If $\mathrm{f}(\mathrm{x})$ is a twice differentiable function such that $\mathrm{f}(\mathrm{a})=0, \mathrm{f}(\mathrm{b})=2, \mathrm{f}(\mathrm{c})=-1, \mathrm{f}(\mathrm{d})=2, \mathrm{f}(\mathrm{e})=0$, where $\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}<\mathrm{e}$, then the minimum number of zeroes of $\mathrm{g}(\mathrm{x})=\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}+\mathrm{f}^{\prime \prime}(\mathrm{x}) \mathrm{f}(\mathrm{x})$ in the interval $[\mathrm{a}, \mathrm{e}]$ is

JEE Advanced · 2006 · 09 Apr 2006 · Application of Derivatives

f (x)

is a twice differentiable function such that

f (a) = 0, f (b) = 2, f (c) = - 1, f (d) = 2, f (e) = 0

, where

a < b < c < d < e

, then the minimum number of zeroes of

g (x) = (f^{'} (x))^{2} + f^{''} (x) f (x)

in the interval

[a, e]

Answer: 6

Note that

g (x) = (f^{'} (x))^{2} + f^{''} (x) f (x) = \frac{d}{d x} (f (x) \cdot f^{'} (x))

.

Let

h (x) = f (x) \cdot f^{'} (x)

. Then

g (x) = h^{'} (x)

.

The zeros of

h (x)

occur when either

f (x) = 0

f^{'} (x) = 0

.

From the given values:

$f (x) = 0$ at $x = a$ and $x = e$ , and by Rolle's theorem, at least 2 more points in between (since f changes sign twice: 0→2→-1→2→0). So minimum 4 zeros of $f (x) = 0$ .
By Rolle's theorem on the intervals where f is non-zero, $f^{'} (x) = 0$ has at least 3 zeros (between b and c, between c and d, and possibly others).

h (x) = 0

has at least 7 zeros in [a,e].

By Rolle's theorem, between any two consecutive zeros of

h (x)

, there is at least one zero of

h^{'} (x) = g (x)

.

With 7 zeros of h(x), there are at least 6 zeros of g(x).

The answer is 6.

JEE Advanced · 2006 · 09 Apr 2006 · Numerical answer