Continuous practice using ML Aggarwal Class 12 ISC Solutions Chapter 5 Continuity and Differentiability Ex 5.14 can lead to a stronger grasp of mathematical concepts.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.14

Question 1.
Examine the continuity the function f(x) = $$\left\{\begin{array}{cc} |x| & , \quad x \leq 0 \\ x & , 0<x<1 \\ 2-x & , 1 \leq x \leq 2 \\ 3 x-5, & x>2 \end{array}\right.$$ at each of the points 0, 1, 2.
Solution:
Continuity at x = 0:
L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ |x|
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ – x = 0

Continuity at x = 1:
$$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ x = 1
$$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 2 – x
= 2 – 1 = 1
also f(1) = 2 – 1 = 1
∴ $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = f(1)
∴ f is continuous at x = 1.

Continuity at x = 2:
L.H.L. = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ 2 – x
= 2 – 2
= 0
R.H.L. = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ 3x – 5
= 6 – 5 = 1
∴ $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) ≠ $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x)
Thus, f(x) is discontinuous at x = 2.

Question 2.
Determine the constant k so that the function f(x) = may be continuous.
Solution:

Also f(2) = k
Now, the function f(x) is continuous at x = 2.
if $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ f(2) = if $$\frac{1}{8}$$ = k.

Question 3.
Examine for continuity and differentiability, each of the following functions:
(i) f(x) = $$\left\{\begin{array}{cc} x \sin \frac{1}{x}, & x<0 \\ 0 & , x \geq 0 \end{array}\right.$$ at x = 0 (ii) f(x) = $$\left\{\begin{array}{cc} |x| \sin \frac{1}{x}, & x>0 \\ 0, & x \leq 0 \end{array}\right.$$ at x = 0
(iii) f(x) = $$\left\{\begin{array}{cc} (x-c)^2 \cos \frac{1}{x-c}, & x \neq c \\ 0 & , x=c \end{array}\right.$$ at x = c.
Solution:
(i) Continuity at x = 0:
$$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(0) = 0
$$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ x sin $$\frac{1}{x}$$
Let g(x) = x
and h(x) = sin $$\frac{1}{x}$$
$$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ g(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ x = 0
and sin $$\frac{1}{x}$$ is bounded in the deleted ngd of 0.
[∵ – 1 ≤ sin t ≤ 1 ∀ t ∈ R]
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ g(x) h(x) = 0
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ x sin $$\frac{1}{x}$$ = 0
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = 0
∴ $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = 0,
also f(0) = 0
Thus $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x0 = f(0)
Hence, f is continuous at x = 0.

Differentiability at x = 0:
Lf'(0) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ $$\frac{f(x)-f(0)}{x-0}$$

∴ f(x) is not differentiable at x = 0.

(ii) Continuity at x = 0:
$$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ |x| sin $$\frac{1}{x}$$
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ x sin $$\frac{1}{x}$$
Let g(x) = x and
h(x) = sin $$\frac{1}{x}$$
$$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ g(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ x = 0
and sin $$\frac{1}{x}$$ is boundede in the deleted ngd of 0.
[∵|sin t| ≤ 1 ∀ t ∈ R]
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ g(x) h(x) = 0
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ x sin $$\frac{1}{x}$$ = 0
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = 0
⇒ $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = 0
$$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ 0 = 0
and f(0) = 0
∴ $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = f(0)
Thus f(x) is continuous at x = 0

Differentiability at x = 0:
Rf'(0) = $$\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$$
= $${Lt}_{x \rightarrow 0^{+}} \frac{x \sin \frac{1}{x}-0}{x}$$
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ sin $$\frac{1}{x}$$, which does not exists.
[since sin $$\frac{1}{x}$$ be a real number oscillating between – 1 and 1]
If x = $$\frac{1}{2 n \pi+\frac{\pi}{2}}$$, n ∈ N, n → ∞ ⇒ x → 0
∴ sin $$\frac{1}{x}$$ = sin (2nπ + $$\frac{\pi}{2}$$)
= sin $$\frac{\pi}{2}$$ = 1
If x = $$\frac{1}{2 n \pi-\frac{\pi}{2}}$$, n ∈ N, n → ∞ ⇒ x → 0
∴ sin $$\frac{1}{x}$$ = sin (2nπ – $$\frac{\pi}{2}$$)
= sin (- $$\frac{\pi}{2}$$)
= – 1
∴ Rf'(0) does not exists.
Thus f is not differentiable at x = 0.

