Continuity and Differentiability

Given $f(x) = 2x^2 - 1$

Value of function at $x = 3$: $f(3) = 2(3)^2 - 1 = 18 - 1 = 17$

Limit as $x \to 3$: $\lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 17$

Since $\lim_{x \to 3} f(x) = f(3)$, the function is continuous at $x = 3$.

At $x = 2$, $f(2) = 2(2) + 3 = 7$.

Left Hand Limit (LHL): $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (2x + 3) = 2(2) + 3 = 7$.

Right Hand Limit (RHL): $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (2x - 3) = 2(2) - 3 = 1$.

Since $\text{LHL} \neq \text{RHL}$, the limit does not exist at $x = 2$. Therefore, $f(x)$ is discontinuous at $x = 2$.

Let $c$ be any real number. Then, we check for three cases: $c < 5, c = 5, c > 5$.

Case 1 ($c < 5$): $f(x) = -(x - 5)$. Limit: $\lim_{x \to c} f(x) = -(c - 5) = f(c)$. Continuous.

Case 2 ($c > 5$): $f(x) = x - 5$. Limit: $\lim_{x \to c} f(x) = c - 5 = f(c)$. Continuous.

Case 3 ($c = 5$): $f(5) = 0$. $\text{LHL} = \lim_{x \to 5^-} -(x - 5) = 0$. $\text{RHL} = \lim_{x \to 5^+} (x - 5) = 0$. Since $\text{LHL} = \text{RHL} = f(5)$, it is continuous at $x = 5$.

Hence, $f(x)$ is continuous for all $x \in \mathbb{R}$.

Let $c$ be any integer. Then $g(c) = c - [c] = c - c = 0$.

$\text{LHL}: \lim_{x \to c^-} (x - [x])$. As $x \to c^-$, $[x] = c - 1$. So, $\text{LHL} = c - (c - 1) = 1$.

$\text{RHL}: \lim_{x \to c^+} (x - [x])$. As $x \to c^+$, $[x] = c$. So, $\text{RHL} = c - c = 0$.

Since $\text{LHL} \neq \text{RHL}$, the limit does not exist at integral points. Hence, $g(x)$ is discontinuous at all integers.

$f(0) = 0 + 1 = 1$.

$\text{LHL}: \lim_{x \to 0^-} \frac{\sin x}{x} = 1$ (Standard Limit).

$\text{RHL}: \lim_{x \to 0^+} (x + 1) = 0 + 1 = 1$.

Since $\text{LHL} = \text{RHL} = f(0) = 1$, the function is continuous at $x = 0$.

$f(0) = 1$.

$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ (Standard limit formula).

Since $\lim_{x \to 0} f(x) = f(0) = 1$, the function is continuous at $x = 0$.

Since $f(x)$ is continuous at $x = \frac{\pi}{2}$, $\lim_{x \to \pi/2} f(x) = f(\frac{\pi}{2}) = 3$.

$\lim_{x \to \pi/2} \frac{k \cos x}{\pi - 2x}$. Let $x = \frac{\pi}{2} - h$. As $x \to \frac{\pi}{2}$, $h \to 0$.

$\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} - h)}{\pi - 2(\frac{\pi}{2} - h)} = \lim_{h \to 0} \frac{k \sin h}{2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2} (1) = \frac{k}{2}$.

Therefore, $\frac{k}{2} = 3 \implies k = 6$.

For $f(x)$ to be continuous, it must be continuous at $x=2$ and $x=10$.

At $x=2$: $\text{LHL} = 5$, $\text{RHL} = \lim_{x \to 2^+} (ax + b) = 2a + b$. So, $2a + b = 5$ (Eq 1).

At $x=10$: $\text{LHL} = \lim_{x \to 10^-} (ax + b) = 10a + b$, $\text{RHL} = 21$. So, $10a + b = 21$ (Eq 2).

Subtract Eq 1 from Eq 2: $8a = 16 \implies a = 2$.

Substitute $a=2$ in Eq 1: $2(2) + b = 5 \implies 4 + b = 5 \implies b = 1$.

Therefore, $a = 2, b = 1$.

$f(0) = \frac{2(0)+1}{0-1} = -1$.

$\text{RHL} = -1$.

