Q6: lim_x → 0 frac(x+1)^5-1x

Given: lim_x → 0 frac(x+1)^5-1x Solution: Let y = x+1. As x → 0, we have y → 1. Also, x = y-1. Substituting these into the limit expression: lim_y → 1 y^5 - 1/y-1 This is in the form lim_y → a y^n - a^n/y-a = na^n-1. Here, n=5 and a=1. lim_y → 1 y^5 - 1^5/y-1 = 5(1)^5-1 = 5(1)^4 = 5 Final Answer: 5

Q10: lim_z → 1 fracz^1/3-1z^1/6-1

Given: lim_z → 1 fracz^1/3-1z^1/6-1 Solution: If we substitute z=1, we get the indeterminate form 0/0. Let z = y^6. As z → 1, we have y → 1. Substituting into the expression: lim_y → 1 frac(y^6)^1/3-1(y^6)^1/6-1 = lim_y → 1 y^2-1/y-1 Factorizing the numerator: lim_y → 1 (y-1)(y+1)/y-1 Since y → 1,…

Q12: lim_x → -2 frac1/x+1/2x+2

Given: lim_x → -2 frac1/x+1/2x+2 Solution: If we substitute x=-2, we get the indeterminate form 0/0. Let's simplify the expression first. frac1/x+1/2x+2 = frac2+x/2xx+2 = x+2/2x(x+2) Since x → -2, x ≠ -2, we can cancel the (x+2) term. lim_x → -2 1/2x = 1/2(-2) = -1/4 Final Answer: -1/4

Q13: lim_x → 0 sin ax/bx

Given: lim_x → 0 sin ax/bx Solution: We use the standard limit lim_θ → 0 sin θ/θ = 1. lim_x → 0 sin ax/bx = lim_x → 0 sin ax/ax · ax/bx As x → 0, ax → 0. Let θ = ax. = left( lim_ax → 0 sin ax/ax right) · a/b = (1) · a/b = a/b Final Answer: a/b

Q14: lim_x → 0 sin ax/sin bx, a, b ≠ 0

Given: lim_x → 0 sin ax/sin bx, a, b ≠ 0 Solution: We use the standard limit lim_θ → 0 sin θ/θ = 1. Divide the numerator and denominator by x. lim_x → 0 fracsin ax/xsin bx/x Now, manipulate the expressions to fit the standard limit form. = lim_x → 0 fracsin ax/ax · asin bx/bx · b = fraca · lim_x →…

Q15: lim_x → π sin (π-x)/π(π-x)

Given: lim_x → π sin (π-x)/π(π-x) Solution: Let y = π - x. As x → π, we have y → 0. Substituting into the expression: lim_y → 0 sin y/π y = 1/π lim_y → 0 sin y/y Using the standard limit lim_y → 0 sin y/y = 1. = 1/π · 1 = 1/π Final Answer: 1/π

Q18: lim_x → 0 ax+x cos x/b sin x

Given: lim_x → 0 ax+x cos x/b sin x Solution: Factor out x from the numerator: lim_x → 0 x(a+cos x)/b sin x Rearrange the terms: = lim_x → 0 left( x/sin x right) · a+cos x/b = left( lim_x → 0 x/sin x right) · left( lim_x → 0 a+cos x/b right) We know lim_x → 0 sin x/x = 1, so lim_x → 0 x/sin x = 1.…

Q19: lim_x → 0 x sec x

Given: lim_x → 0 x sec x Solution: Rewrite sec x as 1/cos x. lim_x → 0 x · 1/cos x = lim_x → 0 x/cos x Substitute x=0: = 0/cos(0) = 0/1 = 0 Final Answer: 0

Limits and Derivatives Class 11 Mathematics NCERT Solutions

Q1EXERCISE 12.1

Evaluate the following limits in Exercises 1 to 22. $\lim_{x \rightarrow 3} x+3$

Solution

Given:

\lim_{x \rightarrow 3} x+3

Solution: The given limit is of a polynomial function. Hence, the limit can be found by direct substitution of the value of

x

.

\lim_{x \rightarrow 3} (x+3) = 3 + 3 = 6

Final Answer: 6

Q2EXERCISE 12.1

$\lim_{x \rightarrow \pi} \left(x-\frac{22}{7}\right)$

Solution

Given:

\lim_{x \rightarrow \pi} \left(x-\frac{22}{7}\right)

Solution: The given limit is of a polynomial function. Hence, the limit can be found by direct substitution of the value of

x

.

\lim_{x \rightarrow \pi} \left(x-\frac{22}{7}\right) = \pi - \frac{22}{7}

Final Answer:

\pi - \frac{22}{7}

Q3EXERCISE 12.1

$\lim_{r \rightarrow 1} \pi r^{2}$

Solution

Given:

\lim_{r \rightarrow 1} \pi r^{2}

Solution: The given limit is of a polynomial function in

r

. Hence, the limit can be found by direct substitution of the value of

r

.

\lim_{r \rightarrow 1} \pi r^{2} = \pi (1)^2 = \pi

Final Answer:

\pi

Q4EXERCISE 12.1

$\lim_{x \rightarrow 4} \frac{4x+3}{x-2}$

Solution

Given:

\lim_{x \rightarrow 4} \frac{4x+3}{x-2}

Solution: The given function is a rational function. We can find the limit by substituting

x=4

, as the denominator is not zero.

\lim_{x \rightarrow 4} \frac{4x+3}{x-2} = \frac{4(4)+3}{4-2} = \frac{16+3}{2} = \frac{19}{2}

Final Answer:

\frac{19}{2}

Q5EXERCISE 12.1

$\lim_{x \rightarrow -1} \frac{x^{10}+x^{5}+1}{x-1}$

Solution

Given:

\lim_{x \rightarrow -1} \frac{x^{10}+x^{5}+1}{x-1}

Solution: The given function is a rational function. We can find the limit by substituting

x=-1

, as the denominator is not zero.

\lim_{x \rightarrow -1} \frac{x^{10}+x^{5}+1}{x-1} = \frac{(-1)^{10}+(-1)^{5}+1}{(-1)-1} = \frac{1-1+1}{-2} = -\frac{1}{2}

Final Answer:

-\frac{1}{2}

Q6EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}$

Solution

Given:

\lim_{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}

Solution: Let

y = x+1

. As

x \rightarrow 0

, we have

y \rightarrow 1

. Also,

x = y-1

. Substituting these into the limit expression:

\lim_{y \rightarrow 1} \frac{y^5 - 1}{y-1}

This is in the form

\lim_{y \rightarrow a} \frac{y^n - a^n}{y-a} = na^{n-1}

. Here,

n=5

and

a=1

.

\lim_{y \rightarrow 1} \frac{y^5 - 1^5}{y-1} = 5(1)^{5-1} = 5(1)^4 = 5

Final Answer: 5

Q7EXERCISE 12.1

$\lim_{x \rightarrow 2} \frac{3x^2-x-10}{x^2-4}$

Solution

Given:

\lim_{x \rightarrow 2} \frac{3x^2-x-10}{x^2-4}

Solution: If we substitute

x=2

, we get the indeterminate form

\frac{0}{0}

. Let's factorize the numerator and the denominator. Numerator:

3x^2-x-10 = 3x^2 - 6x + 5x - 10 = 3x(x-2) + 5(x-2) = (3x+5)(x-2)

Denominator:

x^2-4 = (x-2)(x+2)

So the expression becomes:

\lim_{x \rightarrow 2} \frac{(3x+5)(x-2)}{(x-2)(x+2)}

Since

x \rightarrow 2

,

x \neq 2

, so we can cancel the

(x-2)

term.

\lim_{x \rightarrow 2} \frac{3x+5}{x+2} = \frac{3(2)+5}{2+2} = \frac{6+5}{4} = \frac{11}{4}

Final Answer:

\frac{11}{4}

Q8EXERCISE 12.1

$\lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}$

Solution

Given:

\lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}

Solution: If we substitute

x=3

, we get the indeterminate form

\frac{0}{0}

. Let's factorize the numerator and the denominator. Numerator:

x^4-81 = (x^2-9)(x^2+9) = (x-3)(x+3)(x^2+9)

Denominator:

2x^2-5x-3 = 2x^2 - 6x + x - 3 = 2x(x-3) + 1(x-3) = (2x+1)(x-3)

So the expression becomes:

\lim_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(2x+1)(x-3)}

Since

x \rightarrow 3

,

x \neq 3

, so we can cancel the

(x-3)

term.

\lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{2x+1} = \frac{(3+3)(3^2+9)}{2(3)+1} = \frac{6(9+9)}{6+1} = \frac{6 \times 18}{7} = \frac{108}{7}

Final Answer:

\frac{108}{7}

Q9EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{ax+b}{cx+1}$

Solution

Given:

\lim_{x \rightarrow 0} \frac{ax+b}{cx+1}

Solution: The given function is a rational function. We can find the limit by substituting

x=0

, as the denominator is not zero.

\lim_{x \rightarrow 0} \frac{ax+b}{cx+1} = \frac{a(0)+b}{c(0)+1} = \frac{b}{1} = b

Final Answer: b

Q10EXERCISE 12.1

$\lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

Solution

Given:

\lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}

Solution: If we substitute

z=1

, we get the indeterminate form

\frac{0}{0}

. Let

z = y^6

. As

z \rightarrow 1

, we have

y \rightarrow 1

. Substituting into the expression:

\lim_{y \rightarrow 1} \frac{(y^6)^{1/3}-1}{(y^6)^{1/6}-1} = \lim_{y \rightarrow 1} \frac{y^2-1}{y-1}

Factorizing the numerator:

\lim_{y \rightarrow 1} \frac{(y-1)(y+1)}{y-1}

Since

y \rightarrow 1

,

y \neq 1

, we can cancel the

(y-1)

term.

\lim_{y \rightarrow 1} (y+1) = 1+1 = 2

Final Answer: 2

Q11EXERCISE 12.1

$\lim_{x \rightarrow 1} \frac{ax^2+bx+c}{cx^2+bx+a}, a+b+c \neq 0$

Solution

Given:

\lim_{x \rightarrow 1} \frac{ax^2+bx+c}{cx^2+bx+a}, a+b+c \neq 0

Solution: The given function is a rational function. We check the denominator at

x=1

:

c(1)^2 + b(1) + a = c+b+a

. Since it is given that

a+b+c \neq 0

, the denominator is not zero. We can find the limit by direct substitution.

