OP Malhotra Class 12 Maths Solutions Chapter 10 Mean Value Theorems Ex 10(b)

Utilizing OP Malhotra Maths Class 12 Solutions Chapter 10 Mean Value Theorems Ex 10(b) as a study aid can enhance exam preparation.

S Chand Class 12 ICSE Maths Solutions Chapter 10 Mean Value Theorems Ex 10(b)

Question 1.
(i) f(x) = x(x – 2) in [1, 2]
(ii) f(x) = x² – 2x + 4 in [1, 5].
Solution:
(i) Given f(x) = x (x – 2) in [1, 2]
since f(x) be polynomial in x and hence continuous on [1, 2]
Now f'(x) = 2x – 2 which exists for all x ∈ (1, 2)
∴ f(x) be derivable on (1, 2)
Thus by Lagrange’s mean value theorem ∃ atleast one real number c∈(l, 2)
s.t. \(\frac{f(2)-f(1)}{2-1}\) = f(c)
⇒ f(2) – f(1) = f(c) … (i)
Now f(1) = 1 (1 -2) = – 1;
f(2) = 0 & f(x) = 2x – 2
∴ from (i); 0 – (-1) = 2c -2
⇒ 2c = 3 ⇒ c = \(\frac { 3 }{ 2 }\) ∈(1, 2)
∴ L.M.V. is verified.

(ii) 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
⇒ c = 3
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.
f(x) = x² + x – 1 in the interval [0, 4].
Solution:
Given f (x) = x² + x – 1
since f (x) be polynomial function in x and hence everywhere continuous and differentiable. Thus f (x) is continuous on [0, 4] and differentiable on (0, 4).
Hence both conditions of L.M.V theorem are satisfied. So ∃ atleast one real number c ∈ (0, 4) such that f’ (c) = \(\frac{f(4)-f(0)}{4-0}\) …(1)
We have f (x) = x² + x – 1
∴ f’ (x) = 2x + 1
f(0) = 0 + 0 – 1 = – 1 ;
f(4) = 16 + 4- 1 = 19
∴ from (1) ; 2c + 1 = \(\frac{19-(-1)}{4-0}\) = 5
⇒ 2c = 4 ⇒ c = 2 ∈ (0, 4)
s.t. f'(c) = \(\frac{f(4)-f(0)}{4-0}\)
Hence L.M.V theorem is verified.

Question 3.
f(x) = 2x² – 10x + 29 in [2, 7].
Solution:
Given fix) = 2x² – 10x + 29 in [2, 7]
since f(x) is polynomial in x and hence continuous on [2, 7]
Now f'(x) = 4x – 10 which exists for all x ∈ (2, 7)
Here, f(2) = 8 – 20 + 29 = 17;
f(7) = 98 – 70 + 29 = 57
Thus both conditions of L.M.V. are satisfied so 3 atleast one real no. c ∈ (2, 7)
s.t. \(\frac{f(7)-f(2)}{7-2}\) = f'(c)
⇒ \(\frac{57-17}{7-2}\) = 4c – 10 ⇒ \(\frac { 40 }{ 5 }\) = 4c – 10
⇒ 4c = 18 ⇒ c = \(\frac { 9 }{ 2 }\) ∈ (2, 7)
Thus, L.M.V. is verified.

Question 4.
f(x) = x – \(\frac { 1 }{ x }\); x ∈ [1, 3].
Solution:
Given f(x) = x – \(\frac { 1 }{ x }\); x ∈ [1, 3]
∴ f(x) = \(\frac{x^2-1}{x}\) which is a rational function and hence continuous on [1, 3] since x ≠ 0.
Now f'(x) = 1 + \(\frac { 1 }{ x² }\) which exists for all x∈(1, 3)
Thus, f(x) is derivable on (1, 3).
Now f(3) = 3 – \(\frac { 1 }{ 3 }\) = \(\frac { 8 }{ 3 }\); f(1) = 1 – \(\frac { 1 }{ 1 }\) = 0
Thus both conditions of L.M.V. are satisfied so ∃ atleast one real number c ∈ (1, 3) such that

Thus L.M.V. is verified.