(iii) Continuity at x = c:
Let g(x) = (x – c)2 ;
h(x) = cos $$\frac{1}{x-c}$$
$$\underset{x \rightarrow c}{\mathrm{Lt}}$$ g(x) = $$\underset{x \rightarrow c}{\mathrm{Lt}}$$ (x – c)2 = 0
and h(x) = cos $$\frac{1}{x-c}$$ is bounded in the ddeleted ngd of c.
[∵ – 1 ≤ cos $$\frac{1}{x}$$ ≤ 1 ∀ x ∈ R]
∴ $$\underset{x \rightarrow c}{\mathrm{Lt}}$$ g(x) h(x) = 0
⇒ $$\underset{x \rightarrow c}{\mathrm{Lt}}$$ (x – c)2 cos $$\frac{1}{x-c}$$ = 0
⇒ $$\underset{x \rightarrow c}{\mathrm{Lt}}$$ f(x) = 0 ;
also f(c) = 0
∴ $$\underset{x \rightarrow c}{\mathrm{Lt}}$$ f(x) = f(c)
Thus f is continuous at x = c.

Differentiability at x = c:
Lf'(c) = $$\ {Lt}_{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}$$
= $$\mathrm{Lt}_{h \rightarrow 0^{+}} \frac{f(c-h)-f(c)}{c-h-c}$$
= $$\ {Lt}_{h \rightarrow 0^{+}} \frac{(c-h-c)^2 \cos \left(\frac{-1}{h}\right)-0}{-h}$$
[put x = c – h as x → c ⇒ h → 0+]
= $$\ {Lt}_{h \rightarrow 0^{+}} \frac{h^2 \cos \frac{1}{h}}{h}$$
= $$\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}$$ h cos $$\frac{1}{h}$$ = 0
[Here g(h) = h
and $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ g(h) = $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h = 0
and f(h) = cos $$\frac{1}{h}$$ is bounded in the deleted ngd of 0.
∴ $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ g(h) f(h) = $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ h cos $$\frac{1}{h}$$ = 0]
Rf'(c) = $$\ e{Lt}_{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c}$$
= $$\ {Lt}_{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{c+h-c}$$
[put x = c + h as x → c+
⇒ h → 0+]
= $$\ {Lt}_{h \rightarrow 0^{+}} \frac{(c+h-c)^2 \cos \frac{1}{c+h-c}-0}{c+h-c}$$
= $$\ {Lt}_{h \rightarrow 0^{+}} \frac{h^2 \cos \frac{1}{h}-0}{h}$$
= $$\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}$$ h cos $$\frac{1}{h}$$ = 0
∴ Lf'(c) = R f'(c)
Thus f(x) is also differentiable at x = c.
Hence f(x) is continuous as well as differentiable at x = c.

Question 4.
Prove that the derivative of an odd function is always an even function.
Solution:
Let f(x) be an odd function
∴ f(- x) = – f(x) …………….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
f'(- x) $$\frac{d}{d x}$$ (- x) = – $$\frac{d}{d x}$$ f(x)
⇒ f'(- x) (- 1) = – f'(x)
⇒ f'(- x) = f'(x)
Thus f'(x) be an even function.
Hence, the deerivative of an add function is always an even function.

