If $$f (x) =\displaystyle |\log _{10}x|$$ then at $$x = 1 $$

f is continuous but not differentiable

If $${ x }^{ y }={ e }^{ x-y }$$ then $$\cfrac { dy }{ dx } $$ is equal to

$$\cfrac { \log { x } }{ \log { (x-y) } } $$

$$\cfrac { { e }^{ x } }{ { x }^{ x-y } } $$

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

$$\cfrac { 1 }{ y } -\cfrac { 1 }{ x-y } $$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 3

$f (x) =\displaystyle |\log _{10}x|$ then at

$x = 1$

0%
f is not continuous
0%
f is continuous but not differentiable
0%
f is differentiable
0%
the derivative is 1

Explanation

$\displaystyle f(x)=|\log _{10}x|$
So clearly from the graph it is continuous but not differentiable at

$x = 1$
(

$\displaystyle \because$ it is point of sharpe edge)

$\sqrt { x+y } +\sqrt { y-x } =c,$ then

$\displaystyle \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } }$ is equal to

0%
$\displaystyle \frac { 2 }{ c }$
0%
$\displaystyle-\frac { 2 }{ { c }^{ 2 } }$
0%
$\displaystyle\frac { 2 }{ { c }^{ 2 } }$
0%
none of these

Explanation

$\sqrt { x+y } +\sqrt { y-x } =c$ ...(1)

Rationalising we get

$\dfrac{(y+x)-(y-x)}{\sqrt{x+y}-\sqrt{y-x}}=c$

We know that

$(y+x)-(y-x)=2x$

we get

$\displaystyle \sqrt { x+y } -\sqrt { y-x } =\frac { 2x }{ c }$ ...(2)

From (1) and (2)

$\displaystyle 2\sqrt { y+x } =c+\frac { 2x }{ c }$

$\displaystyle \Rightarrow 4\left( y+x \right) ={ c }^{ 2 }+\frac { 4{ x }^{ 2 } }{ { c }^{ 2 } } +4x$

$\displaystyle \Rightarrow 4y={ c }^{ 2 }+\frac { 4{ x }^{ 2 } }{ { c }^{ 2 } } \Rightarrow 4\frac { dy }{ dx } =4.\frac { 2x }{ { c }^{ 2 } }$

$\displaystyle \Rightarrow \frac { dy }{ dx } =\frac { 2x }{ { c }^{ 2 } } \Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { 2 }{ { c }^{ 2 } }$

$f(x)=\begin{cases} a{ x }^{ 2 }-b,\quad if\quad \left| x \right| <-1 \\ -\cfrac { 1 }{ \left| x \right| } if\quad \quad \left| x \right| \ge -1 \end{cases}$ is differential at

$x=1$ . Find the values of

$a$ and

$b$

0%
$a=1/2$ ; $b=3/2$
0%
$a=1/2$ ; $b=-3/2$
0%
$a=-1/2$ ; $b=3/2$
0%
$a=-1/2$ ; $b=-3/2$

Explanation

$f(x)=\begin{cases} a{ x }^{ 2 }-b\quad ;\quad \quad -1<x<1 \\ \cfrac { -1 }{ x } \quad \quad ;\quad \quad x\ge 1 \\ \cfrac { 1 }{ x } \quad \quad ;\quad \quad x\le -1 \end{cases}$
Now

$f(x)$ is differentiable at

$x=1$

$\Rightarrow 2ax=\cfrac{1}{{x}^{2}}$ at

$x=1$

$\Rightarrow a=\cfrac{1}{2}$
Also

$f(x)$ is continuous at

$x=1$

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

$\Rightarrow b=\cfrac{3}{2}$

Find

$\displaystyle \frac{dy}{dx}$ if

$y=x^{x}$

0%
$\displaystyle x^{x}\left ( lnx+1 \right )$
0%
$\displaystyle x^{x}\left ( lnx-1 \right )$
0%
$\displaystyle x .x^{x-1}$
0%
$\displaystyle x^{x-1}\left ( lnx+1 \right )$

Explanation

$y = x^x$
Take log both sides,

$\log y = x\log x$
Now differentiate both side w.r.t

$x$

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

$\Rightarrow \cfrac{dy}{dx} =x^x(1+\log x)$

Find $\displaystyle \frac { dy }{ dx }$ at $t =\displaystyle \frac { \pi }{ 4 }$ if $y =\displaystyle \cos ^{ 4 }{ t }$ & $x =\displaystyle \sin ^{ 4 }{ t }$ .

