Let f(x) be a polynomial satisfying f(0) = 2, f'(0) = 3 and f''(x) = f(x). Then f(4) is equal to
The general solution of \(f'' = f\) is \(f(x) = Ae^x + Be^{-x}\). (No non-zero polynomial satisfies \(f'' = f\), so "polynomial" in the question should be read as "function".)
\(f(0) = A + B = 2\) and \(f'(0) = A - B = 3\). So \(A = \frac{5}{2}\), \(B = -\frac{1}{2}\).
\(f(4) = \frac{5}{2}e^4 - \frac{1}{2}e^{-4} = \frac{5e^8 - 1}{2e^4}\)
Answer: (B)
Reinforce your conceptual grasp with related questions from MAH MCA CET 2019:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →