Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x ∈ R. If | f(x), f'(x); f'(x), f''(x) | = 0, for all x∈R, then the value of f(1) lies in the interval :

Solution

Maths sir · 4 steps

Expand the determinant and you get a quotient-rule numerator, so the ratio f'/f is constant.

1. Expand the determinant. The condition reads f(x)f''(x) - (f'(x))² = 0 for every x.
2. Spot the quotient rule. Since f'(x) ≠ 0, we may divide by (f')². The quotient rule gives (f/f')' = ((f')² - f f'')/(f')². By the condition this is 0, so f/f' is a constant.
3. Fix the constant. At x = 0, f(0)/f'(0) = 1/2. So f'(x) = 2f(x) for all x. This is a linear equation whose solution is f(x) = Ce²ˣ, and f(0) = 1 gives C = 1.

   f(x) = e²ˣ ⇒ f(1) = e²

4. So. e² ≈ 2.718² ≈ 7.39, which lies in (6, 9). The answer is (d).

The idea to keep. When you see ff'' - (f')², think of the quotient rule: it is the numerator of (f'/f)' or of (f/f')'. Setting it to zero means the ratio is constant, which gives an exponential function.

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