CBSE Class 12 Math 2013 Solved Paper

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Question : 16 of 29
 
Marks: +1, -0
Show that the function f (x) = |x - 3| , x ∊ R, is continuous but not differentiable at x = 3.
OR
If x = a sin t and y = a (cost+logtant2) find d2ydx2
Solution:
f (x) = |x - 3| = {3−xx<3x−3x≥3
Let c be a real number.
Case I: c < 3. Then f(c) = 3 – c.
limx→c f (x) = limx→c (3 - x) = 3 - x
Since, limx→c f (x) = f (c), f is continuous at all negative real numbers.
Case II: c = 3. Then f(c) = 3 – 3 = 0
limx→c f (x) = limx→c f (x) (x - 3) = 3 - 3 = 0
Since, limx→c f (x) f (x) = f (3), f is continuous at x = 3.
Case III: c > 3. Then f(c) = c – 3.
limx→c f (x) = limx→c (x - 3) = x - 3
Since, limx→c f (x) = f (c), f is continuous at all positive real numbers
Therefore, f is continuous function.
Now, we need to show that f(x) = |x - 3|, x ∊ R is not differentiable at x = 3.
Consider the left hand limit of f at x = 3
limh→0− f(3+h)−f(3)h = limh→0− |3+h−3|−|3−3|h = limh→0− |h|−0h = limh→0− −hh = - 1
h < 0 ⇒ |h| = - h
Consider the right hand limit of f at x = 3
limh→0+ f(3+h)−f(3)h limh→0+ |3+h−3|−|3−3|h = limh→0+ |h|−0h = limh→0+ hh = 1
h > 0 ⇒ |h| = h
Since the left and right hand limits are not equal, f is not differentiable at x = 3.
OR
y = a (cost+logtant2) find d2ydx2
⇒ dydt = a |ddt+cost+ddt(logtant2)|
= a |−sint+cott2×sec3t2×12|
= a |−sint+12sint2cost2|
= a (−sint+1sint) = a (−sin2t+1sint) = a cos2tsint
x = a sin t
dxdt = a ddt sin t = a cos t
∴ dydx = (dydx)(dxdt) = (acos2tsint)acost = costsint = cot t
d2ydx2 = - cosec2tdtdx = - cosec2t×1acost = - 1asin2tcost
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