Question 5.
If 2 f(x) + 3 f(- x) = x2 + x + 1, find f'(1).
Solution:
Given 2 f(x) + 3 f(- x) = x2 + x + 1 ……………..(1)
Changing x to – x in eqn. (1) ; we have
2 f(- x) + 3 f(x) = x2 – x + 1 ………….(2)
Multiply eqn. (1) by 2 and eqn. (2) by 3 and subtracting ; we have
– 5 f(x) = 2x2 + 2x + 2 – 3x2 + 3x – 3
⇒ – 5 f(x) = – x2 + 5x – 1
⇒ f(x) = $$\frac{1}{5}$$ [x2 – 5x + 1]
Diff. both sides w.r.t. x ; we have
f'(x) = $$\frac{1}{5}$$ (2x – 5)
⇒ f'(1) = $$\frac{1}{5}$$ (2 – 5)
= – $$\frac{3}{5}$$

Question 6.
Differentiate the following functions w.r.t. x :
(i) tan-1 $$\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)$$
(ii) tan-1 $$\left(\frac{\left(3 a^2 x-x^3\right)}{a\left(a^2-3 x^2\right)}\right)$$
Solution:
(i) Let y = tan-1 $$\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)$$
put √x = tan θ.
⇒ θ = tan-1 x
Thus from (1) ; we have
y = tan-1 $$\left\{\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right\}$$
⇒ y = tan-1 (tan 3θ) = 3θ = tan-1 √x
Diff. both sides w.r.t. x, we get
$$\frac{d y}{d x}=\frac{3}{1+(\sqrt{x})^2} \frac{d}{d x} \sqrt{x}$$
= $$\frac{3}{(1+x)} \frac{1}{2 \sqrt{x}}$$
= $$\frac{3}{2 \sqrt{x}(1+x)}$$

(ii) Let y = tan-1 $$\left(\frac{\left(3 a^2 x-x^3\right)}{a\left(a^2-3 x^2\right)}\right)$$
⇒ y = tan-1 $$\left\{\frac{\frac{3 x}{a}-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2}\right\}$$ ……(1)
put $$\frac{x}{a}$$ = tan θ
⇒ θ = tan-1 $$\frac{x}{a}$$
∴ from (1) ;
y = tan-1 $$\left\{\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right\}$$
⇒ y = tan-1 (tan 3θ)
⇒ y = 3θ
= 3 tan-1 $$\frac{x}{a}$$
Diff. both sides w.r.t. x, we have
$$\frac{d y}{d x}=3 \cdot \frac{1}{1+\left(\frac{x}{a}\right)^2} \cdot \frac{1}{a}$$
= $$\frac{3 a}{a^2+x^2}$$

Question 7.
Differentiate tan-1 $$\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$$ w.r.t. tan-1 ax.
Solution:
Let y = tan-1 $$\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$$ ………..(1)
and z = tan-1 ax ………..(2)
So we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$.
put ax = tan θ
i.e. θ = tan-1 ax
∴ from (1) ;
y = tan-1 $$\left(\frac{\sec \theta-1}{\tan \theta}\right)$$

Question 8.
If y = $$\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{1+\ldots \infty}}}}$$, prove that $$\frac{d y}{d x}=\frac{(1+y) \cos x+y \sin x}{1+2 y+\cos x-\sin x}$$.
Solution:
Given y = $$\frac{\sin x}{1+\frac{\cos x}{1+y}}$$
⇒ y = $$\frac{(1+y) \sin x}{1+y+\cos x}$$
⇒ y + y2 + y cos x = (1 + y) sin x ……….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
$$\frac{d y}{d x}$$ + 2y $$\frac{d y}{d x}$$ + cos x $$\frac{d y}{d x}$$ + y (- sin x) = (1 + y) cos x + sin x $$\frac{d y}{d x}$$
⇒ (1 + 2y + cos x – sin x) $$\frac{d y}{d x}$$ = (1 + y) cos x + y sin x
⇒ $$\frac{d y}{d x}=\frac{(1+y) \cos x+y \sin x}{1+2 y+\cos x-\sin x}$$.

