JEE Advanced (Single Correct MCQs): Differentiation Chapter MCQs & solutions

We have  y² = P (x), ... (1)
where P(x) is a polynomial of degree 3 and hence thrice differentiable. Differentiating (1) w.r. to x, we get

 ......(2)

Again differentiating with respect to x, we get

Again differentiating w.r. to x, we get 
= P '''(x) P (x) + P ''(x) P '(x)- P '(x) P ''(x)
= P '''(x) P(x)

Let f (x) = ax² + bx + c

∴ a > 0 and D < 0
⇒ a > 0 and b² – 4ac < 0 ..... (1)
Now, g (x) = f (x) + f ' (x) + f '' (x)
= ax² + bx + c + 2ax + b + 2a
= ax² + (2a + b) x + (2a + b + c)
Here, D = (2a + b)² – 4a(2a + b + c)
= 4a² + b² + 4ab – 8a² – 4ab – 4ac
= b2 – 4a² – 4ac = – 4a² + b² – 4ac
= (–ve) + (–ve) = –ve [Using eq. (1)] Also a > 0 from (1)

y = (sin x) tan x ⇒ log y= tan x. log sinx
Differentiating w.r.t. x,

x² + y² = 1

⇒ 2x + 2yy' = 0

⇒ x + yy'= 0
⇒ 1+ yy'' + (y')² = 0

⇒ yy'' + (y')²+1 = 0

log (x + y) = 2xy when x = 0 then y = 1
Differentiating w.r.t. x

Let us  consider the function g (x) = f (x) – x2  so that
g (1) = f (1) – 12 = 1 – 1 = 0
g (2) = f (2) – 22 = 4 – 4 = 0
g (3) = f (3) – 32 = 9 – 9 = 0
Since f (x) is twice differentiable we can say g (x) is continuous and differentiable everywhere and
g (1) = g (2) = g (3) = 0
∴ By Rolle's theorem, g' (c) = 0 for some c ∈ (1,2)
and g '(d) =0 for some d ∈ (2,3)
Again by Rolle's theorem,
g ''(e) =0 for some e ∈ (c,d) ⇒ e ∈ (1,3)
⇒ f ''(e) - 2 = 0 or f ''(e) = 2 for some x ∈ (1, 3)
f ''(x) = 2 for some x∈ (1,3)

Given that g(x) = log f (x) ⇒ g (x + 1) = log f (x + 1)
⇒ g (x + 1) = log x f (x)  [∴ f (x + 1) = x f (x)]
⇒ g(x + 1) = logx + log f (x) ⇒ g(x + 1) – g(x) = log(x)

Adding all the above equations, we get

f(x) = x³ + 3x + 2
⇒ f '(x) = 3x2 + 3
Also
f(0) = 2,
f (1) = 6,
f(2) = 16,
f(3) = 38,
f(6) = 236
And
g(f(x)) = x
⇒ g(2) = 0,
g(6) = 1,
g(16) = 2,
g (38) = 3,
g (236) = 6

(a) g (f(x)) = x
⇒ g' (f(x)). f ¢(x) = 1
For
g'(2), f(x) = 2
⇒ x = 0
∴ Putting x = 0,
we get g'(f (0)) f '(0) = 1
⇒ g'(2) = 1/3

(b) h (g(g(x))) = x
⇒ h'(g(g(x))). g'(g(x)). g'(x) = 1 For h'(1),
we need g (g(x)) = 1
⇒ g (x) = 6
⇒x = 236
∴ Putting x = 236, we get

(c) h [g(g(x))] = x
For h (0), g (g(x)) = 0
⇒ g(x) = 2
⇒ x = 16
∴ Putting x = 16, we get
h(g(g(16))) = 16
⇒ h(0) = 16

(d) h[g(g(x))] = x
For h(g(3)), we need g(x) = 3
⇒ x = 38
∴ Putting x = 38,
we get
h [g(g(38))] = 38
⇒ h (g(3)) = 38