Question 5.
f(x) = (x – 1)(x – 2)(x – 3) in [1, 4].
Solution:
Given f(x) = (x – 1) (x – 2) (x – 3) in [1, 2]
⇒ f(x) = (x + 1) (x² – 5x + 6)
= x³ – 6x² + 11x – 6
Which is polynomial in x and hence continuous on [1, 2]
further f'(x) = 3x² – 12x + 11
which exists for all x∈(1, 4)
∴ f(x) is derivable on (1, 4).
further, f(4) = 3.2.1. = 6 & f(1) = 0
so both conditions of L.M.V. theorem are satisfied so ∃atleast one real no c∈ (1, 4)

Clearly c = 3 ∈(1, 4) [Here c = 1 ∉ (1, 4)
Thus, L.M.V. Theorem is verified.

Question 6.
f(x) = logx in [1, e].
Solution:
Given f(x) = logx in [i.e.]
Clearly lagarithmic function is continuous in its domain.
∴ f(x) is continuous on [1, e].
Now, f'(x) = \(\frac { 1 }{ x }\) which is exists for all x ∈ (1, e)
∴ f(x) is derivable on (1, e.).
Further, f(e) = loge = 1;
f(l) = log 1 = 0
Thus, both conditions of L.M.V. theorem are satisfied so ∃ atleast one real number c∈(1, e)
s.t. \(\frac{f(e)-f(1)}{e-1}\) = f'(c)
⇒ \(\frac{1-0}{e-1}=\frac{1}{c}\) ⇒ c = e – 1 ∈(1, e.) [∵ e > 2 ⇒ e – 1 > 1]
Thus Lagrange’s mean value theorem is verified.

Question 7.
f(x) = e^x in [0, 1].
Solution:
Given f(x) = e^x in [0, 1]
Clearly exponential function is continuous and differentiable every where in R.
Thus f (x) is continuous on [0, 1] and derivable on (0, 1).
Now f'(x) = e^x ; f(1) = e;
f(0) = e⁰ = 1
Thus both conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real no. c∈(0, 1)
s.t. \(\frac{f(1)-f(0)}{1-0}\) = f'(c)
⇒ \(\frac{e-1}{1}\) = e^c ⇒ c = log (e – 1)∈(0, 1)
[∵ 1 < e – 1 < e ⇒ 0 < log (e – 1) < 1]
Thus L.M.V. theorem is verified.

Question 8.
f(x) = [x] in [-1, 1],
Solution:
Given f (x) = [x] in [-1, 1]
Clearly, we know that inegral part function is discontinous for all integral values.
∴ f(x) is discontinous at x = – 1, 0, 1
Thus f is not continuous on [-1, 1]
Therefore L.M.V. theorem is not applicable.

Question 9.
(i) Find c so that f'(c) = \(\frac{f(6)-f(4)}{6-4}\), where f(x) = \(\sqrt{x+2}\) and c ∈ (4, 6).
(ii) Given f(x) = x^3/2 , find the value of c ∈ (0,1) such that f'(c) = \(\frac{f(1)-f(0)}{1-0}\).
Solution:

Question 10.
(i) Find a point on the graph of y = x³, where tangent is parallel to the chord joining (1, 1) and (3, 27).
(ii) Use L.M.V. to determine the point on the curve y = x³ – 3x, where the tangent to the curve is parallel to the chord joining (1, -2) and (2, 2).
Solution:
(i) Given y = f(x) = x³
Since f(x) be polynomial in x so it continuous in [1, 3].
Also f'(x) = 3x² which is exists for all x∈(1, 3)
∴ f(x) is derivable on (1, 3).
Now f(1) = 1; f(3) = 3³ = 27
Thus both conditions of Lagrange’s mean value theorem are satisfied so ∃ atleast one real no. c∈(1, 3)

Thus, there exists a point \(\left(\sqrt{\frac{13}{3}}, \frac{13}{3} \sqrt{\frac{13}{3}}\right)\) on given curve where the tangent is parallel to the chord joining the points (1, 1) and (3, 27).

