Mean Value Theorems Class 12 OP Malhotra Exe-10A ISC Maths Solutions Ch-10 Solutions. In this article you would learn about Rolle’s Theorem and Geometrical Intrepretation. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Mean Value Theorems Class 12 OP Malhotra Exe-10A ISC Maths Solutions Ch-10

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-10	Mean Value Theorems
Writer	OP Malhotra
Exe-10(A)	Rolle’s Theorem and Geometrical Intrepretation

Rolle’s Theorem and Geometrical Intrepretation

Mean Value Theorems Class 12 OP Malhotra Exe-10A ISC Maths Solutions Ch-10 Solutions

Que-1: f(x) = x² – x – 6 on [-2, 3].

Sol: Since f(x) be polynomial in x and hence continuous and differentiable on R.
Thus f(x) be continuous on [- 2, 3]
and f(x) be differentiable on (- 2, 3).
& f(-2) = (-2)² + 2 – 6 = 0;
f(3) = 3² – 3 – 6 = 0
∴ f(-2) = f(3)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real number c ∈ (-2, 3)
s.t. f(c) = 0
Now, f'(x) = 2x – 1 ⇒ f'(c) = 2c – 1
So f'(c) = 0 ⇒ 2c – 1 = 0
⇒ c = 1/2 ∈ (- 2, 3)
Hence Rolle’s theorem is verified

Que-2: f(x) = x²-6x + 5, in the interval [1, 5],

Sol: Given f(x) = x² – 6x + 5 in [1, 5]
∴ f’ (x) = 2x – 6
which exists for all x ∈ R
Thus f is derivable on (1, 5), Since every derivable function is continuous
f(x) is continuous on [1, 5].
Now f(1) = 1 – 6 + 5 = 0;
f(5) = 25 – 30 + 5 = 0
∴ f(1) = f(5)
So all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real no c ∈ (1, 5)
s.t f(c) = 0 ⇒ 2c – 6 = 0 ⇒ c = 3 ∈ (1, 5)
Thus Rolle’s theorem is verified.

Que-3: f(x) = x² -5x + 6, 2 ≤ x ≤ 3.

Sol: Given f(x) = x² – 5x + 6, 2 ≤ x ≤ 3
Since f(x) be polynomial in x and hence continuous and differentiable for all x ∈ R.
Thus f(x) is continuous on [2, 3]
and diffemtiable in (2, 3)
Now, f(2) = 4 – 10 + 6 = 0;
f(3) = 9 – 15 + 6 = 0
∴ f(2) = f(3)
Thus all the three conditions of Roll’s theorem are satisfied So ∃ atleast one real numbers c ∈ (2, 3)
s.t. f'(c) = 0
Now f'(x) = 2x – 5 ∴ f'(c) = 0
⇒ 2c – 5 = 0 ⇒ c = 5/2 ∈(2, 3)
Thus, Rolle’s theorem is verified.

Que-4: y = 16 – x², x ∈ [-1, 1].

Sol: Given y = f(x) = 16 – x², x ∈ [-1, 1]
∴ f'(x) = – 2x which exists eveiywhere on R.
Thus f(x) is continuous on [-1, 1]
Si, f(x) is differentiable on (-1, 1)
Since f(x) be polynomial in x and every polynomial function is differentiable.
Also, f(-1) = f(1)
Thus, all the three conditions of Rolle’s
theorem are satisfied So ∃ atleast one real no. c ∈ (-1, 1)
s.t. f(c) = 0 ⇒ – 2c = 0
⇒ c = 0 ∈ (-1, 1)
Thus, Rolle’s theorem is verified.

Que-5: f(x) = x(x – 3)² in the interval [0, 3].

Sol: Given f(x) = x (x – 3)² in [0, 3]
Since f(x) is polynomial in x hence continuous in [0, 3]
Also, f'(x) = (x – 3)² + 2x (x – 3)
= (x – 3) (3x – 3) = 3x² – 12x + 9
which exists for all x ∈ R
Thus f(x) is derivable on (0, 3).
Now f(0) = 0 = f(3)
Now, all the three conditions of Rolle’s
theorem are satisfied so ∃, atleast one real number c ∈ (0, 3)
s.t. f'(c) = 0 ⇒ 3c² – 12c + 9 = 0
⇒ 3(c² – 4c + 3) = 0
⇒ (c – 1) (c -3) = 0
⇒ c = 1, 3
but c = 3 ∉ (0, 3) and c = 1 ∈(0, 3)
Thus, Rolle’s theorem is verified.

Que-6: f(x) = x(x – 3)² on [2, 4].

Sol: Given f(x) = (x – 2) (x – 3) (x – 4)
= (x – 2) (x² – 7x + 12)
∴ f(x) = x³ – 9x² + 26x – 24
Since f(x) is polynomial in x and hence continuous on [2, 4]
Now f'(x) = 3x² – 18x + 26 which exists ∀x ∈ R
∴ f(x) be derivable on (2, 4)
Now f(2) = f(4) = 2
Thus, all the three conditions of Rolle’s theorem are satisfied ∃ so atleast one real no. c ∈ (2, 4)
s.t. f(c) = 0 ⇒ 3c² – 18c + 26 = 0
∴ c = 18±√(324-312)/6 = 18±2√3/6
= 3±1/√3 ∈(2,4)
Thus Rolle’s theorem is verified.