$\text{LHL} = \lim_{x \to 0^-} \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x}$. Rationalizing the numerator:

$= \lim_{x \to 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})}$

$= \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{2k}{1 + 1} = k$.

For continuity, $\text{LHL} = f(0) \implies k = -1$.

Since $f(x)$ is continuous at $x=0$, $k = \lim_{x \to 0} \frac{1-\cos 4x}{8x^2}$.

Using $1-\cos \theta = 2\sin^2(\theta/2)$, we get $1-\cos 4x = 2\sin^2(2x)$.

$k = \lim_{x \to 0} \frac{2\sin^2(2x)}{8x^2} = \lim_{x \to 0} \frac{2}{8} \left(\frac{\sin(2x)}{x}\right)^2$.

Multiply and divide inside the square by 2: $k = \frac{1}{4} \lim_{x \to 0} \left(\frac{\sin(2x)}{2x} \cdot 2\right)^2 = \frac{1}{4} \cdot (1 \cdot 2)^2 = \frac{4}{4} = 1$.

Therefore, $k = 1$.

For continuity at $x=3$, $\text{LHL} = \text{RHL} = f(3)$.

$\text{LHL} = \lim_{x \to 3^-} (ax + 1) = 3a + 1$.

$\text{RHL} = \lim_{x \to 3^+} (bx + 3) = 3b + 3$.

$f(3) = 3a + 1$.

Equating LHL and RHL: $3a + 1 = 3b + 3 \implies 3a - 3b = 2 \implies a - b = \frac{2}{3}$.

This is the required relationship between $a$ and $b$.

To check differentiability at $x=1$, we evaluate LHD and RHD.

$f(1) = |1-1| = 0$.

$\text{LHD} = \lim_{h \to 0^-} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0^+} \frac{|(1-h)-1| - 0}{-h} = \lim_{h \to 0^+} \frac{|-h|}{-h} = \lim_{h \to 0^+} \frac{h}{-h} = -1$.

$\text{RHD} = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{|(1+h)-1| - 0}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.

Since $\text{LHD} \neq \text{RHD}$, the function is not differentiable at $x=1$.

$f(x) = \begin{cases} -x^2, & \text{if } x < 0 \\ x^2, & \text{if } x \ge 0 \end{cases}$

Continuity at $x=0$: $\text{LHL} = \lim_{x \to 0} (-x^2) = 0$, $\text{RHL} = \lim_{x \to 0} (x^2) = 0$, $f(0) = 0$. It is continuous.

Differentiability at $x=0$:

$\text{LHD} = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{-(-h)^2 - 0}{-h} = \lim_{h \to 0^+} \frac{-h^2}{-h} = \lim_{h \to 0^+} h = 0$.

$\text{RHD} = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0$.

Since $\text{LHD} = \text{RHD}$, $f(x)$ is differentiable at $x=0$.

We know that a function must be continuous to be differentiable. In Q4, we proved that $f(x) = [x]$ is discontinuous at all integer points.

Since differentiability implies continuity, the contrapositive is true: if a function is not continuous at a point, it cannot be differentiable at that point.

Therefore, $f(x) = [x]$ is not differentiable at integer points.

If $f(x)$ is differentiable at $x=1$, it MUST be continuous at $x=1$.

Continuity at $x=1$: $\text{LHL} = \text{RHL} \implies 1^2 + 3(1) + a = b(1) + 2 \implies 4 + a = b + 2 \implies b - a = 2$ (Eq 1).

Differentiability at $x=1$: $\text{LHD} = \text{RHD}$.

$\text{LHD}$ is the derivative of the left branch at $x=1$: $\frac{d}{dx}(x^2+3x+a) = 2x+3$. At $x=1$, $\text{LHD} = 2(1)+3 = 5$.

$\text{RHD}$ is the derivative of the right branch at $x=1$: $\frac{d}{dx}(bx+2) = b$. So, $\text{RHD} = b$.

Equating LHD and RHD: $b = 5$.

Substitute $b=5$ in Eq 1: $5 - a = 2 \implies a = 3$.

Thus, $a = 3, b = 5$.

Let $y = \sin(x^2 + 5)$. Let $u = x^2 + 5$, then $y = \sin(u)$.

By chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \cos(u) \cdot \frac{d}{dx}(x^2 + 5) = \cos(x^2 + 5) \cdot (2x)$.

