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Discuss the continuity and differentiability of the function y = f(x) defined parametrically; x = 2t – |t-1| and y = 2t2 + t|t|.

Solution :

Here x = 2t – |t-1| and y = 2t2 + t|t|.

Now when t < 0;

x = 2t – {-(t-1)} = 3t – 1 and y = 2t2t2 = t2 = y = 19(x+1)2

= when 0 t < 1

x = 2t – {-(t-1)} = 3t – 1 and y = 2t2t2 = 3t2 = y = 13(x+1)2

when t 1;

x = 2t – (t-1) = t + 1 and y = 2t2 + t2 = 3t2 = y = 3(x+1)2

Thus, y = f(x) = 19(x+1)2, x < -1

y = f(x) = 13(x+1)2, -1x < 2

y = f(x) = 3(x+1)2, x 2

We have to check differentiability at x = -1 and 2.

Differentiabilty at x = -1;

LHD = f'(1) = limh0f(1h)f(1)h = limh0 19(1h+1)20h = 0

RHD = f'(1)+ = limh0f(1+h)f(1)h = limh0 13(1+h+1)20h = 0

Hence f(x) is differentiable at x = -1

  continuous at x = -1.

To check differentiability at x = 2;

LHD = f'(2) = limh013(2h+1)23h = 2

RHD = f'(2)+ = limh03(2+h1)23h = 6

Hence f(x) is not differentiable at x = 2.

But continuous at x = 2, because LHD and RHD both are finite.

f(x) is continuous for all x and differentiable for all x, except x = 2.

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