Exemplar 12 3 — Class 11 NCERT Maths Exercise Solutions

Derivatives using First Principle

Differentiate the function with respect to $x$ using first principle: $\cos(x^2 + 1)$

Let $f(x) = \cos(x^2 + 1)$.

By the first principle of derivative, $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{\cos((x+h)^2 + 1) - \cos(x^2 + 1)}{h}$

Using the identity $\cos C - \cos D = -2 \sin\frac{C+D}{2} \sin\frac{C-D}{2}$:

Here, $C = (x+h)^2 + 1 = x^2 + 2xh + h^2 + 1$ and $D = x^2 + 1$.

$\frac{C+D}{2} = \frac{2x^2 + 2xh + h^2 + 2}{2} = x^2 + xh + \frac{h^2}{2} + 1$

$\frac{C-D}{2} = \frac{2xh + h^2}{2} = \frac{h(2x+h)}{2}$

Substituting these back: $f'(x) = \lim_{h \to 0} \frac{-2 \sin(x^2 + xh + \frac{h^2}{2} + 1) \sin(\frac{h(2x+h)}{2})}{h}$

Multiply and divide by $\frac{2x+h}{2}$: $f'(x) = -2 \lim_{h \to 0} \sin(x^2 + xh + \frac{h^2}{2} + 1) \cdot \lim_{h \to 0} \frac{\sin(\frac{h(2x+h)}{2})}{\frac{h(2x+h)}{2}} \cdot \frac{2x+h}{2}$

As $h \to 0$, the first term becomes $\sin(x^2 + 1)$, the limit $\frac{\sin \theta}{\theta} \to 1$, and the last term becomes $\frac{2x}{2} = x$.

$f'(x) = -2 \sin(x^2 + 1) \cdot 1 \cdot x = -2x \sin(x^2 + 1)$

$-2x \sin(x^2 + 1)$

Differentiate the function with respect to $x$ using first principle: $\frac{ax + b}{cx + d}$

Let $f(x) = \frac{ax + b}{cx + d}$.

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{a(x+h) + b}{c(x+h) + d} - \frac{ax + b}{cx + d}}{h}$

$f'(x) = \lim_{h \to 0} \frac{(ax + ah + b)(cx + d) - (ax + b)(cx + ch + d)}{h(cx + ch + d)(cx + d)}$

Expanding the numerator: $(acx^2 + adx + achx + adh + bcx + bd) - (acx^2 + achx + adx + bcx + bch + bd)$

$= acx^2 + adx + achx + adh + bcx + bd - acx^2 - achx - adx - bcx - bch - bd$

$= adh - bch = h(ad - bc)$

Substituting back into the limit: $f'(x) = \lim_{h \to 0} \frac{h(ad - bc)}{h(cx + ch + d)(cx + d)}$

$f'(x) = \lim_{h \to 0} \frac{ad - bc}{(cx + ch + d)(cx + d)}$

As $h \to 0$, $ch \to 0$: $f'(x) = \frac{ad - bc}{(cx + d)(cx + d)} = \frac{ad - bc}{(cx + d)^2}$

$\frac{ad - bc}{(cx + d)^2}$

Differentiate the function with respect to $x$ using first principle: $x^{\frac{2}{3}}$

Let $f(x) = x^{\frac{2}{3}}$.

$f'(x) = \lim_{h \to 0} \frac{(x+h)^{\frac{2}{3}} - x^{\frac{2}{3}}}{h}$

Factor out $x^{\frac{2}{3}}$ from the first term: $f'(x) = \lim_{h \to 0} \frac{x^{\frac{2}{3}}(1 + \frac{h}{x})^{\frac{2}{3}} - x^{\frac{2}{3}}}{h} = x^{\frac{2}{3}} \lim_{h \to 0} \frac{(1 + \frac{h}{x})^{\frac{2}{3}} - 1}{h}$

Using the binomial expansion $(1+y)^n \approx 1 + ny$ for small $y$:

Here $y = \frac{h}{x}$ and $n = \frac{2}{3}$. So, $(1 + \frac{h}{x})^{\frac{2}{3}} \approx 1 + \frac{2}{3}\frac{h}{x}$

$f'(x) = x^{\frac{2}{3}} \lim_{h \to 0} \frac{(1 + \frac{2}{3}\frac{h}{x}) - 1}{h}$

$f'(x) = x^{\frac{2}{3}} \lim_{h \to 0} \frac{\frac{2}{3}\frac{h}{x}}{h} = x^{\frac{2}{3}} \cdot \frac{2}{3x}$

$f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} = \frac{2}{3} x^{-\frac{1}{3}}$

$\frac{2}{3} x^{-\frac{1}{3}}$

Differentiate the function with respect to $x$ using first principle: $x \cos x$

Let $f(x) = x \cos x$.

$f'(x) = \lim_{h \to 0} \frac{(x+h)\cos(x+h) - x \cos x}{h}$

$f'(x) = \lim_{h \to 0} \frac{x \cos(x+h) + h \cos(x+h) - x \cos x}{h}$

Group terms with $x$: $f'(x) = \lim_{h \to 0} \frac{x(\cos(x+h) - \cos x) + h \cos(x+h)}{h}$

$f'(x) = x \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} + \lim_{h \to 0} \cos(x+h)$

Using $\cos C - \cos D = -2 \sin\frac{C+D}{2} \sin\frac{C-D}{2}$:

$\cos(x+h) - \cos x = -2 \sin(x + \frac{h}{2}) \sin(\frac{h}{2})$

$f'(x) = x \lim_{h \to 0} \frac{-2 \sin(x + \frac{h}{2}) \sin(\frac{h}{2})}{h} + \cos x$

$f'(x) = x \lim_{h \to 0} -\sin(x + \frac{h}{2}) \cdot \frac{\sin(h/2)}{h/2} + \cos x$

As $h \to 0$, $\sin(x + h/2) \to \sin x$ and $\frac{\sin(h/2)}{h/2} \to 1$.

$f'(x) = x (-\sin x)(1) + \cos x = \cos x - x \sin x$

$\cos x - x \sin x$

Evaluation of Limits

Evaluate the limit: $\lim_{y \to 0} \frac{(x+y)\sec(x+y) - x \sec x}{y}$

Let $f(x) = x \sec x$. The given limit is the definition of the derivative $f'(x)$ with respect to $x$ (where $y$ acts as $h$).

$f'(x) = \frac{d}{dx}(x \sec x)$

Using the product rule $(uv)' = u'v + uv'$:

$u = x, u' = 1$ and $v = \sec x, v' = \sec x \tan x$

$f'(x) = 1 \cdot \sec x + x \cdot (\sec x \tan x)$

$f'(x) = \sec x + x \sec x \tan x$

$f'(x) = \sec x (1 + x \tan x)$

$\sec x (1 + x \tan x)$

Evaluate the limit: $\lim_{x \to 0} \frac{x[\sin(\alpha + \beta)x + \sin(\alpha - \beta)x + \sin(2\alpha x)]}{\cos(2\beta x) - \cos(2\alpha x)}$

Let the expression be $L$.

Numerator part inside brackets: $\sin(\alpha + \beta)x + \sin(\alpha - \beta)x + \sin(2\alpha x)$

Using $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$:

$= 2 \sin(\alpha x) \cos(\beta x) + 2 \sin(\alpha x) \cos(\alpha x) = 2 \sin(\alpha x) [\cos(\beta x) + \cos(\alpha x)]$

Denominator: $\cos(2\beta x) - \cos(2\alpha x)$

Using $\cos C - \cos D = -2 \sin\frac{C+D}{2} \sin\frac{C-D}{2}$:

$= -2 \sin\frac{(2\beta + 2\alpha)x}{2} \sin\frac{(2\beta - 2\alpha)x}{2} = -2 \sin((\alpha + \beta)x) \sin(-(\alpha - \beta)x) = 2 \sin((\alpha + \beta)x) \sin((\alpha - \beta)x)$

Now substitute back into the limit: $L = \lim_{x \to 0} \frac{x \cdot 2 \sin(\alpha x) [\cos(\beta x) + \cos(\alpha x)]}{2 \sin((\alpha + \beta)x) \sin((\alpha - \beta)x)}$

$L = \lim_{x \to 0} \frac{x \sin(\alpha x) [\cos(\beta x) + \cos(\alpha x)]}{\sin((\alpha + \beta)x) \sin((\alpha - \beta)x)}$

Rearrange terms: $L = \lim_{x \to 0} \frac{x \cdot \sin(\alpha x)}{\sin((\alpha + \beta)x) \sin((\alpha - \beta)x)} \cdot [\cos(\beta x) + \cos(\alpha x)]$

Multiply and divide by appropriate constants to use $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$L = \lim_{x \to 0} \frac{x \cdot \frac{\sin(\alpha x)}{\alpha x} \cdot \alpha x}{\frac{\sin((\alpha + \beta)x)}{(\alpha + \beta)x} \cdot (\alpha + \beta)x \cdot \frac{\sin((\alpha - \beta)x)}{(\alpha - \beta)x} \cdot (\alpha - \beta)x} \cdot [\cos(\beta x) + \cos(\alpha x)]$

$L = \frac{1 \cdot \alpha x^2}{1 \cdot (\alpha + \beta)x \cdot 1 \cdot (\alpha - \beta)x} \cdot [\cos 0 + \cos 0]$

$L = \frac{\alpha x^2}{(\alpha^2 - \beta^2)x^2} \cdot 2 = \frac{2\alpha}{\alpha^2 - \beta^2}$

$\frac{2\alpha}{\alpha^2 - \beta^2}$

Evaluate the limit: $\lim_{x \to \frac{\pi}{4}} \frac{\tan^3 x - \tan x}{\cos(x + \frac{\pi}{4})}$

Let $L = \lim_{x \to \frac{\pi}{4}} \frac{\tan x (\tan^2 x - 1)}{\cos(x + \frac{\pi}{4})}$

Using $\tan^2 x - 1 = \frac{\sin^2 x}{\cos^2 x} - 1 = \frac{\sin^2 x - \cos^2 x}{\cos^2 x} = \frac{-\cos 2x}{\cos^2 x}$:

$L = \lim_{x \to \frac{\pi}{4}} \frac{\tan x (-\cos 2x)}{\cos^2 x \cos(x + \frac{\pi}{4})}$

As $x \to \frac{\pi}{4}$, $\tan x \to 1$ and $\cos^2 x \to \frac{1}{2}$.