\lim_{x \rightarrow 1} \frac{ax^2+bx+c}{cx^2+bx+a} = \frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a} = \frac{a+b+c}{c+b+a} = 1

Final Answer: 1

Q12EXERCISE 12.1

$\lim_{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

Solution

Given:

\lim_{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}

Solution: If we substitute

x=-2

, we get the indeterminate form

\frac{0}{0}

. Let's simplify the expression first.

\frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \frac{\frac{2+x}{2x}}{x+2} = \frac{x+2}{2x(x+2)}

Since

x \rightarrow -2

,

x \neq -2

, we can cancel the

(x+2)

term.

\lim_{x \rightarrow -2} \frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}

Final Answer:

-\frac{1}{4}

Q13EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{\sin ax}{bx}$

Solution

Given:

\lim_{x \rightarrow 0} \frac{\sin ax}{bx}

Solution: We use the standard limit

\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1

.

\lim_{x \rightarrow 0} \frac{\sin ax}{bx} = \lim_{x \rightarrow 0} \frac{\sin ax}{ax} \cdot \frac{ax}{bx}

As

x \rightarrow 0

,

ax \rightarrow 0

. Let

\theta = ax

.

= \left( \lim_{ax \rightarrow 0} \frac{\sin ax}{ax} \right) \cdot \frac{a}{b}

= (1) \cdot \frac{a}{b} = \frac{a}{b}

Final Answer:

\frac{a}{b}

Q14EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{\sin ax}{\sin bx}, a, b \neq 0$

Solution

Given:

\lim_{x \rightarrow 0} \frac{\sin ax}{\sin bx}, a, b \neq 0

Solution: We use the standard limit

\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1

. Divide the numerator and denominator by

x

.

\lim_{x \rightarrow 0} \frac{\frac{\sin ax}{x}}{\frac{\sin bx}{x}}

Now, manipulate the expressions to fit the standard limit form.

= \lim_{x \rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot a}{\frac{\sin bx}{bx} \cdot b} = \frac{a \cdot \lim_{x \rightarrow 0} \frac{\sin ax}{ax}}{b \cdot \lim_{x \rightarrow 0} \frac{\sin bx}{bx}}

As

x \rightarrow 0

, both

ax \rightarrow 0

and

bx \rightarrow 0

.

= \frac{a \cdot 1}{b \cdot 1} = \frac{a}{b}

Final Answer:

\frac{a}{b}

Q15EXERCISE 12.1

$\lim_{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$

Solution

Given:

\lim_{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}

Solution: Let

y = \pi - x

. As

x \rightarrow \pi

, we have

y \rightarrow 0

. Substituting into the expression:

\lim_{y \rightarrow 0} \frac{\sin y}{\pi y} = \frac{1}{\pi} \lim_{y \rightarrow 0} \frac{\sin y}{y}

Using the standard limit

\lim_{y \rightarrow 0} \frac{\sin y}{y} = 1

.

= \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}

Final Answer:

\frac{1}{\pi}

Q16EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{\cos x}{\pi-x}$

Solution

Given:

\lim_{x \rightarrow 0} \frac{\cos x}{\pi-x}

Solution: The function is well-defined at

x=0

. We can find the limit by direct substitution.

\lim_{x \rightarrow 0} \frac{\cos x}{\pi-x} = \frac{\cos(0)}{\pi-0} = \frac{1}{\pi}

Final Answer:

\frac{1}{\pi}

Q17EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{\cos 2x-1}{\cos x-1}$

Solution

Given:

\lim_{x \rightarrow 0} \frac{\cos 2x-1}{\cos x-1}

Solution: We use the identity

1 - \cos \theta = 2 \sin^2(\theta/2)

. So,

\cos 2x - 1 = -(1 - \cos 2x) = -2 \sin^2(x)

. And,

\cos x - 1 = -(1 - \cos x) = -2 \sin^2(x/2)

. The expression becomes:

\lim_{x \rightarrow 0} \frac{-2 \sin^2 x}{-2 \sin^2 (x/2)} = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sin^2 (x/2)}

We can rewrite this as:

= \lim_{x \rightarrow 0} \left( \frac{\sin x}{\sin (x/2)} \right)^2

Now, we divide the numerator and denominator inside the bracket by

x

.

= \left( \lim_{x \rightarrow 0} \frac{\frac{\sin x}{x}}{\frac{\sin (x/2)}{x}} \right)^2 = \left( \lim_{x \rightarrow 0} \frac{\frac{\sin x}{x}}{\frac{\sin (x/2)}{x/2} \cdot \frac{1}{2}} \right)^2

Using the standard limit

\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1

.

= \left( \frac{1}{1 \cdot \frac{1}{2}} \right)^2 = (2)^2 = 4

Final Answer: 4

Q18EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{ax+x \cos x}{b \sin x}$

Solution

Given:

\lim_{x \rightarrow 0} \frac{ax+x \cos x}{b \sin x}

Solution: Factor out

x

from the numerator:

\lim_{x \rightarrow 0} \frac{x(a+\cos x)}{b \sin x}

Rearrange the terms:

= \lim_{x \rightarrow 0} \left( \frac{x}{\sin x} \right) \cdot \frac{a+\cos x}{b}

= \left( \lim_{x \rightarrow 0} \frac{x}{\sin x} \right) \cdot \left( \lim_{x \rightarrow 0} \frac{a+\cos x}{b} \right)

We know

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

, so

\lim_{x \rightarrow 0} \frac{x}{\sin x} = 1

.

= (1) \cdot \frac{a+\cos(0)}{b} = \frac{a+1}{b}

Final Answer:

\frac{a+1}{b}

Q19EXERCISE 12.1

$\lim_{x \rightarrow 0} x \sec x$

Solution

Given:

\lim_{x \rightarrow 0} x \sec x

Solution: Rewrite

\sec x

as

\frac{1}{\cos x}

.

\lim_{x \rightarrow 0} x \cdot \frac{1}{\cos x} = \lim_{x \rightarrow 0} \frac{x}{\cos x}

Substitute

x=0

:

= \frac{0}{\cos(0)} = \frac{0}{1} = 0

Final Answer: 0

Q20EXERCISE 12.1

$\lim_{x \rightarrow 0} \frac{\sin ax+bx}{ax+\sin bx} a, b, a+b \neq 0$

Solution

Given:

\lim_{x \rightarrow 0} \frac{\sin ax+bx}{ax+\sin bx} a, b, a+b \neq 0

Solution: If we substitute

x=0

, we get the indeterminate form

\frac{0}{0}

. Divide the numerator and the denominator by

x

.

\lim_{x \rightarrow 0} \frac{\frac{\sin ax}{x}+b}{\frac{ax}{x}+\frac{\sin bx}{x}} = \lim_{x \rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot a+b}{a+\frac{\sin bx}{bx} \cdot b}

Using the standard limit

\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1

.

= \frac{\left( \lim_{x \rightarrow 0} \frac{\sin ax}{ax} \right) \cdot a+b}{a+\left( \lim_{x \rightarrow 0} \frac{\sin bx}{bx} \right) \cdot b} = \frac{1 \cdot a+b}{a+1 \cdot b} = \frac{a+b}{a+b}

Since

a+b \neq 0

, the value is 1.

Final Answer: 1

Q21EXERCISE 12.1

$\lim_{x \rightarrow 0} (\csc x - \cot x)$

Solution

Given:

\lim_{x \rightarrow 0} (\csc x - \cot x)

Solution: Rewrite in terms of

\sin x

and

\cos x

.

\lim_{x \rightarrow 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) = \lim_{x \rightarrow 0} \frac{1-\cos x}{\sin x}

If we substitute

x=0

, we get the indeterminate form

\frac{0}{0}

. Use the identity

1 - \cos x = 2 \sin^2(x/2)

and

\sin x = 2 \sin(x/2) \cos(x/2)

.

= \lim_{x \rightarrow 0} \frac{2 \sin^2(x/2)}{2 \sin(x/2) \cos(x/2)} = \lim_{x \rightarrow 0} \frac{\sin(x/2)}{\cos(x/2)} = \lim_{x \rightarrow 0} \tan(x/2)

Substitute

x=0

:

= \tan(0) = 0

Final Answer: 0

Q22EXERCISE 12.1

$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}$

Solution

Given:

\lim_{x \rightarrow \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}

Solution: Let

y = x - \frac{\pi}{2}

. As

x \rightarrow \frac{\pi}{2}

, we have

y \rightarrow 0

. Also,

x = y + \frac{\pi}{2}

. Substituting into the expression:

\lim_{y \rightarrow 0} \frac{\tan 2(y + \frac{\pi}{2})}{y} = \lim_{y \rightarrow 0} \frac{\tan(2y + \pi)}{y}

We know that

\tan(\pi + \theta) = \tan \theta

.

= \lim_{y \rightarrow 0} \frac{\tan 2y}{y}

Rewrite in terms of the standard limit

\lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1

.

= \lim_{y \rightarrow 0} \frac{\tan 2y}{2y} \cdot 2

Let

z=2y

. As

y \rightarrow 0

,

z \rightarrow 0

.

= 2 \lim_{z \rightarrow 0} \frac{\tan z}{z} = 2 \cdot 1 = 2

Final Answer: 2

Q23EXERCISE 12.1

Find $\lim_{x \rightarrow 0} f(x)$ and $\lim_{x \rightarrow 1} f(x)$ , where $f(x)=\begin{cases}2 x+3, & x \leq 0 \ 3(x+1), & x>0\end{cases}$

Solution

Given:

f(x)=\begin{cases}2 x+3, & x \leq 0 \ 3(x+1), & x>0\end{cases}

Part 1: Find $\lim_{x \rightarrow 0} f(x)$ To find the limit at

x=0

, we need to evaluate the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL).

LHL at

x=0

:

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (2x+3) = 2(0) + 3 = 3

RHL at

x=0

:

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} 3(x+1) = 3(0+1) = 3

Since LHL = RHL = 3, the limit exists and

\lim_{x \rightarrow 0} f(x) = 3

.

Part 2: Find $\lim_{x \rightarrow 1} f(x)$ For

x \rightarrow 1

, the values of

x

are in the neighborhood of 1, which means

x > 0

. So we use the function definition

f(x) = 3(x+1)

.

\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x+1) = 3(1+1) = 3(2) = 6

Final Answer:

\lim_{x \rightarrow 0} f(x) = 3

\lim_{x \rightarrow 1} f(x) = 6

Q24EXERCISE 12.1

Find $\lim_{x \rightarrow 1} f(x)$ , where $f(x)= \begin{cases}x^{2}-1, & x \leq 1 \ -x^{2}-1, & x>1\end{cases}$

Solution

Given:

f(x)= \begin{cases}x^{2}-1, & x \leq 1 \ -x^{2}-1, & x>1\end{cases}

To Find:

\lim_{x \rightarrow 1} f(x)

Solution: To find the limit at

x=1

, we need to evaluate the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL).