0%
1
0%
0
0%
-1
0%
4

Explanation

$\cfrac{dx}{dt} = 4\sin^3t .\cos t$ and

$\cfrac{dy}{dt} =-4\cos^3t. \sin t$

$\Rightarrow \cfrac{dy}{dx} = -\cfrac{\cos^2t}{\sin^2t}$
Thus at

$t = \cfrac{\pi}{4}$ ,

$\cfrac{dy}{dx} = -1$

Differentiate $\displaystyle { x }^{ \ln x }$ with respect to $\ln x$ .

0%
2 $\displaystyle \left( { x }^{ lnx } \right) \left( lnx \right)$
0%
$\displaystyle -\left( { lnx }^{ lnx } \right) \left( lnx \right)$
0%
$\displaystyle \left( { x }^{ lnx } \right) \left( lnx \right)/x$
0%
$\displaystyle -\left( { lnx }^{ lnx } \right) \left( lnx \right)/x$

Explanation

Let

$y = x^{\ln x}$ and

$z = \ln x\Rightarrow x = e^z$ we have to find

$\cfrac{dy}{dz}$

$\Rightarrow y = (e^z)^z = e^{z^2}$

$\Rightarrow \cfrac{dy}{dz} = e^{z^2}.(2z) = x^{\ln x}.(2\ln x)$

$\displaystyle \frac{d}{dx}\left ( x^{\log x} \right )$ is equal to

0%
$\displaystyle 2x^{\log x-1}\log x$
0%
$\displaystyle x^{\log x-1}$
0%
$\dfrac 23 (\log x)$
0%
$\displaystyle x^{\log x-1}.\log x$

Explanation

$\displaystyle \dfrac { d }{ dx } \left( x^{ \log x } \right) =\dfrac { d }{ dx } \left( { e }^{ \log { x } \log { x } } \right)$

$\displaystyle =\dfrac { d }{ dx } \left( { \left( \log { x } \right) }^{ 2 } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$

$\displaystyle =2\log { x } \dfrac { d }{ dx } \left( \log { x } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$

$\displaystyle ={ e }^{ { \left( \log { x } \right) }^{ 2 } }2\log { x } \left( \dfrac { d }{ dx } \left( \log { x } \right) \right)$

$\displaystyle =\dfrac { 1 }{ x } 2\log { x } { e }^{ { \left( \log { x } \right) }^{ 2 } }=2x^{ \log { x } -1 }\log { x }$

$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$ , then

$y'(0)$ is equal to

0%
$-1$
0%
$1$
0%
$2$
0%
Non existant

Explanation

$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$
Take log both sides,

$\displaystyle \log y = \frac{1}{3}\log(1+3x)+\frac{1}{4}\log(1+4x)+\frac{1}{5}\log(1+5x)-\frac{1}{7}\log(1+7x)-\frac{1}{8}\log(1+8x)$
Differentiating both side w.r.t

$x$

$\displaystyle \frac {1}{y}\times \frac {dy}{dx}=\frac {1}{(1+3x)}+\frac {1}{1+4x}+\frac {1}{1+5x}-\frac {1}{1+7x}-\frac {1}{1+8x}$
at

$x=0, y=1$ ,

$\therefore \displaystyle \frac {dy}{dx}=1+1+1-1-1=1$

$f(x)$ and

$g(x)$ are both continuous at

$x= c$ then which of the following is/are always continuous at

$x=c$ ?

0%
$f(x) + g(x)$
0%
$(f(x) - g(x))\times f(x)$
0%
$g(x)\times f(x)$
0%
$\dfrac{f(x) - g(x)}{g(x)}$

Explanation

$f(x)$ and

$g(x)$ are continuous at

$x=c$ .

$f+g,f-g,f \times g,$ are continuous

$(f-g) \times f$ is also continuous.

But

$\frac {f-g}{g}$ may not be continuous because

$g(c)$ should not be equal to zero and should be given in question.

without that condition, we cannot ascess anything.
For example: Let

$f = 2x^2$ and

$g = x-3$ then both

$f$ and

$g$ are continuous at

$3$ but

$\dfrac{2x^2 - (x-3)}{x-3}$ is not continuous at

$x=3$ as denominator is

$0$ in this situation.

