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

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

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

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

If $$f (x)$$ is differentiable everywhere, then:

$$ f | f |$$ is not differentiable at some point

If $$f(x)=x^{n}\sin \dfrac {1}{x},f(0)=0$$

$$f'(x)$$ is discountinuous at $$x=0$$ if $$n=0$$

$$f$$ is continuous differentiable at $$x=0$$ for any $$n< 2, n\in N$$

$$f$$ is continuous differentiable for all $$x$$ for any $$n\ge 2, n\in N$$

$$f$$ is discountinuous at $$x=0$$ if $$n=0$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 7

$x=\sqrt { { a }^{ \sin ^{ -1 }{ t } } } ;\quad y=\sqrt { { a }^{ \cos ^{ -1 }{ t } } }$ , then

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

0%
True
0%
False

$f(x)=\dfrac {1-\cos(1-\cos x)}{x^4}$ is continuous at $x= 0$ , then $f(0)=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{8}$

Explanation

$f(x)=\dfrac {1-\cos(1-\cos x)}{x^4}$

It is given that the function

$f$ is continuous at

$x = 0.$

Then,

$\displaystyle\lim_{x\rightarrow 0}f(x)=f(0)$

$f(x)=\displaystyle\lim_{x\rightarrow 0}\dfrac {1-\cos(1-\cos x)}{x^4}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{1-\cos\left(2\left(\sin\dfrac{x}{2}\right)^2\right)}{x^4}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{x^4}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{\left(\sin\dfrac{x}{2}\right)^4}\dfrac{\left(\sin\dfrac{x}{2}\right)^4}{x^4}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{\left(\sin\dfrac{x}{2}\right)^4}\dfrac{\left(\sin\dfrac{x}{2}\right)^4}{\left(\dfrac{x}{2}\right)^42^4}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{1}{2^3}\dfrac{\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{\left(\sin\dfrac{x}{2}\right)^4}\displaystyle\lim_{x\rightarrow 0}\dfrac{\left(\sin\dfrac{x}{2}\right)^4}{\left(\dfrac{x}{2}\right)^4}$

$=\dfrac{1}{8}(1)(1)$

$=\dfrac{1}{8}$

$\therefore$

$f(0)=\dfrac{1}{8}$

Let

$f(x)$ be a function defined on

$(-a,a)$ with

$a > 0$ . Amuse that

$f(x)$ is continuous at

$x=0$ and

$\underset{x\to 0}{\lim}\dfrac {f(x)-f(kx)}{x}=\alpha$ , where

$k \in (0,1)$ then compute

$f'(0^{+})$ and

$f'(0^{-})$ , and comment upon the differentiablity of

$f$ at

$x=0$ ? Denote

$\alpha$ .

0%
$\underset{x\to o}{\lim}\dfrac{f(x) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$
0%
$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{kx} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$
0%
$\underset{x\to 0}{\lim}\dfrac{f(x) - f(kx)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$
0%
$\underset{x\to o}{\lim}\dfrac{f(kx) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$

Explanation

$\underset{x \to 0^-}{\lim} f(x) = \underset{x\to 0}{\lim} + f(x)$

$\underset{x \to 0^-}{\lim} f'(x) = \underset{x\to 0}{\lim} + f'(x)$

$f'(0^=) = f'(0^-)$

$\underset{x\to 0}{\lim}\dfrac {f(x)-f(kx)}{x}=\alpha$ ...(1)

$f'(0^+)= \underset{x\to 0}{\lim} \dfrac{f(x) - f(0)}{x}$

$f' (0^-) = \underset{x\to 0}{\lim} \dfrac{f(x) - f(0)}{-x}$
From eqn (1)

$\underset{x\to 0}{\lim} \dfrac{f(x)+f(0) - f(o) - f(kx)}{x}$

$\underset{x\to o}{\lim}\dfrac{f(x) - f(0)}{x} \cdot \dfrac{f(kx) - f(0)}{kx} \times k = \alpha$

$x=a\left(\cos t+\log \tan\dfrac{t}{2}\right), y=a\sin t$ , then evaluate

$\dfrac{d^2y}{dx^2}$ at

$t=\dfrac{\pi}{3}$ .

