CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 4 - MCQExams.com

$$\sec\left( \dfrac { x+y }{ x-y }  \right) =a$$ $$\longrightarrow (1)$$

g is inverse of f 

$$\textbf{Step -1: Differentiating with respect to x }$$

                  $$\text{Given,}$$  $${ y=x \tan y }$$

                  $${\Rightarrow\tan \ y=\dfrac{y}{x} \qquad\rightarrow (1)}$$

                  $$\text{Now,}$$ 

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}=1\times\tan y+x\sec ^{2}y\dfrac{\mathrm{d} y}{\mathrm{d} x}$$           $$\textbf{[ Using Product rule of derivative ]}$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}[1-x\sec ^{2}y]=\tan y$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\tan y}{1-x\sec ^{2}y}\rightarrow (2)$$

$$\textbf{Step -2: Find the value of }$$$$\mathbf{ sec ^{2}y.}$$

                  $$\text{We know, }$$  $$\sec ^{2}y-\tan ^{2}y=1$$

                  $$\Rightarrow\sec ^{2}y=1+\tan ^{2}y$$

                  $$\text{Putting the value from eq(1),}$$ 

                  $$\Rightarrow\sec ^{2}y=1+\dfrac{y^{2}}{x^{2}}$$               $$\qquad\rightarrow (3)$$

$$\textbf{Step -3: Prove the desired.}$$

                  $$\text{From eq(2),}$$ 

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\tan y}{1-x\sec ^{2}y}$$

                  $$\Rightarrow\mathrm{\dfrac{dy}{dx}}=\dfrac{\tan y}{1-x(1+\dfrac{y^{2}}{x^{2}})}$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}$$ $$=\dfrac{x\tan y}{x-x^{2}-y^{2}}$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}$$ $$=\dfrac{y}{x-x^{2}-y^{2}}$$                      $$\textbf{[From eq(1)]}$$ 

                  $$\therefore\ \text{If }$$ $$y=x\tan y\text{ then, }\dfrac{\mathrm{d} y}{\mathrm{d} x} \text{ = } \dfrac{ y}{x-x^{2}-y^{2}}.$$

$$\textbf{Hence, the correct answer is option B.}$$

$$\log _{ 10 }{ ({ { x }^{ 2 }-{ y }^{ 2 } }/{ { x }^{ 2 }+{ y }^{ 2 }) } } =2\\ \Rightarrow \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } ={ 10 }^{ 2 }\\ \Rightarrow 99{ x }^{ 2 }=-101{ y }^{ 2 }$$

$$f(x)=\cfrac { \log { (1+2x) } \sin { { x }^{ 0 } }  }{ { x }^{ 2 } } $$ for $$x\ne 0$$

Let $$x=\sin\theta $$

Given, $$x=\sec { \theta  } -\cos { \theta  } $$ and $$y=\sec ^{ n }{ \theta  } -\cos ^{ n }{ \theta  } $$

Continuity at $$x = 0$$
$$\displaystyle \lim_{x\rightarrow 0^{-}} f(x) = \displaystyle \lim_{x\rightarrow 0} x = 0$$
$$\displaystyle \lim_{x\rightarrow 0^{+}}f(x) = 0$$
$$f(0) = 0$$
$$\therefore$$ continuous at $$x = 0$$
For differentiablility 
$$f'(x) =\begin{cases} 1,\ x\leq 0 \\ 0,\ x>0\end{cases}$$
$$\therefore$$ not differentiable
It has sharp edge at $$x = 0$$
$$\therefore$$ not differentiable but continuous.

Since, $$f$$ is continuous at $$x=0$$

If $$-1 \leq x \leq 1$$, then $$0 \leq x \sin \pi x \leq \displaystyle\frac{1}{2}$$
$$\therefore f(x)=[x \sin \pi x]=0,$$ for $$-1 \leq x \leq 1$$
If $$1 < x < 1+h$$, where h is a small positive real number, then
$$\pi < \pi x < \pi +\pi h$$
$$\Rightarrow -1 < \sin \pi x< 0$$
$$\Rightarrow -1 < x \sin \pi x < 0$$
$$\therefore f(x)=[x\sin \pi x]=-1$$ in the right neighbourhood of $$x=1$$.
Thus, $$f(x)$$ is constant and equal to zero in $$[-1, 1]$$ and so $$f(x)$$ is differentiable and hence continuous on $$(-1, 1)$$.
At $$x=1, f(x)$$ is discontinuous because $$\displaystyle\lim_{x\rightarrow 1^-}f(x)=0$$ and $$\displaystyle\lim_{x\rightarrow 1^+}f(x)=-1$$
Hence, $$f(x)$$ is not differentiable at $$x=1$$.

Let $$f(x) = \sin\left| x \right| -\left| x \right|$$ 

$${ 2 }^{ x }+{ 2 }^{ y }={ 2 }^{ x+y }$$

Differentiating both sides

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

So option $$C$$ is correct.

Given, $$f\left( x \right) =\begin{cases} \dfrac { 1-\cos { 4x }  }{ 8{ x }^{ 2 } } &,x\neq 0 \\ k &,x=0 \end{cases}$$
LHL $$=\displaystyle\lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right)  } =\displaystyle\lim _{ h\rightarrow 0 }{ f\left( 0-h \right)  } =\displaystyle\lim _{ h\rightarrow 0 }{ f\left( h \right)  } $$
$$=\displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { 1-\cos { 4h }  }{ 8{ h }^{ 2 } }  } $$ ........ $$\dfrac{0}{0}$$ form
$$=\displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { 4\sin { 4h }  }{ 16h }  }$$ ........ (using L'Hospital's rule)

Since, $$f\left( x \right) $$ is continuous at $$x=0$$, therefore
$$\displaystyle\lim _{ x\rightarrow 0 }{ f\left( x \right)  } =f\left( 0 \right) \Rightarrow \displaystyle\lim _{ x\rightarrow 0 }{ { x }^{ n }\sin { \dfrac { 1 }{ x }  }  } =0,\forall n> 0$$
$$f\left( x \right) $$ is differentiable at $$x=0$$, if $$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) -f\left( 0 \right)  }{ x-0 }  } $$ exists finitely.
$$\Rightarrow \displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) -f\left( 0 \right)  }{ x-0 }  } $$ exists finitely
$$\Rightarrow \displaystyle\lim _{ x\rightarrow 0 }{ { x }^{ n-1 } } \sin { \dfrac { 1 }{ x }  } $$ exists finitely
$$\Rightarrow n-1 > 0$$    $$\Rightarrow $$   $$n > 1$$
Hence, $$f\left( x \right) $$ is continuous but not differentiable at $$x=0$$, if $$n\in \left( 0,1 \right) $$.

Solution:

Given, $$u = \tan^{-1} \left \{\dfrac {\sqrt {1 - x^{2}} - 1}{x}\right \}$$
and $$v = \sin^{-1} x$$
Put $$x = \sin \theta \Rightarrow \theta = \sin^{-1}x$$, then
$$u = \tan^{-1} \left \{\dfrac {\cos \theta - 1}{\sin \theta}\right \} (\because 1 - \cos^{2} \theta = \sin^{2} \theta )$$
$$= -\tan^{-1}\left \{\dfrac {1 - \cos \theta}{\sin \theta}\right \}$$
$$= -\tan^{-1} \left \{\dfrac {2\sin^{2} \theta /2}{2\sin \theta/2 \cdot \cos \theta /2}\right \}$$
$$= -\tan^{-1} \left (\tan \dfrac {\theta}{2}\right )$$
$$= -\dfrac {1}{2} \theta = -\dfrac {1}{2} \sin^{-1}x$$
Therefore, $$\dfrac {du}{dx} = -\dfrac {1}{2\sqrt {1 - x^{2}}}$$
and $$\dfrac {dv}{dx} = \dfrac {1}{\sqrt {1 - x^{2}}}$$
Thus $$ \dfrac {du}{dv} = \dfrac {du}{dx}\times \dfrac {dx}{dv} = -\dfrac {1}{2\sqrt {1 - x^{2}}} \times \sqrt {1 - x^{2}}$$
$$= -\dfrac {1}{2}$$

We have $${ 2 }^{ x }+{ 2 }^{ y }={ 2 }^{ x+y }$$

Given, $$x=a\cos^3\theta, y=a\sin ^3\theta$$

Given curve is $$y^{2} = px^{3} + q .... (i)$$
On differentiating w.r.t. $$x$$, we get
$$2y \dfrac {dy}{dx} = 3px^{2} \Rightarrow \dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$$
At $$(2, 3), \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$$
Since, line $$y = 4x - 5$$ is a tangent to the given curve.
Therefore, $$\text{slope} = 4 = 2p$$
$$\Rightarrow p = 2$$
Now, put the value of $$p, x$$ and $$y$$ in Eq. (i), we get
$$(3)^{2} = (2)(2)^{3} + q$$
$$\Rightarrow 9 = 16 + q$$
$$\Rightarrow q = -7$$
Hence, $$p + q = 2 - 7 = -5$$.