(ii) Let y = f (x) = x³ – 3x
Here we discuss the applicability of L.M.V. theorem in [1, 2].
Since f (x) is polynomial in x
∴ f (x) is continuous in [1, 2]
Further, f(x) = 3x² – 3 exists ∀ x ∈ (1, 2).
∴ f(x) is derivable in (1, 2)
and f(1) = 1 – 3 = – 2 and f(2) = 8 – 6 = 2
Thus all the two conditions of L.M.V. theorem are satisfied.
∴ ∃ atleast one real number c ∈ (1,2) s.t.

at which tangents is parallel to the chord joining (1, -2) and (2, 2) by using geometrical interpretation of L.M.V. Theorem.

Question 11.
Explain why Lagrange’s Mean Value theorem is not applicable to the following functions:
(i) f(x) = \(\left\{\begin{array}{c}
|x| \text { if } x \neq 0 \\
0 \text { if } x=0
\end{array}, x \in[-1,1]\right.\)
(ii) f(x) = |x| in [- 1, 2]
Solution:
(i) Given f(x) = \(\left\{\begin{array}{c}
|x| \text { if } x \neq 0 \\
0 \text { if } x=0
\end{array}, x \in[-1,1]\right.\)
Lf'(0) = \(\underset{x \rightarrow 0^{-}}{\mathbf{L t}}\)\(\frac{f(x)-f(0)}{x-0}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\)\(\frac{-x-0}{x-0}\)
& Rf'(0) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\frac{x-0}{x-0}\) = 1
∴ Lf'(0) ≠ Rf'(0)
Thus f is not differentiable at x = 0 ∈ (-1, 1)
Thus f is not derivable on (-1, 1).
∴ L.M.V. theorem is not applicable.

(ii) Given f (x) = | x | in [- 1, 1]

∴ f'(x) does not exists at x = 0 ∈ (-1, 1)
Hence f is not derivable in (-1, 1)
∴ L.M.V. theorem is not applicable.

Question 12.
f(x) = x(1 – logx); x > 0, show that (a-b)logc = b(1-logb)-a(1-loga); 0 < a < b.
Solution:
Given f(x) = x( 1 – log x) ; x > 0 in [a, b] where 0 < a<b.
Since polynomial function is continuous & differentiable everywhere while logarithmic function is continuous & differentiable in its domain.
Thus f(x) is continuous in [a, b] & differentiable for all x > 0 and in [a, b] where 0 < a < b
Thus both conditions of L.M.V. theorem are satisfied so ∃ atleast one real no c ∈ (a, b)

Example

Question 1.
Find ‘c’ of the Lagrange’s Mean Value Theorem,
if f(x) = x² – 3x – 1 ∈ \(\left[\frac{-11}{7}, \frac{13}{7}\right]\)
Solution:
Given f(x) = x² – 3x – 1 ∈ \(\left[\frac{-11}{7}, \frac{13}{7}\right]\)
since f(x) be polynomial in x and hence continuous in \(\left[\frac{-11}{7}, \frac{13}{7}\right]\)0
∴ f be derivable on \(\left(\frac{-11}{7}, \frac{13}{7}\right)\)

Question 2.
Verify Rolle’s Theorem for the function
f(x) = log\(\frac{x^2+a b}{x(a+b)}\), x ∈ [a, b]
Solution:
Given f(x) = log\(\left(\frac{x^2+a b}{x(a+b)}\right)\)
= log (x² + ab) – log (a + b) – log x
Since log x is continuous for all x > 0
∴ f(x) is continuous in [a, b]
Now f(x) = \(\frac{2 x}{x^2+a b}-\frac{1}{x}=\frac{2 x^2-x^2-a b}{x\left(x^2+a b\right)}\)
= \(\frac{x^2-a b}{x\left(x^2+a b\right)}\)
which exists for all x ∈ (a, b)
∴ f(x) is derviable on (a, b).
Further, f(a) = log\(\left(\frac{a^2+a b}{a(a+b)}\right)\) – log1 = 0
& f(b) = log\(\left(\frac{b^2+a b}{b(a+b)}\right)\) = log1 = 0
∴ f(a) = f(b)
Thus all the three conditions of Rolle’s theorem are satisfied so, ∃ atleast one real number c∈(a, b) s.t. f(c) = 0
\(\frac{c^2-a b}{c\left(c^2+a b\right)}\) ⇒ c² – ab = 0
⇒ c = ± \(\sqrt{a b}\)
but c = – \(\sqrt{a b}\) ∉ (a, b)
since GM. of a, b .i.e. \(\sqrt{a b}\) lies between a & b
c = \(\sqrt{a b}\) ∈ (a, b)
Thus Rolle’s theorem is satisfied.

Question 3.
Examine the validity and conclusion of Lagrange’s Mean Value Theorem for the function f(x) = x(x – 1)(x – 2) for every x ∈ \(\left[0, \frac{1}{2}\right]\).
Solution:
Given f(x) = x (x – 1) (x – 2)
= x³ – 3x² + 2x in \(\left[0, \frac{1}{2}\right]\)
Since f(x) be polynomial in x and hence continuous in \(\left[0, \frac{1}{2}\right]\)
Further f'(x) = 3x² – 6x + 2, which exists for all x ∈\(\left(0, \frac{1}{2}\right)\)
∴ f is derivable on \(\left(0, \frac{1}{2}\right)\)
Thus both conditions of Lagrange’s mean value theorem are satisfied so ∃ atleast are real number c ∈ \(\left(0, \frac{1}{2}\right)\) s.t.

Thus L.M.V. is satisfied.

Question 4.
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).
Solution:
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 is 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 = x² – 6x + 1 where the tangent is parallel to the chord joining the points (1, – 4) and (3, – 8).

Question 5.
Examine the validity and conclusion of Rolle’s Theorem for the f(x) = e^x sin x in x ∈[0, π].
Solution:
Given f (x) = e^x sin x
since, e^x and sin x are differentiable and continuous everywhere. Therefore, product of two functions i.e. f (x) = e^x sin x is continuous on [0, π] and differentiable on (0, π).
Now f(0) = 0 ; f(π) = e^π x 0 = 0
∴ f(0) = f(π)
So, all the three conditions of Rolle’s theorem are satisfied. Now we want to show that, ∃ atleast one real number c ∈ (0, π) such that f'(c) = 0.
We have f(x) = e^x sin x
∴ f (x) = e^x cos x + sin x e^x = e^x (cos x + sin x)
Now f’ (c) = 0 ⇒ (e^c cos c + sin c) = 0
⇒ cos c + sin c = 0 [∵ e^c > 0]
⇒ tan c = – 1 = tan \(\left(\frac{3 \pi}{4}\right)\)
⇒ c = \(\frac { 3π }{ 4 }\) ∈(0, π) such that f’ (c) = 0
Hence Rolle’s theorem verified.

Question 6.
Verify Rolle’s theorem for the function
f(x) = e^2x(sin 2x – cos 2x) in \(\left[\frac{\pi}{8}, \frac{5 \pi}{8}\right]\).
Solution:
Given f (x) = e^-x sin x
Clearly e^x is continuous and derivable everywhere and sinx is also continuous and differentiable in its domain.
Also product of two continuous functions is continuous.
Thus f (x) be continuous in [0, π]
and f(x) be derivable in (0, π).
Also f(0) = e^-0 sin 0 = 0 ;
f(π) = e^-π sin π = e^-π x 0 = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied. So ∃ atleast one real number c ∈ (0, π) such that f’ (c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
f’ (x) = e^-x cos x + sin x e^-x (- 1)
⇒ f’ (x) = e^-x (cos x – sin x)
Now f’ (c) = 0 ⇒ e^-c (cos x – sin c) = 0
⇒ cos c – sin c = 0 (∵e^-c > 0)
⇒ tan c = 1 = tan \(\frac { π }{ 4 }\)
⇒ c = nπ + \(\frac { π }{ 4 }\) ∀ n ∈ I
but c ∈ (0, π) ∴ c = \(\frac { π }{ 4 }\)
Thus there exists \(\frac { π }{ 4 }\) ∈ (0, π) s.t f’ (c) = 0
i.e. f'(\(\frac { π }{ 4 }\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac { π }{ 4 }\)

