Interactive OP Malhotra Maths Class 12 Solutions Chapter 10 Mean Value Theorems Chapter Test engage students in active learning and exploration.
S Chand Class 12 ICSE Maths Solutions Chapter 10 Mean Value Theorems Chapter Test
Verify Rolle’s Theorem for the function:
Question 1.
(i) f(x) = x² -5x + 4 on [1, 4].
Solution:
f (x) = x² – 5x + 4 ; since f (x) is polynomial in x.
∴ It is continuous in [1, 4] and derivable in (1, 4). also f(1) = 1 – 5 + 4 = 0
and f(4) = 16 – 20 + 4 = 0
∴ f(1) = f(4)
∴ all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈ (1, 4) s.t.
f'(c) = 0 i.e. 2c – 5 = 0 ⇒ c = \(\frac { 5 }{ 2 }\) ∈ (1, 4)
Hence, Rolle’s theorem is verified and c = \(\frac { 5 }{ 2 }\).
Question 2.
f(x) = (x – 1)(x – 2)² on [0, 2π].
Solution:
Given, f(x) = (x – 1) (x – 2)² = (x – 1) (x² – 4x + 4) = (x³ – 5x² + 8x – 4) in [1, 2]
Since f(x) is polynomial in x
∴ continuous in [1, 2] as f'(x) = 3x² – 10x + 8 exists ∀ x ∈ (1, 2)
∴ f is derivable in (1, 2),
also f(1) = f(2) = 0
Thus all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no c ∈ (1,2) s.t. f'(c) = 0
i.e. 3c² – 10c + 8 = 0
i.e. c = \(\frac{10 \pm 2}{6}=2, \frac{4}{3}\)
but c = 2 ∉ (1,2) ∴ c = \(\frac { 4 }{ 3 }\) ∈ (1, 2).
Hence Rolle’s theorem is verified.
Question 3.
f (x) = cos x + sin x in [0, 2π].
Solution:
Given f(x) = cos x + sin x in [0, 2π]
since sine & cosine function both are continuous eveiywhere & sum of two continuous functions is also continuous.
Thus f is continuous on [0, 2π]
Further f(x) = – sin x + cos x which exists ∀ x ∈ R
∴ f(x) is derivable on (0, 2π)
Also, f(0) = 1 + 0 = 1; f(2π) = 1 + 0 = 1
∴ f(0) = f(2π)
Thus all the three conditions of Rolle’s theorem are satisfied so∃ atleast one real number c ∈(0, 2π) s.t. f'(c) = 0
∴ – sin c + cos c = 0 ⇒ tan c = 1
But c ∈ (0, 2π)
∴ c = \(\frac{\pi}{4}, \frac{5 \pi}{4} \in(0,2 \pi)\)
Thus Rolle’s theorem is verified
∴ c = \(\frac{\pi}{4}, \frac{5 \pi}{4}\)
Question 4.
Using Rolle’s Theorem, find points on the curve y = 16 – x², x∈ [- 1, 1], where tangent is parallel to x-axis.
Solution:
Given y = f(x) = 16 – x², x∈[-1, 1]
since f(x) is polynomial in x and hence continuous on [-1, 1].
Also, f(x) = -2x which exists for all X ∈ (-1, 1)
∴ f(x) is derivable on (-1, 1).
Also, f(-1) = 16 – (-1)² = 15;
f(1) = 16 – 1² = 15
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)
∴ Rolle’s theorem is verified.
Question 5.
Verify Lagrange’s 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 6.
Verify Mean Value Theorem, if
f (x) = x³ – 5x² – 3x in the interval [a, b] where a = 1, 6 = 3. Find c ∈ (1, 3) for which f’ (c) = 0.
Solution:
Given f(x) = x³ – 5x² – 3x in [1, 3]
Here f(x) be polynomial in x and hence continuous in [1, 3].
also f(x) = 3x² – 10x – 3 which exists for all x∈R
∴ f(x) is devirable on (1, 3)
Further f(1) = 1 – 5 – 3 = – 7;
f(3) = 27 – 45 – 9 = – 27
Thus both conditions of lagrange’s mean value theorem are satisfied so ∃ atleast one real number c ∈(1, 3)
s.t. \(\frac{f(3)-f(1)}{3-1}\) = f'(c)
⇒ \(\frac{-27+7}{2}\) = 3c² – 10c – 3
⇒ – 10 = 3c² – 10c – 3
⇒ 3c² – 10c + 7 = 0 ⇒ c = 1, \(\frac { 7 }{ 3 }\)
but c∈(1, 3)
∴ c = \(\frac { 7 }{ 3 }\)∈(1, 3)
Thus L.M.V. theorem is verified.
Question 7.
Verify Lagrange’s Mean Value Theorem for the function f (x) = x (x – 2) on [1, 3].
Solution:
Given f(x) = x (x – 2) = x² – 2x in [1, 3]
since f(x) is polynomial in x and hence continuous on [1, 3].
also, f(x) = 2x – 2 which exists for all x∈R
∴ f(x) is derivable on (1, 3)
Also f(3) = 3(3 – 2) = 3;
f(1) = 1, lagrange’s (1 – 2) = – 1
Thus both conditions of lagrange’s mean value theorem are satisfied so ∃ atleast one real number c ∈ (1, 3)
s.t. f'(c) = \(\frac{f(3)-f(1)}{3-1}\)
⇒ 2c – 2 = \(\frac{3-(-1)}{3-1}\) = 2
⇒ 2c = 4 ⇒ c = 2 ∈ (1, 3)
Thus LM.V. theorem is verified & c = 2
Question 8.
Using Lagrange’s Mean Value Theorem, find a point on the parabola y = (x + 3)², where the tangent is parallel to the chord joining (-3, 0) and (-4, 1).
Solution:
Let f(x) = y = (x + 3)² in [-4, -3]
Clearly f(x) be polynomial in * and hence continuous everywhere in R. Thus f(x) in continuous on [-4, -3].
Also, f'(x) = 2(x + 3) which is exists for all x∈R
∴ f(x) is derivable on (-4, -3)
Now, f(-4) = 1; f(-3) = 0
Thus both conditions of L.M.V. theorem satisfied so ∃ atleast one real number c∈(-4, -3)
s.t. \(\frac{f(-3)-f(-4)}{-3-(-4)}\) = f'(c)
⇒ \(\frac{0-(1)}{-3+4}\) = 2 (c + 3)
⇒ – 1 = 2c + 6 c ⇒ c = – \(\frac { 7 }{ 2 }\) ∈ (- 4, – 3)
∴ L.M.V. theorem is verified.
when x = – \(\frac { 7 }{ 2 }\) ∴ y = (-\(\frac { 7 }{ 2 }\) + 3)² = \(\frac { 1 }{ 4 }\)
so required point an curve be (-\(\frac { 7 }{ 2 }\), \(\frac { 1 }{ 4 }\))
where tangent is parallel to the chord joining the point (-3, 0) & (-4, 1).