0%
$\cfrac{4\sqrt3}{a}$
0%
$\cfrac{\sqrt3}{a}$
0%
$\cfrac{8\sqrt3}{a}$
0%
$\cfrac{8\sqrt3}{5a}$

Explanation

$x = a(cost+logtan\cfrac{t}{2})$

$1 = a(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}}) \cfrac{dt}{dx}$

$\cfrac{dt}{dx} = \cfrac{1}{a(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}})}$

$\cfrac{dy}{dt} = acost$

$\cfrac{dy}{dx} = \cfrac{cost}{(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}})}$

$\cfrac{dy}{dx} = \cfrac{cost}{(-sint+\cfrac{1}{2sin\cfrac{t}{2}cos\cfrac{t}{2}})}$

$\cfrac{dy}{dx} = \cfrac{cost}{(-sint+\cfrac{1}{sint})}$

$\cfrac{dy}{dx} = \cfrac{costsint}{1-sin^2t}$

$\cfrac{dy}{dx} = tant$

$\cfrac{d^2y}{dx^2} = sec^2t \cfrac{dt}{dx}$

$t = \cfrac{\pi}{3}$

$\cfrac{d^2y}{dx^2} = \cfrac{4}{a(-sint+\cfrac{sec^2\cfrac{t}{2}}{2tan\cfrac{t}{2}})} = \cfrac{8\sqrt3}{a}$

Let

$f(x)$ be a continuous funct and

$g(x)$ be a discontinuous

$funct^{n}$ , then

$f(x) + g(x)$ is continuous

$funct^{n}$ . ? it is .

0%
True
0%
False

$x=\cos ec \theta-\sin \theta,y=\cos ec^{n}\theta-\sin^{n}\theta$ then

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

0%
$n^{2}$
0%
$2n^{2}$
0%
$3n^{2}$
0%
$4n^{2}$

Explanation

$x = cosec \theta - sin\theta$

$x^2+4 = (cosec\theta - sin\theta)^2+4$

$= (cosec\theta + sin\theta)^2$

$y = cosec^n \theta - sin^n \theta$

$y^2+4 = (cosec^n \theta + sin^n \theta)^2$

$\dfrac{dy}{d\theta} = n. cosec^{n} \theta (-cosec \theta cot \theta)+ nsin^{n-1}\theta. cos\theta.$

$= n cot \theta (cosec^n\theta+sin^n \theta)$

$\dfrac{dx}{d\theta} = cot\theta cosec\theta-cos \theta$

$= -cot\theta = (cosec \theta + sin\theta )$

$\dfrac{dy}{dx} = \dfrac{ncot \theta}{-cot \theta} (\dfrac{cosec^n \theta + sin^n \theta}{ cosec\theta + sin\theta})$

$(\dfrac{dy}{dx})^{2} = \dfrac{n^2(cosec^n \theta + sin^n \theta )^{2}}{(-1)^2 (cosec \theta + sin \theta)^{2}}$

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

$(x^2+4) (\dfrac{dy}{dx})^2 - x^{2}y^{2} = 4x^{2}$

option D

1142314_1143747_ans_072ae63c054d4bf8a86e6bec41ff3e5d.jpg

$x=a\sec^{2}\theta$ ,

$y=a\tan^{3}\theta$ then

$\dfrac {d^{3}y}{dx^{3}}$

0%
$\dfrac {-3}{8a^{2}}\cot^{3}\theta$
0%
` $\dfrac {3}{8a^{2}}\cot^{3}\theta$
0%
$3\sec^{2}\theta \tan \theta$
0%
$\dfrac {3}{4a^{2}}\cot^{3}\theta$

Explanation

$\alpha =a\sec^2 \theta , \ y=a\tan 3 \theta ,\ \dfrac {d^3y}{dx^3}=?$

$\dfrac {d^3y}{dx^3}=\dfrac {}d^3\theta{d\theta ^3} \dfrac {d^3\theta}{dx^3}$

$\dfrac {dy}{d\theta} =\dfrac {d}{d\theta} (a\tan ^3\theta) =a.3\tan^2 \theta \sec^2 \theta$

$\dfrac {d^2 y}{d\theta ^2}=a^3 (\sec^2 \theta. \tan \theta .\sec^2 \theta +\tan^2 \theta .2\sec \theta (\sec \theta \tan \theta))$

