Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x⁶ is unity and it has extrema at x = -1 and x = 1. If lim _(x → 0) f(x)/x³ = 1, then 5.f(2) is equal to _________.

Solution

Maths sir · 4 steps

The limit fixes the bottom four coefficients, and the two extrema give the two you still need.

1. Read the limit. For f(x)/x³ to approach 1 as x → 0, the constant, x and x² terms of f must be zero, and the coefficient of x³ must be 1. The leading coefficient is also 1, so only the x⁵ and x⁴ coefficients are unknown.

   f(x) = x⁶ + a x⁵ + b x⁴ + x³, f'(x) = 6x⁵ + 5a x⁴ + 4b x³ + 3x²

2. Use the extrema. f'(1) = 0 gives 6 + 5a + 4b + 3 = 0, so 5a + 4b = -9. f'(-1) = 0 gives -6 + 5a - 4b + 3 = 0, so 5a - 4b = 3.
3. Solve and check. Adding the two equations gives 10a = -6, so a = -3/5, and then b = -3/2. As a check, f'(x) = 3x²(2x-1)(x-1)(x+1). The sign of f' changes at ± 1, so both points are true extrema.
4. So. f(2) = 64 - 96/5 - 24 + 8 = 144/5, so 5f(2) = 144.

The common slip: 28.8. Stopping at f(2) gives 28.8. The question asks for 5f(2), which is 144.

The idea to keep. A limit like lim(x→ 0) f(x)/xⁿ = L tells you the lowest-power terms of a polynomial for free. Each extremum then gives one equation f'(c) = 0.

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