Question 7.
Verify Lagrange’s Mean Value Theorem for the function
f(x) = 3x² – 5x + 1 defined in the interval [2, 5].
Solution:
Given f(x) = 3x² – 5x + 1 in [2, 5]
Since f(x) is a polynomial in x and hence continuous in [2, 5]
Further, f(x) = 6x – 5 which exists for all x∈(2, 5)
∴ f'(x) is derivable on (2, 5)
Also, f(5) = 75 – 25 + 1 = 51 ;
f(2) = 12 – 10 + 1 = 3
Thus both conditions of L.M.V. theorem are satisfied so ∃ atleast one real no. c∈(2, 5)
S.t. \(\frac{f(5)-f(2)}{5-2}\) = f'(c)
⇒ \(\frac{51-3}{5-2}\) = 6c – 5 ⇒ \(\frac { 48 }{ 3 }\) = 6c – 5
⇒ 21 = 6c ⇒ c = \(\frac { 7 }{ 2 }\)
∴ c = \(\frac { 7 }{ 2 }\)∈(2, 5)
Thus, L.M.V. theorem is satisfied.

Question 8.
Use Lagrange’s Mean Value Theorem to determine a point P on the curve y = \(\sqrt{x-2}\) defined in the interval [2, 3] where the tangent is parallel to the chord joining the end points on the curve.
Solution:
Given f(x) = e^x cos x in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Since cosine and exponential function are continuous everywhere.
Also product of two continuous functions is continuous.and
∴ f (x) is continuous in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and derivable in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Also f’ (x) = e^x cos x – e^x sin x
Also, f(-\(\frac { π }{ 2 }\)) = 0 = f(\(\frac { π }{ 2 }\))
Therefore all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast are real number c ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
s.t. f'(c) = 0, i.e. e^c (cos c – sin c) = 0
⇒ cos c – sin c = 0 [ ∵ e^c ≠ 0]
⇒ tan c = 1 ⇒ c = \(\frac { π }{ 4 }\) ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
∴ Rolle’s theorem is verified.

Question 9.
It is given that Rolle’s Theorem holds good for the function
f(x) = x³ + ax² + bx, x ∈ [1, 2] at the point x = \(\frac { 4 }{ 3 }\). Find the values of a and b.
Solution:
It is given that Rolle’s theorem holds for the function f (x) = x³ + bx² + cx defined on [1, 2] with c = 4/3.
∴ f(1) = f(2) ⇒ 1 + b + c = 23 + 4b + 2c
⇒ 3b + c = – 7 …(1)
and f’ (c) = 0 ⇒ f’ (4/3) = 0 …(2)
since f (x) = x³ + bx² + cx
⇒ f’ (x) = 3x² + 2bx + c
∴ from (2); 3 x \(\frac { 16 }{ 9 }\) + \(\frac { 8 }{ 3 }\)b + c = o
⇒ 8b + 3c = – 16 …(3)
On solving (1) and (3); we get
b = – 5 and c = 8.

