Continuity and Differentiability Test - 11

We know that, if  f and g be continuous functions, then

(A) f + g is continuous

(B) f - g is continuous.

(C) fg is continuous.

(D) \(\frac{f}{g} \) is continuous at these points, where g(x) \(\neq\) 0.

Here, \(\frac{g(x)}{f(x)} = \frac{\frac{x^2}{2} + 1}{2x} = \frac{x^2 + 2}{4x}\)

Which is discontinuous at x = 0

We have, f(x) = \(\frac{4 - x^2}{4x - x^3}\) = \(\frac{(4 - x^2)}{x(4 - x^2)}\)

= \(\frac{(4 - x^2)}{x(2^2 - x^2)}\) = \(\frac{4 - x^2}{x(2 + x)(2 - x)}\)

Clearly, f(x) is discontinuous at exactly three points x = 0, x = -2 and x = 2.

We have, f(x) = |2x - 1| sin x

At x = \(\frac{1}{2}\), f(x) is not differentiable

Hence, f(x) is differentiable in R - {\({\frac{1}{2}}\)}

\(\because\) Rf'(\(\frac{1}{2}\)) = \(\lim \limits_{h \to 0} \frac{{f(\frac{1}{2} + h) - f(\frac{1}{2})}}{h}\)

= \(\lim \limits_{h \to 0} \frac{|2(\frac{1}{2} + h) - 1| sin (\frac{1}{2} + h) - 0}{h}\)

= \(\lim \limits_{h \to 0} \frac{|2h|.sin(\frac{1 + 2h}{2})}{h} = 2.sin \frac{1}{2}\)

and Lf'\((\frac{1}{2}) = \lim \limits_{h \to 0} \frac{f(\frac{1}{2} - h) - f(\frac{1}{2})}{-h}\)

= \(\lim \limits_{h \to 0} \frac{|2(\frac{1}{2} - h)^{-1}|sin(\frac{1}{2} - h) - 0}{-h}\)

= \(\lim \limits_{h \to 0} \frac{|0 - 2h| sin(\frac{1}{2} - h)}{-h}\) = -2sin\((\frac{1}{2})\)

\(\because\) Rf'\((\frac{1}{2})\) \(\neq\) Lf' \((\frac{1}{2})\)

So, f(x) is not differentiable at x = \(\frac{1}{2}\)

We have, f(x) = cot x is continuous in R - {x = n \(\pi\) : n \(\in\) Z}.

Since, f(x) = cot x = \(\frac{cos x}{sin x}\) [since, sin x = 0 at n\(\pi\), n \(\in\) Z]

Hence, f(x) = cot x is discontinuous on the set {x = n\(\pi\) : n \(\in\) Z}

Let u(x) = |x| and v(x) = \(e^x\)

\(\therefore\) f(x) = vou(x) = v[u(x)]

= v|x| = \(e^{|x|}\)

Since, u(x) and v(x) are both continuous functions.

So, f(x) is also continuous function but u(x) = |x| is not differentiable at x = 0, whereas v(x) = \(e^x\) is differentiable at everywhere.

Hence, f(x) is continuous everywhere but not differentiable at x = 0.

\(\because\) f(x) = \(x^2 sin(\frac{1}{x}), where \; x \neq 0\)

Hence, value of the function f at x = 0, so that it is continuous at x = 0, is 0.

We have, f(x) = \(\begin{cases} mx + 1, & \quad \text{if } x \leq \frac{\pi}{2}\\ sin x + n, & \quad \text{if } x > \frac{\pi}{2} \end{cases}\) is continuous at x = \(\frac{\pi}{2}\)

\(\therefore\) LHL = \(\lim \limits_{x \to \frac{\pi}{2}}(mx + 1)\) = \(\lim \limits_{h \to 0}[m(\frac{\pi}{2} - h) + 1] = \frac{m \pi}{2} + 1\)

and RHL = \(\lim \limits_{x \to \frac{\pi}{2}}\)(sin x + n)  \(= \lim \limits_{x \to h}[sin(\frac{\pi}{2} + h) + n]\)

= \( \lim \limits_{x \to h} cos h + n = 1 + n\)

\(\therefore\) LHL = RHL[ to be continuous at x = \(\frac{\pi}{2}\)]

⇒ m. \(\frac{\pi}{2}\) + 1 = n + 1

\(\therefore\) n = m.\(\frac{\pi}{2}\)

We have, f(x) = \(|sin x|\)

Let \(f(x) = vou(x) = v[u(x)][where, u(x) = sin x \) and v(x) = |x|]

= v(sin x) = |sin x|

where, u(x) and v(x) are both continuous.

Hence, f(x) = vo u(x) is also a continuous function but v(x) is not differentiable at x = 0

So, f(x) is not differentiable where sin x = 0 

⇒ \(x = n \pi, n \in Z\)

Hence, f(x) is continuous everywhere but not differentiable at \(x = n \pi, n \in Z\)

We have, y = log\((\frac{1 - x^2}{1 + x^2})\)

\(\therefore \frac{dy}{dx} = \frac{1}{\frac{1 - x^2}{1 + x^2}}.\frac{d}{dx}(\frac{1 - x^2}{1 + x^2})\)

= \(\frac{(1 + x^2)}{(1 - x^2)}.\frac{(1 + x^2).(-2x)-(1 - x^2).2x}{(1 + x^2)^2}\)

\(= \frac{-2x[1 + x^2 + 1 - x^2]}{(1 - x^2).(1 + x^2)}\)

= \(\frac{-4x}{1 - x^4}\)

