_{ 0 }$$ and $$f\left( { x }_{ 0 } \right) =g\left( { x }_{ 0 } \right) , f^{ ' }\left( x \right) >g^{ ' }\left( x \right) for\ x>{ x }_{ 0 }$$ then

A
$$f\left( x \right) < g\left( x \right)$$ for some $$x>{ x }_{ 0 }$$

B
$$f\left( x \right) =g\left( x \right)$$ for some $$x>{ x }_{ 0 }$$

C
$$f\left( x \right) >g\left( x \right)$$ for all $$x>{ x }_{ 0 }$$

D
None of these
Question 7
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Let g (x), $$x \geq 0$$, be a non-negative continuous function and let $$F(x) = \int_{0}^{x} f(t) dt$$, $$x \geq 0$$. If for some c > 0, f(x) $$\leq$$ c F(x) for all $$x \geq 0$$, then

A
$$f(x) = 0, \forall x \leq 1$$

B
$$f(x) = 0, \forall x \geq 1$$

C
$$f(x) = 1, \forall x \geq 1$$

D
$$f(x) = -1, \forall x \geq 1$$

Solution

$$f\left(x\right)\ge 0$$ for $$x\ge 0$$
$$F\left(x\right)=\int_{0}^{x}{f\left(t\right)dt}$$ for all $$x\ge 0$$
As $$c<0,f\left(x\right)\le cF\left(x\right)$$ for $$x\ge 0$$
$$f\left(x\right)=0\Rightarrow {F}^{\prime}{\left(x\right)}=f\left(x\right)$$ for $$x\ge 0$$
Since $$f\left(x\right)\le cF{\left(x\right)}$$ since $${F}^{\prime}{\left(x\right)}=f\left(x\right)$$
$$\Rightarrow {e}^{-cx}{F}^{\prime}{\left(x\right)}-c{e}^{-cx}F{\left(x\right)}\le 0$$ by multiplying by $${e}^{-cx}$$ where $${e}^{-cx}>0$$
When $$x\rightarrow \infty, {e}^{-cx}\rightarrow 0$$
$$\Rightarrow \dfrac{d}{dx}{\left({e}^{-cx}F\left(x\right)\right)}\le 0$$ for all $$x\ge 0$$
$$\Rightarrow {e}^{-cx}F\left(x\right)$$ is a decreasing function for $$x\ge 0$$
Let $$g\left(x\right)={e}^{-cx}F\left(x\right)$$ be a decreasing function.
$$\Rightarrow g\left(0\right)\ge g\left(x\right)$$ for $$x\ge 0$$
$$\Rightarrow 1\times F\left(0\right)\ge {e}^{-cx}F\left(x\right)$$
Given:$$F{\left(0\right)}=\int_{0}^{0}{F\left(t\right)dt}=0$$
$$\Rightarrow {e}^{-cx}F\left(x\right)\le 0$$
$$\Rightarrow F\left(x\right)\le 0$$ and $${e}^{-cx}\ge 0$$
$$\Rightarrow c>0$$
$$\Rightarrow cF\left(x\right)\le 0$$
$$\Rightarrow f\left(x\right)\le cF\left(x\right)\le 0$$
Since $$f\left(x\right)$$ is non-negative $$f\left(x\right)\le 0\Rightarrow f\left(x\right)=0$$
Question 8
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If the function $$f:\left[ 0,8 \right] \rightarrow R$$ is differentiable, then for $$0<a,b<2,\int _{ 0 }^{ 8 }{ f(t) } dt$$ is equal to

A
$$2\left[ { \alpha }^{ 3 }f({ \alpha }^{ 2 })+{ \beta }^{ 2 }f({ \beta }^{ 2 }) \right] $

B
$$3\left[ { \alpha }^{ 3 }f({ \alpha }^{ })+{ \beta }^{ 2 }f({ \beta }^{ }) \right] $$

C
$$3\left[ { \alpha }^{ 2 }f({ \alpha }^{ 3 })+{ \beta }^{ 2 }f({ \beta }^{ 3 }) \right] $$