LHL at

x=1

:

\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^-} (x^2-1) = (1)^2 - 1 = 1 - 1 = 0

RHL at

x=1

:

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (-x^2-1) = -(1)^2 - 1 = -1 - 1 = -2

Since LHL

\neq

RHL (

0 \neq -2

), the limit of the function as

x

approaches 1 does not exist.

Final Answer:

\lim_{x \rightarrow 1} f(x)

does not exist.

Q25EXERCISE 12.1

Evaluate $\lim_{x \rightarrow 0} f(x)$ , where $f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \ 0, & x=0\end{cases}$

Solution

Given:

f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \ 0, & x=0\end{cases}

To Find:

\lim_{x \rightarrow 0} f(x)

Solution: We need to evaluate the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) at

x=0

. Recall the definition of the absolute value function:

|x| = x

if

x \geq 0

, and

|x| = -x

if

x < 0

.

LHL at

x=0

: For

x \rightarrow 0^-

,

x < 0

, so

|x| = -x

.

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{|x|}{x} = \lim_{x \rightarrow 0^-} \frac{-x}{x} = \lim_{x \rightarrow 0^-} -1 = -1

RHL at

x=0

: For

x \rightarrow 0^+

,

x > 0

, so

|x| = x

.

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{|x|}{x} = \lim_{x \rightarrow 0^+} \frac{x}{x} = \lim_{x \rightarrow 0^+} 1 = 1

Since LHL

\neq

RHL (

-1 \neq 1

), the limit of the function as

x

approaches 0 does not exist.

Final Answer:

\lim_{x \rightarrow 0} f(x)

does not exist.

Q26EXERCISE 12.1

Find $\lim_{x \rightarrow 0} f(x)$ , where $f(x)=\begin{cases}\frac{x}{|x|}, & x \neq 0 \ 0, & x=0\end{cases}$

Solution

Given:

f(x)=\begin{cases}\frac{x}{|x|}, & x \neq 0 \ 0, & x=0\end{cases}

To Find:

\lim_{x \rightarrow 0} f(x)

Solution: This function is very similar to the one in the previous question. We evaluate the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) at

x=0

. Recall:

|x| = x

if

x > 0

, and

|x| = -x

if

x < 0

.

LHL at

x=0

: For

x \rightarrow 0^-

,

x < 0

, so

|x| = -x

.

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{x}{|x|} = \lim_{x \rightarrow 0^-} \frac{x}{-x} = \lim_{x \rightarrow 0^-} -1 = -1

RHL at

x=0

: For

x \rightarrow 0^+

,

x > 0

, so

|x| = x

.

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{x}{|x|} = \lim_{x \rightarrow 0^+} \frac{x}{x} = \lim_{x \rightarrow 0^+} 1 = 1

Since LHL

\neq

RHL (

-1 \neq 1

), the limit of the function as

x

approaches 0 does not exist.

Final Answer:

\lim_{x \rightarrow 0} f(x)

does not exist.

Q27EXERCISE 12.1

Find $\lim_{x \rightarrow 5} f(x)$ , where $f(x)=|x|-5$

Solution

Given:

f(x)=|x|-5

To Find:

\lim_{x \rightarrow 5} f(x)

Solution: The function

f(x)=|x|-5

is a continuous function for all real numbers. Therefore, the limit at any point is equal to the value of the function at that point. We can find the limit by direct substitution.

\lim_{x \rightarrow 5} f(x) = \lim_{x \rightarrow 5} (|x|-5) = |5| - 5 = 5 - 5 = 0

Alternatively, we can check LHL and RHL. LHL: For

x \rightarrow 5^-

,

x

is slightly less than 5 but positive, so

|x|=x

.

\lim_{x \rightarrow 5^-} (x-5) = 5-5=0

. RHL: For

x \rightarrow 5^+

,

x

is slightly more than 5, so

|x|=x

.

\lim_{x \rightarrow 5^+} (x-5) = 5-5=0

. Since LHL = RHL = 0, the limit is 0.

Final Answer: 0

Q28EXERCISE 12.1

Suppose $f(x)= \begin{cases}a+b x, & x<1 \ 4, & x=1 \ b-a x, & x>1\end{cases}$ and if $\lim_{x \rightarrow 1} f(x)=f(1)$ what are possible values of $a$ and $b$ ?

Solution

Given:

f(x)= \begin{cases}a+b x, & x<1 \ 4, & x=1 \ b-a x, & x>1\end{cases}

Also given,

\lim_{x \rightarrow 1} f(x) = f(1)

.

Solution: From the definition of the function,

f(1) = 4

. Therefore, we have

\lim_{x \rightarrow 1} f(x) = 4

. For the limit to exist, the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) must be equal to each other and to the value of the limit.

LHL at

x=1

:

\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^-} (a+bx) = a+b(1) = a+b

RHL at

x=1

:

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (b-ax) = b-a(1) = b-a

Since the limit exists and is equal to 4, we must have: LHL = RHL = 4 So, we have a system of two linear equations:

$a+b = 4$
$b-a = 4$

Adding the two equations:

(a+b) + (b-a) = 4+4

2b = 8

b = 4

Substitute

b=4

into the first equation:

a+4 = 4

a = 0

Final Answer: The possible values are

a=0

and

b=4

.

Q29EXERCISE 12.1

Let $a_{1}, a_{2}, \ldots, a_{n}$ be fixed real numbers and define a function $f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)$ What is $\lim _{x \rightarrow a_{1}} f(x)$ ? For some $a \neq a_{1}, a_{2}, \ldots, a_{n}$ , compute $\lim _{x \rightarrow a} f(x)$

Solution

Given:

f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)

Part 1: Find $\lim _{x \rightarrow a_{1}} f(x)$

Solution: The function

f(x)

is a polynomial function. The limit of a polynomial function at any point can be found by direct substitution.

\lim_{x \rightarrow a_1} f(x) = f(a_1)

f(a_1) = (a_1 - a_1)(a_1 - a_2) \ldots (a_1 - a_n)

f(a_1) = 0 \cdot (a_1 - a_2) \ldots (a_1 - a_n) = 0

Therefore,

\lim_{x \rightarrow a_1} f(x) = 0

.

Part 2: Compute $\lim _{x \rightarrow a} f(x)$ for some $a \neq a_{1}, a_{2}, \ldots, a_{n}$

Solution: Since

f(x)

is a polynomial, we can again find the limit by direct substitution.

\lim_{x \rightarrow a} f(x) = f(a)

f(a) = (a - a_1)(a - a_2) \ldots (a - a_n)

Since

a \neq a_i

for any

i=1, 2, \ldots, n

, none of the factors are zero.

Final Answer:

\lim_{x \rightarrow a_1} f(x) = 0

\lim_{x \rightarrow a} f(x) = (a - a_1)(a - a_2) \ldots (a - a_n)

Q30EXERCISE 12.1

If $f(x)=\begin{cases}|x|+1, & x<0 \ 0, & x=0 \ |x|-1, & x>0\end{cases}$ . For what value (s) of $a$ does $\lim_{x \rightarrow a} f(x)$ exists?

Solution

Given:

f(x)=\begin{cases}|x|+1, & x<0 \ 0, & x=0 \ |x|-1, & x>0\end{cases}

Solution: Let's analyze the function definition. For

x<0

,

|x| = -x

, so

f(x) = -x+1

. For

x>0

,

|x|=x

, so

f(x) = x-1

. The function can be rewritten as:

f(x)=\begin{cases}-x+1, & x<0 \ 0, & x=0 \ x-1, & x>0\end{cases}

The function is defined by polynomials for

x<0

and

x>0

. Therefore, the limit will exist for all values of

a

except possibly at

x=0

, where the definition of the function changes.

Let's check the limit at

a=0

. LHL at

x=0

:

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (-x+1) = -0+1 = 1

RHL at

x=0

:

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} (x-1) = 0-1 = -1

Since LHL

\neq

RHL (

1 \neq -1

),

\lim_{x \rightarrow 0} f(x)

does not exist.

For any other real number

a \neq 0

: Case 1:

a < 0

. Then in the neighborhood of

a

,

f(x) = -x+1

. This is a polynomial, so the limit exists.

\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} (-x+1) = -a+1

Case 2:

a > 0

. Then in the neighborhood of

a

,

f(x) = x-1

. This is a polynomial, so the limit exists.

\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} (x-1) = a-1

Thus, the limit exists for all real numbers

a

except for

a=0

.

Final Answer: The limit

\lim_{x \rightarrow a} f(x)

exists for all

a \in \mathbb{R} - \{0\}

.

Q31EXERCISE 12.1

If the function $f(x)$ satisfies $\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi$ , evaluate $\lim_{x \rightarrow 1} f(x)$

Solution

Given:

\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi

To Find:

\lim_{x \rightarrow 1} f(x)

Solution: We are given that the limit of the quotient exists and is a finite number,

\pi

. Let's look at the limit of the denominator as

x \rightarrow 1

.

\lim_{x \rightarrow 1} (x^2 - 1) = 1^2 - 1 = 0

Since the overall limit is finite and the denominator approaches 0, the numerator must also approach 0. If the numerator approached a non-zero number, the limit of the fraction would be infinite, which contradicts the given information. Therefore, we must have:

\lim_{x \rightarrow 1} (f(x)-2) = 0

Using the algebra of limits:

\lim_{x \rightarrow 1} f(x) - \lim_{x \rightarrow 1} 2 = 0

\lim_{x \rightarrow 1} f(x) - 2 = 0

\lim_{x \rightarrow 1} f(x) = 2

Final Answer: 2

Q32EXERCISE 12.1

If $f(x)=\begin{cases}m x^{2}+n, & x<0 \ n x+m, & 0 \leq x \leq 1 \ n x^{3}+m, & x>1\end{cases}$ . For what integers $m$ and $n$ does both $\lim_{x \rightarrow 0} f(x)$ and $\lim_{x \rightarrow 1} f(x)$ exist?

Solution

Given:

f(x)=\begin{cases}m x^{2}+n, & x<0 \ n x+m, & 0 \leq x \leq 1 \ n x^{3}+m, & x>1\end{cases}

Condition 1: $\lim_{x \rightarrow 0} f(x)$ exists For the limit to exist at

x=0

, the LHL must equal the RHL.

LHL at

x=0

:

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (mx^2+n) = m(0)^2 + n = n

RHL at

x=0

:

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} (nx+m) = n(0) + m = m

For the limit to exist, LHL = RHL, which implies:

n = m

Condition 2: $\lim_{x \rightarrow 1} f(x)$ exists For the limit to exist at

x=1

, the LHL must equal the RHL.

LHL at

x=1

:

\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^-} (nx+m) = n(1) + m = n+m

RHL at

x=1

:

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (nx^3+m) = n(1)^3 + m = n+m

In this case, LHL = RHL is always true (

n+m = n+m

). This means the limit at

x=1

exists for any values of

m

and

n

.

Conclusion: For both limits to exist, we only need to satisfy the condition from the limit at

x=0

, which is

m=n

. Since the question asks for integers

m

and

n

, the condition is that

m

and

n

can be any pair of equal integers.

Final Answer: Both limits exist for any integers

m

and

n

such that

m=n

.

Q1EXERCISE 12.2

Find the derivative of $x^2-2$ at $x=10$ .

Solution

Given: Function

f(x) = x^2-2

.

To Find: The derivative of

f(x)

at

x=10

, i.e.,

f'(10)

.

Solution: First, find the derivative of

f(x)

with respect to

x

.

f'(x) = \frac{d}{dx}(x^2 - 2)

Using the power rule and the fact that the derivative of a constant is 0:

f'(x) = 2x^{2-1} - 0 = 2x

Now, evaluate the derivative at

x=10

.

f'(10) = 2(10) = 20

Final Answer: 20

Q2EXERCISE 12.2

Find the derivative of $x$ at $x=1$ .

Solution

Given: Function

f(x) = x

.

To Find: The derivative of

f(x)

at

x=1

, i.e.,

f'(1)

.

Solution: First, find the derivative of

f(x)

with respect to

x

.

f'(x) = \frac{d}{dx}(x) = 1

The derivative is a constant function. Now, evaluate the derivative at

x=1

.

f'(1) = 1

Final Answer: 1

Q3EXERCISE 12.2

Find the derivative of $99x$ at $x=100$ .

Solution

Given: Function

f(x) = 99x

.

To Find: The derivative of

f(x)

at

x=100

, i.e.,

f'(100)

.

Solution: First, find the derivative of

f(x)

with respect to

x

.

f'(x) = \frac{d}{dx}(99x) = 99 \frac{d}{dx}(x) = 99(1) = 99

The derivative is a constant function. Now, evaluate the derivative at

x=100

.

f'(100) = 99

Final Answer: 99

Q4EXERCISE 12.2

Find the derivative of the following functions from first principle.

(i)

$x^3-27$

(ii)

$(x-1)(x-2)$

(iii)

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

(iv)

$\frac{x+1}{x-1}$

Solution

Solution using the First Principle: The derivative of a function

f(x)

is given by:

f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}

(i) $f(x) = x^3 - 27$

f'(x) = \lim_{h \rightarrow 0} \frac{((x+h)^3 - 27) - (x^3 - 27)}{h}

= \lim_{h \rightarrow 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 27 - x^3 + 27}{h}

= \lim_{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h}

= \lim_{h \rightarrow 0} \frac{h(3x^2 + 3xh + h^2)}{h}

= \lim_{h \rightarrow 0} (3x^2 + 3xh + h^2) = 3x^2 + 3x(0) + (0)^2 = 3x^2

(ii) $f(x) = (x-1)(x-2) = x^2 - 3x + 2$

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

= \lim_{h \rightarrow 0} \frac{x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2}{h}

= \lim_{h \rightarrow 0} \frac{2xh + h^2 - 3h}{h}

= \lim_{h \rightarrow 0} \frac{h(2x + h - 3)}{h}

= \lim_{h \rightarrow 0} (2x + h - 3) = 2x + 0 - 3 = 2x - 3

(iii) $f(x) = \frac{1}{x^2}$

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

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{x^2 - (x+h)^2}{x^2(x+h)^2} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x+h)^2} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{-2xh - h^2}{x^2(x+h)^2} \right)

= \lim_{h \rightarrow 0} \frac{-h(2x + h)}{h \cdot x^2(x+h)^2}

= \lim_{h \rightarrow 0} \frac{-(2x + h)}{x^2(x+h)^2} = \frac{-(2x+0)}{x^2(x+0)^2} = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}

(iv) $f(x) = \frac{x+1}{x-1}$

f'(x) = \lim_{h \rightarrow 0} \frac{\frac{(x+h)+1}{(x+h)-1} - \frac{x+1}{x-1}}{h}

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{(x+h-1)(x-1)} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{(x^2-x+hx-h+x-1) - (x^2+hx-x+x+h-1)}{(x+h-1)(x-1)} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{(x^2+hx-h-1) - (x^2+hx+h-1)}{(x+h-1)(x-1)} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{x^2+hx-h-1 - x^2-hx-h+1}{(x+h-1)(x-1)} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{-2h}{(x+h-1)(x-1)} \right)

= \lim_{h \rightarrow 0} \frac{-2}{(x+h-1)(x-1)} = \frac{-2}{(x+0-1)(x-1)} = \frac{-2}{(x-1)^2}

Final Answer:

(i)

3x^2

(ii)

2x-3

(iii)

-\frac{2}{x^3}

(iv)

-\frac{2}{(x-1)^2}

Q5EXERCISE 12.2

For the function $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$ Prove that $f^{\prime}(1)=100 f^{\prime}(0)$ .

Solution

Given:

f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1

To Prove:

f^{\prime}(1)=100 f^{\prime}(0)

Proof: First, we find the derivative of

f(x)

,

f'(x)

.

f'(x) = \frac{d}{dx} \left( \frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1 \right)

Using the power rule

\frac{d}{dx}(x^n) = nx^{n-1}

for each term:

f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \ldots + \frac{2x^1}{2} + 1 + 0

f'(x) = x^{99} + x^{98} + \ldots + x + 1

Now, we evaluate

f'(1)

and

f'(0)

. For

x=1

:

f'(1) = (1)^{99} + (1)^{98} + \ldots + (1) + 1

This is a sum of 1, repeated 100 times (from power 0 to 99).

f'(1) = \underbrace{1+1+\ldots+1}_{100 \text{ terms}} = 100

For

x=0

:

f'(0) = (0)^{99} + (0)^{98} + \ldots + (0) + 1

f'(0) = 0 + 0 + \ldots + 0 + 1 = 1

Now, we check the condition to be proved:

f'(1) = 100

100 f'(0) = 100 \times 1 = 100

Since

f'(1) = 100

and

100 f'(0) = 100

, we have

f'(1) = 100 f'(0)

.

Hence Proved.

Q6EXERCISE 12.2

Find the derivative of $x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n}$ for some fixed real number $a$ .

Solution

Given: Function

f(x) = x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n}

, where

a

is a fixed real number.

To Find: The derivative of

f(x)

, i.e.,

f'(x)

.

Solution: We differentiate the function term by term using the power rule

\frac{d}{dx}(x^k) = kx^{k-1}

. Note that

a, a^2, \ldots, a^n

are constants.

f'(x) = \frac{d}{dx} (x^{n}) + \frac{d}{dx} (a x^{n-1}) + \frac{d}{dx} (a^{2} x^{n-2}) + \ldots + \frac{d}{dx} (a^{n-1} x) + \frac{d}{dx} (a^{n})

f'(x) = nx^{n-1} + a \cdot (n-1)x^{n-2} + a^2 \cdot (n-2)x^{n-3} + \ldots + a^{n-1} \cdot (1) + 0

Final Answer: The derivative is

nx^{n-1} + (n-1)ax^{n-2} + (n-2)a^2x^{n-3} + \ldots + a^{n-1}

.

Q7EXERCISE 12.2

For some constants $a$ and $b$ , find the derivative of

(i)

$(x-a)(x-b)$

(ii)

$(ax^2+b)^2$

(iii)

$\frac{x-a}{x-b}$

Solution

Solution:

(i) $f(x) = (x-a)(x-b)$ We use the product rule:

(uv)' = u'v + uv'

. Let

u = x-a

and

v = x-b

. Then

u' = 1

and

v' = 1

.

f'(x) = (1)(x-b) + (x-a)(1)

= x-b + x-a = 2x - a - b

Alternatively, we can first expand the expression:

f(x) = x^2 - (a+b)x + ab

. Then differentiate:

f'(x) = 2x - (a+b)

.

(ii) $f(x) = (ax^2+b)^2$ First, we expand the expression:

f(x) = (ax^2)^2 + 2(ax^2)(b) + b^2 = a^2x^4 + 2abx^2 + b^2

Now, we differentiate term by term:

f'(x) = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + \frac{d}{dx}(b^2)

= a^2(4x^3) + 2ab(2x) + 0

= 4a^2x^3 + 4abx = 4ax(ax^2+b)

(iii) $f(x) = \frac{x-a}{x-b}$ We use the quotient rule:

(\frac{u}{v})' = \frac{u'v - uv'}{v^2}

. Let

u = x-a

and

v = x-b

. Then

u' = 1

and

v' = 1

.

f'(x) = \frac{(1)(x-b) - (x-a)(1)}{(x-b)^2}

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

Final Answer:

(i)

2x - a - b

(ii)

4a^2x^3 + 4abx

or

4ax(ax^2+b)

(iii)

\frac{a-b}{(x-b)^2}

Q8EXERCISE 12.2

Find the derivative of $\frac{x^{n}-a^{n}}{x-a}$ for some constant $a$ .

Solution

Given: Function

f(x) = \frac{x^{n}-a^{n}}{x-a}

, where

a

is a constant.

To Find: The derivative of

f(x)

, i.e.,

f'(x)

.

