Differentiation MCQ (With Solution) - 2 (Competition Level 1) - JEE MCQ

 To resolve y into partial fractions we put x –1 = z. Then 
 arranging the numerator and the denominator both in ascending powers of z.
Now dividing the numerator 1 by the denominator –1 + z till z³ is a common factor in the remainder, we get 

 Now differentiate n times and you will get the required result. 

on resolving into partial fractions. 
Now differentiating both sides (n-1) times w.r.t. 'x' , we have 

∴ sin ∞ and cos ∞ lies between –1 to 1.  
for p ≥ 5, RHL = 0  
LHL = 0 
 V.F.  = 0 
 for p ∈ [5, ∞), f ′′ (x) is continuous .
 (C) [5, ∞) (D) none of these

Clearly x² + y² = a² and y(π/4) = 

f(x) = x² + 2x + C, but f(0) = 6  
So, f(x) = x² + 2x + 6. 

Put x = 0
Then k = 0
f (x) = f ′(x)

 Since the given fraction is not a proper one, therefore we should first divide the numerator by the denominator before resolving it into particle fractions. Here we observe orally that the quotient will be 1. so let 


Now differentiating both sides n times, we get 

Taking suitable intervals we understand that function is not differentiable at  x = –1, 0, 1. 

We have y = (sin^–1x)²
therefore, 

now using the standard formula Dⁿsin(ax+(B), we get