Answer: $2x \cos(x^2 + 5)$.

Let $y = \cos(\sqrt{x})$.

$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$

$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \left(\frac{1}{2\sqrt{x}}\right)$.

Answer: $-\frac{\sin(\sqrt{x})}{2\sqrt{x}}$.

Let $y = 2(\cot(x^2))^{1/2}$.

$\frac{dy}{dx} = 2 \cdot \frac{1}{2}(\cot(x^2))^{-1/2} \cdot \frac{d}{dx}(\cot(x^2))$

$= \frac{1}{\sqrt{\cot(x^2)}} \cdot (-\text{cosec}^2(x^2)) \cdot \frac{d}{dx}(x^2)$

$= \frac{-\text{cosec}^2(x^2)}{\sqrt{\cot(x^2)}} \cdot 2x = \frac{-2x \text{ cosec}^2(x^2)}{\sqrt{\cot(x^2)}}$.

Let $y = \cos x^3 \cdot (\sin(x^5))^2$. Use product rule: $(uv)' = u'v + uv'$.

$y' = [\frac{d}{dx}(\cos x^3)] \cdot \sin^2(x^5) + \cos x^3 \cdot [\frac{d}{dx}(\sin^2(x^5))]$

$= [-\sin x^3 \cdot 3x^2] \cdot \sin^2(x^5) + \cos x^3 \cdot [2\sin(x^5) \cdot \cos(x^5) \cdot 5x^4]$

$= -3x^2 \sin x^3 \sin^2(x^5) + 10x^4 \cos x^3 \sin(x^5) \cos(x^5)$.

Let $y = e^{\sin^{-1}x}$. Using the chain rule:

$\frac{dy}{dx} = e^{\sin^{-1}x} \cdot \frac{d}{dx}(\sin^{-1}x)$

$\frac{dy}{dx} = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}$.

Answer: $\frac{e^{\sin^{-1}x}}{\sqrt{1 - x^2}}$.

Differentiating both sides w.r.t $x$:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$

$2 + 3\frac{dy}{dx} = \cos y \frac{dy}{dx}$

Rearranging terms with $\frac{dy}{dx}$ to one side:

$\cos y \frac{dy}{dx} - 3\frac{dy}{dx} = 2$

$\frac{dy}{dx} (\cos y - 3) = 2 \implies \frac{dy}{dx} = \frac{2}{\cos y - 3}$.

Differentiating both sides w.r.t $x$:

$(x\frac{dy}{dx} + y \cdot 1) + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$

$x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y$

$\frac{dy}{dx}(x + 2y - 1) = \sec^2 x - y$

$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$.

Differentiating w.r.t $x$:

$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)$

$2\sin y \cos y \frac{dy}{dx} - \sin(xy) \cdot \frac{d}{dx}(xy) = 0$

$\sin(2y) \frac{dy}{dx} - \sin(xy) \cdot (x\frac{dy}{dx} + y) = 0$

$\sin(2y) \frac{dy}{dx} - x\sin(xy) \frac{dy}{dx} - y\sin(xy) = 0$

$\frac{dy}{dx}(\sin(2y) - x\sin(xy)) = y\sin(xy) \implies \frac{dy}{dx} = \frac{y\sin(xy)}{\sin(2y) - x\sin(xy)}$.

Take natural log on both sides: $y \log x = x \log y$.

Differentiating both sides w.r.t $x$ using product rule:

$y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} = x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y \cdot 1$

$\frac{dy}{dx} \left(\log x - \frac{x}{y}\right) = \log y - \frac{y}{x}$

$\frac{dy}{dx} \left(\frac{y \log x - x}{y}\right) = \frac{x \log y - y}{x}$

$\frac{dy}{dx} = \frac{y(x \log y - y)}{x(y \log x - x)}$.

Method 1: Direct chain rule is tedious. We use substitution.

Let $x = \sin \theta \implies \theta = \sin^{-1} x$.

$y = \sin^{-1}(2 \sin \theta \sqrt{1 - \sin^2 \theta}) = \sin^{-1}(2 \sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta)$.

Assuming principal values, $y = 2\theta = 2\sin^{-1} x$.

$\frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1} x) = \frac{2}{\sqrt{1 - x^2}}$.

Let $x = \cos \theta \implies \theta = \cos^{-1} x$.