Solution: We use the quotient rule:

(\frac{u}{v})' = \frac{u'v - uv'}{v^2}

. Let

u = x^n - a^n

and

v = x-a

. Then

u' = nx^{n-1}

and

v' = 1

.

f'(x) = \frac{(nx^{n-1})(x-a) - (x^n - a^n)(1)}{(x-a)^2}

= \frac{nx^n - nax^{n-1} - x^n + a^n}{(x-a)^2}

= \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2}

Final Answer:

\frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2}

Q9EXERCISE 12.2

Find the derivative of

(i)

$2x - \frac{3}{4}$

(ii)

$(5x^3+3x-1)(x-1)$

(iii)

$x^{-3}(5+3x)$

(iv)

$x^5(3-6x^{-9})$

(v)

$x^{-4}(3-4x^{-5})$

(vi)

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

Solution

Solution:

(i) $f(x) = 2x - \frac{3}{4}$

f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}(\frac{3}{4}) = 2 - 0 = 2

(ii) $f(x) = (5x^3+3x-1)(x-1)$ Using the product rule

(uv)' = u'v + uv'

. Let

u = 5x^3+3x-1

and

v = x-1

. Then

u' = 15x^2+3

and

v' = 1

.

f'(x) = (15x^2+3)(x-1) + (5x^3+3x-1)(1)

= 15x^3 - 15x^2 + 3x - 3 + 5x^3 + 3x - 1

= 20x^3 - 15x^2 + 6x - 4

(iii) $f(x) = x^{-3}(5+3x) = 5x^{-3} + 3x^{-2}$

f'(x) = 5(-3x^{-4}) + 3(-2x^{-3}) = -15x^{-4} - 6x^{-3} = -\frac{15}{x^4} - \frac{6}{x^3}

(iv) $f(x) = x^5(3-6x^{-9}) = 3x^5 - 6x^{-4}$

f'(x) = 3(5x^4) - 6(-4x^{-5}) = 15x^4 + 24x^{-5} = 15x^4 + \frac{24}{x^5}

(v) $f(x) = x^{-4}(3-4x^{-5}) = 3x^{-4} - 4x^{-9}$

f'(x) = 3(-4x^{-5}) - 4(-9x^{-10}) = -12x^{-5} + 36x^{-10} = -\frac{12}{x^5} + \frac{36}{x^{10}}

(vi) $f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1}$ We differentiate each term separately using the quotient rule. For the first term,

\frac{d}{dx}(\frac{2}{x+1}) = \frac{(0)(x+1) - 2(1)}{(x+1)^2} = -\frac{2}{(x+1)^2}

. For the second term,

\frac{d}{dx}(\frac{x^2}{3x-1}) = \frac{(2x)(3x-1) - (x^2)(3)}{(3x-1)^2} = \frac{6x^2-2x-3x^2}{(3x-1)^2} = \frac{3x^2-2x}{(3x-1)^2}

.

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

Final Answer:

(i)

2

(ii)

20x^3 - 15x^2 + 6x - 4

(iii)

-15x^{-4} - 6x^{-3}

(iv)

15x^4 + 24x^{-5}

(v)

-12x^{-5} + 36x^{-10}

(vi)

-\frac{2}{(x+1)^2} - \frac{3x^2-2x}{(3x-1)^2}

Q10EXERCISE 12.2

Find the derivative of $\cos x$ from first principle.

Solution

Given: Function

f(x) = \cos x

.

To Find: The derivative of

f(x)

from the first principle.

Solution: By definition of the derivative (first principle):

f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}

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

Using the trigonometric identity

\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)

:

f'(x) = \lim_{h \rightarrow 0} \frac{-2 \sin\left(\frac{x+h+x}{2}\right) \sin\left(\frac{x+h-x}{2}\right)}{h}

= \lim_{h \rightarrow 0} \frac{-2 \sin\left(\frac{2x+h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}

= \lim_{h \rightarrow 0} - \sin\left(x+\frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}

= \left( \lim_{h \rightarrow 0} - \sin\left(x+\frac{h}{2}\right) \right) \cdot \left( \lim_{h \rightarrow 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)

As

h \rightarrow 0

,

h/2 \rightarrow 0

. Using the standard limit

\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1

.

= -\sin(x+0) \cdot 1

= -\sin x

Final Answer: The derivative of

\cos x

is

-\sin x

.

Q11EXERCISE 12.2

Find the derivative of the following functions:

(i)

$\sin x \cos x$

(ii)

$\sec x$

(iii)

$5 \sec x+4 \cos x$

(iv)

$\csc x$

(v)

$3 \cot x+5 \csc x$

(vi)

$5 \sin x-6 \cos x+7$

(vii)

$2 \tan x-7 \sec x$

Solution

Solution:

(i) $f(x) = \sin x \cos x$ Using the product rule

(uv)' = u'v + uv'

. Let

u = \sin x

and

v = \cos x

. Then

u' = \cos x

and

v' = -\sin x

.

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

= \cos^2 x - \sin^2 x = \cos 2x

(ii) $f(x) = \sec x = \frac{1}{\cos x}$ Using the quotient rule

(\frac{u}{v})' = \frac{u'v - uv'}{v^2}

. Let

u=1

and

v=\cos x

. Then

u'=0

and

v'=-\sin x

.

f'(x) = \frac{(0)(\cos x) - (1)(-\sin x)}{(\cos x)^2} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x

(iii) $f(x) = 5 \sec x + 4 \cos x$

f'(x) = 5 \frac{d}{dx}(\sec x) + 4 \frac{d}{dx}(\cos x)

= 5(\sec x \tan x) + 4(-\sin x) = 5 \sec x \tan x - 4 \sin x

(iv) $f(x) = \csc x = \frac{1}{\sin x}$ Using the quotient rule. Let

u=1

and

v=\sin x

. Then

u'=0

and

v'=\cos x

.

f'(x) = \frac{(0)(\sin x) - (1)(\cos x)}{(\sin x)^2} = \frac{-\cos x}{\sin^2 x} = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\csc x \cot x

(v) $f(x) = 3 \cot x + 5 \csc x$ First, find the derivative of

\cot x = \frac{\cos x}{\sin x}

.

\frac{d}{dx}(\cot x) = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x

. Now, differentiate the given function:

f'(x) = 3(-\csc^2 x) + 5(-\csc x \cot x)

= -3 \csc^2 x - 5 \csc x \cot x

(vi) $f(x) = 5 \sin x - 6 \cos x + 7$

f'(x) = 5 \frac{d}{dx}(\sin x) - 6 \frac{d}{dx}(\cos x) + \frac{d}{dx}(7)

= 5(\cos x) - 6(-\sin x) + 0 = 5 \cos x + 6 \sin x

(vii) $f(x) = 2 \tan x - 7 \sec x$ First, find the derivative of

\tan x = \frac{\sin x}{\cos x}

.

\frac{d}{dx}(\tan x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x

. Now, differentiate the given function:

f'(x) = 2(\sec^2 x) - 7(\sec x \tan x)

Final Answer:

(i)

\cos 2x

(ii)

\sec x \tan x

(iii)

5 \sec x \tan x - 4 \sin x

(iv)

-\csc x \cot x

(v)

-3 \csc^2 x - 5 \csc x \cot x

(vi)

5 \cos x + 6 \sin x

(vii)

2 \sec^2 x - 7 \sec x \tan x

Q1Miscellaneous Exercise on Chapter 12

Find the derivative of the following functions from first principle:

(i)

$-x$

(ii)

$(-x)^{-1}$

(iii)

$\sin (x+1)$

(iv)

$\cos \left(x-\frac{\pi}{8}\right)$

Solution

Solution using the First Principle:

f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}

(i) $f(x) = -x$

f'(x) = \lim_{h \rightarrow 0} \frac{-(x+h) - (-x)}{h} = \lim_{h \rightarrow 0} \frac{-x-h+x}{h} = \lim_{h \rightarrow 0} \frac{-h}{h} = -1

(ii) $f(x) = (-x)^{-1} = -\frac{1}{x}$

f'(x) = \lim_{h \rightarrow 0} \frac{-\frac{1}{x+h} - \left(-\frac{1}{x}\right)}{h} = \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{1}{x} - \frac{1}{x+h} \right)

= \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{x+h-x}{x(x+h)} \right) = \lim_{h \rightarrow 0} \frac{1}{h} \frac{h}{x(x+h)} = \lim_{h \rightarrow 0} \frac{1}{x(x+h)} = \frac{1}{x^2}

(iii) $f(x) = \sin(x+1)$

f'(x) = \lim_{h \rightarrow 0} \frac{\sin(x+h+1) - \sin(x+1)}{h}

Using

\sin C - \sin D = 2\cos(\frac{C+D}{2})\sin(\frac{C-D}{2})

:

= \lim_{h \rightarrow 0} \frac{2\cos(\frac{x+h+1+x+1}{2})\sin(\frac{x+h+1-x-1}{2})}{h}

= \lim_{h \rightarrow 0} \frac{2\cos(\frac{2x+2+h}{2})\sin(\frac{h}{2})}{h} = \lim_{h \rightarrow 0} \cos\left(x+1+\frac{h}{2}\right) \frac{\sin(h/2)}{h/2}

= \cos(x+1) \cdot 1 = \cos(x+1)

(iv) $f(x) = \cos\left(x-\frac{\pi}{8}\right)$

f'(x) = \lim_{h \rightarrow 0} \frac{\cos(x+h-\frac{\pi}{8}) - \cos(x-\frac{\pi}{8})}{h}

Using

\cos C - \cos D = -2\sin(\frac{C+D}{2})\sin(\frac{C-D}{2})

:

= \lim_{h \rightarrow 0} \frac{-2\sin(\frac{x+h-\pi/8+x-\pi/8}{2})\sin(\frac{x+h-\pi/8-x+\pi/8}{2})}{h}

= \lim_{h \rightarrow 0} \frac{-2\sin(x-\frac{\pi}{8}+\frac{h}{2})\sin(\frac{h}{2})}{h} = \lim_{h \rightarrow 0} -\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right) \frac{\sin(h/2)}{h/2}

= -\sin\left(x-\frac{\pi}{8}\right) \cdot 1 = -\sin\left(x-\frac{\pi}{8}\right)

Final Answer:

(i)

-1

(ii)

\frac{1}{x^2}

(iii)

\cos(x+1)

(iv)

-\sin\left(x-\frac{\pi}{8}\right)

Q2Miscellaneous Exercise on Chapter 12

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): 2. $(x+a)$

Solution

Given:

f(x) = x+a

Solution:

f'(x) = \frac{d}{dx}(x+a) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1

Final Answer: 1

Q3Miscellaneous Exercise on Chapter 12

$(p x+q)\left(\frac{r}{x}+s\right)$

Solution

Given:

f(x) = (px+q)(\frac{r}{x}+s)

Solution: Using the product rule

(uv)' = u'v + uv'

. Let

u = px+q

and

v = \frac{r}{x}+s = rx^{-1}+s

. Then

u' = p

and

v' = -rx^{-2} = -\frac{r}{x^2}

.

f'(x) = (p)\left(\frac{r}{x}+s\right) + (px+q)\left(-\frac{r}{x^2}\right)

= \frac{pr}{x} + ps - \frac{prx}{x^2} - \frac{qr}{x^2}

= \frac{pr}{x} + ps - \frac{pr}{x} - \frac{qr}{x^2} = ps - \frac{qr}{x^2}

Final Answer:

ps - \frac{qr}{x^2}

Q4Miscellaneous Exercise on Chapter 12

$(a x+b)(c x+d)^{2}$

Solution

Given:

f(x) = (ax+b)(cx+d)^2

Solution: First, expand

(cx+d)^2 = c^2x^2+2cdx+d^2

.