Question 9.
If y = (logcos x sin x) (logsin x cos x)-1, prove that $$\frac{d y}{d x}$$π/4 = – 8 log2 e.
Solution:

Question 10.
Differentiate xx sin-1 √x w.r.t. x.
Solution:
Let y = xx sin-1 √x
Diff. both sides w.r.t. x, we have
$$\frac{d y}{d x}$$ = xx $$\frac{1}{\sqrt{1-(\sqrt{x})^2}} \frac{1}{2 \sqrt{x}}$$ + sin-1 √x $$\frac{d}{d x}$$ xx
= $$\frac{x^x}{\sqrt{1-x}} \frac{1}{2 \sqrt{x}}$$ + sin-1 √x $$\frac{d}{d x}$$ elog xx
= $$\frac{x^x}{2 \sqrt{x-x^2}}$$ + sin-1 √x $$\frac{d}{d x}$$ elog xx
= $$\frac{x^x}{2 \sqrt{x-x^2}}$$ + sin-1 √x ex log x [x × $$\frac{1}{x}$$ + log x . 1]
= $$\frac{x^x}{2 \sqrt{x-x^2}}$$ + sin-1 √x elog xx [1 + log x]
= $$\frac{x^x}{2 \sqrt{x-x^2}}$$ + sin-1 √x . xx (1 + log x).

Question 11.
If x = sin t and y = sin 2t, prove that
(i) (1 – x2) ($$\frac{d y}{d x}$$)2 = 4 (1 – y2)
(ii) (1 – x2) $$\frac{d^2 y}{d x^2}$$ – x $$\frac{d y}{d x}$$ + 4y = 0.
Solution:
(i) Given x = sin t ………..(1)
and y = sin 2t ………..(2)
Diff. eqn. (1) and (2) w.r.t. t ; we have
$$\frac{d x}{d t}$$ = cos t ;
$$\frac{d y}{d t}$$ = 2 cos 2t
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{2 \cos 2 t}{\cos t}$$ ………..(3)
On squaring both sides of eqn. (3) ; we have
cos2 t ($$\frac{d y}{d x}$$)2 = 4 cos2 2t
⇒ (1 – sin2 t) ($$\frac{d y}{d x}$$)2 = 4 [1 – sin2 2t]
⇒ (1 – x2) ($$\frac{d y}{d x}$$)2 = 4 [1 – y2] ………..(4)
Diff. eqn. (4) w.r.t. x ; we have
(1 – x2) 2 $$\frac{d y}{d x}$$ $$\frac{d^2 y}{d x^2}$$ + ($$\frac{d y}{d x}$$)2 (- 2x) = – 8y $$\frac{d y}{d x}$$
Dividing throughout by 2 $$\frac{d y}{d x}$$ ≠ 0 ; we get
(1 – x2) $$\frac{d^2 y}{d x^2}$$ – x $$\frac{d y}{d x}$$ = – 4y
⇒ (1 – x2) $$\frac{d^2 y}{d x^2}$$ – x $$\frac{d y}{d x}$$ + 4y = 0.

Question 12.
If y1/m + y-1/m = 2x, prove that (x2 – 1) y2 + xy1 = m2y.
Solution:
Given y1/n + y-1/n = 2x
⇒ y1/n + $$\frac{1}{y^{1 / n}}$$ = 2x
⇒ (y1/n)2 – 2xy1/n + 1 = 0 ; …………(1)
putting y1/n = t in eqn. (1) ; we get
⇒ t2 – 2xt + 1 = 0
⇒ t = $$\frac{2 x \pm \sqrt{4 x^2-4}}{2}$$
⇒ t = x ± $$\sqrt{x^2-1}$$

Case I:
When y1/n = x + $$\sqrt{x^2-1}$$
⇒ y = (x + $$\sqrt{x^2-1}$$)n …………(2)
Diff. eqn. (2) w.r.t. x, we have
∴ y1 = n (x + $$\sqrt{x^2-1}$$)n-1 $$\frac{d}{d x}$$ [x + $$\sqrt{x^2-1}$$]
∴ y1 = n (x + $$\sqrt{x^2-1}$$)n-1 $$\left[1+\frac{x}{\sqrt{x^2-1}}\right]$$
= $$\frac{n\left(x+\sqrt{x^2-1}\right)^{n-1}\left[x+\sqrt{x^2-1}\right]}{\sqrt{x^2-1}}$$
⇒ $$\sqrt{x^2-1}$$ y1 = ny ………..(3) [using (2)]
Again diff. both sides of eqn. (3) w.r.t. x, we have
⇒ $$\sqrt{x^2-1}$$ y2 + $$\frac{y_1}{\sqrt{x^2-1}}$$ × x = ny1
⇒ (x2 – 1) y2 + xy1 = n2y [using (3)]

