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

$$\displaystyle f(x)=|\log _{10}x|$$
So clearly from the graph it is continuous but not differentiable at $$x = 1$$
($$\displaystyle \because $$ it is point of sharpe edge)

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

$$f(x)=\begin{cases} a{ x }^{ 2 }-b\quad ;\quad \quad -1<x<1 \\ \cfrac { -1 }{ x } \quad \quad ;\quad \quad x\ge 1 \\ \cfrac { 1 }{ x } \quad \quad ;\quad \quad x\le -1 \end{cases}$$
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 $$x=1$$
$$\Rightarrow a{(1)}^{2}-b=-1$$
$$\Rightarrow b=\cfrac{3}{2}$$

$$y = x^x$$
Take log both sides,
$$\log y = x\log x$$
Now differentiate both side w.r.t $$x$$
$$\Rightarrow \cfrac{1}{y}\cfrac{dy}{dx} = 1+\log x$$

$$\cfrac{dx}{dt} = 4\sin^3t .\cos t$$ and $$\cfrac{dy}{dt} =-4\cos^3t. \sin t$$
$$\Rightarrow \cfrac{dy}{dx} = -\cfrac{\cos^2t}{\sin^2t}$$
Thus at $$t = \cfrac{\pi}{4}$$,  $$\cfrac{dy}{dx} = -1$$

Let $$y = x^{\ln x}$$ and $$z = \ln x\Rightarrow x = e^z$$ we have to find $$\cfrac{dy}{dz}$$
$$\Rightarrow  y = (e^z)^z = e^{z^2}$$
$$\Rightarrow \cfrac{dy}{dz} = e^{z^2}.(2z) = x^{\ln x}.(2\ln x)$$

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

$$f(x)$$ and $$g(x)$$ are continuous at $$x=c$$.

Given, $$y=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\sin { x }  }{ 1+\sin { x }  }  }  } $$
                 $$=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\cos { \left( \dfrac { \pi  }{ 2 } -x \right)  }  }{ 1+\cos { \left( \dfrac { \pi  }{ 2 } -x \right)  }  }  }  } $$
                 $$=\tan ^{ -1 }{ \left| \tan { \left( \dfrac { \pi  }{ 4 } -\dfrac { x }{ 2 }  \right)  }  \right|  } $$
                 $$=\dfrac { \pi  }{ 4 } -\dfrac { x }{ 2 } $$                 $$\left[ \because x=\dfrac { \pi  }{ 6 }  \right] $$
$$\Rightarrow \dfrac { dy }{ dx } =-\dfrac { 1 }{ 2 } $$

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

Given, $$ y=1-\cos { \theta  } $$

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

Let $$y=\log_{10}x$$ and $$z =\log_x10$$

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

We have, $$x=a\cos^3\theta$$ and $$y=a\sin^3\theta$$

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

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

At $$x=0$$, for the function to be continuous $$b=-1$$

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.

$$\displaystyle \lim_{ x\rightarrow \frac { \pi  }{ 2 }  }f(x)=\lim_{x\rightarrow \frac{\pi}{2}}\dfrac { \alpha \cos x }{ \pi -2x }$$ which is in $$\dfrac{0}{0}$$ form

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