$y = \sec^{-1}\left(\frac{1}{2\cos^2 \theta - 1}\right) = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) = \sec^{-1}(\sec 2\theta)$.

Since $0 < x < \frac{1}{\sqrt{2}}$, $0 < \cos \theta < \frac{1}{\sqrt{2}} \implies \frac{\pi}{4} < \theta < \frac{\pi}{2} \implies \frac{\pi}{2} < 2\theta < \pi$.

This falls in the principal range of $\sec^{-1}$, so $y = 2\theta = 2\cos^{-1} x$.

$\frac{dy}{dx} = 2 \cdot \left(\frac{-1}{\sqrt{1 - x^2}}\right) = \frac{-2}{\sqrt{1 - x^2}}$.

Let $x = \tan \theta \implies \theta = \tan^{-1} x$.

$y = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right) = \tan^{-1}(\tan 3\theta)$.

Assuming principal values, $y = 3\theta = 3\tan^{-1} x$.

$\frac{dy}{dx} = \frac{3}{1 + x^2}$.

Let $x = \tan \theta \implies \theta = \tan^{-1} x$.

$y = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta} - 1}{\tan \theta}\right) = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$

$y = \tan^{-1}\left(\frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}}\right) = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right)$

Use half angle formulas: $1 - \cos \theta = 2\sin^2(\theta/2)$, $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$

$y = \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan^{-1}(\tan(\theta/2)) = \frac{\theta}{2} = \frac{1}{2}\tan^{-1} x$.

$\frac{dy}{dx} = \frac{1}{2(1 + x^2)}$.

Let $x = \tan \theta \implies \theta = \tan^{-1} x$.

$y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos 2\theta)$.

Assuming principal values, $y = 2\theta = 2\tan^{-1} x$.

$\frac{dy}{dx} = \frac{2}{1 + x^2}$.

Taking log on both sides: $\log y = \log(x^x) = x \log x$.

Differentiate w.r.t $x$:

$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1$

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

$\frac{dy}{dx} = y(1 + \log x) = x^x(1 + \log x)$.

Let $y = u + v$, where $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$. Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

For $u$: $\log u = \sin x \log x \implies \frac{1}{u}\frac{du}{dx} = \sin x \cdot \frac{1}{x} + \log x \cdot \cos x$

$\frac{du}{dx} = x^{\sin x} \left(\frac{\sin x}{x} + \cos x \log x\right)$.

For $v$: $\log v = \cos x \log(\sin x) \implies \frac{1}{v}\frac{dv}{dx} = \cos x \cdot \frac{1}{\sin x} \cdot \cos x + \log(\sin x) \cdot (-\sin x)$

$\frac{dv}{dx} = (\sin x)^{\cos x} (\cot x \cos x - \sin x \log(\sin x))$.

$\frac{dy}{dx} = x^{\sin x} \left(\frac{\sin x}{x} + \cos x \log x\right) + (\sin x)^{\cos x} (\cot x \cos x - \sin x \log(\sin x))$.

Let $y = \left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)^{1/2}$.

Take log: $\log y = \frac{1}{2} [\log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) - \log(x-5)]$.

Differentiate: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$.

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$.

We can rewrite the equation as: $y = \sqrt{\sin x + y}$.

Squaring both sides: $y^2 = \sin x + y$.

Differentiating both sides w.r.t $x$: $2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}$.

$2y\frac{dy}{dx} - \frac{dy}{dx} = \cos x \implies \frac{dy}{dx}(2y - 1) = \cos x$.

$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$.

$\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}$.

$\frac{dx}{d\theta} = a(1 - \cos \theta) = a(2\sin^2(\theta/2))$.

$\frac{dy}{d\theta} = a(-\sin \theta) = -a(2\sin(\theta/2)\cos(\theta/2))$.

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-2a\sin(\theta/2)\cos(\theta/2)}{2a\sin^2(\theta/2)} = -\cot(\theta/2)$.

$\frac{dy}{dt} = a \cos t$.

$\frac{dx}{dt} = a \left(-\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}\right) = a \left(-\sin t + \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2}\right)$

$= a \left(-\sin t + \frac{1}{2\sin(t/2)\cos(t/2)}\right) = a \left(-\sin t + \frac{1}{\sin t}\right) = a \left(\frac{1 - \sin^2 t}{\sin t}\right) = \frac{a \cos^2 t}{\sin t}$.

