Continuity and Differentiability Test

Result

Continuity and Differentiability Test - 27

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Differentiate $$\log(1+x^{2})$$ with respect $$\tan^{-1}x$$.

A
$$2x$$

B
$$x^2$$

C
$$x$$

D
$$x^3$$

Solution

Let $$y=\log(1+x^{2})$$
$$y=\tan^{-1}(x)$$
To find $$\dfrac{dy}{dz}$$
first we find $$\dfrac{dy}{dx}=\dfrac{d}{dx}(\log(1+x^{2})=\dfrac{2x}{1+x^{2}}$$
Next we find
$$\dfrac{dz}{dx}=\dfrac{d}{dx}(\tan^{-1}(x))=\dfrac{1}{1+x^{2}}$$
Now $$\dfrac{dy}{dz}=\dfrac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}}=\dfrac{\dfrac{2x}{1+x^{2}}}{\dfrac{1}{1+x^{2}}}=2x$$
Question 2
1 / -0

An angle $$\theta$$ through which a pulley turns with time $$'t'$$ is completed by $$\theta = t^{2} + 3t - 5\ sq. cms/ min$$. Then the angular velocity for $$t = 5\ sec$$.

A
$$5^{c}/ sec$$

B
$$13^{c}/ sec$$

C
$$23^{c}/ sec$$

D
$$35^{c}/ sec$$

Solution
Question 3
1 / -0

If $$f\left( x \right) = \left( {x - a} \right)g\left( x \right)$$ and $$g\left( x \right)$$ is continuous $$x = a$$ then $${f^{'}}\left( 1 \right) = $$

A
$$g\left( 1 \right)$$

B
$${g^{'}}\left( 1 \right)$$

C
$$ - {g^{'}}\left( 1 \right)$$

D
$${g^{'}}\left( { - 1} \right)$$

Solution

From given identity ,
$$f(a)=0$$
$$f(a+h)=(a+h-a)g(a+h)=hg(a+h)$$
$$f(a-h)=(a-h-a)g(a-h)=-hg(a-h)$$

$$f'(a)= \displaystyle \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\dfrac{hg(a+h)-f(a)}{h}$$ .......(RHD,Similiarly LHD can be worked out )
Since $$g(x)$$ is continuous at $$x=a$$ , we can write above equation as
$$f'(a)=\displaystyle \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\dfrac{hg(a)-f(a)}{h}=\lim_{h\to 0}\dfrac{hg(a)}{h}=g(a)$$

So finally we arrive at $$f'(a)=g(a)$$
So $$f'(1)=g(1)$$
Question 4
1 / -0

If the following function is continuous at $$x=0$$, find the value of $$k$$:
$$f\left( x \right) = \left\{ {\begin{array}{ccccccccccccccc}{\dfrac{{\sin \frac{{3x}}{2}}}{x}}&{,x \ne 0}\\k&{,x = 0}\end{array}} \right.$$

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

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

C
$$\dfrac{4}{3}$$

D
$$\dfrac{3}{2}$$

Solution
Question 5
1 / -0

Let $$g\left(x\right)=\dfrac{f\left(x\right)}{x+1}$$ where $$f\left(x\right) $$ is differentiable on $$\left[0,5\right]$$ such that $$f\left(0\right)=4,f\left(5\right)=-1$$. There exists $$c\in \left(0,5\right)$$ such that $$g^{ ' }\left( c \right) $$ is ?

A
$$-\dfrac{1}{6}$$

B
$$\dfrac{1}{6}$$

C
$$-\dfrac{5}{6}$$

D
$$-1$$

Solution

$$g\left( x \right) =\cfrac { f\left( x \right) }{ x+1 } $$
$$\Rightarrow g(0)=\cfrac { 4 }{ 1 } =4;g\left( 5 \right)=\cfrac { -1 }{ 6 } $$
By Lagrange's mean value theorem,
$$g'\left( c \right)=\cfrac { g\left( 5 \right)-g\left( 0 \right) }{ 5-0 } =\cfrac { \cfrac { -1 }{ 6 } -4 }{ 5 } $$
$$=\cfrac { -25 }{ 30 } =\cfrac { -5 }{ 6 } $$
Question 6
1 / -0

Let $$f(x)=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$$.
The value of $$\lim _{ h\rightarrow 0 }{ \dfrac { f\left( 1-h \right) -f\left( 1 \right) }{ { h }^{ 3 }+3h } } $$ is

A
$$\dfrac{50}{3}$$

B
$$\dfrac{22}{3}$$

C
$$13$$

D
None of these
Question 7
1 / -0

If $$y=y(x)$$ and it follows the relation $$e^{xy^{2}}+y\cos(x^{2})=5$$ then $$y'(0)$$ is equal to

A
$$4$$

B
$$-16$$

C
$$-4$$

D
$$16$$

Solution
Question 8
1 / -0

A function $$f:R\rightarrow R$$ is such that $$f(1)=3$$ and $$f'(1)=6$$. Then $$\rightarrow { \displaystyle \lim _{ x\rightarrow 0 }{ \left[ \dfrac { f\left( 1+x \right) }{ f\left( 1 \right) } \right] } }^{ 1/x }=$$ ?

A
$$1$$

B
$$e^{2}$$

C
$$e^{1/2}$$

D
$$e^{3}$$

Solution
Question 9
1 / -0

If $$f$$ is a real valued differentiable function satisfying $$\left| f\left( x \right) -f\left( y \right) \right| \le { \left( x-y \right) }^{ 2 },x,y\epsilon R\ and f\left( 0 \right) ,\ then f\left( 1 \right)$$ equals ?

A
$$-1$$

B
$$0$$

C
$$2$$

D
$$1$$

Solution
Question 10
1 / -0

Solve:
$$\dfrac { d } { d x } \tan ^ { - 1 } ( \sec x + \tan x )$$

A
$$1$$

B
$$1 / 2$$

C
$$\cos x$$

D
sec $$x$$

Solution

$$\dfrac d{dx} \tan ^{-1} (\sec x+\tan x)$$

$$\implies \dfrac 1{1+(\sec x+\tan x)^2}\dfrac d{dx}(\sec x+\tan x)$$

$$\implies \dfrac 1{1+(\sec x+\tan x)^2}(\sec x\tan x+\sec ^2x)$$

$$\implies \dfrac 1{1+\sec ^2 x+\tan ^2x +2\sec x\tan x}(\sec x\tan x+\sec ^2x)$$

$$\implies \dfrac 1{2\sec^2x+2\sec x \tan x}(\sec x\tan x+\sec ^2x)$$

$$\implies \dfrac 12 $$