Case – II:
When y1/n = x – $$\sqrt{x^2-1}$$
⇒ y = [x – $$\sqrt{x^2-1}$$]n ………..(4)
Diff. eqn. (4) w.r.t. x; we have
∴ y1 = n (x – $$\sqrt{x^2-1}$$)n-1 $$\left(1-\frac{x}{\sqrt{x^2-1}}\right)$$
⇒ $$\sqrt{x^2-1}$$ y1 = – ny
Again diff. eqn. (5) w.r.t. x ; we have
⇒ $$\sqrt{x^2-1}$$ y2 + xy1 = n2y [using eqn. (5)]

Question 13.
If x = et sin t and y = et cos t, then prove that (x + y)2 $$\frac{d^2 y}{d x^2}$$ = 2 (x $$\frac{d y}{d x}$$ – y).
Solution:
Given, x = et sin t …………(1)
y = et cos t ………..(2)
Diff. eqn. (1) and (2) both sides w.r.t. t ; we have
$$\frac{d x}{d t}$$ = et cos t + et sin t
= et (cos t + sin t) ………….(3)
and $$\frac{d x}{d t}$$ = et (- sin t) + cos t et
= et (cos t – sin t)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{e^t(\cos t-\sin t)}{e^t(\cos t+\sin t)}$$
= $$\frac{\cos t-\sin t}{\cos t+\sin t}$$ ………….(4)
Diff. eqn. (4) both sides w.r.t. x, we have

Question 14.
Using Rolle’s theorem, show that between any two distinct real roots of the equation ax3 + bx2 + cx + d = 0, there is atleast one real root of the equation 3ax2+ 2bx + c = 0.
Solution:
Let f(x) = ax3 + bx2 + cx + d
Let α and β be the distinct real rootsof the equation
f(x) = 0 and let α < β.
∴ f (α) = 0 = (β)
Since f(x) be a polynomial function and hence f(x) be continuous and differentiable ∀ x ∈ R.
Thus f(x) is continuous in [α, β].
f(x) is derivable in (α, β).
Also f(α) = f(β)
Thus, all the three conditions of Rolle’s theorem are satisfied so 3 atleast one real number y ∈ (α, β)
s.t f’ (γ) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
∴f’(x) = 3ax2 + 2bx + c
Now f’(y) = 0
⇒ 3ay2 + 2by + c = 0
⇒ γ be the root of the equation 3ax2 + 2bx + c = 0
Hence, between any two distinct real roots of the equation f(x)= 0, there is atleast one real root of the equation 3ax2 + 2bx + c = 0.

Question 15.
Discuss the applicability of Rolle’s theorem for the following functions in the indicated intervals :
(i) f(x) = $$\begin{cases}x+3, & x \leq 2 \\ 7-x, & x>2\end{cases}$$ in [- 3, 7]
(ii) f(x) = $$\left\{\begin{array}{cc} x^2+1, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2 \end{array}\right.$$ in [0, 2] (NCERT Exampler)
Solution:
(i) Given, f(x) = $$\begin{cases}x+3, & x \leq 2 \\ 7-x, & x>2\end{cases}$$
$$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ 7 – x
= 7 – 2 = 5
$$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ x + 3
= 2 + 3 = 5
also f(2) = 2 + 3 = 5
∴ $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = f(2)
Thus f(x) is continuous at x = 2.