Derivative of sin x w.r.t. cos x is

0%
cos x
0%
cot x
0%
- cot x
0%
tan x

Explanation

Let

$cos\,x = t$

than,

$dt = -sinx\,dx...(1)$

let

$y = sinx$

$dy = cosx\,dx...(2)$

then (2)

$\div$ (1)

$\dfrac{dy}{dx} = \dfrac{cosx\,dx}{-sinx\,dx} = -cotx$

1117048_460516_ans_5edfb8e232f74270a3462850d30914c0.JPG

$y=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\sin { x } }{ 1+\sin { x } } } }$ , then the value of

$\dfrac { dy }{ dx }$ at

$x=\dfrac { \pi }{ 6 }$ is

0%
$-\dfrac { 1 }{ 2 }$
0%
$\dfrac { 1 }{ 2 }$
0%
$1$
0%
$-1$

Explanation

Given,

$y=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\sin { x } }{ 1+\sin { x } } } }$

$=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\cos { \left( \dfrac { \pi }{ 2 } -x \right) } }{ 1+\cos { \left( \dfrac { \pi }{ 2 } -x \right) } } } }$

$=\tan ^{ -1 }{ \left| \tan { \left( \dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right) } \right| }$

$=\dfrac { \pi }{ 4 } -\dfrac { x }{ 2 }$

$\left[ \because x=\dfrac { \pi }{ 6 } \right]$

$\Rightarrow \dfrac { dy }{ dx } =-\dfrac { 1 }{ 2 }$

$x=a\, cos^3\theta$ and

$y=a\, sin^3\theta$ , then

$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}$ is equal to

0%
$|sec \,\theta |$
0%
$sec^2\theta$
0%
$sec\, \theta$
0%
$| tan\, \theta|$

Explanation

$x=a\cos^3\theta$ ,

$y=0\sin^3\theta$

diff wrt

$'\theta'$

$\dfrac{dx}{d\theta}=3a\cos^2\theta(-\sin\theta)$

$\dfrac{dy}{d\theta}=3a\sin^2\theta(+\cos \theta)$

$\dfrac{dy}{dx}=\dfrac{3a\sin^2\theta \cos\theta}{-3a\cos^2\theta \sin\theta}=-\dfrac{\sin\theta}{\cos\theta}=-\tan\theta$

$\therefore 1+\left(\dfrac{dy}{dx}\right)^2=1+\tan^2\theta =\sec^2\theta$

$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\sqrt{\sec^2\theta}$ .

$=|\sec \theta|$

$x=\cos^3\theta$ and

$y=\sin^3\theta$ , then

$1+\left(\displaystyle\frac{dy}{dx}\right)^2$ is equal to:

0%
$\tan^2\theta$
0%
$\cot^2\theta$
0%
$\sec^2\theta$
0%
$\text{cosec}^2\theta$

Explanation

$\Rightarrow \dfrac{dx}{d\theta}=3\cos^2\theta(-\sin\theta)=-3\cos^2\theta\sin\theta$ ------(1)

and

$\Rightarrow \dfrac{dy}{d\theta}=3\sin^2\theta\cos\theta$ -----(2)

Hence,

$1+\left( \dfrac{dy}{dx}\right)^2=1+\left( \dfrac{dy/d\theta}{dx/d\theta}\right)^2=1+\dfrac{\sin^2\theta}{\cos^2\theta}=\sec^2\theta$

$y=1-\cos { \theta } ,x=1-\sin { \theta }$ , then

$\cfrac { dy }{ dx }$ at

$\theta =\cfrac { \pi }{ 4 }$ is

0%
$-1$
0%
$1$
0%
$\cfrac{1}{2}$
0%
$\cfrac{1}{\sqrt{2}}$

Explanation

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

$\therefore { \left[ \cfrac { dy }{ dx } \right] }_{ \theta =\frac { \pi }{ 4 } }=-\tan { \cfrac { \pi }{ 4 } } =-1$

$y = (1+x) (1+x^2)(1+x^4)......(1+x^{2n})$ then the value of

$\begin{pmatrix}\dfrac{dy}{dx}\end{pmatrix}$ at

$x=0$ is

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

Explanation

We have,

$y=(1+x)(1+x^2)(1+x^4).......(1+x^4)$

take natural logarithm both sides

$\ln y=\ln(1+x)+\ln(1+x^2)+\ln(1+x^4)+................+\ln(1+x^{2n})$

Now differentiate both sides w.r.t.