D
$$3\left[ { \alpha }^{ 2 }f({ \alpha }^{ 2 })+{ \beta }^{ 2 }f({ \beta }^{ 2 }) \right] $$
Question 9
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A point where function $$f(x)$$ is not continuous where $$f(x)=\left[ \sin { \left[ x \right] } \right] $$ in $$\left( 0,2\pi \right) $$; is ($$\left[ \ast \right] $$ denotes greatest integer $$\le x$$)

A
$$(3,0)$$

B
$$(2,0)$$

C
$$(1,0)$$

D
$$(4,-1)$$
Question 10
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Consider $$f(x)=\lim _{ n\rightarrow \infty }{ \cfrac { { x }^{ n }-\sin { { x }^{ n } } }{ { x }^{ n }+\sin { { x }^{ n } } } } $$ for $$x>0,x\neq 1,f(1)=0$$ then

A
$$f$$ is continuous at $$x=1$$

B
$$f$$ has a discontinuity at $$x=1$$

C
$$f$$ has an infinite or oscillatory discontinuity at $$x=1$$

D
$$f$$ has a removal type of discontinuity at $$x=1$$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 39

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Continuity and Differentiability Test - 39

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SHARING IS CARING

Question 1

$$\sin t=\displaystyle\frac{2a^2+b^2}{3ab}$$

$$\cos t=-\left [\displaystyle\frac{a^2+2b^2}{3ab}\right]$$

$$\tan t=a/b$$

None of the above

Solution

Question 2

$$f(4)<8$$

$$f(4)\geq 8$$

$$f(4) > 12$$

$$f(4) >8$$

Solution

Question 3

$$\dfrac {1}{3}$$

$$\dfrac {1}{4}$$

$$\dfrac {7}{12}$$

$$\dfrac {5}{12}$$

Solution

Question 4

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

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

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

$$0$$

Solution

Question 5

$$\cfrac { 6 }{ 1+4{ x }^{ 2 } } $$

$$\cfrac { 3 }{ 1+4{ x }^{ 2 } } $$

$$\cfrac { 5 }{ 1+{ 25x }^{ 2 } } $$

$$\dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} - \dfrac{1.5}{(1+2.25x^{2})}$$

Solution

Question 6

$$f\left( x \right) < g\left( x \right)$$ for some $$x>{ x }_{ 0 }$$

$$f\left( x \right) =g\left( x \right)$$ for some $$x>{ x }_{ 0 }$$

$$f\left( x \right) >g\left( x \right)$$ for all $$x>{ x }_{ 0 }$$

None of these

Question 7

$$f(x) = 0, \forall x \leq 1$$

$$f(x) = 0, \forall x \geq 1$$

$$f(x) = 1, \forall x \geq 1$$

$$f(x) = -1, \forall x \geq 1$$

Solution

Question 8

$$2\left[ { \alpha }^{ 3 }f({ \alpha }^{ 2 })+{ \beta }^{ 2 }f({ \beta }^{ 2 }) \right] $

$$3\left[ { \alpha }^{ 3 }f({ \alpha }^{ })+{ \beta }^{ 2 }f({ \beta }^{ }) \right] $$

$$3\left[ { \alpha }^{ 2 }f({ \alpha }^{ 3 })+{ \beta }^{ 2 }f({ \beta }^{ 3 }) \right] $$

$$3\left[ { \alpha }^{ 2 }f({ \alpha }^{ 2 })+{ \beta }^{ 2 }f({ \beta }^{ 2 }) \right] $$

Question 9

$$(3,0)$$

$$(2,0)$$

$$(1,0)$$

$$(4,-1)$$

Question 10

$$f$$ is continuous at $$x=1$$

$$f$$ has a discontinuity at $$x=1$$

$$f$$ has an infinite or oscillatory discontinuity at $$x=1$$

$$f$$ has a removal type of discontinuity at $$x=1$$

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