$=3a (2\tan \theta \sec^4\theta +2\tan^3 \theta \sec^2\theta )$

$\dfrac {dy}{d\theta^2} =60 [(\tan \theta .4\sec^3\theta \sec \theta +\sec^4 \theta .\sec^2 \theta) +(\tan^3\theta .2\sec \theta \sec \theta \tan \theta +\sec^2 \theta .3\tan^2 \theta \sec^2 \theta)]$

$=60 [5\sec^4\theta +\sec^6 \theta +2\tan^4 \theta \sec^2\theta +3\tan^2 \theta \sec^2 \theta]$

$=60[7\sec^4\theta \tan^2 \theta +2\tan^4 \theta \sec^2 \theta +\sec^6 \theta]$

$\dfrac {dx}{d\theta} \Rightarrow \ \dfrac {d}{d\theta}(a\sec^2 \theta)=a.2\sec \theta \sec \theta \tan \theta$

$=2a(\sec^2\theta \tan \theta)$

$\dfrac {d^2x}{d\theta} =2a(\tan \theta .2\sec \theta \tan \theta +\sec^2 \theta -\sec^2 \theta )$

$=2a((2\tan^2 \theta \sec^2 \theta)+\sec ^4 \theta)$

$\dfrac {d^3 x}{d\theta ^3}=2a[(2\tan^2\theta 32\sec \theta \sec \theta \tan \theta+2\sec^2 \theta +2\sec^2\theta 2\tan \theta \sec^2 \theta )+ 4\sec^3\theta \sec \theta \tan \theta]$

$=2a [4\tan^3\theta \sec^2\theta +4\sec \theta \sec^4 \theta+4\tan \theta \sec^3\theta]$

$=2a.4 [\tan^3\theta \sec^2\theta +2\tan \theta \sec^4 \theta]$

$\dfrac {d^3y}{dx^3}\ \Rightarrow \ \dfrac {6a (7\sec^4 \theta \tan^2 \theta +2\tan^4 \theta \sec^2\theta +\sec^6 \theta)}{8a (\tan \theta \sec^2 \theta [\tan^2\theta +2\sec^2 \theta])}$

$\Rightarrow \ \dfrac {3}{4} \dfrac {\sec^2\theta [7 \sec^2 \theta \tan ^2 \theta + 2\tan ^4 \theta +\sec^4 \theta]}{\tan \theta \sec^4 \theta (\tan ^2\theta +2 \sec^2 \theta)}$

$\Rightarrow \ (Since\ 1+\tan^2\theta =\sin^2\theta)$

$\Rightarrow \ \dfrac {3}{4} \left [\dfrac {7(1+\tan^2\theta) (\tan^2 \theta)+2\tan^2 \theta +(1+\tan^2\theta)}{\tan \theta (\tan \theta+2(1+\tan \theta))}\right]$

on further simplyfying we get

$\Rightarrow \ 3\sec^2 \theta \tan \theta$

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?

0%
$8$
0%
$1$
0%
$-1$
0%
None

Explanation

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

Let

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

$f$ is

0%
continuous at $x = \dfrac{3}{2}$
0%
continuous at $x = 3$
0%
differentiable at $x = \dfrac{3}{2}$
0%
differentiable at $x = 0$

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

0%
Differentiable at $x = 0$
0%
Differentiable at $x = 2$
0%
Differentiable at $x = 1$ and $x = 2$
0%
Not differentiable at $x = 0$

Range of function

$f(x)-g(x)$ in [-

$\pi, \pi$ ] is-

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

Which of the following is NOT CORRECT-

0%
$b=d$
0%
$a=d$
0%
$a\neq b$
0%
$c\neq d$

$f\left( x \right) = \left( {x - a} \right)g\left( x \right)$ and

$g\left( x \right)$ is continuous

$x = a$ then

${f^{'}}\left( 1 \right) =$

0%
$g\left( 1 \right)$
0%
${g^{'}}\left( 1 \right)$
0%
$- {g^{'}}\left( 1 \right)$
0%
${g^{'}}\left( { - 1} \right)$

Explanation

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

$f'(a)=g(a)$

$f'(1)=g(1)$

Function

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

0%
continuous & differentiable at $x=0$
0%
continuous but not differentiable at $x=0$
0%
discontinuous at $x=0$
0%
non differentiable because it is discontinuous at $x=0$

Differentiate

$\log(1+x^{2})$ with respect

$\tan^{-1}x$ .