Differentiability at x = 2:
Lf'(2) = $$\ {Lt}_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}$$
= $$\ {Lt}_{x \rightarrow 2^{-}} \frac{(x+3)-(2+3)}{x-2}$$
= $$\ {Lt}_{x \rightarrow 2^{-}} \frac{x-2}{x-2}$$ = 1

and Rf'(2) = $$\ {Lt}_{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}$$
= $$\ {Lt}_{x \rightarrow 2^{+}} \frac{7-x-(2+3)}{x-2}$$
= $$\ {Lt}_{x \rightarrow 2^{+}} \frac{2-x}{x-2}$$ = – 1
∴ Lf'(2) ≠ Rf'(2)
Hence f(x) is not derivable at x = 2 ∈ (- 3, 7)
∴ f(x) is not derivable in (- 3, 7)
So condition (ii) of Rolle’s theorem is not satisfied.
Hence Rolle’s theorem is not applicable to the function f(x) in [- 3, 7]

(ii) Given f(x) = $$\left\{\begin{array}{cc} x^2+1 ; & 0 \leq x \leq 1 \\ 3-x & ; 1 \end{array}\right.$$ in [0, 2]

Derivability at x = 1:
Rf'(1) = $$\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$$
= $$\ {Lt}_{x \rightarrow 1^{+}} \frac{3-x-(1+1)}{x-1}$$
= $$\ {Lt}_{x \rightarrow 1^{+}} \frac{3-x-2}{x-1}$$
= $$\ {Lt}_{x \rightarrow 1^{+}} \frac{1-x}{x-1}$$ = – 1

Lf'(1) = $$\ {Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$$
= $$\ {Lt}_{x \rightarrow 1^{-}} \frac{\left(x^2+1\right)-(1+1)}{x-1}$$
= $$\ {Lt}_{x \rightarrow 1^{-}} \frac{x^2-1}{x-1}$$
= $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ x + 1
= 1 + 1 = 2
∴ Lf'(1) ≠ Rf'(1)
Thus, f(x) is not derivable at x I E (0, 2)
Hence fis not derivable in (0, 2)
So condition (ii) of RoBe’s theorem is not satisfied.
Hence Rolle’s theorem is not applicable to the function f(x) in [0, 2].

Question 16.
Discuss the applicability of Lagrange’s mean value theorem for the following functions in the indicated intervals.
(i) f(x) = $$\left\{\begin{array}{rll} 2+x^3 & , \text { if } & x \leq 1 \\ 3 x & \text {, if } & x>1 \end{array}\right.$$ on [- 1, 2]
(ii) f(x) = $$\begin{cases}\frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}$$ in [- 1, 1]
(iii) f(x) = $$\left\{\begin{array}{cl} 3 x+2 & , x<2 \\ 14-3 x & , 2 \leq x \end{array}\right.$$ in [- 2, 6]
(iv) f(x) = $$\left\{\begin{array}{cc} x \sin \frac{1}{x}, & x \neq 0 \\ 0 & , x=0 \end{array}\right.$$ in [- 1, 1]
Solution:
(i) Given $$\left\{\begin{array}{rll} 2+x^3 & , \text { if } & x \leq 1 \\ 3 x & \text {, if } & x>1 \end{array}\right.$$ on [- 1, 2]
Clearly f(x) is polynomial in x,
also $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 3x
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 3 (1 + h) = 3
and $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ (2 + x3)
= $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ [2 + (1 – h)3]
= 2 + (1 – 0)3 = 3
∴ $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = f(1)
i.e. f(x) is continuous at x = 1 and hence f(x) is continuous in [- 1, 2]

at x = 1:

∴ Lf’(1) = Rf’(1)
∴ f(x) is differentiable at x = 1
Hencef is derivable on (- 1, 2).
Further f(- 1) = 2 + (- 1)3 = 1
and f(2) = 3 × 2 = 6
Thus all the two conditions of L.M.V. theorem are satisfied
∴ ∃ atleast our real number c ∈ (- 1, 2).
s.t. f'(c) = $$\frac{f(b)-f(a)}{b-a}$$
Also f'(c) = $$\left\{\begin{array}{cc} 3 x^2, & x \leq 1 \\ 3, & x>1 \end{array}\right.$$

When x ≤ 1 :
i.e. 3c2 = $$\frac{6-1}{2-(-1)}=\frac{5}{3}$$
c = $$\pm \frac{\sqrt{5}}{3}$$
but c = – $$\frac{\sqrt{5}}{3}$$ ∈ (- 1, 2)

When x > 1 ; f'(c) = 3
i.e. 3 = $$\frac{6-1}{2-(-1)}$$
⇒ 3 = $$\frac{5}{3}$$ which is impossible
Hence L.M.V. theorem is applicable and c = $$\frac{\sqrt{5}}{3}$$.

