Continuity and Differentiability Test - 38

Let $$ y= \dfrac{1}{2x^{2}-1} $$

$$\displaystyle y=(\tan x)^{x}$$

For $$f(x)$$ is continuous at $$x = \dfrac {\pi}{2}$$, we must have
$$\displaystyle \lim_{x\rightarrow \frac {\pi}{2}} f(x) = f\left (\dfrac {\pi}{2}\right )$$
$$\Rightarrow \displaystyle \lim_{x\rightarrow \frac {\pi}{2}} \dfrac {1 - \sin x}{(\pi - 2x)^{2}} \cdot \dfrac {\log \sin x}{\log (1 + \pi^{2} - 4\pi x + 4x^{2})} = k$$
$$\Rightarrow \displaystyle \lim_{h\rightarrow 0} \dfrac {1 - \cos h}{4h^{2}} \cdot \dfrac {\log \cos h}{\log (1 + 4h^{2})} = k$$
$$\Rightarrow \displaystyle \lim_{h\rightarrow 0} \dfrac {1 - \cos h}{4h^{2}} \times \dfrac {\log(1 + \cos h - 1)}{\cos h - 1} \times \dfrac {4h^{2}}{\log (1 + 4h^{2})} \times \dfrac {\cos h - 1}{4h^{2}} = k$$
$$\Rightarrow -\displaystyle \lim_{h\rightarrow 0} \left (\dfrac {1 - \cos h}{4h^{2}}\right )^{2} \times \dfrac {\log (1 + (\cos h - 1))}{\cos h - 1}\times \dfrac {4h^{2}}{\log (1 + 4h^{2})} = k$$
$$\Rightarrow -\dfrac {1}{64} \displaystyle \lim_{h\rightarrow 0} \left (\dfrac {\sin (\frac{h}{2})}{\frac{h}{2}}\right )^{4} \dfrac {\log (1 + (\cos h - 1))}{\cos h - 1}\times \dfrac {4h^{2}}{\log (1 + 4h^{2})} = k$$
$$\Rightarrow -\dfrac {1}{64} = k\Rightarrow k = -\dfrac {1}{64}$$.

We have to find the derivative of $$tan^{-1} \dfrac{cos x-sin x}{cos x+sin x}$$ with respect to $$x$$.

Let $$x=a\cos ^{ 2 }{ \theta  } +b\sin ^{ 2 }{ \theta  } $$
Here $$ a-x=a-a\cos ^{ 2 }{ \theta  } -b\sin ^{ 2 }{ \theta  } =\left( a-b \right) \sin ^{ 2 }{ \theta  } $$
and $$x-b=a\cos ^{ 2 }{ \theta  } -b\sin ^{ 2 }{ \theta  } -b=\left( a-b \right) \cos ^{ 2 }{ \theta  } $$
and $$ y=\left( a-b \right) \sin { \theta  } \cdot \cos { \theta  } -\left( a-b \right) \tan ^{ -1 }{ \tan { \theta  }  } $$
$$=\dfrac { a-b }{ 2 } \sin { 2\theta  } -\left( a-b \right) \theta $$
Therefore, $$ \dfrac { dy }{ dx } =\dfrac { { dy }/{ d\theta  } }{ { dx }/{ d\theta  } } =\dfrac { \left( a-b \right) \cos { 2\theta  } -\left( a-b \right)  }{ \left( b-a \right) \sin { 2\theta  }  } $$
$$=\dfrac { 1-\cos { 2\theta  }  }{ \sin { \theta  }  } =\tan { \theta  } =\sqrt { \dfrac { a-x }{ x-b }  } $$

Given, $$f(x) = \left\{\begin{matrix}2a - x, &if\ -a < x < a \\ 3x - 2a, &if\ a \leq x \end{matrix}\right.$$
At $$x = a$$,
$$LHL = \displaystyle \lim_{x\rightarrow a^{-}} f(x) = \displaystyle \lim_{x\rightarrow a} (2a - x)$$
$$= 2a - a = a\ RHL$$
$$= \displaystyle \lim_{x\rightarrow a^{+}} f(x) = \displaystyle \lim_{x\rightarrow a}(3x - 2a)$$
$$= 3(a) - 2a = a$$
and $$f(a) = 3(a) - 2a = a$$
$$\therefore LHL = RHL = f(a)$$
Hence, it is continuous at $$x = a$$
Again, at $$x = a$$
$$LHD = \displaystyle \lim_{h\rightarrow 0}\dfrac {f(a - h) - f(a)}{-h}$$
$$= \displaystyle \lim_{h\rightarrow 0} = \dfrac {2a - (a - h) - a}{-h} =-1$$
and $$RHD = \displaystyle \lim_{h\rightarrow 0} \dfrac {f(a + h) - f(a)}{h}$$
$$= \displaystyle \lim_{h\rightarrow 0} \dfrac {3(a + h) - 2a - a}{h} = 3$$
$$\therefore LHD \neq RHD$$
Hence, it is not differentiable at $$x = a$$.

Given, $$x = A\cos 4t + B\sin 4t$$
On differentiating w.r.t., $$t$$, we get
$$\dfrac {dx}{dt} = -4A\sin 4t + 4B\cos 4t$$
Again, differentiating w.r.t. $$t$$, we get
$$\dfrac {d^{2}x}{dt^{2}} = -6A\cos 4t - 16B \sin 4t$$
$$\Rightarrow \dfrac {d^{2}x}{dt^{2}} = -16(x)$$.

We have
$$\displaystyle\lim _{ x\rightarrow { 2 }^{ - } }{ f(x) } =\lim _{ h\rightarrow 0 }{ f(2-h) } =\lim _{ h\rightarrow 0 }{ { (2-h) }^{ 2 }+{ (-2+h) }^{ 2 } } $$
$$\displaystyle\Rightarrow \lim _{ x\rightarrow { 2 }^{ - } }{ f(x) } =3+{ (-2) }^{ 2 }=7$$
$$\displaystyle\lim _{ x\rightarrow { 2 }^{ + } }{ f(x) } =\lim _{ h\rightarrow 0 }{ f(2+h) } =\lim _{ h\rightarrow 0 }{ { (2+h) }^{ 2 }+{ (-2-h) }^{ 2 } } $$
$$\displaystyle\lim _{ x\rightarrow { 2 }^{ + } }{ f(x) } =4+{ (-3) }^{ 2 }=13\quad $$
clearly, $$\displaystyle\lim _{ x\rightarrow { 2 }^{ - } }{ f(x) } \neq \lim _{ x\rightarrow { 2 }^{ + } }{ f(x) } $$
so, $$f(x)$$ is discontinuous at $$x=2$$
Consequently, it is not differentiable at $$x=2$$

let $$ \tan^{-1}x = t$$ and $$\dfrac{\tan^{-1}x}{1+\tan^{-1}x} = y$$