0%
$2x$
0%
$x^2$
0%
$x$
0%
$x^3$

Explanation

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

$\dfrac{dy}{dz}=\dfrac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}}=\dfrac{\dfrac{2x}{1+x^{2}}}{\dfrac{1}{1+x^{2}}}=2x$

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

$t = 5\ sec$ .

0%
$5^{c}/ sec$
0%
$13^{c}/ sec$
0%
$23^{c}/ sec$
0%
$35^{c}/ sec$

If the following function is continuous at

$x=0$ , find the value of

$k$ :

$f\left( x \right) = \left\{ {\begin{array}{ccccccccccccccc}{\dfrac{{\sin \frac{{3x}}{2}}}{x}}&{,x \ne 0}\\k&{,x = 0}\end{array}} \right.$

0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{3}{2}$

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 ?

0%
$-\dfrac{1}{6}$
0%
$\dfrac{1}{6}$
0%
$-\dfrac{5}{6}$
0%
$-1$

Explanation

$g\left( x \right) =\cfrac { f\left( x \right) }{ x+1 }$

$\Rightarrow g(0)=\cfrac { 4 }{ 1 } =4;g\left( 5 \right)=\cfrac { -1 }{ 6 }$

By Lagrange's mean value theorem,

$g'\left( c \right)=\cfrac { g\left( 5 \right)-g\left( 0 \right) }{ 5-0 } =\cfrac { \cfrac { -1 }{ 6 } -4 }{ 5 }$

$=\cfrac { -25 }{ 30 } =\cfrac { -5 }{ 6 }$

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

0%
$\dfrac{50}{3}$
0%
$\dfrac{22}{3}$
0%
$13$
0%
None of these

$f$ is differentiable on

$(a,b)$ and if

$f(a)=f(b)=0$ then for any

$\alpha$ there is an

$x\in (a,b)$ such that

$\alpha f(x)+f'(x)-1=0$

0%
True
0%
False

A function

$f:R\rightarrow R$ is such that

$f(1)=3$ and

$f'(1)=6$ . Then

$\rightarrow { \displaystyle \lim _{ x\rightarrow 0 }{ \left[ \dfrac { f\left( 1+x \right) }{ f\left( 1 \right) } \right] } }^{ 1/x }=$ ?

0%
$1$
0%
$e^{2}$
0%
$e^{1/2}$
0%
$e^{3}$

$y=y(x)$ and it follows the relation

$e^{xy^{2}}+y\cos(x^{2})=5$ then

$y'(0)$ is equal to

0%
$4$
0%
$-16$
0%
$-4$
0%
$16$

$f$ 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 ?

0%
$-1$
0%
$0$
0%
$2$
0%
$1$

$f (x)$ is differentiable everywhere, then:

0%
$| f |$ is differentiable everywhere
0%
$| { f } | ^ { 2 }$ is differentiable everywhere
0%
$f | f |$ is not differentiable at some point
0%
$f + | f |$ is differentiable everywhere

Explanation

Given:

$f(x)$ is differentiable everywhere

$|f|$ is differentiable everywhere

Let

$f(x)=x$

$\Rightarrow |f(x)|=|x|$

$\therefore |f|$ is not differentiable everywhere

$|f|^2$ is differentiable everywhere

Let

$f(x)=x$

$\Rightarrow |f(x)|^2=x^2$

$f(x)=e^x$

$|f(x)|^2=e^{2x}$

Differentiable everywhere

$f|f|$ is not differentiable at same points

$f(x)=e^x\Rightarrow |f|=e^x$

$f|f|=e^{2x}\rightarrow$ differentiable everywhere

$\Rightarrow$ This option is wrong

$f+|f|$ is differentiable everywhere

$f(x)=x$ ,

$|f|=|x|$

$f+|f|=\left\{\begin{matrix} 0, x < 0\\ 2x, x\geq 0\end{matrix}\right\}$ not

differentiable at

$x=0$

Options (ii), (iii), (iv) are wrong.

$\Rightarrow$ (ii) is correct.

