Differentiation - Model Problems - 7

1. There is exists a function f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x , and

a. f’’(x)>0 for all x
b. -1c. -2≤f’’(x) ≤-1 or all x
d. f’’(x)<-2 for all x

(JEE, 1982)

Answer: (a)

Reason:

x² -x+1 = (x – ½)² + ¾ can be a solution to the f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x as f(0) = 1 and f’(0) = -1

f’’(x) = 2 and hence f’’(x)>0 for all x.

We may assume f(x) = e^-x and get a similar conclusion. But e^-x may tend to zero as x tends to infinity. Hence f(x) = x² -x+1 is more appropriate function.

2. If y = f[(2x-1)/( x²+1)] and f’(x) = sin x², then dy/dx = --------------.
(JEE 1982)

Answer: dy/dx = sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²

Solution: f’(x) = sin x²
=> f’(t) = sin t²
Given problem is visualized as
y = f(t)
t = [(2x-1)/( x²+1)]
dy/dx = dy/dt*dt/dx
=> sin t² [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2x²+2 –(4x²-2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²