$x$

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+......+\dfrac{2nx^{2n-1}}{1+x^{2n}}$

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

Now substitute

$x=0$ to get the required value

$\Rightarrow \dfrac{dy}{dx}\bigg|_{x=0}=1(1+0+0+......+0)=1$

Note that at

$x=0$ , value of

$y$ is also

$1$

The differential coefficient of

$\log _{ 10 }{ x }$ with respect to

$\log _{ x }{ 10 }$ is

0%
$1$
0%
$-{ \left( \log _{ 10 }{ x } \right) }^{ 2 }\quad$
0%
${ \left( \log _{ x }{ 10} \right) }^{ 2 }\quad$
0%
$\cfrac { { x }^{ 2 } }{ 100 }$

Explanation

Let

$y=\log_{10}x$ and

$z =\log_x10$

We have to find the value of

$\dfrac{dy}{dz}$

Now using base change formula

$yz =\dfrac{\log x}{\log 10}\cdot \dfrac{\log 10}{\log x}=1$

$\Rightarrow y=\dfrac{1}{z}$

$\Rightarrow \dfrac{dy}{dz}=-\dfrac{1}{z^2}=-y^2=-(\log_{10}x)^2$

Since

$\dfrac{1}{z}=y$

The function represented by the following graph is.

0%
Differentiable but not continuous $x=1$
0%
Neither continuous nor Differentiable at $x=1$
0%
Continuous but not Differentiable at $x=1$
0%
Continuous but Differentiable at $x=1$

Explanation

Since there is no discontinuity in the graph so clearly function is continuous,

but is not differentiable at

$x=1$ , because graph of the function has a sharp point or kink.

Note: The point on the graph of any function where it has a sharp corner or kink, function is not differentiable .

$x = a \cos^{3}\theta, y = a\sin^{3}\theta$ , then

$1 + \left (\dfrac {dy}{dx}\right )^{2}$ is _____

0%
$\tan \theta$
0%
$\tan^{2}\theta$
0%
$\sec^{2}\theta$
0%
$1$

Explanation

We have,

$x=a\cos^3\theta$ and

$y=a\sin^3\theta$

Differentiate both w.r.t.

$\theta$

We get:

$\dfrac{dx}{d\theta}=-3a\cos^2\theta\sin\theta$ and

$\dfrac{dy}{d\theta}=3a\sin^2\theta\cos\theta$

Thus

$\dfrac{dy}{dx}=\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\tan\theta$

Therefore

$1+\left(\dfrac{dy}{dx}\right)^2=1+\tan^2\theta=\sec^2\theta$

${ x }^{ y }={ e }^{ x-y }$ then

$\cfrac { dy }{ dx }$ is equal to

0%
$\cfrac { \log { x } }{ \log { (x-y) } }$
0%
$\cfrac { { e }^{ x } }{ { x }^{ x-y } }$
0%
$\cfrac { \log { x } }{ { (1+\log { x } ) }^{ 2 } }$
0%
$\cfrac { 1 }{ y } -\cfrac { 1 }{ x-y }$
0%
$\dfrac{y(x -y)}{x^2}$

Explanation

We have

$x^y=e^{x-y}$

Take natural logarithm both sides

$\Rightarrow y\ln x=x-y$ ....(1)

Differentiate both sides w.r.t.

$x$

$\Rightarrow \dfrac{y}{x}+\ln x\dfrac{dy}{dx}=1-\dfrac{dy}{dx}$

$\Rightarrow \dfrac{dy}{dx}(\ln x+1)=1-\dfrac{y}{x}=\dfrac{x-y}{x}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{x-y}{x(\ln x+1)}=\dfrac{x-y}{x(\frac{x-y}{y}+1)}$ , use (1)

$=\dfrac{y(x-y)}{x^2}$

$x = ct$ and

$y = \dfrac {c}{t}$ , find

$\dfrac {dy}{dx}$ at

$t = 2$

0%
$\dfrac {1}{4}$
0%
$4$
0%
$-\dfrac {1}{4}$
0%
$0$

Explanation

We have,

$x=ct\Rightarrow \dfrac{dx}{dt}=c$

and

$y=\dfrac{c}{t}\Rightarrow \dfrac{dy}{dt}=-\dfrac{c}{t^2}$

$\therefore \dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\dfrac{1}{t^2}$

Hence at

$t=2$

$\dfrac{dy}{dx}=-\dfrac{1}{2^2}=-\dfrac{1}{4}$

Given a function

$f(x) = \left\{\begin{matrix}-1 & if & x \leq 0\\ ax + b & if & 0 < x < 1\\ 1 & if & x \geq 1\end{matrix}\right.$ where

$a, b$ are constants. The function is continuous everywhere.

