Match the following items of Column I with Column II–
Column-I | Column-II |
(i) f(x) is an odd differentiable function on (– ∞ , ∞) such that f ′(3) = 2, then f ′(3) + f ′(– 3) is | [A] 0 |
(ii) If f(x) = , then f ′(1) is | [B] 1 |
(iii) The number of solutions of = 2 is | [C] 2 |
(iv) The number of real roots of the equation(sin 2x) (cos 2x) = (2x + 2–x) is | [D] 4 |
Text Solution
Verified by Experts(i) [C]; (ii) [B]; (iii) [A]; (iv) [A]
Ans.
(i) [C]
(ii) [B]
(iii) [A]
(iv) [A]
Sol. (i) f(x) is odd ∴ f(– x) = – f(x)
f ′ (–x) = –f ′ (x) ⇒ f ′ (–x) = –f(x)
f ′ (–3) = f ′ (3) =2
∴ f ′ (–3) + f ′ (3) = 4
(ii) y = x y taking log both side, we get
log y = y log x
diff. w.r.to x
y ′ = 
+ y ′ log x
⇒ f ′ (1) = 1
(iii) 10(x
2 + 1) = (x – 3) 2 ⇒ 9x
2 + 6x + 1 = 0
⇒ x = 
= 
But x = 
is not a solution.
∴ Number of solution is 0.
(iv) sin 2 (2 x ) = 
Since, sin2(2x) < 1 but 
≥ 1
by AM ≥ GM.
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