\(\because y = (sin x + y)^{1/2}\)

\(\therefore \frac{dy}{dx} = \frac{1}{2} (sin x + y)^{-1/2}\). \(\frac{d}{dx}(sin x + y)\)

⇒ \(\frac{dy}{dx} = \frac{1}{2}.\frac{1}{(sin x + y)^{1/2}}\)(cos x + \(\frac{dy}{dx})\)

⇒ \(\frac{dy}{dx} = \frac{1}{2y}\)(cos x + \(\frac{dy}{dx})\)[\(\because(sin x + y)^{1/2} = y]\)

⇒ \(\frac{dy}{dx}(1 - \frac{1}{2y}) = \frac{cos x}{2y}\)

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

Let \(u = cos^{-1}(2x^2 - 1)\) and \(v = cos^{-1}x\)

\(\therefore \frac{du}{dx}\) = \(\frac{+ -1}{\sqrt{1 - (2x^2 - 1)^2}}.4x\) = \(\frac{-4x}{\sqrt{1 -(4x^4 + 1 - 4x^2)}}\)

\(\frac{du}{dx}\) = \(\frac{ -1}{\sqrt{1 - (2x^2 - 1)^2}}.4x\) = \(\frac{-4x}{\sqrt{1 -(4x^4 + 1 - 4x^2)}}\)

= \(\frac{-4x}{\sqrt{-4x^4 + 4x^2}} = \frac{-4x}{\sqrt{4x^2(1 - x^2)}}\)

= \(\frac{-2}{\sqrt{1 - x^2}}\)

and \(\frac{dv}{dx} = \frac{-1}{\sqrt{1 - x^2}}\)

\(\therefore \frac{du}{dv} = \frac{du/dx}{dv/dx} =\frac{\frac{-2}{\sqrt{1 - x^2}}}{\frac{-1}{\sqrt{1 - x^2}}}\)

= 2

We have, x = \(t^2, y = t^3\),

\(\therefore \frac{dx}{dt}\) = 2t and \(\frac{dy}{dt} = 3t^2\)

\(\therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t}\) = \(\frac{3}{2} t\)

On further differentiating w.r.t. x, we get

\(\frac{d^2y}{dx^2} = \frac{3}{2}.\frac{d}{dt}t.\frac{dt}{dx}\)

= \(\frac{3}{2}.\frac{1}{2t}[\because \frac{dt}{dx} = \frac{1}{2t}]\)

= \(\frac{3}{4t}\)

\(\because\) f'(c) = 0 [\(\because \) f'(x) = \(3x^2 - 3]\)

⇒ \(3 c^2\) - 3 = 0

⇒ \(c^2 = \frac{3}{3} = 1\)

⇒ c = \(\pm\)1, where 1 \(\in\) \((0, \sqrt 3)\)

\(\therefore\) c = 1

\(\because \) f'(c) = \(\frac{f(b) - f(a)}{b -a}\)

⇒ 1 - \(\frac{1}{c^2} = \frac{[3 + \frac{1}{3}] - [1 + \frac{1}{1}]}{3 - 1}\) \([\because f'(x) = 1 - \frac{1}{x^2} \) and b = 3, a = 1]

⇒ \(\frac{c^2 - 1}{c^2} = \frac{\frac{10}{3} - 2}{2}\)

⇒ \(\frac{c^2 - 1}{c^2} = \frac{4}{3 \times 2} = \frac{2}{3}\)

⇒ 3\((c^2 - 1) = 2c^2\)

⇒ \(3c^2 - 2c^2 = 3\)

⇒ \(c^2 = 3 \) ⇒ c = \(\pm \sqrt 3\)

\(\because c = \sqrt 3 \in\) (1, 3)

Statement is true, because differentiable function is always continuous.

= x \(\leq x^2\) ⇒ x(1 - x) \(\leq 0\)

= x(x - 1) \(\geq 0\)

= x \(\leq 0\) or x \(\geq 1\)

\(\therefore\) h(x) = \(\begin{cases} x : & \quad x \leq 0\\ x^2: & \quad 0 < x < 1\\x : & \quad x \geq 1 \end{cases}\)

h(x) is continuous for every x but not differentiable at x = 0 and 1. 

Also h'(x) = \(\begin{cases} 1 & \quad x < 0\\ not \;exists & \quad x = 0\\2x & \quad 0 < x < 1\\ not \; exists& \quad x = 1\\1 & \quad x > 1 \end{cases}\)

\(\therefore\) h'(x) = 1 for all x > 1

Here f(2) = 0

\(\lim\limits_{x \to 2-} f(x) = \lim\limits_{h \to 0} f(2 - h)\)

= \(\lim\limits_{h \to 0} |2 - h - 2| = 0\)

\(\lim\limits_{x \to 2-} f(x)\)  \(= \lim\limits_{h \to 0} f(2 - h)\)

 \(= \lim\limits_{h \to 0} |2 - h - 2| = 0\)

Hence it is continuous at x = 2.

\(\lim\limits_{x \to 0} f(x) = f(0)\)

\(\lim\limits_{x \to 0} f(x) = (1 + x)^{1/x} = e\)

(i) When \(0 \leq x < 1\)

f(x) doesn't exist as [x] = 0 here.

(ii) Also \(\lim \limits_{x \to 1+}\) f(x) and \(\lim \limits_{x \to 1-}\) f(x) does not exist. 

Hence f(x) is discontinuous at all integers and also in (0, 1).

Obviously \(\lim \limits_{x \to b}f(x) = f(b) = 0\)