Que-7: f(x) = x³ – 7x² + 16x – 12 on [2, 3],

Sol: Given f(x) = x³ – 7x² + 16x – 12 on [2, 3]
Since f(x) be polynomial in x and hence continuous in [2, 3]
Also, f'(x) = 3x² – 14x + 16 which exists for all x ∈ R
∴ f(x) is differentiable on (2, 3)
Now f(2) = 8 – 28 + 32 – 12 = 0;
f(3) = 27 – 63 + 48 – 12 = 0
∴ f(2) = f(3)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one rreal number c ∈ (2, 3)
s.t. f ‘(c) = 0 ⇒ 3c² – 14c + 16 = 0
∴ c = 14±√(196-192)/6
= 14+2/6 = 8/3,2
Now c = 8/3 ∈ (2, 3) [∵ c = 2 ∉ (2, 3)]
Therefore Rolle’s theorem is verified.

Que-8: f(x) = 4sinx, x ∈ [0, π]

Sol: Given f(x) n since, x ∈ [0, π]
Since stine function is continous and differentiable energy where in its domain.
∴ f(x) is continous on [0, π] and derivable on (0, π)
Now f(0) = 4 x 0 = 0;
f(x) = sin π = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied so ∃ one real no. C∈(0, π)
s.t. f'(c) = 0 4 cosx = 0
⇒ cos x = 0 ⇒ x = π/2 ∈ (0, π)
Thus Rolle’s theorem is neasified & c = π/2.

Que-9: f(x) = sin2x, x ∈ [0 , π/2].

Sol: Given f (x) = sin 2x in [0 , π/2]
Since sine function is continuous and derivable everywhere
∴ f(x) is continuous in [0 , π/2] and derivable in ([0 , π/2).
Also, f (0) = 0 = f(π/2)
∴ All the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈(0 , π/2) s.t. f'(c) = 0
i.e. 2 cos 2c = 0 ⇒ cos 2c = 0
⇒ 2c = π/2 ⇒ c = π/4 ∈ (0 , π/2)
Hence Rolle’s theorem is verified and c = π/4.

Que-10: f(x) = sin² x, 0 ≤ x ≤ π.

Sol: Given f(x) = sin²x; 0 ≤ x ≤ π
since sine function and cosine function both are continuous & differentiable everywhere in its domain.
Also, f'(x) = 2 sin x cos x = sin 2x
which exists for all x ∈ [0, π]
Thus f(x) is continuous on [0, π]
& f(x) is differentiable on (0, π)
Now f(0) = sin² 0 = 0; f(π) sin²π = 0
∴ f(0) = f(π)
Thus, all the theree conditions of Rolle’s theorem are satisfied so ∃ atleast one real so c ∈ (0, π) s.t. f'(c) = 0
⇒ sin 2c = 0
⇒ x = 0, π, 2π ⇒ c = 0, π/2 , π
since c = 0, π ∉ (0, π)
but c = π/2 (0, π)
Thus Rolle’s theorem is verified.

Que-11: f(x) = e^xcosx, x ∈ [-π/2,π/2].

Sol: Given f(x) = e^xcosx, x ∈ [-π/2,π/2]
Since cosine and exponential function are continuous everywhere.
∴ f (x) is continuous in [-π/2,π/2] and derivable in [-π/2,π/2] .
Also, f (-π/2) = 0 = f(π/2)
Therefore all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast are real number c ∈ (0 , π/2) 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 = π/4 ∈ (-π/2,π/2)
∴ Rolle’s theorem is verified.

Que-12: f(x) = sin x/e^x, x ∈ [0, π].

Sol: Given f (x) = sin x/e^x
since sin x and e^x are continuous and differentiable everywhere.
∴ quotient function f (x) is also continuous and differentiable everywhere.
∴ f (x) is continuous on [0, π] and differentiable on (0, π).
Now f(0) = 0/e⁰ = 0/1 = 0;
f(π) = sin π/e^π = 0
∴ f(0) = f(π)
Hence, all the three conditions of Rolle’s theorem are satisfied. Now we want to show that ∃ atleast one real number
c ∈ (0, π) s.t. f’ (c) = 0
we have f (x) = sin x/e^x
⇒ f'(x) =(e^xcos x – sin xe^x)/e^2x
∴ f’ (c) = 0 ⇒ e^c (cos c – sin c) = 0
⇒ cos c – sin c = 0 [∵ e^c > 0]
⇒ tan c = 1 ⇒ c = π/4 ∈ (0, π)
such that f’ (c) = 0.
Hence Rolle’s theorem verified.

Que-13: f(x) = x(x + 3)e^-x/2 defined in the interval [-3, 0].