Question 10.
Using Rolle’s Theorem, find a point on the curve sin x + cos x – 1 = 0, x ∈ \(\left[0, \frac{\pi}{2}\right]\) where the tangent is parallel to the x-axis.
Solution:
Let f(x) = sinx + cosx – 1, x ∈ \(\left[0, \frac{\pi}{2}\right]\)
Since sine & cosine function is differentiable in its domain. Sum of two differentiable function is differentiable. Further f'(x) = cosx + sinx which exists for x ∈ R
∴ f(x) is derivable in \(\left(0, \frac{\pi}{2}\right)\) & hence continuous in \(\left[0, \frac{\pi}{2}\right]\).
Also f(0) = 0 + 1 – 1 = 0;
& f\(\frac { π }{ 2 }\) = 1 + 0 – 1 = 0;
∴ f(0) = f(\(\frac { π }{ 2 }\))
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast are real number c ∈ \(\left(0, \frac{\pi}{2}\right)\)
s.t. f(c) = 0 ⇒ cosc – sin c = 0
⇒ tan c = 1 ⇒ c = \(\frac { π }{ 4 }\) ⇒ c = \(\frac { π }{ 4 }\) ∈ (0, \(\frac { π }{ 2 }\))
So, ∃ c = \(\frac { π }{ 4 }\) ∈ (0, \(\frac { π }{ 2 }\)) s.t. f'(\(\frac { π }{ 4 }\)) = 0
i.e., the tangent to a curve is parallel x-axis.
When x = \(\frac { π }{ 4 }\)
∴ y = sin \(\frac { π }{ 4 }\) + cos \(\frac { π }{ 4 }\) – 1
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1=\sqrt{2}\) – 1
Hence there exists a point \(\left(\frac{\pi}{4}, \sqrt{2}-1\right)\) on the curve where tangent is parallel to x-axis.

Question 11.
Verify Lagrange’s mean value theorem for the function f (x) = sin x – sin 2x in the interval [0, π].
Solution:
f (x) = sin x – sin 2x on [0, 2π]
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. f.
∴∃ 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 = \(\frac{0-0}{2 \pi}/latex] = 0
⇒ cos c – 2 cos 2c = 0
⇒ cos c – 2 (2 cos² c – 1) = 0
⇒ 4 cos² c – cos c – 2 = 0
∴ cos c = [latex]\frac{1 \pm \sqrt{33}}{8}\)
∴ c = cos^-1\(\left(\frac{1 \pm \sqrt{33}}{8}\right) \in[0,2 \pi]\)
[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 12.
Verify Rolle’s theorem for
f(x) = log[(x² + ab) /(a + b)x] in [a b]; 0 ∉ [a, b].
Solution:
Given f(x) = log\(\left(\frac{x^2+a b}{x(a+b)}\right)\)
= log (x² + ab) – log (a + b) – log x
Since log x is continuous for all x > 0
∴ f(x) is continuous in [a, b]
Thus f'(x) = \(\frac{2 x}{x^2+a b}-\frac{1}{x}=\frac{2 x^2-x^2-a b}{x\left(x^2+a b\right)}\)
= \(\frac{x^2-a b}{x\left(x^2+a b\right)}\)
which exists for all x ∈ (a, b)
∴ f(x) is derviable on (a, b).
Further, f(a) = log\(\left(\frac{a^2+a b}{a(a+b)}\right)\) = log1 = 0
f(b) = log\(\left(\frac{b^2+a b}{b(a+b)}\right)\) = log1 = 0
∴ f(a) = f(b)
Thus all the three conditions of Rolle’s theorem are satisfied so, ∃ atleast one real number c ∈ (a, b) s.t. f ’(c) = 0
\(\frac{c^2-a b}{c\left(c^2+a b\right)}\) = 0 ⇒ c² – ab = 0
⇒ c = ± \(\sqrt{ab}\)
but c = – \(\sqrt{ab}\) ∉ (a, b)
since GM. of a, b .i.e. \(\sqrt{ab}\) lies between a & b.
∴ c = \(\sqrt{ab}\) ∈ (a, b)
Thus Rolle’s theorem is satisfied.

