ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.13

Effective ML Aggarwal Class 12 Solutions ISC Chapter 5 Continuity and Differentiability Ex 5.13 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.13

Verify Lagrange’s mean value theorem for the following (1 to 4) functions in the given intervals. Also find ‘c’ of this theorem :

Question 1.
(i) f(x) = x² in [2, 4] (NCERT)
(ii) f(x) = x² – 4x – 3 in [1, 4] (NCERT)
(iii) f(x) = x + \(\frac {1}{x}\) in [1, 3]. [ISC 2019]
Solution:
(i) Given f(x) = x² is continuous on [2, 4]
and derivable on (2, 4) as f(x) is polynomial in x which is continuous and differentiable everywhere.
∴ f’(x) = 2x
and f(2) = 4;
f (4) = 16
Thus both the conditions of L.M.V are satisfied.
∴ ∃ atleast one real number c ∈ (2, 4) s.t.
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
⇒ 2c = \(\frac{16-4}{4-2}\)
Hence L.M.V theorem is verified and c = 3.

(ii) Given f(x) = x² – 4x – 3
∴ f(x) is continuous in [1, 4]
[∵ f(x) is polynomial in x]
also f’(x) = 2x – 4 exists ∀ x ∈ (1, 4)
∴ f(x) is derivable in (1, 4).
Now, f( 1) = – 6
and f(4) = 3
∴ Both the conditions of L.M.V. theorem are satiafied.
∴ ∃ atleast one real no. c ∈ (1, 4)
s.t. f'(c) = \(\frac{f(4)-f(1)}{4-1}\)
2c – 4 = \(\frac{-3-(-6)}{4-1}\) = 1
c = \(\frac{5}{2}\) ∈ (1, 4).
∴ L.M.V theorem is verified and c = \(\frac{5}{2}\).

(iii) Given f(x) = x + \(\frac {1}{x}\)
∴ f'(x) = 1 – \(\) which exists in [1, 3]
Thus, f(x) is continuous in [1, 3] and derivable in (1, 3)
Now f(1) = 1 + \(\frac{1}{1}\) = 2 ;
f(3) = 3 + \(\frac{1}{3}\)
= \(\frac{10}{3}\)
Therefore both conditions of Lagrange’s mean value theorem are satisfied so ∃ atleast one real number c ∈ (1, 3) such that
\(\frac{f(3)-f(1)}{3-1}\) = f'(c)
⇒ \(\frac{\frac{10}{3}-2}{2}=1-\frac{1}{c^2}\)
⇒ \(\frac{2}{3}=1-\frac{1}{c^2}\)
⇒ \(\frac{1}{c^2}=1-\frac{2}{3}=\frac{1}{3}\)
⇒ c² = 3
⇒ c = ± √3
But c ∈ (1, 3)
∴ c = √3 ∈ (1, 3)
Thus ∃ c = √3 ∈ (1, 3) s.t.
f'(c) = \(\frac{f(3)-f(c)}{3-1}\)
Hence, Lagrange’s mean value theorem is verified.

Question 1 (old).
(iii) f(x) = x² – 2x + 4 on [1, 5]
Solution:
Given f(x) = x² – 2x + 4
since polynomial function is everywhere continuous and differentiable.
Thus f (x) is continuous on [1, 5] and differentiable on (1, 5).
So both conditions of L.M.V theorem are satisfied.
So ∃ atleast one real number c ∈ (1, 5) such that
f'(c) = \(\frac{f(5)-f(1)}{5-1}\) ……….(1)
since f(x) = x² – 2x + 4
∴ f’(x) = 2x – 2
f(1) = 1 – 2 + 4 = 3 ;
f(5) = 25 – 10 + 4 = 19
∴ from (1) ;
2c – 2 = \(\frac{19-3}{4}\)
= \(\frac{16}{4}\) = 4
Thus c = 3 ∈ (1, 5) s.t.
f’ (c) = \(\frac{f(5)-f(1)}{5-1}\)
Thus, L.M.V Theorem is verified.

Question 2.
(i) f(x) = x³ – 2x² – x + 3 on [0, 1] (NCERT Exampler)
(ii) f(x) = (x – 4) (x – 6) (x – 8) on [4, 8].
Solution:
(i) Given f(x) = x³ – 2x² – x + 3 in [0, 1]
Since f(x) is polynomial in x.
∴ f(x) is continuous in [0, 1] and derivable in (0, 1).
Now f'(x) = 3x² – 4x – 1
and f'(0) = 3
and f(1) = 1 – 2 – 1 + 3 = 1
∴ Both the conditions of L.M.V. theorem are satisfied.
∴ ∃ atleast one real number c ∈ (0, 1) s.t.
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
i.e. 3c² – 4c – 1 = \(\frac{1-3}{1-0}\)
= – 2
⇒ 3c² – 4c + 1 = 0
⇒ c = 1, \(\frac{1}{3}\)
but c = 1 ∉ (0, 1)
∴ c = \(\frac{1}{3}\) ∈ (0, 1)
Hence L.M.V. theorem is applicable and c = \(\frac{1}{3}\).

