Notice Board: Due to unavoidable circumstances, Ctanujit Classes will be closed till 31st May, 2018. Application for ISI 2019 Entrance exam study material & Online classes will start from 1st June, 2018

IIT JEE SOLVED PAPERS ON DIFFERENTIATIONS

Here is some previous year IIT - JEE solved problems on differentiation


Q. [IIT-JEE 2006] If f″(x)= -f(x), where f(x) is a continuous double differentiable function and g(x)=f′(x);
If F(x)= (f(x/2))2 + (g(x/2))2 and F(5)=5, then F(10) is
(a)  0      (b) 5     (c) 10     (d) 25
Solution :(b)Here F′(x)= f(x/2)g(x/2) – g(x/2)f(x/2)
F′(x)=0 . Thus, F(x) is a constant. Now, F(5)=5 ⇒ F(10)=5.
Q. [IIT-JEE 2004] If y is a function of x and log(x+y)=2xy, then the value of y′(0) is equal to
(a) 1     (b) -1     (c) 2     (d) 0
Solution : (a) Differentiating this equation log(x+y)=2xy w.r.t. x,         we get  (1+y′(x)/(x+y) = 2(xy′(x) + y).
Now when x=0, we get  y=1.
Putting these, we have y′(0)=1.
Q. [IIT-JEE 2000] If x2 + y2 = 1, then
(a) yy″ - 2(y′)2 + 1 = 0                 (b) yy″ + (y′)2 + 1 = 0
(c) yy″ + (y′)2 - 1 = 0                   (d) yy″ + 2(y′)2 + 1 = 0
Solution :(b) Given x2 + y2 = 1
Differentiating w.r.t. x, we have x + yy′ = 0
Differentiating w.r.t. x again, we have 1 + yy″ + (y′)2 = 0.