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JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2006 · Mathematics · Application of Derivatives

If f(x) is twice differentiable function such that f(a)=0, f(b)=2,...

If

f(x)

is 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 zeros of

g(x)=\left\{f^{\prime}(x)\right\}^2+f^{\prime \prime}(x) \cdot f(x)

in the interval

[a, e]

is

Correct answer

6

Step-by-step solution

Let,

g(x)=\frac{d}{d x} \left[f(x) \cdotf^{\prime}(x)\right]

to get the zero of

g(x)

we take function

h(x)=f(x) \cdotf^{\prime}(x)

between any two roots of

h(x)

there lies atleast one root of

h^{\prime}(x)=0

\begin{aligned} & \Rightarrow \quad g(x)=0 \\ & \Rightarrow \quad h(x)=0 \\ & \Rightarrow \quad f(x)=0 \text{ or } f^{\prime}(x)=0 \\ & \text{ If } f(x)=0 \text{ has }4 \text{ minimum solutions, }\\ & f^{\prime}(x)=0 \text{ has }3 \text{ minimum solutions, }\\ & h(x)=0 \text{ has }7 \text{ minimum solutions, then }\\ & h^{\prime}(x)=g(x)=0 \text{ has }6 \text{ minimum solutions. }\\ & \end{aligned}

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