(ii) Given f(x) = (x – 4) (x – 6) (x – 8) …………..(1)
Clcarly f(x) be a polynomial function hence continuous and differentiable eveiy here.
∴ f(x) is continuous in [4, 8] and derivable in (4, 8).
Thus, both conditions of Lagrange’s mean value Theorem are satisfied.
So ∃ exists at least one real number c ∈ (4, 8)
s.t f'(c) = \(\frac{f(8)-f(4)}{8-4}\) …………(2)
Duff. (1) both sides w.r.t. x, we have
Since f(x) = x³ – 18x² + 104x – 192
f’(x) = 3x² – 36x + 104
also, f(8) = 0; f(4) = 0
∴ from (2) ;
3c² – 36c + 104 = \(\frac{0-0}{4}\) = 0
⇒ c = \(\frac{36 \pm \sqrt{1296-1248}}{6}\)
= \(\frac{36 \pm \sqrt{48}}{6}\)
⇒ c = \(\frac{36 \pm 4 \sqrt{3}}{6}\)
= 6 ± \(\frac{2}{3}\) √3.

Question 3.
(i) f(x) = sin x in [0, \(\frac{\pi}{2}\)]
(ii) f(x) = x – 2 sin x in [- π, π]
Solution:
(i) Given f(x) = sin x …………….. (1)
Clearly f (x) be continuous and differentiable in its domain.
∴ f be continuous in [0, \(\frac{\pi}{2}\)] and derivable in (o, \(\frac{\pi}{2}\)).
Thus both conditions of Lagrange’s mean value theorem are satisfied for function f in [0, \(\frac{\pi}{2}\)]
Then ∃ atleast one real number c ∈ (0, \(\frac{\pi}{2}\))
s.t. f'(c) = \(\frac{f\left(\frac{\pi}{2}\right)-f(0)}{\frac{\pi}{2}-0}\) ………..(2)
Hence f'(x) = cos x ;
f(\(\frac{\pi}{2}\)) = 1 ;
f(0) = 0
∴ from (2) ;
cos c = \(\frac{1-0}{\frac{\pi}{2}-0}=\frac{2}{\pi}\)
⇒ c = cos^-1 \(\left(\frac{2}{\pi}\right)\)
So ∃ a real number c = cos^-1 \(\frac{2}{\pi}\) ∈ (0, \(\frac{\pi}{2}\))
s.t. f'(c) = \(\frac{f\left(\frac{\pi}{2}\right)-f(0)}{\frac{\pi}{2}-0}\)
Hence lagrange’s mean value Theorem is verified
and c = cos^-1 \(\left(\frac{2}{\pi}\right)\)
[When 0 < c < \(\frac{\pi}{2}\)
⇒ 0 < cos c < 1
⇒ 0 < \(\frac{2}{\pi}\) < 1, which is true]

(ii) Given f(x) = x – 2 sin x ……………(1)
Clearly polynomial function and trigonometric function are continuous and derivable everywhere.
Thus f(x) is continuous in [- π, π] and derivable on (- π, π).
So both conditions of lagrange’s mean value are satisfied.
So ∃ atleast one real number e E (- π, π).
s.t. f'(c) = \(\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}\) ……….(2)
Diff. w.r.t. x ; we get
f'(x) = 1 – 2 cos x ;
f(π) = π – 2 sin π
= π – 0 = π
f(- π) = – π – 2 sin (- π) = – π
∴ from (2) ;
1 – 2 cos c = \(\frac{\pi-(-\pi)}{\pi-(-\pi)}\) = 1
⇒ – 2 cos c = 0
⇒ cos c = 0
⇒ c = (2n + 1) \(\frac{\pi}{2}\) ∀ n ∈ I
But c ∈ (- π, π)
∴ c = \(\frac{\pi}{2}\), – \(\frac{\pi}{2}\)
Thus ∃ two real numbers \(\frac{\pi}{2}\) and – \(\frac{\pi}{2}\)
s.t. f'(\(\frac{\pi}{2}\)) = \(\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}\)
and f'(- \(\frac{\pi}{2}\)) = \(\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}\)
∴ Lagrange mean value theorem is verified and c = ± \(\frac{\pi}{2}\).

