Match the column:
Column-I | Column-II |
(i) Let g(x) be a polynomial of degree two and let f(x) be defined by f(x) = The value of continuous function f(x) at x =–4 if f(1) = f(–1) and f(–2) = 0 | [A] 6 |
(ii) Given that f(x) = ,g(0) = g'(0) = 0 and f(x) is continuous at x = 0, the value of f '(0) is | [B] |
(iii) If [x] denotes the integral part of x and f(x) = then the number of points where f(x) is discontinuous in [0, 5] | [C] 3 |
(iv) If [x] denotes the integral part of x then\ =(for n > 1) | [D] 0 |
Text Solution
Verified by Experts(i) [C]; (ii) [D]; (iii) [A]; (iv) [C]
Ans.
(i) [C]
(ii) [D]
(iii) [A]
(iv) [C]
Sol.
(i) Let g(x) = ax 2 + bx + c
f(0 – ) = f(0) = c
f(0 + ) = 
= 0
⇒ c = 0 ….(i)
⇒ 
= f(–1)
⇒ 
= a – b
⇒ a – b = 
= 
= 
….(ii)
Also, f(–2) = 0 ⇒ 4a – 2b = 0
⇒ 2a – b = 0 ….(iii)
From Eqs.(ii) and (iii), we get
a = 
b = 
f(–4) = g(–4) = 16a – 4b = 
(ii) Since, f(x) = 
, g(0) = g '(0) = 0
and f(x) is continuous at x = 0
f(0 + ) = 
= 
= 0
Now f '(0 + ) = 
= 
= 
= g'(0) = 0
and f(0 – ) = 
= 
= 
= g '(0) = 0
⇒ f '(0 + ) = f '(0 – ) = 0
(iii) f(x) = 
⇒ f(x) = 
as sin π [x + 1] = 0
f(I + ) = 
= 
= 
….(i)
f(I – ) = 
= 
….(ii)
f(I) = 
…..(iii)
From Eqs.(i), (ii) and (iii) f(x) is discontinuous at all integral points.
⇒ Number of points of discontinuity of f(x) in
[0, 5] is 6.
(iv) When n > 1, x → 
, we have sin x > sin n x.
So, 
= 
= 
= 3.
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