Consider the function
where P(x) is a polynomial such that P'' (x) is always a constant and P(3) = 9. If f(x) is continuous at x = 2, then P(5) is equal to _____________.f(x) = P(x)/sin(x-2), x ≠ 2 ; = 7, x = 2

Solution

Maths sir · 4 steps

The denominator vanishes at x = 2, so P(2) must be 0 too. Then the limit gives P'(2) = 7.

1. Shape of P. P''(x) is a constant, so P has degree at most 2. Write P(x) = ax² + bx + c.
2. Numerator must vanish. As x → 2, sin(x-2) → 0. For the limit to be the finite value 7, the numerator must also go to 0, so P(2) = 0. That means (x-2) is a factor: P(x) = (x-2)(ax + b).

   lim(x → 2) (x-2)(ax+b)/sin(x-2) = lim(x → 2) (x-2)/sin(x-2) × (ax+b) = 2a + b

3. Two equations. Continuity at x = 2 gives 2a + b = 7. The condition P(3) = 9 gives (3-2)(3a + b) = 9, so 3a + b = 9. Subtracting, a = 2 and then b = 3.
4. So. P(x) = (x-2)(2x+3), and P(5) = 3 × 13 = 39.

The idea to keep. If a limit of the form P(x)/sin(x-a) is finite, the numerator must vanish at x = a. Factor out (x-a) and the limit becomes P'(a).

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