1379344_1216266_ans_eb40e952ab6c4ce9ae4f053c2f911bd9.JPG

Let

$f$ differentiable at

$x=0$ and

$f'\left( 0 \right) = 1$ . Then

$\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( h \right) - f\left( { - 2h} \right)}}{h} =$ =

0%
3
0%
2
0%
1
0%
6

Explanation

$\begin{matrix} { \lim }_{ h\to 0 }\dfrac { { F\left( h \right) -f\left( { -2h } \right) } }{ h } \\ \Rightarrow { \lim }_{ h\to 0 }\dfrac { { f'\left( h \right) +2f'\left( { -2h } \right) } }{ 1 } \, \, \left[ { \because f\left( 0 \right) =1 } \right] \\ \to f'\left( 0 \right) +2f'\left( 0 \right) \\ =1+2+3=6 \\ \end{matrix}$

Let

$f(x)$ be differentiable function such that

$f\left(\dfrac{x+y}{1-xy}\right)=f(x)+f(y) \forall x$ and

$y$ . If

$\underset { x\rightarrow 0 }{ lt } \dfrac { f\left( x \right) }{ x } =\dfrac { 1 }{ 3 }$ then

$f(1)$ equals

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{12}$
0%
$\dfrac{1}{8}$

Explanation

We have,

$f\left( {\dfrac{{x + y}}{{1 - xy}}} \right) = f\left( x \right) + f\left( y \right)\,\,\,\forall {x^2}y$

$\Rightarrow f\left( x \right) = k{\tan ^{ - 1}}\left( x \right)$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{k{{\tan }^{ - 1}}x}}{x} = \dfrac{1}{3}$

$k = \dfrac{1}{3}$

$f\left( x \right) = \dfrac{1}{3}{\tan ^{ - 1}}x$

$f\left( 1 \right) = \dfrac{1 }{{12}}$

Hence, this is the answer.

$f(x)=x^{n}\sin \dfrac {1}{x},f(0)=0$

0%
$f$ is continuous differentiable at $x=0$ for any $n< 2, n\in N$
0%
$f$ is continuous differentiable for all $x$ for any $n\ge 2, n\in N$
0%
$f$ is discountinuous at $x=0$ if $n=0$
0%
$f'(x)$ is discountinuous at $x=0$ if $n=0$

$x=\dfrac{e^t+e^{-t}}{2}, y=\dfrac{e^t-e^{-t}}{2}$ , then

$\dfrac{dx}{dy}=$

0%
$-\dfrac{x}{y}$
0%
$\dfrac{y}{x}$
0%
$\dfrac{x}{y}$
0%
$-\dfrac{y}{x}$

Explanation

$x=\dfrac{e^t+e^{-t}}{2},$

$y=\dfrac{e^t-e^{-t}}{2}$

$\Rightarrow \dfrac{dx}{dt}=\dfrac{e^t-e^{-t}}{2}=y,\dfrac{dy}{dt}=\dfrac{e^t+e^{-t}}{2}=x$

$\Rightarrow \dfrac{dx}{dy}=\dfrac{dx}{dt}\times \dfrac{dt}{dy}$

$=y\times \dfrac{1}{x}$

$=\dfrac{y}{x}$

Hence, the answer is

$\dfrac{y}{x}.$

$y=a^{{a}^{{x}}}$ , then

$\dfrac {dy}{dx}=$

0%
$y.a^{x}{(\log a)}^{2}$
0%
$y.a^{x}.\log a$
0%
${(y.a^{x})}^{2}$
0%
$(y.a^{x})$

Explanation

$\quad y={ a }^{ { x }^{ y } }$

taking logarithm on both sides

$\log { y } ={ x }^{ y }\log { a }$

again we take logarithm on both sides

$\log { \log { y } } =y\left( \log { x } \right) +\log { \log { a } }$

taking derivative on both sides w.r. t

$x$

$\cfrac { 1 }{ \log { y } } .\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =\cfrac { dy }{ dx } \log { x } +y.\cfrac { 1 }{ x } +0$

$\Rightarrow \cfrac { dy }{ dx } \left( \cfrac { 1 }{ y\log { y } } -\log { x } \right) =\cfrac { y }{ x } \Rightarrow \cfrac { dy }{ dx } \left( \cfrac { 1-y\log { x } .\log { y } }{ y\log { y } } \right) =\cfrac { y }{ x }$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { { y }^{ 2 }\log { y } }{ x\left( 1-y\log { x } .\log { y } \right) }$

Solve:

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

0%
$1$
0%
$1 / 2$
0%
$\cos x$
0%
sec $x$

Explanation

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

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

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

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

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

$\implies \dfrac 12$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 7 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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