Let f : ℝ → ℝ be a polynomial function of degree four having extreme values at x = 4 and x = 5 . If lim(x → 0) f(x)/x² = 5 , then f(2) is equal to :

Solution

Maths sir · 4 steps

The limit tells you the first three coefficients, so only two unknowns are left, and the extrema find them.

1. Read the limit. For f(x)/x² to tend to 5, the constant and x terms of f must be zero and the x² coefficient must be 5. So f(x) = ax⁴ + bx³ + 5x².
2. Use the extrema. f'(x) = 4ax³ + 3bx² + 10x = x(4ax² + 3bx + 10). The quadratic in the bracket must vanish at x = 4 and x = 5, so it is a multiple of (x-4)(x-5).

   4ax² + 3bx + 10 = 1/2(x² - 9x + 20)

3. Match coefficients. The constant terms give the factor 1/2, because 20 × 1/2 = 10. Then 4a = 1/2 gives a = 1/8, and 3b = -9/2 gives b = -3/2.
4. So. f(2) = 1/8(16) - 3/2(8) + 5(4) = 2 - 12 + 20 = 10, which is option (b).

The idea to keep. A limit like f(x)/xⁿ → L as x → 0 fixes the lowest coefficients of a polynomial: all terms below xⁿ vanish and the xⁿ coefficient is L. Then use the extrema as roots of f'.

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