Let f: R → R be defined as :
f(x)= (a-b cos 2 x)/x² ; x<0 ; ; x²+c x+2 ; 0 ≤ x ≤ 1 ; ; 2 x+1 ; x>1
If f is continuous everywhere in R and m is the number of points where f is NOT differential then m+a+b+c equals :

Solution

Maths sir · 5 steps

Use continuity to fix a, b and c first, then test both joins for differentiability.

1. Continuity at 0. As x → 0⁻ the denominator x² → 0, so the numerator must also go to 0. That needs a - b = 0, so a = b. Then the left piece becomes a(1-cos 2x)/x² = (2a sin² x)/x² → 2a, and this must equal f(0) = 2.

   2a = 2 ⇒ a = b = 1

2. Continuity at 1. The middle piece gives f(1) = 1 + c + 2 and the right piece tends to 2(1) + 1 = 3. So 3 + c = 3, which gives c = 0.
3. Check x = 0. For x < 0, f(x) = (2 sin² x)/x² = 2 - 2x²/3 + …, so the left derivative is lim (f(x) - 2)/x = lim (-2x/3) = 0. The right derivative is 2x + c = 0 at x = 0. They match, so f is differentiable at 0.
4. Check x = 1. The left derivative is 2x = 2 at x = 1 and the right derivative of 2x + 1 is 2. They match, so f is differentiable at 1 too. Each piece is smooth on its own, so m = 0.
5. So. m + a + b + c = 0 + 1 + 1 + 0 = 2, which is option (d).

The idea to keep. A 0/0 form must appear for a finite limit, which gives the first equation. And never assume a join is a corner: compute the left and right derivatives and compare them.

Answer from the NTA final answer key. Solution by PrepWiser.