f(x) = (ax+b)(c^2x^2+2cdx+d^2)

. Using the product rule

(uv)' = u'v + uv'

. Let

u = ax+b

and

v = c^2x^2+2cdx+d^2

. Then

u' = a

and

v' = 2c^2x+2cd

.

f'(x) = (a)(c^2x^2+2cdx+d^2) + (ax+b)(2c^2x+2cd)

= ac^2x^2+2acdx+ad^2 + (2ac^2x^2+2acdx+2bc^2x+2bcd)

= 3ac^2x^2 + 4acdx + 2bc^2x + ad^2 + 2bcd

= 3ac^2x^2 + 2(2ad+bc^2)x + ad^2 + 2bcd

Final Answer:

3ac^2x^2 + 2(2ad+bc^2)x + ad^2 + 2bcd

Q5Miscellaneous Exercise on Chapter 12

$\frac{a x+b}{c x+d}$

Solution

Given:

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

Solution: Using the quotient rule

(\frac{u}{v})' = \frac{u'v - uv'}{v^2}

. Let

u=ax+b

and

v=cx+d

. Then

u'=a

and

v'=c

.

f'(x) = \frac{(a)(cx+d) - (ax+b)(c)}{(cx+d)^2}

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

Final Answer:

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

Q6Miscellaneous Exercise on Chapter 12

$\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$

Solution

Given:

f(x) = \frac{1+\frac{1}{x}}{1-\frac{1}{x}}

Solution: First, simplify the expression.

f(x) = \frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}

Now, use the quotient rule. Let

u=x+1

and

v=x-1

. Then

u'=1

and

v'=1

.

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

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

Final Answer:

-\frac{2}{(x-1)^2}

Q7Miscellaneous Exercise on Chapter 12

$\frac{1}{a x^{2}+b x+c}$

Solution

Given:

f(x) = \frac{1}{ax^2+bx+c}

Solution: Using the quotient rule with

u=1

and

v=ax^2+bx+c

. Then

u'=0

and

v'=2ax+b

.

f'(x) = \frac{(0)(ax^2+bx+c) - (1)(2ax+b)}{(ax^2+bx+c)^2}

= \frac{-(2ax+b)}{(ax^2+bx+c)^2}

Final Answer:

-\frac{2ax+b}{(ax^2+bx+c)^2}

Q8Miscellaneous Exercise on Chapter 12

$\frac{a x+b}{p x^{2}+q x+r}$

Solution

Given:

f(x) = \frac{ax+b}{px^2+qx+r}

Solution: Using the quotient rule. Let

u=ax+b

and

v=px^2+qx+r

. Then

u'=a

and

v'=2px+q

.

f'(x) = \frac{(a)(px^2+qx+r) - (ax+b)(2px+q)}{(px^2+qx+r)^2}

= \frac{apx^2+aqx+ar - (2apx^2+aqx+2bpx+bq)}{(px^2+qx+r)^2}

= \frac{apx^2+aqx+ar - 2apx^2-aqx-2bpx-bq}{(px^2+qx+r)^2}

= \frac{-apx^2 - 2bpx + ar-bq}{(px^2+qx+r)^2}

Final Answer:

\frac{-apx^2 - 2bpx + ar-bq}{(px^2+qx+r)^2}

Q9Miscellaneous Exercise on Chapter 12

$\frac{p x^{2}+q x+r}{a x+b}$

Solution

Given:

f(x) = \frac{px^2+qx+r}{ax+b}

Solution: Using the quotient rule. Let

u=px^2+qx+r

and

v=ax+b

. Then

u'=2px+q

and

v'=a

.

f'(x) = \frac{(2px+q)(ax+b) - (px^2+qx+r)(a)}{(ax+b)^2}

= \frac{2apx^2+2bpx+aqx+bq - (apx^2+aqx+ar)}{(ax+b)^2}

= \frac{2apx^2+2bpx+aqx+bq - apx^2-aqx-ar}{(ax+b)^2}

= \frac{apx^2+2bpx+bq-ar}{(ax+b)^2}

Final Answer:

\frac{apx^2+2bpx+bq-ar}{(ax+b)^2}

Q10Miscellaneous Exercise on Chapter 12

$\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$

Solution

Given:

f(x) = \frac{a}{x^4} - \frac{b}{x^2} + \cos x = ax^{-4} - bx^{-2} + \cos x

Solution: Differentiate term by term.

f'(x) = a(-4x^{-5}) - b(-2x^{-3}) + (-\sin x)

= -4ax^{-5} + 2bx^{-3} - \sin x

= -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x

Final Answer:

- \frac{4a}{x^5} + \frac{2b}{x^3} - \sin x

Q11Miscellaneous Exercise on Chapter 12

$4 \sqrt{x}-2$

Solution

Given:

f(x) = 4\sqrt{x} - 2 = 4x^{1/2} - 2

Solution:

f'(x) = 4 \cdot \frac{d}{dx}(x^{1/2}) - \frac{d}{dx}(2)

= 4 \left( \frac{1}{2}x^{1/2 - 1} \right) - 0

= 2x^{-1/2} = \frac{2}{\sqrt{x}}

Final Answer:

\frac{2}{\sqrt{x}}

Q12Miscellaneous Exercise on Chapter 12

$(a x+b)^{n}$

Solution

Given:

f(x) = (ax+b)^n

Solution: Using the product rule and induction (or the chain rule, which is more direct). Let's use the product rule for

n=2,3

to see the pattern. For

n=2

,

f(x) = (ax+b)(ax+b)

,

f'(x) = a(ax+b) + (ax+b)a = 2a(ax+b)

. For

n=3

,

f(x) = (ax+b)^2(ax+b)

,

f'(x) = [2a(ax+b)](ax+b) + (ax+b)^2(a) = 2a(ax+b)^2 + a(ax+b)^2 = 3a(ax+b)^2

. The pattern suggests the derivative is

n \cdot a \cdot (ax+b)^{n-1}

. This is a direct application of the chain rule: Let

u=ax+b

, then

f(u)=u^n

.

f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = nu^{n-1} \cdot a = na(ax+b)^{n-1}

.

Final Answer:

na(ax+b)^{n-1}

Q13Miscellaneous Exercise on Chapter 12

$(a x+b)^{n}(c x+d)^{m}$

Solution

Given:

f(x) = (ax+b)^n (cx+d)^m

Solution: Using the product rule

(uv)' = u'v + uv'

. Let

u = (ax+b)^n

and

v = (cx+d)^m

. From the previous question,

u' = na(ax+b)^{n-1}

and

v' = mc(cx+d)^{m-1}

.

f'(x) = [na(ax+b)^{n-1}](cx+d)^m + (ax+b)^n[mc(cx+d)^{m-1}]

We can factor out common terms

(ax+b)^{n-1}

and

(cx+d)^{m-1}

.

f'(x) = (ax+b)^{n-1}(cx+d)^{m-1} [na(cx+d) + mc(ax+b)]

= (ax+b)^{n-1}(cx+d)^{m-1} [nacx + nad + mcax + mcb]

= (ax+b)^{n-1}(cx+d)^{m-1} [(na+mc)ax + (nad+mcb)]

Final Answer:

(ax+b)^{n-1}(cx+d)^{m-1} [na(cx+d) + mc(ax+b)]

Q14Miscellaneous Exercise on Chapter 12

$\sin (x+a)$

Solution

Given:

f(x) = \sin(x+a)

Solution: Using the sum formula for sine:

f(x) = \sin x \cos a + \cos x \sin a

. Since

a

is a constant,

\cos a

and

\sin a

are constants.

f'(x) = \frac{d}{dx}(\sin x \cos a) + \frac{d}{dx}(\cos x \sin a)

= (\cos x)(\cos a) + (-\sin x)(\sin a)

= \cos x \cos a - \sin x \sin a

Using the formula for

\cos(A+B)

:

= \cos(x+a)

Final Answer:

\cos(x+a)

Q15Miscellaneous Exercise on Chapter 12

$\csc x \cot x$

Solution

Given:

f(x) = \csc x \cot x

Solution: Using the product rule

(uv)' = u'v + uv'

. Let

u = \csc x

and

v = \cot x

. We know

u' = -\csc x \cot x

and

v' = -\csc^2 x

.

f'(x) = (-\csc x \cot x)(\cot x) + (\csc x)(-\csc^2 x)

= -\csc x \cot^2 x - \csc^3 x

= -\csc x (\cot^2 x + \csc^2 x)

Using the identity

\cot^2 x = \csc^2 x - 1

:

= -\csc x ((\csc^2 x - 1) + \csc^2 x) = -\csc x (2\csc^2 x - 1)

Final Answer:

-\csc x (\cot^2 x + \csc^2 x)

Q16Miscellaneous Exercise on Chapter 12

$\frac{\cos x}{1+\sin x}$

Solution

Given:

f(x) = \frac{\cos x}{1+\sin x}

Solution: Using the quotient rule. Let

u = \cos x

and

v = 1+\sin x

. Then

u' = -\sin x

and

v' = \cos x

.

f'(x) = \frac{(-\sin x)(1+\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2}

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

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

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

Final Answer:

-\frac{1}{1+\sin x}

Q17Miscellaneous Exercise on Chapter 12

$\frac{\sin x+\cos x}{\sin x-\cos x}$

Solution

Given:

f(x) = \frac{\sin x+\cos x}{\sin x-\cos x}

Solution: Using the quotient rule. Let

u = \sin x+\cos x

and

v = \sin x-\cos x

. Then

u' = \cos x - \sin x

and

v' = \cos x + \sin x

.

f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}

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

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

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

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

Final Answer:

\frac{-2}{(\sin x - \cos x)^2}

Q18Miscellaneous Exercise on Chapter 12

$\frac{\sec x-1}{\sec x+1}$

Solution

Given:

f(x) = \frac{\sec x-1}{\sec x+1}

Solution: First, simplify by converting sec to cos.

f(x) = \frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1} = \frac{\frac{1-\cos x}{\cos x}}{\frac{1+\cos x}{\cos x}} = \frac{1-\cos x}{1+\cos x}

Now use the quotient rule. Let

u = 1-\cos x

and

v = 1+\cos x

. Then

u' = \sin x

and

v' = -\sin x

.

f'(x) = \frac{(\sin x)(1+\cos x) - (1-\cos x)(-\sin x)}{(1+\cos x)^2}

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

= \frac{2\sin x}{(1+\cos x)^2}

Final Answer:

\frac{2\sin x}{(1+\cos x)^2}

Q19Miscellaneous Exercise on Chapter 12

$\sin ^{n} x$

Solution

Given:

f(x) = \sin^n x = (\sin x)^n

Solution: Using the chain rule, let

u = \sin x

, so

f(u) = u^n

.

f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = (nu^{n-1}) \cdot (\cos x)

= n(\sin x)^{n-1} \cos x = n \sin^{n-1} x \cos x

Final Answer:

n \sin^{n-1} x \cos x

Q20Miscellaneous Exercise on Chapter 12

$\frac{a+b \sin x}{c+d \cos x}$

Solution

Given:

f(x) = \frac{a+b \sin x}{c+d \cos x}

Solution: Using the quotient rule. Let

u = a+b\sin x

and

v = c+d\cos x

. Then

u' = b\cos x

and

v' = -d\sin x

.

f'(x) = \frac{(b\cos x)(c+d\cos x) - (a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}

= \frac{bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x}{(c+d\cos x)^2}

= \frac{bc\cos x + ad\sin x + bd(\cos^2 x + \sin^2 x)}{(c+d\cos x)^2}

= \frac{bc\cos x + ad\sin x + bd}{(c+d\cos x)^2}

Final Answer:

\frac{bc\cos x + ad\sin x + bd}{(c+d\cos x)^2}

Q21Miscellaneous Exercise on Chapter 12

$\frac{\sin (x+a)}{\cos x}$

Solution

Given:

f(x) = \frac{\sin (x+a)}{\cos x}

Solution: Using the quotient rule. Let

u = \sin(x+a)

and

v = \cos x

. Then

u' = \cos(x+a)

and

v' = -\sin x

.

f'(x) = \frac{(\cos(x+a))(\cos x) - (\sin(x+a))(-\sin x)}{(\cos x)^2}

= \frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x}

Using the identity

\cos(A-B) = \cos A \cos B + \sin A \sin B

, with

A=x+a

and

B=x

.

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

Final Answer:

\frac{\cos a}{\cos^2 x}

Q22Miscellaneous Exercise on Chapter 12

$x^{4}(5 \sin x-3 \cos x)$

Solution

Given:

f(x) = x^4(5 \sin x - 3 \cos x)

Solution: Using the product rule. Let

u = x^4

and

v = 5\sin x - 3\cos x

. Then

u' = 4x^3

and

v' = 5\cos x - 3(-\sin x) = 5\cos x + 3\sin x

.

f'(x) = (4x^3)(5\sin x - 3\cos x) + (x^4)(5\cos x + 3\sin x)

= 20x^3\sin x - 12x^3\cos x + 5x^4\cos x + 3x^4\sin x

= x^3[(20+3x)\sin x + (5x-12)\cos x]

Final Answer:

x^3[(20+3x)\sin x + (5x-12)\cos x]

Q23Miscellaneous Exercise on Chapter 12

$\left(x^{2}+1\right) \cos x$

Solution

Given:

f(x) = (x^2+1)\cos x

Solution: Using the product rule. Let

u = x^2+1

and

v = \cos x

. Then

u' = 2x

and

v' = -\sin x

.

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

= 2x\cos x - (x^2+1)\sin x

Final Answer:

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

Q24Miscellaneous Exercise on Chapter 12

$\left(a x^{2}+\sin x\right)(p+q \cos x)$

Solution

Given:

f(x) = (ax^2+\sin x)(p+q\cos x)

Solution: Using the product rule. Let

u = ax^2+\sin x

and

v = p+q\cos x

. Then

u' = 2ax+\cos x

and

v' = -q\sin x

.

f'(x) = (2ax+\cos x)(p+q\cos x) + (ax^2+\sin x)(-q\sin x)

= 2apx + 2aqx\cos x + p\cos x + q\cos^2 x - aqx^2\sin x - q\sin^2 x

Final Answer:

(2ax+\cos x)(p+q\cos x) - q\sin x(ax^2+\sin x)

Q25Miscellaneous Exercise on Chapter 12

$(x+\cos x)(x-\tan x)$

Solution

Given:

f(x) = (x+\cos x)(x-\tan x)

Solution: Using the product rule. Let

u = x+\cos x

and

v = x-\tan x

. Then

u' = 1-\sin x

and

v' = 1-\sec^2 x

.

f'(x) = (1-\sin x)(x-\tan x) + (x+\cos x)(1-\sec^2 x)

Using the identity

1-\sec^2 x = -\tan^2 x

.

f'(x) = (1-\sin x)(x-\tan x) - (x+\cos x)\tan^2 x

Final Answer:

(1-\sin x)(x-\tan x) - (x+\cos x)\tan^2 x

Q26Miscellaneous Exercise on Chapter 12

$\frac{4 x+5 \sin x}{3 x+7 \cos x}$

Solution

Given:

f(x) = \frac{4x+5\sin x}{3x+7\cos x}

Solution: Using the quotient rule. Let

u = 4x+5\sin x

and

v = 3x+7\cos x

. Then

u' = 4+5\cos x

and

v' = 3-7\sin x

.

f'(x) = \frac{(4+5\cos x)(3x+7\cos x) - (4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}

= \frac{(12x+28\cos x+15x\cos x+35\cos^2 x) - (12x-28x\sin x+15\sin x-35\sin^2 x)}{(3x+7\cos x)^2}

= \frac{12x+28\cos x+15x\cos x+35\cos^2 x - 12x+28x\sin x-15\sin x+35\sin^2 x}{(3x+7\cos x)^2}

= \frac{15x\cos x+28x\sin x+28\cos x-15\sin x+35(\cos^2 x+\sin^2 x)}{(3x+7\cos x)^2}

= \frac{15x\cos x+28x\sin x+28\cos x-15\sin x+35}{(3x+7\cos x)^2}

Final Answer:

\frac{(15x+28)\cos x + (28x-15)\sin x + 35}{(3x+7\cos x)^2}

Q27Miscellaneous Exercise on Chapter 12

$\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$

Solution

Given:

f(x) = \frac{x^2 \cos(\frac{\pi}{4})}{\sin x}

Solution: Note that

\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}

is a constant. Let

k = \cos(\frac{\pi}{4})

. So,

f(x) = k \frac{x^2}{\sin x}

. Using the quotient rule for

\frac{x^2}{\sin x}

. Let

u = x^2

and

v = \sin x

. Then

u' = 2x

and

v' = \cos x

.

f'(x) = k \left( \frac{(2x)(\sin x) - (x^2)(\cos x)}{(\sin x)^2} \right)

= \frac{\cos(\frac{\pi}{4}) (2x\sin x - x^2\cos x)}{\sin^2 x}

= \frac{x\cos(\frac{\pi}{4}) (2\sin x - x\cos x)}{\sin^2 x}

Final Answer:

\frac{x\cos(\frac{\pi}{4}) (2\sin x - x\cos x)}{\sin^2 x}

Q29Miscellaneous Exercise on Chapter 12

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

Solution

Given:

f(x) = (x+\sec x)(x-\tan x)

Solution: Using the product rule. Let

u = x+\sec x

and

v = x-\tan x

. Then

u' = 1+\sec x \tan x

and

v' = 1-\sec^2 x

.

f'(x) = (1+\sec x \tan x)(x-\tan x) + (x+\sec x)(1-\sec^2 x)

Using the identity

1-\sec^2 x = -\tan^2 x

.

f'(x) = (1+\sec x \tan x)(x-\tan x) - (x+\sec x)\tan^2 x

Final Answer:

(1+\sec x \tan x)(x-\tan x) - (x+\sec x)\tan^2 x

Q30Miscellaneous Exercise on Chapter 12

$\frac{x}{\sin ^{n} x}$

Solution

Given:

f(x) = \frac{x}{\sin^n x}

Solution: Using the quotient rule. Let

u = x

and

v = \sin^n x

. Then

u' = 1

and

v' = n\sin^{n-1} x \cos x

(from question 19).

f'(x) = \frac{(1)(\sin^n x) - (x)(n\sin^{n-1} x \cos x)}{(\sin^n x)^2}

= \frac{\sin^n x - nx\sin^{n-1} x \cos x}{\sin^{2n} x}

Factor out

\sin^{n-1} x

from the numerator:

= \frac{\sin^{n-1} x (\sin x - nx\cos x)}{\sin^{2n} x}

= \frac{\sin x - nx\cos x}{\sin^{n+1} x}

Final Answer:

\frac{\sin x - nx\cos x}{\sin^{n+1} x}

Limits and DerivativesClass 11 Mathematics NCERT Solutions

Evaluate the following limits in Exercises 1 to 22. lim⁡x→3x+3\lim_{x \rightarrow 3} x+3limx→3​x+3

Solution

lim⁡x→π(x−227)\lim_{x \rightarrow \pi} \left(x-\frac{22}{7}\right)limx→π​(x−722​)

Solution

lim⁡r→1πr2\lim_{r \rightarrow 1} \pi r^{2}limr→1​πr2

Solution

lim⁡x→44x+3x−2\lim_{x \rightarrow 4} \frac{4x+3}{x-2}limx→4​x−24x+3​

Solution

lim⁡x→−1x10+x5+1x−1\lim_{x \rightarrow -1} \frac{x^{10}+x^{5}+1}{x-1}limx→−1​x−1x10+x5+1​

Solution

lim⁡x→0(x+1)5−1x\lim_{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}limx→0​x(x+1)5−1​

Solution

lim⁡x→23x2−x−10x2−4\lim_{x \rightarrow 2} \frac{3x^2-x-10}{x^2-4}limx→2​x2−43x2−x−10​

Solution

lim⁡x→3x4−812x2−5x−3\lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}limx→3​2x2−5x−3x4−81​

Solution

lim⁡x→0ax+bcx+1\lim_{x \rightarrow 0} \frac{ax+b}{cx+1}limx→0​cx+1ax+b​

Solution

lim⁡z→1z13−1z16−1\lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}limz→1​z61​−1z31​−1​

Solution

lim⁡x→1ax2+bx+ccx2+bx+a,a+b+c≠0\lim_{x \rightarrow 1} \frac{ax^2+bx+c}{cx^2+bx+a}, a+b+c \neq 0limx→1​cx2+bx+aax2+bx+c​,a+b+c=0

Solution

lim⁡x→−21x+12x+2\lim_{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}limx→−2​x+2x1​+21​​

