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Continuity and Differentiability

Long Answer Type

611.

Find all points of discontinuity of f, where f is defined as following:

$f (x) = \{\begin{cases} \begin{matrix} |x| + 3, x \leq - 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$

612.

$Find \frac{dy}{dx}, if y = {(cosx)}^{x} + {(sinx)}^{\frac{1}{x}}$

613.

Find the value of ‘a’ for which the function f defined as

$f (x) = \{\begin{cases} a \sin \frac{π}{2} (x + 1), x \leq 0 \\ \frac{\tan x - \sin x}{x^{3}}, x > 0 \end{cases}$

is continuous at x = 0.

614.

Differentiate $X^{x \cos x} + \frac{x^{2} + 1}{x^{2} - 1} w . r . t . x$

615.
If $x = a (θ - \sin θ), y = (1 + \cos θ), find \frac{d^{2} y}{{dx}^{2}}$

$x = a (θ - \sin θ), y = a (1 + \cos θ) Differentiating x and y w . r . t . θ, \frac{dx}{dθ} = a (1 - \cos θ) . . . . . . . . . (i) \frac{dy}{dθ} = - a \sin θ . . . . . . . . . . (ii) Dividing (2) by (1), \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} = \frac{- a \sin θ}{a (1 - \cos θ)}$

$\Rightarrow \frac{dy}{dx} = \frac{- sinθ}{1 - \cos θ} \Rightarrow \frac{dy}{dx} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} \Rightarrow \frac{dy}{dx} = \frac{- \cos \frac{θ}{2}}{\sin \frac{θ}{2}} \Rightarrow \frac{dy}{dx} = - \cot \frac{θ}{2}$

Differentiating w.r.t. x,

$\frac{d}{dx} (\frac{dy}{dx}) = \frac{d}{dθ} (\frac{dy}{dx}) x \frac{dθ}{dx} \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dθ} (\frac{dy}{dx}) x \frac{dθ}{dx} \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dθ} (- \cot \frac{θ}{2}) x \frac{dθ}{dx} . . . . [From equation (iii)] \frac{d^{2} y}{{dx}^{2}} = - (- {cosec}^{2} \frac{θ}{2} x \frac{1}{2}) x \frac{dθ}{dx} = \frac{1}{2} {cosec}^{2} \frac{θ}{2} x \frac{1}{(\frac{dx}{dθ})}$

$= \frac{1}{2} {cosec}^{2} \frac{θ}{2} x \frac{1}{a (1 - \cos θ)} . . . . . . . . [From equation (i)] = \frac{{cosec}^{2} \frac{θ}{2}}{2 a (1 - \cos θ)} = \frac{{cosec}^{2} \frac{θ}{2}}{2 a (2 \sin^{2} \frac{θ}{2})} = \frac{1}{4 a} x {cosec}^{4} \frac{θ}{2}$