Question

If f(x) is a differentiable function and g(x)is a double differentiable function such that |f(x)|≤1 and f'(x)=g(x). If f2(0)+g2(0)=9. Prove that there exists some c∈(–3,3)such that g(c).g''(c)<0.

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Answer 1

Answer:

Solution:

TO prove: g(x).g''(x)<0 for some c∈(–3,3)

This means that one of g(x) or g"(x) is negative is negative

Let's assume that both are positive.

Since, |f(x)|≤1

$\Rightarrow (x) \in [-1,1]\\\\\Rightarrow f^2 (x)\leq 1 , f^2 (0) + g^2 (0) =9\\\\\Rightarrow g(0) \leq 9$

Given: f(x)' = g(x)

$\Rightarrow \int_{-3}^{x} f (x) dx = \int_{-3}^{x} g(x)dx$

$\Rightarrow f (x) =\int_{-3}^{x} g(x)dx+f(-3)$_____(since f(-3) can take minimum value at -1)

$since,$ $g(x)> 0$

=> curve is opening upwards for any g(x) satisfying the conditions

case 1: g(x) is decreasing

$\int_{-3}^{x} g(x)dx > 3^* 2\sqrt{2} = 6 \sqrt{2}$ $(Area$ $of$ $rectangle)$

case 2: g(x) is increasing

$\int_{-3}^{x} g(x)dx > 3^* 2\sqrt{2} = 6 \sqrt{2}$ $( Area$ $of$ $rectangle)$

case 1: g(x) is take minimum value at x = 0

$\int_{-3}^{x} g(x)dx > 3^* 2\sqrt{2} = 6 \sqrt{2}$  $(Area$ $of$ $rectangle)$

$\Rightarrow f(x) > 6\sqrt{2} - 1$

But f(x) cant be greater than 1

This shows the contradiction which means our assumption is wrong

=> g(x) and g"(x) positive at the same time

similarly, they cant be negative at the same time

Hence,there exist c ∈ ( -3 , 3 ) such that g(c)g" (c) < 0

Hence, proved

$\bigstar$ Hope its help u $\bigstar$