$\frac{dy}{dx} = \frac{a \cos t}{\frac{a \cos^2 t}{\sin t}} = \tan t$.

Let $u = \sin^2 x$ and $v = e^{\cos x}$. We need to find $\frac{du}{dv}$.

$\frac{du}{dx} = 2\sin x \cos x = \sin 2x$.

$\frac{dv}{dx} = e^{\cos x} (-\sin x) = -\sin x e^{\cos x}$.

$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2\sin x \cos x}{-\sin x e^{\cos x}} = \frac{-2\cos x}{e^{\cos x}}$.

$\frac{dy}{dx} = 2x + 3$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x + 3) = 2$.

$y_1 = \frac{dy}{dx} = -3\sin(\log x) \cdot \frac{1}{x} + 4\cos(\log x) \cdot \frac{1}{x}$.

Multiply by $x$: $x y_1 = -3\sin(\log x) + 4\cos(\log x)$.

Differentiate again w.r.t $x$:

$x \cdot y_2 + y_1 \cdot 1 = -3\cos(\log x) \cdot \frac{1}{x} - 4\sin(\log x) \cdot \frac{1}{x}$.

Multiply by $x$ again: $x^2 y_2 + x y_1 = -(3\cos(\log x) + 4\sin(\log x)) = -y$.

Therefore, $x^2 y_2 + x y_1 + y = 0$.

$y_1 = mAe^{mx} + nBe^{nx}$.

$y_2 = m^2Ae^{mx} + n^2Be^{nx}$.

Substitute in LHS: $(m^2Ae^{mx} + n^2Be^{nx}) - (m+n)(mAe^{mx} + nBe^{nx}) + mn(Ae^{mx} + Be^{nx})$

$= m^2Ae^{mx} + n^2Be^{nx} - m^2Ae^{mx} - mnBe^{nx} - mnAe^{mx} - n^2Be^{nx} + mnAe^{mx} + mnBe^{nx}$

All terms cancel out. LHS = 0 = RHS.

$\frac{dx}{d\theta} = -a\sin \theta$, $\frac{dy}{d\theta} = b\cos \theta \implies \frac{dy}{dx} = \frac{b\cos \theta}{-a\sin \theta} = -\frac{b}{a}\cot \theta$.

Now, differentiate $\frac{dy}{dx}$ w.r.t $x$ (NOT $\theta$!):

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{b}{a}\cot \theta\right) = \frac{d}{d\theta}\left(-\frac{b}{a}\cot \theta\right) \cdot \frac{d\theta}{dx}$.

$\frac{d^2y}{dx^2} = \left(-\frac{b}{a}(-\text{cosec}^2 \theta)\right) \cdot \frac{1}{-a\sin \theta} = \frac{b}{a}\text{cosec}^2 \theta \cdot \frac{-1}{a\sin \theta} = -\frac{b}{a^2}\text{cosec}^3 \theta$.

1. $f(x)$ is a polynomial, so it is continuous in $[-4, 2]$.

2. $f(x)$ is differentiable in $(-4, 2)$.

3. $f(-4) = (-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0$. $f(2) = (2)^2 + 2(2) - 8 = 4 + 4 - 8 = 0$. So, $f(-4) = f(2)$.

All conditions of Rolle's theorem are satisfied. There must exist $c \in (-4, 2)$ such that $f'(c) = 0$.

$f'(x) = 2x + 2$. $f'(c) = 2c + 2 = 0 \implies c = -1$.

Since $c = -1 \in (-4, 2)$, Rolle's theorem is verified.

1. $f(x)$ is a polynomial, so it is continuous in $[1, 4]$.

2. $f(x)$ is differentiable in $(1, 4)$.

There must exist $c \in (1, 4)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

$f(4) = 4^2 - 4(4) - 3 = -3$. $f(1) = 1^2 - 4(1) - 3 = -6$.

$f'(x) = 2x - 4 \implies f'(c) = 2c - 4$.

$2c - 4 = \frac{-3 - (-6)}{4 - 1} = \frac{3}{3} = 1$.

$2c - 4 = 1 \implies 2c = 5 \implies c = 2.5$.

Since $c = 2.5 \in (1, 4)$, MVT is verified.