(ii) as x → 0, $$\frac{1}{x}$$ → ∞
as x → 0+, $$\frac{1}{x}$$ → ∞
Thus f(x) = $$\frac{1}{x}$$ oscillating between – ∞ to + ∞.
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) does not exist.
Thus f(x) is discontinuous at x = 0 ∈ [- 1, 1]
Hence lagrange mean value theorem is not applicable.

(iii) Given f(x) = $$\begin{cases}3 x+2 & ; \quad x<2 \\ 14-3 x & ; \quad 2 \leq x\end{cases}$$ in [- 2, 6]
L.H. Limit = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ 3x + 2
= 6 + 2 = 8
R.H.Limit = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ (14 – 3x)
= 14 – 6 = 8
and f(2) = 8
∴ f is continuous at x = 2.

Differentiability at x = 2:

∴ Lf'(2) ≠ Rf'(2)
Thus f(x) is not derivable at x = 2 ∈ (- 2, 6)
Hence f(x) is not derivable in (- 2, 6).
Thus lagrange’s mean value theorem is not applicable for f(x) in [- 2, 6].

(iv) f'(0) = $$\ {Lt}_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{x \sin \frac{1}{x}-0}{x-0}$$
= $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ sin $$\frac{1}{x}$$,
which does not exists
[Since sin $$\frac{1}{x}$$ be a real number oscillating between – 1 and 1]
at x = $$\frac{1}{2 n \pi+\frac{\pi}{2}}$$, n ∈ N, n → ∞
⇒ x → 0
∴ sin $$\frac{1}{x}$$ = sin (2nπ + $$\frac{\pi}{2}$$)
= siin $$\frac{\pi}{2}$$ = 1
at x = $$\frac{1}{2 n \pi+\frac{\pi}{2}}$$, n ∈ N, n is large
⇒ x → 0
∴ sin $$\frac{1}{x}$$ = sin (2nπ – $$\frac{\pi}{2}$$)
= sin $$\frac{-\pi}{2}$$ = – 1
Thus f(x)is not derivable at x ∈ (- 1, 1)
Hencef(x) is not derivable in (- 1, 1).
Thus the condition (ii) of lagrange’s mean values is not satisfied.
Thus, the lagrange’s mean value theorem is not applicable for the function f(x) in [- 1, 1].

Question 17.
Using Lagrange’s Mean Value Theorem, show that | sin α – sin β | ≤ |α – β|.
Solution:
Two cases arise.

Case – I:
When α = β
∴ | sin α – sin β | = | sin α – sin α |
= |0| = 0
|α – β| = |α – α|
= |0| = 0
Thus, | sin α – sin β | = |α – β|

Case – II:
When α ≠ β, let α < β
Let f(x) = sin x ………….(1)
which is continuous and differentiable everywhere.
Thus f(x) is continuous in [α, β] and f(x) is derivable in (α, β).
So both the conditions of lagrange’s mean value theorem are satisfied.
Then ∃ atleast one real number c ∈ (α, β)
s.t. f'(c) = $$\frac{f(\beta)-f(\alpha)}{\beta-\alpha}$$ …………..(2)
Diff. eqn. (1) w.r.t. x ; we have
f'(x) = cos x
Now f'(c) = cos c ;
f(β) = sin β ;
f(α) = sin α
∴ from (2) ;
$$\frac{\sin \beta-\sin \alpha}{\beta-\alpha}$$ = cos c
⇒ $$\frac{\sin \beta-\sin \alpha}{\beta-\alpha}$$ = |cos c| ≤ 1
[∵ – 1 ≤ cos x ≤ 1 ∀ x ∈ R]
⇒ |sin α – sin β| ≤ |β – α|
= |- (α – β)|
= |α – β|
[∵ |- x| = |x| ∀ x ∈ R]
So, combining two cases, we have |sin α – sin β| ≤ |α – β|.