Solution

lim⁡x→0sin⁡axbx\lim_{x \rightarrow 0} \frac{\sin ax}{bx}limx→0​bxsinax​

Solution

lim⁡x→0sin⁡axsin⁡bx,a,b≠0\lim_{x \rightarrow 0} \frac{\sin ax}{\sin bx}, a, b \neq 0limx→0​sinbxsinax​,a,b=0

Solution

lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}limx→π​π(π−x)sin(π−x)​

Solution

lim⁡x→0cos⁡xπ−x\lim_{x \rightarrow 0} \frac{\cos x}{\pi-x}limx→0​π−xcosx​

Solution

lim⁡x→0cos⁡2x−1cos⁡x−1\lim_{x \rightarrow 0} \frac{\cos 2x-1}{\cos x-1}limx→0​cosx−1cos2x−1​

Solution

lim⁡x→0ax+xcos⁡xbsin⁡x\lim_{x \rightarrow 0} \frac{ax+x \cos x}{b \sin x}limx→0​bsinxax+xcosx​

Solution

lim⁡x→0xsec⁡x\lim_{x \rightarrow 0} x \sec xlimx→0​xsecx

Solution

lim⁡x→0sin⁡ax+bxax+sin⁡bxa,b,a+b≠0\lim_{x \rightarrow 0} \frac{\sin ax+bx}{ax+\sin bx} a, b, a+b \neq 0limx→0​ax+sinbxsinax+bx​a,b,a+b=0

Solution

lim⁡x→0(csc⁡x−cot⁡x)\lim_{x \rightarrow 0} (\csc x - \cot x)limx→0​(cscx−cotx)

Solution

lim⁡x→π2tan⁡2xx−π2\lim_{x \rightarrow \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}limx→2π​​x−2π​tan2x​

Solution

Find lim⁡x→0f(x)\lim_{x \rightarrow 0} f(x)limx→0​f(x) and lim⁡x→1f(x)\lim_{x \rightarrow 1} f(x)limx→1​f(x), where f(x)={2x+3,x≤0 3(x+1),x>0f(x)=\begin{cases}2 x+3, & x \leq 0 \ 3(x+1), & x>0\end{cases}f(x)={2x+3,​x≤0 3(x+1),​x>0​

Solution

Find lim⁡x→1f(x)\lim_{x \rightarrow 1} f(x)limx→1​f(x), where f(x)={x2−1,x≤1 −x2−1,x>1f(x)= \begin{cases}x^{2}-1, & x \leq 1 \ -x^{2}-1, & x>1\end{cases}f(x)={x2−1,​x≤1 −x2−1,​x>1​

Solution

Evaluate lim⁡x→0f(x)\lim_{x \rightarrow 0} f(x)limx→0​f(x), where f(x)={∣x∣x,x≠0 0,x=0f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \ 0, & x=0\end{cases}f(x)={x∣x∣​,​x=0 0,​x=0​

Solution

Find lim⁡x→0f(x)\lim_{x \rightarrow 0} f(x)limx→0​f(x), where f(x)={x∣x∣,x≠0 0,x=0f(x)=\begin{cases}\frac{x}{|x|}, & x \neq 0 \ 0, & x=0\end{cases}f(x)={∣x∣x​,​x=0 0,​x=0​

Solution

Find lim⁡x→5f(x)\lim_{x \rightarrow 5} f(x)limx→5​f(x), where f(x)=∣x∣−5f(x)=|x|-5f(x)=∣x∣−5

Solution

Suppose f(x)={a+bx,x<1 4,x=1 b−ax,x>1f(x)= \begin{cases}a+b x, & x<1 \ 4, & x=1 \ b-a x, & x>1\end{cases}f(x)={a+bx,​x<1 4,​x=1 b−ax,​x>1​ and if lim⁡x→1f(x)=f(1)\lim_{x \rightarrow 1} f(x)=f(1)limx→1​f(x)=f(1) what are possible values of aaa and bbb?

Solution

Solution

If f(x)={∣x∣+1,x<0 0,x=0 ∣x∣−1,x>0f(x)=\begin{cases}|x|+1, & x<0 \ 0, & x=0 \ |x|-1, & x>0\end{cases}f(x)={∣x∣+1,​x<0 0,​x=0 ∣x∣−1,​x>0​. For what value (s) of aaa does lim⁡x→af(x)\lim_{x \rightarrow a} f(x)limx→a​f(x) exists?

Solution

If the function f(x)f(x)f(x) satisfies lim⁡x→1f(x)−2x2−1=π\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pilimx→1​x2−1f(x)−2​=π, evaluate lim⁡x→1f(x)\lim_{x \rightarrow 1} f(x)limx→1​f(x)

Solution

Solution

Find the derivative of x2−2x^2-2x2−2 at x=10x=10x=10.

Solution

Find the derivative of xxx at x=1x=1x=1.

Solution

Find the derivative of 99x99x99x at x=100x=100x=100.

Solution

Find the derivative of the following functions from first principle.

x3−27x^3-27x3−27

(x−1)(x−2)(x-1)(x-2)(x−1)(x−2)

1x2\frac{1}{x^2}x21​

x+1x−1\frac{x+1}{x-1}x−1x+1​

Solution

For the function f(x)=x100100+x9999+…+x22+x+1f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1f(x)=100x100​+99x99​+…+2x2​+x+1 Prove that f′(1)=100f′(0)f^{\prime}(1)=100 f^{\prime}(0)f′(1)=100f′(0).

Solution

Find the derivative of xn+axn−1+a2xn−2+…+an−1x+anx^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n}xn+axn−1+a2xn−2+…+an−1x+an for some fixed real number aaa.

Solution

For some constants aaa and bbb, find the derivative of

Evaluate the following limits in Exercises 1 to 22. $\lim_{x \rightarrow 3} x+3$

$\lim_{x \rightarrow \pi} \left(x-\frac{22}{7}\right)$

$\lim_{r \rightarrow 1} \pi r^{2}$

$\lim_{x \rightarrow 4} \frac{4x+3}{x-2}$

$\lim_{x \rightarrow -1} \frac{x^{10}+x^{5}+1}{x-1}$

$\lim_{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}$

$\lim_{x \rightarrow 2} \frac{3x^2-x-10}{x^2-4}$

$\lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}$

$\lim_{x \rightarrow 0} \frac{ax+b}{cx+1}$

$\lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

$\lim_{x \rightarrow 1} \frac{ax^2+bx+c}{cx^2+bx+a}, a+b+c \neq 0$

$\lim_{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

$\lim_{x \rightarrow 0} \frac{\sin ax}{bx}$

$\lim_{x \rightarrow 0} \frac{\sin ax}{\sin bx}, a, b \neq 0$

$\lim_{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$

$\lim_{x \rightarrow 0} \frac{\cos x}{\pi-x}$

$\lim_{x \rightarrow 0} \frac{\cos 2x-1}{\cos x-1}$

$\lim_{x \rightarrow 0} \frac{ax+x \cos x}{b \sin x}$

$\lim_{x \rightarrow 0} x \sec x$

$\lim_{x \rightarrow 0} \frac{\sin ax+bx}{ax+\sin bx} a, b, a+b \neq 0$

$\lim_{x \rightarrow 0} (\csc x - \cot x)$

$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}$

Find $\lim_{x \rightarrow 0} f(x)$ and $\lim_{x \rightarrow 1} f(x)$ , where $f(x)=\begin{cases}2 x+3, & x \leq 0 \ 3(x+1), & x>0\end{cases}$

Find $\lim_{x \rightarrow 1} f(x)$ , where $f(x)= \begin{cases}x^{2}-1, & x \leq 1 \ -x^{2}-1, & x>1\end{cases}$

Evaluate $\lim_{x \rightarrow 0} f(x)$ , where $f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \ 0, & x=0\end{cases}$

Find $\lim_{x \rightarrow 0} f(x)$ , where $f(x)=\begin{cases}\frac{x}{|x|}, & x \neq 0 \ 0, & x=0\end{cases}$

Find $\lim_{x \rightarrow 5} f(x)$ , where $f(x)=|x|-5$

Suppose $f(x)= \begin{cases}a+b x, & x<1 \ 4, & x=1 \ b-a x, & x>1\end{cases}$ and if $\lim_{x \rightarrow 1} f(x)=f(1)$ what are possible values of $a$ and $b$ ?

If $f(x)=\begin{cases}|x|+1, & x<0 \ 0, & x=0 \ |x|-1, & x>0\end{cases}$ . For what value (s) of $a$ does $\lim_{x \rightarrow a} f(x)$ exists?

If the function $f(x)$ satisfies $\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi$ , evaluate $\lim_{x \rightarrow 1} f(x)$

Find the derivative of $x^2-2$ at $x=10$ .

Find the derivative of $x$ at $x=1$ .

Find the derivative of $99x$ at $x=100$ .

$x^3-27$

$(x-1)(x-2)$

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

$\frac{x+1}{x-1}$

For the function $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$ Prove that $f^{\prime}(1)=100 f^{\prime}(0)$ .

Find the derivative of $x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n}$ for some fixed real number $a$ .

For some constants $a$ and $b$ , find the derivative of

$(x-a)(x-b)$

$(ax^2+b)^2$

$\frac{x-a}{x-b}$

Find the derivative of $\frac{x^{n}-a^{n}}{x-a}$ for some constant $a$ .

$2x - \frac{3}{4}$

$(5x^3+3x-1)(x-1)$

$x^{-3}(5+3x)$

$x^5(3-6x^{-9})$

$x^{-4}(3-4x^{-5})$

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

Find the derivative of $\cos x$ from first principle.

$\sin x \cos x$

$\sec x$

$5 \sec x+4 \cos x$

$\csc x$

$3 \cot x+5 \csc x$

$5 \sin x-6 \cos x+7$

$2 \tan x-7 \sec x$

$-x$

$(-x)^{-1}$

$\sin (x+1)$

$\cos \left(x-\frac{\pi}{8}\right)$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): 2. $(x+a)$

$(p x+q)\left(\frac{r}{x}+s\right)$

$(a x+b)(c x+d)^{2}$

$\frac{a x+b}{c x+d}$

$\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$

$\frac{1}{a x^{2}+b x+c}$

$\frac{a x+b}{p x^{2}+q x+r}$

$\frac{p x^{2}+q x+r}{a x+b}$

$\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$

$4 \sqrt{x}-2$

$(a x+b)^{n}$

$(a x+b)^{n}(c x+d)^{m}$

$\sin (x+a)$

$\csc x \cot x$

$\frac{\cos x}{1+\sin x}$

$\frac{\sin x+\cos x}{\sin x-\cos x}$

$\frac{\sec x-1}{\sec x+1}$

$\sin ^{n} x$