$L = \frac{1 \cdot (-1)}{1/2} \lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{\cos(x + \frac{\pi}{4})} = -2 \lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{\cos(x + \frac{\pi}{4})}$

This is a $0/0$ form. Using L'Hopital's Rule:

Differentiate numerator: $\frac{d}{dx}(\cos 2x) = -2 \sin 2x$

Differentiate denominator: $\frac{d}{dx}(\cos(x + \frac{\pi}{4})) = -\sin(x + \frac{\pi}{4})$

$L = -2 \lim_{x \to \frac{\pi}{4}} \frac{-2 \sin 2x}{-\sin(x + \frac{\pi}{4})}$

$L = -2 \frac{-2 \sin(\pi/2)}{-\sin(\pi/2)} = -2 \frac{-2(1)}{-(1)} = -2(2) = -4$

$-4$

Evaluate the limit: $\lim_{x \to \pi} \frac{1 - \sin(\frac{x}{2})}{\cos(\frac{x}{2}) (\cos(\frac{x}{4}) - \sin(\frac{x}{4}))}$

Let $x/2 = y$. As $x \to \pi$, $y \to \pi/2$.

The limit becomes $\lim_{y \to \pi/2} \frac{1 - \sin y}{\cos y (\cos(y/2) - \sin(y/2))}$

We know $1 - \sin y = (\cos(y/2) - \sin(y/2))^2$ and $\cos y = \cos^2(y/2) - \sin^2(y/2) = (\cos(y/2) - \sin(y/2))(\cos(y/2) + \sin(y/2))$.

Substitute these into the expression: $\frac{(\cos(y/2) - \sin(y/2))^2}{(\cos(y/2) - \sin(y/2))(\cos(y/2) + \sin(y/2)) (\cos(y/2) - \sin(y/2))}$

Cancel the common terms $(\cos(y/2) - \sin(y/2))^2$ from numerator and denominator:

$= \frac{1}{\cos(y/2) + \sin(y/2)}$

Now evaluate the limit as $y \to \pi/2$:

$= \frac{1}{\cos(\pi/4) + \sin(\pi/4)} = \frac{1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{1}{\frac{2}{\sqrt{2}}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}$

Show that $\lim_{x \to 4} \frac{|x - 4|}{x - 4}$ does not exist.

Left Hand Limit (LHL) at $x = 4$:

$\lim_{x \to 4^-} \frac{|x - 4|}{x - 4}$. For $x < 4$, $|x - 4| = -(x - 4)$.

LHL $= \lim_{x \to 4^-} \frac{-(x - 4)}{x - 4} = -1$

Right Hand Limit (RHL) at $x = 4$:

$\lim_{x \to 4^+} \frac{|x - 4|}{x - 4}$. For $x > 4$, $|x - 4| = (x - 4)$.

RHL $= \lim_{x \to 4^+} \frac{x - 4}{x - 4} = 1$

Since LHL $\neq$ RHL, the limit does not exist.

$Limit does not exist$

Q10

Let $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x} & x \neq \frac{\pi}{2} \\ 3 & x = \frac{\pi}{2} \end{cases}$. If $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$, find the value of $k$.

Given $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = 3$.

Evaluate the limit: $L = \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$

Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$, $h \to 0$.

$L = \lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)}$

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

$L = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}$

$L = \frac{k}{2} \cdot 1 = \frac{k}{2}$

Since the limit equals 3: $\frac{k}{2} = 3 \implies k = 6$

$k = 6$

Q11

Let $f(x) = \begin{cases} x + 2 & x \le -1 \\ cx^2 & x > -1 \end{cases}$. Find '$c$' if $\lim_{x \to -1} f(x)$ exists.

For the limit to exist at $x = -1$, the Left Hand Limit (LHL) must equal the Right Hand Limit (RHL).

LHL $= \lim_{x \to -1^-} f(x) = \lim_{x \to -1} (x + 2) = -1 + 2 = 1$

RHL $= \lim_{x \to -1^+} f(x) = \lim_{x \to -1} (cx^2) = c(-1)^2 = c$

Equating LHL and RHL: $1 = c \implies c = 1$

$c = 1$

Chapter 12 - Limits and Derivatives

Overview

Derivatives using First Principle

Evaluation of Limits