CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Continuity & Differentiability

Q.2. If y = a(sin θ – θ cos θ) and x = a(cos θ + θ sin θ) , find dy/dx.

Solution :

We have, y = a(sin θ – θ cos θ)
Therefore, dy/dθ = a[cos θ – θ(– sin θ) – cos θ(1)]
= a(cos θ + θ sin θ – cos θ)
= a θ sin θ.
And x = a(cos θ + θ sin θ)
Therefore, dx/dθ = a(– sin θ + θ.cos θ + sin θ.1)
= a(– sin θ + θ cos θ + sin θ)
= aθ cos θ
dy/dx = (dy/dθ)/(dx/dθ)
= (aθ sin θ)/(aθ cos θ)
= tan θ. [Ans.]

5.7. Rolle’s Theorem.

Q.1. Verify Rolle’s Theorem for the following function :
f(x) = sin x + cos x, x ε [0, π/2]

Solution :

We have, f(x) = sin x + cos x, x ε [0, π/2]
Both sin x and cos x are continuous function in [0, π/2] and differentiable in ]0, π/2[.
Therefore, f(x) is continuous in [0, π/2] and differentiable in ]0, π/2[.
Also, f(0) = sin 0 + cos 0 = 1 and f(π/2) = sin π/2 + cos π/2 = 1.
Or, f(0) = f(π/2)
Thus all the conditions of Rolle’s Theorem are satisfiedby the function f(x).
Therefore, there exist c ε ]0, π/2[ such that f’(c) = 0.
Now, f’(x) = cos x – sin x
f’(x) = 0 => cos x – sin x = 0
=> cos x = sin x
=> tan x = 1
=> x = π/4.
Clearly c = π/4 ε ]0, π/2[. [Verified.]

Q.2. Verify Rolle’s Theorem for the following function : f(x) = (x – 1)(x – 2)², [1, 2].

Solution :

We have, f(x) = (x – 1)(x – 2)², x ε [1, 2]
As f(x) being a polynomial function in x is continuous in [1, 2] and differentiable in ]1, 2[.
Also f(1) = 0 = f(2).
Thus we see that all the conditions of Rolle’s Theorem are satisfied.
Hence, there exist c ε ]1, 2[ satisfying f’(c) = 0.
Now, f’(x) = 1.(x – 2)² + (x – 1).2(x – 2)
= (x – 2)[x – 2 + 2x – 2]
= (x – 2)(3x – 4).
And f’(x) = 0 => x = 2, 4/3.
2 does not belongs to ]1, 2[ and 4/3 ε ]1, 2[.
So, the admissible value of c is 4/3 Ans.

5.8. Lagrange’s Mean Value Theorem.

Q.1. Verify Lagrange’s Mean Value Theorem for the following :
f(x) = x² + 2x + 3, [4, 6].

Solution :

We have, f(x) = x² + 2x + 3, x ε [4, 6] f(x) being a polynomial function is both continuous and differentiable everywhere and in particular, in [4, 6].
f(4) = 27 & f(6) = 51.≠ f(4).
Thus conditions of Lagrange’s Mean Value Theorem is satisfied.
Hence, there exist a point c ε ]4, 6[ such that f’(c) = {f(b) – f(a)}/(b – a).
f’(x) = 2x + 2 => f’(c) = 2c + 2.
Therefore, 2c + 2 = {f(6) – f(4)}/(6 – 4)
Or, 1/2 [(62 + 2×6 + 3) – (42 + 2×4 + 3)] = 2c + 2
Or, 2c + 2 = 12 => c = 5 ε ]4, 6[. [Verified.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289