For continuity at x=0, the left-hand limit and right-hand limit must both equal f(0)=b. Using standard limits, we find a2−b2=b from the left and b=52 from the right, giving a2=2512 and a2+b=2524. The correct option is (D).
Concept & Intuition
Continuity at a point means the function’s value equals its limit from both sides. Here, the left piece uses a difference of sines over x2, which suggests using the identity sin2θ=21−cos2θ and the classic limit limt→0t21−cost=21. The right piece involves a subtle combination of a binomial expansion and a small-angle approximation for sine. Matching both limits to b gives two equations that determine a2 and b.
Step-by-step solution
- Left-hand limit (x→0−)
For x<0,
f(x)=x2sin2(ax)−sin2(bx).
Use sin2θ=21−cos2θ:
sin2(ax)−sin2(bx)=21−cos(2ax)−21−cos(2bx)=2cos(2bx)−cos(2ax).
So
f(x)=2x2cos(2bx)−cos(2ax).
Now use the identity cosu−cosv=−2sin2u+vsin2u−v:
cos(2bx)−cos(2ax)=−2sin((a+b)x)sin((b−a)x).
Hence
f(x)=2x2−2sin((a+b)x)sin((b−a)x)=−xsin((a+b)x)⋅xsin((b−a)x).
As x→0, xsin(kx)→k. Therefore
limx→0−f(x)=−(a+b)(b−a)=a2−b2.
For continuity, this must equal f(0)=b:
a2−b2=b⇒a2=b2+b.(1)
- Right-hand limit (x→0+)
For 0<x<π,
f(x)=sin6/5x(x+2x2)1/5−x1/5.
Factor x1/5 from the numerator:
(x+2x2)1/5=x1/5(1+2x)1/5.
So
f(x)=sin6/5xx1/5[(1+2x)1/5−1].
Write sinx∼x as x→0, so sin6/5x∼x6/5. Then
f(x)∼x6/5x1/5[(1+2x)1/5−1]=x(1+2x)1/5−1.
Now use the binomial expansion (1+u)1/5=1+51u+O(u2) for small u:
(1+2x)1/5−1=51(2x)+O(x2)=52x+O(x2).…