Question 4.
(i) f(x) = 2 sin x + sin 2x in [0, π]
(ii) f(x) = sin x – sin 2x on [0, π]
Solution:
(i) Given f(x) = 2 sin x + sin 2x
Since every trigonometric function is continuous and differentiable in its domain.
The sum of two continuous and differentiable functions ¡s continuous and differentiable.
(i) f(x) is continuous in [0, π]
(ii) f(x) is duff. in (0, π)
Thus both conditions of Lagrange’s mean value theorem are satisfied
so ∃ atleast one real no. c ∈ (0, π)
s.t \(\frac{f(\pi)-f(0)}{\pi-0}\) = f’(c) ………….(1)
Here f(π) = 2 sin π + sin 2π = 0
and f(0) = 0 + 0 = 0
and f’(x) = 2 cos x + 2 cos 2x
∴ from (1) ;
\(\frac{0}{\pi}\) = 2 (cos c + cos 2c)
⇒ cos 2c + cos c = 0
⇒ 2 cos \(\frac{3 c}{2}\) cos \(\frac{c}{2}\) = 0
⇒ cos \(\frac{3 c}{2}\) = 0 or
cos \(\frac{c}{2}\) = 0
⇒ \(\frac{3 c}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \ldots\)
or \(\frac{c}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \ldots\)
⇒ c = \(\frac{\pi}{3}, \pi, \frac{5 \pi}{3}, \ldots\)
or c = π, 3π, 5π, ……………
But c ∈ (0, π)
∴ c = \(\frac{\pi}{3}\)
Thus there exists atleast one c = \(\frac{\pi}{3}\) ∈ (0, π)
s.t. f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)

(ii) f(x) = sin x – sin 2x on [0, π]
Since sine function is continuous and derivable everywhere
∴ f is continuous in [0, 2π] and derivable in (0, 2π)
∴ Both the conditions of L.M.V are satisfied.
∴ ∃ atleast one real no c ∈ (0, 2π) s.t.
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
Now f’(c) = cos x – 2 cos 2x ;
f(x) = 0 ;
f(2π) = 0
i.e. cos c – 2 cos 2c = \(\frac{f(2 \pi)-f(0)}{2 \pi-0}\)
i.e. cos c – 2 cos 2c = \(\frac{0-0}{2 \pi}\) = 0
⇒ cos c – 2 cos 2c = 0
⇒ cos c – 2 (2 cos² c – 1) = 0
⇒ 4 cos² c – cos c – 2 = 0
∴ cos c = \(\frac{1 \pm \sqrt{33}}{8}\)
∴ c = cos^-1 \(\left(\frac{1 \pm \sqrt{33}}{8}\right)\) ∈ [0, 2π]
[Since 0 < c < 2π
⇒ – 1 < cos c < 1]
Thus L.M.V. is applicable and
c = cos^-1 \(\left(\frac{1 \pm \sqrt{33}}{8}\right)\)

Question 5.
(i) f(x) = px² + qx + r, p ≠ 0, on [0, 1]
(ii) f(x) = x on [a, b]
(iii) f(x) = (x – 1)^2/3 on [1, 2].
Solution:
(i) Given f(x) = px² + qx + r, p ≠ 0
Sincef(x) be a polynomial function. So it
is continuous and derivable everywhere.
Thus f(x) is continuous in [0, 1] and derivable in (0, 1).
Therefore, both conditions of lagrange’s mean value Theorem are satisfied.
So ∃ atleast one real number c ∈ (0, 1)
s.t. f'(c) = \(\frac{f(1)-f(0)}{1-0}\) ………..(2)
Diff. (1) w.r.t. x, we have
f'(x) = 2px + q ;
f(1) = p + q + r ;
f(0) = r
∴ from (2) ; we have
2pc + q = \(\frac{p+q+r-r}{1}\)
= p + q
⇒ 2pc = p
⇒ c = \(\frac{1}{2}\) ∈ (0, 1) [∵ p ≠ 0]
Thus, ∃ a real number c = \(\frac{1}{2}\) ∈ (0, 1)
such that f'(\(\frac{1}{2}\)) = \(\frac{f(1)-f(0)}{1-0}\)
Hence lagrange’s mean value theorem is verified and c = \(\frac{1}{2}\) .

(ii) Given f(x) = x on [a, b]
Clearly f(x) is polynomial in x.
∴ f is continuous n [a, b] and derivable in (a, b)
s.t. f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
⇒ 1 = \(\frac{b-a}{b-a}\)
⇒ 1 = 1 which is true.
Hence L.M.V. is applicable and ∃ one real number c ∈ (a, b) s.t.
f(c) = \(\frac{f(b)-f(a)}{b-a}\)

(iii) f(x) = (x – 1)^2/3
∴ f'(x) = \(\frac{2}{3(x-1)^{1 / 3}}\)
which does not exist at x = 1 and 1 ∉ (1, 2)
∴ f is derivable in (1, 2) and hence continuous in [1, 2]
also f(1) = 0
and f(2) = 1
∴ Both the conditions of L.M.V. theorem are satisfied.
∴ ∃ atleast one real number c ∈ (1, 2)
s.t. f'(c) = \(\frac{f(2)-f(1)}{2-1}\)
\(\frac{2}{3(c-1)^{1 / 3}}=\frac{1-0}{1}\)
= 1
⇒ (c – 1)^2/3 = \(\frac{2}{3}\)
⇒ c – 1 = \(\frac{8}{27}\)
⇒ c = \(\frac{35}{27}\) ∈ (1, 2)
Thus L.M.V. theorem ia applicable and c = \(\frac{35}{27}\).