Sol: Given f(x) = x (x + 3) e^-x/2 on [-3, 0] Since polynomial function x (x + 3) & exponential function are differential everywhere and product of two differential function is differentiable.
Thus f(x) is continuous on [- 3, 0] and differentiable on (-3, 0)
Now f(- 3) = 0 = f(0)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real no. c∈(-3, 0) s.t. f'(c) = 0
Now f'(x) =

Thus Rolle’s theorem is verified.

Que-14: f(x)=√(4-x²) on [-2,2].

Sol: Given f (x) = √(4-x²) …(1)
Clearly D_f = [- 2, 2],
So f(x) is clearly continuous in its domain
i.e. [- 2, 2]
Also, f’ (x) = 1/2√(4-x²) (- 2x)
= -x/√(4-x²)
Clearly f’ (x) exists in (- 2, 2).
So f(x) is clearly derivable in (- 2, 2).
Also f(- 2) = 0 = f(2)
So all the three conditions of Rolle’s theorem are satisfied. So ∃ atleast one real number c ∈(- 2, 2) s.t f’ (c) = 0
Now, f’ (c) = 0 ⇒ -c/√(4-c²) = 0
⇒ c = 0 ∈ (- 2, 2)
So there exists a real number c ∈ (- 2, 2)
s.t f’ (c) = 0
Hence Rolle’s Theorem verified and c = 0

Que-15: If Rolle’s theorem holds for the function
f(x) = x³ + bx² + ax + 5 on [1, 3] with c = (2+1/√3) , find the values of a and b.

Sol: Given f(x) = x³ + bx² + ax + 5
∴ f'(x) = 3x² + 2bx + a
since Rolle’s theorem hold for f(x) on [1, 3]
∴ f(x) = x³
⇒ 1 + b + a + 5 = 27 + 9b + 3a + 5
⇒ b + a + 6 = 9b + 3a + 32
⇒ 2a + 8b = – 26 ⇒ a + 4b = – 13 …(1)
Now f'(c) = 0 ⇒ 3c² + 2bc + a = 0

Que-16: Apply Rolle’s theorem to find point (or points) on the given curve where the tangent is parallel to the x-axi,
(i) y = x² in [-2, 2]
(ii) y = 16 – x², x ∈ [-1, 1].

Sol: (i) Given y = x² in [- 2, 2]
Since f (x) is polynomial in x
∴ it is continuous and derivable everywhere
∴ f(x) is continuous in [- 2, 2] and derivable in (-2, 2).
Also f (- 2) = 4 = f( 2)
Hence all the three conditions of Rolle’s theorem are satisfied
∴ ∃ atleast one real no. c ∈ (-2, 2) s.t. f'(c) = 0 i.e. c = 0 ∈ (-2, 2). When x = 0, y = 0² = 0
∴ (0, 0) be the required point at which tangent is || to x-axis.
[By Geomatrical interpretation of Rolle’s Theorem]

(ii) Given y = f(x) = 16 – x²
which is polynomial in x and hence differentiable and continuous everywhere. Thus,f(x) is continuous on [- 1, 1] and differentiable on (- 1, 1).
Here f (- 1) = 16 – 1 = 15
and f(1) = 16 – 1 = 15
∴ f(- 1) = f(0)
Thus, all the three conditions of Rolle’s Theorem are satisfied. So ∃ atleast one real number c ∈ (- 1, 1) s.t.f’ (c) = 0
But f’ (c) = 0 ⇒ – 2c = 0 ⇒ c = 0
∴ f(c) = f(0) = 16
Then by the geometrical interpretation of Rolle’s Theorem, (0, 16) be the point on given curve, where tangent is parallel to x- axis.

Que-17: Examine the applicability of Rolle’s theorem on the following functions :
(0 f(x) = √x on [-1, 1]
(ii) f(x) = 1 – (x – 1)^2/3 on [0, 2]
(iii) f (x) = x^2/3 on [- 1, 1].

Sol: (i) Given f(x) = √x in [- 1, 1]
∴ f'(x) = 1/2 (x^{1/2 – 1}) = 1/2√x
which does not exists at x = 0 ∈ (-1, 1)
Thus f(x) is not differentiable at x = 0 ∈ (- 1, 1)
∴ f(x) is not differentiable on (-1, 1)
Thus Rolle’s theorem is not applicable.

(ii) Given f(x) = 1 – (x – 1)^2/3 on [0, 2]
∴ f'(x) = -2/3 (x-1)^{2/3 – 1} = – 2/3(x-1)^1/3
which does not exists at x = 1 ∈ (0, 2).
Thus function f(x) is not derivable at x = 1 ∈ (0, 2)

(iii) Given f (x) = x^2/3
∴ f'(x) = 2/3 (x^-1/3) = 2/3x^-1/3 does not exists at x = 0 ∈ (- 1, 1)
∴function f is not derivable on (0, 2)
Thus, Rolle’s theorem is not applicable.

–: End of Mean Value Theorems Class 12 OP Malhotra Exe-10A ISC Math Ch-10 Solution :–

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