Test: 35 Year JEE Previous Year Questions: Differentiation

f (x) = xⁿ ⇒ f (1) = 1

f(x) = ax² +bx + c
f(1) = f(-1) ⇒ a + b + c = a-b + c or b = 0
∴ f(x) = ax² +c or f '(x) = 2ax

Now f '(a); f '(b); and f '(c) are 2a(a); 2a(b); 2a(c) i.e. 2a², 2ab, 2ac.
⇒ If a,b,c ar e in A.P. th en f '(a); f '(b) and f '(c) are also in A.P.

x² - (a - 2) x -a-1 = 0
⇒ α + b = a- 2 ; α β = -(a+ 1)
α² + β² = (α + β)2 - 2αβ = a² - 2a +6 = (a -1)²+5
For min. value of α² + β² where α is an integer
⇒ a = 1.

Let α , α + 1 be roots
Then α + α + 1 = b = sum of roots α (α + 1) = c
= product of roots
∴ b² - 4c = (2a + 1)² - 4a(a + 1) = 1 

Applying L Hospital rule

∴ f '(x) exist at everywhere.

x^m.yⁿ = ( x+y) ^m+n
⇒ m/nx + n/ny = (m + n)ln(x + y)
Differentiating both sides.

x^2x – 2x^x cot y – 1 = 0
⇒ 2 cot y = x^x –  x ^{– x} ⇒ 2 cot y =

Differentiating both sides with respect to x, we get

where u = x^x ⇒ log u = x log x


Now when x  = 1, x^2x – 2x^x cot y – 1 = 0, gives
1 – 2 cot y – 1 = 0 ⇒ cot y = 0
∴ From equation (i), at x = 1 and cot y = 0, we get

 

= 2 f (2 f (x) + 2) f '(2 f (x))+ 2).(2 f '(x))
⇒ g '(0) = 2 f (2 f (0) + 2). f '( 2 f (0) + 2).2 f '(0)= 4 f (0)(f '(0))² 
= 4(–1)(1)² = – 4

Let y = sec(tan^–1 x) and tan^–1 x = θ.
⇒ x = tan θ

Thus, we have y = sec θ

Since f (x) and g(x) are inverse of each other

⇒ g '(f (x)) =1+x⁵ 

Here x = g(y)

∴ g'(y) = 1+ {g (y)}⁵
⇒ g'( x) = 1+ {g (x)}⁵

Let  f (x) = α log | x | + βx² + x Differentiating both sides,

Since x = –1 and x = 2 are extreme points therefore f ' (x) =0 at these points.
Put x  = –1 and x = 2 in f '(x) , we get

On solving (i) and (ii), we get