Question 6.
Show that the function f(x) = x² – 6x + 1 on [1, 3] satisfies Lagrange’s mean value theorem. Also find the coordinates of a point at which the tangent to the curve represented by the above function is parallel to the chord joining A (1, – 4) and B (3, – 8). (ISC 2004)
(ii) Use Lagrange’s mean valuc theorem to deterniine a point P on the curve y = \(\sqrt{x-2}\) defined in the interval [2, 3] where the tangent ¡s parallel to the chord joining the end points on the curve. (ISC 2008)
Solution:
(i) Given y = f(x)
= x² – 6x + 1
Here we discuss the applicability of L.M.V theorem in [1, 3]
Since f(x) is polynomial in x
∴ it ¡s continuous in [1, 3]
also, f’ (x) = 2x – 6 which is exists ∀ x ∈ (1, 3)
∴ f(x) is derivable in (1, 3)
Now,
f(1) = 1- 6 + 1 = – 4
and f(3) = 9 – 18 + 1 = – 8
Thus all the conditions of L.M.V theorem are satisfied
∴ ∃ atleast one real number c ∈ (1, 3)
s.t. f'(c) = \(\frac{f(3)-f(1)}{3-1}\)
⇒ 2c – 6 = \(\frac{-8-(-4)}{2}\)
= – 2
⇒ 2c = 4
⇒ c = 2 ∈ (1, 3)
i.e.when x = 2
then y = 4 – 12 + 1 = – 7
Hence the required point is (2, – 7).
Thus, there exists a point (2, – 7) on the given curve y = – 6x + 1 where the tangent ¡s parallel to the chord joining the points (1, – 4) and (3, – 8).

(ii) Given y = f(x) = \(\sqrt{x-2}\) ………….(1)
Here we use the lagrange’s mean value theorem for [2, 3]
∴ f'(x) = \(\frac{1}{2 \sqrt{x-2}}\)
and f(2) = 0
and f(3) = 1
Clearly f(x) exists ∀ x ∈ (2, 3)
i.e. f(x) is desirable in (2, 3) and hence continuous in [2, 3].
∴ by L.M.V theorem ∃ atleast one real number x ∈ (2, 3).
s.t. f'(x) = \(\frac{f(3)-f(2)}{3-2}\)
⇒ \(\frac{1}{2 \sqrt{x-2}}=\frac{1-0}{1}\) = 1
⇒ 2 \(\sqrt{x-2}\) = 1
⇒ (x- 2) = \(\frac{1}{4}\)
⇒ x = \(\frac{9}{4}\) ∈ (2, 3)
∴ From (1) ;
y = \(\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
Thus, ∃ a point on the given curve where the tangent is parallel to the chord joining the points on the curve.

Question 7.
What can you say about the applicability of Lagrange’s mean value theorem for the following functions in the indicated intervals ?
(i) f(x) = x^1/3 in [- 1, 2]
(ii) f(x) = |x| in [- 2, 3]
(iii) f(x) = 3 – (2 – x)^2/3 in [0, 3]
Solution:
(i) Given f(x) = x^1/3
∴ f’(x) = \(\frac{1}{3} \frac{1}{x^{2 / 3}}\) does not exists at x = 0 ∈ (- 1, 1)
:. f(x) is not derivable at (- 1, 1)
Hence L.M.V theorem is not applicable.

(ii) Given f(x) = |x|
Clearly f(x) be continuous at x = 0
∴ f be continuous in [- 2, 3].
Thus, derivative of f(x) does not exists at x = 0 ∈ (- 2, 3).
Hence f(x) is not derivable on (- 2, 3).
Hence condition (ii) of lagrange’s mean value is not satisfied byf in [- 2, 3]
Thus, lagrange mean value theorem is not applicable to function fin [- 2, 3].

(ii) Given f(x) = 3 – (2 – x)^2/3 ; x ∈ [0, 3]
∴ f'(x) = – \(\frac{2}{3}\) (2 – x)^-1/3 (- 1)
= \(\frac{2}{3(2-x)^{1 / 3}}\), x ≠ 2
Thus, deerivative of f(x) does not exist at x = 2 ∈ (0, 3)
∴ f is not derivable at x = 2 ∈ (0, 3)
Thus f(x) is not derivable in (0, 3).
Hence lagrange’s mean value theorem is not applicable to function f in [0, 3].