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

Continuity & Differentiability

Q.4. If the following function f(x) is continuous at x = 0, find the value of k :

∫(1 – cos 2x)/2x², x ≠ 0
f(x) |
∫ k, x = 0

Solution :

We have, ∫(1 – cos 2x)/2x², x ≠ 0
f(x) |
∫ k, x = 0
As, f(x) is continuous at x = 0
limx→0–f(x) = limx→0+f(x) = f(0).
limx→0–[(1 – cos 2x)/2x²] = limx→0+ [(1 – cos 2x)/2x²]= k.
Let us consider limx→0+ [(1 – cos 2x)/2x²] = k
Put x = 0 + h, h→0.
limh→0[{1 – cos 2(0 + h)}/2(0 + h)2] = k
Or, limh→0[(1 – cos 2h)/2h²] = k
Or, limh→0[2sin² h/2h²] = k
Or, limh→0(sinh/h)² = k [As, limx→0(sinx/x) = 1.]
Or, (1)2 = k => k = 1. [Ans.]

Q.5. Show that

∫5x – 4, when 0 < x ≤ 1
f(x) = |
∫4x³ – 3x, when 1 < x < 2
is continuous at x = 1.

Solution :

L.H.L. = limx→1–f(x) = limx→1– (5x – 4)
[Put x = 1 – h, h→0.
= limh→0{5(1 – h) – 4}
= limh→0 (5 – 5h – 4)
= limh→0 (1 – 5h)
= 1.
R.H.L. = limx→0+f(x) = limx→0+ (4x³ – 3x)
[Put x = 1 + h, h→0.]
= limh→0{4(1 + h)3 – 3(1 + h)}
= 4(1 + 0)³ – 3(1 + h)}
= 4 – 3
= 1.
f(x) = 5x – 4
Therefore, f(1) = 5(1) – 4
= 5 – 4
= 1.
Therefore, limx→1– f(x) = limx→1+ f(x) = f(1).
Hence, f(x) is continuous at x = 1. [Proved.]

Paper By Mr. M.P.Keshari
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