Given, $$xe^{xy} + ye^{-xy} = \sin^{2}x$$
On differentiating w.r.t. $$x$$, we get
$$e^{xy} + xe^{xy} \left \{x \dfrac {dy}{dx} + y\right \} + \dfrac {dy}{dx} e^{-xy}$$
$$-ye^{-xy} \left \{x \dfrac {dy}{dx} + y\right \} = 2\sin x \cdot \cos x$$
$$\Rightarrow \left \{x^{2} e^{xy} + e^{-xy} - xe^{y} e^{-xy}\right \}\dfrac {dy}{dx} + \left \{e^{xy} + xye^{xy} - y^{2} e^{-xy}\right \} =\sin 2x$$
On putting $$x = 0$$, we get
$$\left \{0 + e^{-0} - 0\right \} \left (\dfrac {dy}{dx}\right )_{(x = 0)} + \left \{e^{0} + 0 - y^{2}e^{-0}\right \} = \sin 0$$
$$\Rightarrow \left [(1)\dfrac {dy}{dx}\right ]_{x = 0} + (1 - y^{2}) = 0$$
$$\Rightarrow \left [\dfrac {dy}{dx}\right ]_{x = 0} = y^{2} - 1$$.

Given, $$ u = 2 (t - \sin t ) $$ and $$ v = 2 (1- \cos t), $$
On differentiating w.r.t. t, we get 
$$ \dfrac {du}{dt} = 2 ( 1- \cos t) $$ and $$ \dfrac {dv}{dt} = 2 ( \sin t ) $$
Therefore, $$  \dfrac {dv}{du} = \dfrac {dv/dt} {du/dt} $$
$$ = \dfrac { 2 ( \sin t)} { 2 (1 - \cos t )} = \dfrac {2 \sin \frac {t}{2} \cos \frac {t}{2}}{2 \sin^2 \frac {t}{2}} $$
$$ \Rightarrow \dfrac {dv}{du} = \cot \dfrac {t}{2} $$
At $$ t = \dfrac { 2 \pi}{3} $$, we have
$$ \dfrac {dv}{du} = \cot \left( \dfrac {2 \pi}{3 \times 2 } \right) = \cot \dfrac {\pi}{3} = \dfrac {1}{\sqrt3} $$

Here, $$x\sin (a + y) + \sin a \cos (a + y) = 0$$     ..... (i)
On differentiating w.r.t. $$x$$, we get
$$\dfrac {d}{dx} [x\sin (a + y)] + \dfrac {d}{dx} [\sin a\cos (a + y)] = 0$$
$$\Rightarrow \left [x\dfrac {d}{dx}\left \{\sin (a + y)\right \} + \sin (a + y) \dfrac {d}{dx}(x)\right ] + \sin a\dfrac {d}{dx} \cos (a + y) = 0$$
(using product rule and chain rule)
$$\Rightarrow \left [x \cos (a + y)\dfrac {d}{dx}(a + y) + \sin (a + y)\right ] + \sin a\left [-\sin (a + y) \dfrac {d}{dx} (a + y)\right ] = 0$$
$$\Rightarrow \left [x \cos (a + y) \left (0 + \dfrac {dy}{dx}\right ) + \sin (a + y)\right ] - \sin a \sin (a + y) \left (0 + \dfrac {dy}{dx}\right ] = 0$$
$$\Rightarrow x\cos (a + y) \dfrac {dy}{dx} + \sin (a + y) - \sin a \sin (a + y) \dfrac {dy}{dx} = 0$$
$$\Rightarrow \dfrac {dy}{dx} [x\cos (a + y) - \sin a \sin (a + y)] = -\sin (a + y)$$
$$\left [from Eq.(i), putting\ x = -\sin a \dfrac {\cos (a + y)}{\sin (a + y)}\right ]$$
$$\Rightarrow \dfrac {dy}{dx}\left [-\sin a\dfrac {\cos^{2}(a + y)}{\sin (a + y)} - \sin a \sin (a + y)\right ]$$
$$= -\sin (a + y)$$
$$\Rightarrow -\dfrac {dy}{dx}\left [\dfrac {\sin a \cos^{2} (a + y) + \sin a \sin^{2} (a + y)}{\sin (a + y)}\right ]$$
$$= -\sin (a + y)$$
$$\Rightarrow \dfrac {dy}{dx} = \sin (a + y) \left [\dfrac {\sin (a + y)}{\sin a \left \{\cos^{2} (a + y) +\sin^{2} (a + y)\right \}}\right ]$$
$$\Rightarrow \dfrac {dy}{dx} = \dfrac {\sin^{2}(a + y)}{\sin a} (\because \sin^{2}\theta + \cos^{2}\theta = 1)$$

Given, $$s=\sec ^{ -1 }{ \left( \cfrac { 1 }{ 2{ x }^{ 2 }-1 }  \right)  } $$