Question 13.
Verify Lagranges’ Mean Value theorem for the function f(x) = \(\sqrt{x^2-x}\) in the interval [1, 4].
Solution:
Given f(x) = \(\sqrt{x^2-x}\)
Since f(x) is defined for x ∈ [1, 4]
∴ f(x) is continuous in [1, 4]
Further, f'(x) = \(\frac { 1 }{ 2 }\)(x² – x)^{\(\frac { -1 }{ 2 }\)} (2x – 1)
= \(\frac{2 x-1}{2 \sqrt{x^2-x}}\)
Which exists for all x ∈ (1, 4)
∴ f(x) is derivable on (1, 4)
Thus both conditions of L.M.V. theorem are satisfied so ∃ atleast one real number c ∈(1, 4)
s.t. \(\frac{f(4)-f(1)}{4-1}=f^{\prime}(c)\)
⇒ \(\frac{\sqrt{4^2-4}-0}{3}=\frac{2 c-1}{2 \sqrt{c^2-c}}\)
⇒ \(\frac{2 \sqrt{3}}{3}=\frac{2 c-1}{2 \sqrt{c^2-c}} \Rightarrow \frac{4}{\sqrt{3}}=\frac{2 c-1}{\sqrt{c^2-c}}\)
On squaring both sides; we have
\(\frac{16}{3}=\frac{(2 c-1)^2}{c^2-c}\)
⇒ 16(c² – c) = 3 (4c² – 4c + 1)
⇒ 4c² – 4c – 3 = 0
⇒ (2c – 3) (2c + 1) = 0
⇒ c = \(\frac { 3 }{ 2 }\), – \(\frac { 1 }{ 2 }\)
but c = – \(\frac { 1 }{ 2 }\) ∉ (1, 4)
∴ c = \(\frac { 3 }{ 2 }\) ∈ (1, 4)
Thus L.M.V. theorem is verified.

Question 14.
Verify Rolle’s theorem for the function f(x) = e^x(sin x – cos x) on \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\).
Solution:
Given f(x) = e^x (sinx – cosx) on \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\) Since exponential functional and sine, cosine functions are continuous every where in R. Also difference of two continuous functions is continous.
∴ sinx – cosx is continuous everywhere.
Also product of two continous functions is also continuous
∴ f(x) = e^x (sinx – cosx) is continuous everywhere and hence f(x) is continuous in \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\).
Also f'(x) = e^x (cosx + sinx) + (sinx – cosx)e^x = 2e^x sinx which exists for all x ∈ R

Thus, Rolle’s theorem is verified.

Question 15.
Verify Lagranges’ Mean Value Theorem for the function f(x) = 2 sin x + sin 2x on [0, π].
Solution:
Given f(x) = 2sinx + sin 2x in [0, π]
Since sine function is continuous every where and sum of two continuous functions is continuous in [0, π].
Also, f'(x) = 2cosx + 2cos2x
which exists for all x∈R
∴ f(x) is derivable on (0, π)
Further, f(0) = 0 ; f(π) = 2 sin π + sin 2π = 0
Thus, both conditions of L.M.V. theorem are satisfied so ∃ atleast one real number c∈(0, π)
s.t. f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)
⇒ 2cosc + 2cos 2c = 0
⇒ 2cos²c + cosc – 1 = 0
⇒ (cosc + 1) (2cosc – 1) = 0
⇒ cosc = – 1 or cos c = \(\frac { 1 }{ 2 }\)
∴ c = π or c = \(\frac { π }{ 3 }\)
but c∈(0, π) ∴ c = π/3 ∈ (0, π)
Thus L.M.V. theorem is verified & c = \(\frac { π }{ 3 }\).

Question 16.
Verify the conditions of Rolle’s Theorem for the following function :
f(x) = log (x² + 2) – log 3 on [- 1, 1].
Solution:
Let f(x) = log (x² + 2) – log 3
Clearly f(x) is continuous in [-1, 1].
Now f(x) = \(\frac{2 x}{x^2+2}\) exists ∀x ∈ (-1, 1)
∴ f is derivable in (-1, 1).
further f (-1) = log 3 – log 3 = 0
and f(1) = log 3 – log 3 = 0
∴ f(-1) = f(1) = 0
Thus all the three conditions of Rolle’s theorem are satisfied
∴ ∃ atleast one real number c ∈ (-1, 1) s.t.f'(c) = 0
i.e. \(\frac{2 c}{c^2+2}\) = 0 ⇒ c = 0 ∈ (-1, 1).
Hence Rolle’s theorem is verified.