What is the value of

$a$ ?

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

Explanation

$x=0$ , for the function to be continuous

$b=-1$

$x=1$ , for the function to be continuous

$a+b=1$

$\Rightarrow$

$a=2$

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?

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

Explanation

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.

Consider the function

$f(x) = \left\{\begin{matrix}\dfrac {\alpha \cos x}{\pi - 2x} & if & x\neq \dfrac {\pi}{2}\\ 3 & if & x = \dfrac {\pi}{2}\end{matrix}\right.$
which is continuous at

$x = \dfrac {\pi}{2}$ , where

$\alpha$ is a constant.

What is the value of

$\alpha$ ?

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

Explanation

$\displaystyle \lim_{ x\rightarrow \frac { \pi }{ 2 } }f(x)=\lim_{x\rightarrow \frac{\pi}{2}}\dfrac { \alpha \cos x }{ \pi -2x }$ which is in

$\dfrac{0}{0}$ form

$=\displaystyle \lim_{x\rightarrow \frac{\pi}{2}}\dfrac { -\alpha \sin x }{ -2 }$

$=\dfrac { \alpha }{ 2 }$

But

$\dfrac { \alpha }{ 2 } =3$ as the function is continuous at

$\dfrac { \pi }{ 2 }$

$\Rightarrow \alpha =6$

If y = cos t and x = sin t, then what is

$\dfrac{dy}{dx}$ equal to ?

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

Explanation

Differentiating

$x,y$ wrt

$t,$ we get

$\dfrac { dx}{ dt } =\dfrac { d }{ dt } (\cos t)=-\sin t$ and

$\dfrac { dx }{ dt } =\dfrac { d }{ dt } (\sin t)=\cos t\\$

$\therefore \dfrac { dy }{ dx } =\dfrac { \left( \dfrac { dy }{ dt } \right) }{ \left( \dfrac { dx }{ dt } \right) } =\dfrac { -\sin t }{ \cos t } =-\dfrac { x }{ y }$

Consider the function

$f(x)=\begin{cases}ax-2 & for & -2 < x < -1 \\ -1 & for & -1\le x\le 1 \\ a+2(x-1)^2 & for & 1 < x < 2\end{cases}$
What is the value of a for which

$f(x)$ is continuous at

$x=-1$ and

$x=1$ ?

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

Explanation

Solution:

$LHL=\lim_{h\rightarrow0}f(-1-h)=\lim_{h\rightarrow0}a(-1-h)-2=-a-2$

$RHL=\lim_{h\rightarrow0}f(-1+h)=\lim_{h\rightarrow0}(-1)=-1$

$x=1$ ,

$LHL=\lim_{h\rightarrow0}f(1-h)=\lim_{h\rightarrow0}(-1)=-1$

$RHL=\lim_{h\rightarrow0}f(1+h)=\lim_{h\rightarrow0}(a+2h^2)=a$

Now, for

$f$ to be continuous at

$x=-1$

we should have

$LHL=RHL=f(-1)$

$\Longrightarrow -a-2=-1$

$\Longrightarrow a=-1$

Also, for

$a=-1$ ,

$f$ is continuous at

$x=1$

Hence,

$a=-1$

Hence, A is the correct option.

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?

0%
$1$ only
0%
$2$ only
0%
Both $1$ and $2$
0%
Neither $1$ nor $2$

Explanation

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

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

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

$x\le 3$

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

$\Rightarrow f\left( x \right)$ is differentiable at

$x=0$

$y={\cot}^{-1}\begin{bmatrix}\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\end{bmatrix}$ , where

$0 < x < \dfrac{\pi}{2}$ , then

$\dfrac{dy}{dx}$ is equal to

0%
$-\dfrac{1}{2}$
0%
2
0%
$\sin x + \cos x$
0%
$\sin x - \cos x$

Explanation

Solution:

$y=\cot^{-1}\left[\cfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$

$=\cot^{-1}\left[\cfrac{\sqrt{\sin^{2}x\cos^{2}x+2\sin \dfrac{x}{2}\cos\dfrac{x}{2}}+\sqrt{\sin^{2}+\cos^{2}x-2\sin \dfrac{x}{2} \cos\dfrac{x}{2}}}{\sqrt{\sin^{2}x+\cos^{2}x+2\sin \dfrac{x}{2}\cos\dfrac{x}{2}}-\sqrt{\sin^{2}x+\cos^{2}x-2\sin \dfrac{x}{2}}\cos\dfrac{x}{2}}\right]$

$=\cot^{-1}\left[\cfrac{\left|\sin \cfrac x2+\cos\cfrac x2\right|+\left|\sin\cfrac x2-\cos\cfrac x2\right|}{\left|\sin\cfrac x2+\cos\cfrac x2\right|-\left|\sin\cfrac x2-\cos\cfrac x2\right|}\right]$

$0<x<\cfrac{\pi}2$

$\therefore y=\cot^{-1}\left[\cfrac{\left(\sin \cfrac x2+\cos\cfrac x2\right)+\left(\sin\cfrac x2-\cos\cfrac x2\right)}{\left(\sin\cfrac x2+\cos\cfrac x2\right)-\left(\sin\cfrac x2-\cos\cfrac x2\right)}\right]$

$=\cot^{-1}\left(\tan\cfrac x2\right)$

$=\cot^{-1}\left[\cot\left(\cfrac{\pi}2-\cfrac x2\right)\right]$

$=\cfrac{\pi}2-\cfrac x2$

$\therefore\cfrac{dy}{dx}=-\cfrac12$

$x^ay^b=(x-y)^{a+b}$ , then the value of

$\dfrac{dy}{dx}-\dfrac{y}{x}$ is equal to

0%
$\dfrac{a}{b}$
0%
$\dfrac{b}{a}$
0%
1
0%
0

Explanation

Solution:

We have,

$x^ay^b=(x-y)^{a+b}$

Taking logarithm on both sides, we get

$a\log x+b\log y=(a+b)\log (x-y)$

Differentiating on both sides, we get

$\cfrac ax+\cfrac by\cfrac{dy}{dx}=(a+b)\cfrac1{x-y}\left(1-\cfrac{dy}{dx}\right)$

or,

$\cfrac{dy}{dx}\left(\cfrac by+\cfrac{a+b}{x-y}\right)=\cfrac{a+b}{x-y}-\cfrac ax$

or,

$\cfrac{dy}{dx}=\cfrac yx.\cfrac{bx+ay}{bx+ay}$

or,

$\cfrac{dy}{dx}-\cfrac yx=0$

Hence, D is the correct option.

Consider the following in respect of the function

$f(x) = | x- 3 |$ :

$f(x)$ is continuous at

$x = 3$

$f(x)$ is differentiable at

$x = 0$ .
Which of the above statements is/are correct ?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

$\displaystyle \lim _{ x\rightarrow { 3 }^{ - } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow { 3 }^{ - } }{ -\left( x-3 \right) } =-\left( 3-3 \right) =0$

$\displaystyle \lim _{ x\rightarrow { 3 }^{ + } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow { 3 }^{ + } }{ \left( x-3 \right) } =3-3=0$

$f\left( 3 \right) =\left| 3-3 \right| =0$

LHL

$=$ RHL

$=f\left( x \right) \Rightarrow$ continuous at

$x=3$

$f'\left( x \right) =\begin{cases} -1;\quad x<3 \\ 0;\quad x=3 \\ 1;\quad x>3 \end{cases}$

$f'\left( x \right) =-1$ for

$x<3$

$\Rightarrow f'\left( x \right) =-1$ for

$x=0$

$\Rightarrow f\left( x \right)$ is differentiable at

$x=0$

$f(x)=\sqrt{25-x^2}$ , then what is

$\displaystyle \lim_{ x\rightarrow 1 } \dfrac{f(x)-f(1)}{x-1}$ equal to?

0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{24}$
0%
$\sqrt{24}$
0%
$-\dfrac{1}{\sqrt{24}}$

Explanation

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

$\displaystyle \lim_{x\rightarrow 1}{\cfrac{f(x)-f(1)}{x-1}}$

$={\cfrac{f'(1)}{1}}=\left|{\cfrac{1(-2x)}{2\sqrt{25-x^2}}}\right|_{at\ x=1}$

$=-\cfrac{1}{\sqrt {24}}$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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