Consider the function f(x)= max x2,(1-x)2, 2 x(1-x) where 0 ≤ x ≤ 1. Let Rolle's Theorem is applicable for f(x) on greatest interval [a, b] then a+b+c is (where c is point such that f prime(c)=0 )

Question

Consider the function f(x)= max x2,(1-x)2, 2 x(1-x) where 0 ≤ x ≤ 1. Let Rolle's Theorem is applicable for f(x) on greatest interval [a, b] then a+b+c is (where c is point such that f prime(c)=0 ) (A) (2/3) (B) (1/3) (C) (1/2) (D) (3/2)

Tardigrade Admin · Accepted Answer

Draw the graph of f1(x)=x2, f2(x)=(1-x)2 f3=2 x(1-x) Now the bold part is the graph of f(x) Hence <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/398c7105ccd3cf9d33a26d47db555897-.png" /> f(x)= begincases (1-x)2 , 0 ≤ x < (1/3) 2 x(1-x) , (1/3) ≤ x ≤ (2/3) x2 , (2/3) < x ≤ 1 endcases Clearly Rolle's theorem is applicable on [(1/3), (2/3)] where f(x)=2 x(1-x) ⇒ f prime(c)=2-4 c=0 ⇒ c=1 / 2 ⇒ a+b+c=(1/3)+(2/3)+(1/2) ⇒ (3/2)

Q. Consider the function $f (x) = max {x^{2}, (1 - x)^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ . Let Rolle's Theorem is applicable for $f (x)$ on greatest interval $[a, b]$ then $a + b + c$ is (where $c$ is point such that $f^{'} (c) = 0$ )

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{2}$

Q. Consider the function f(x)=max{x2,(1−x)2,2x(1−x)} where 0≤x≤1. Let Rolle's Theorem is applicable for f(x) on greatest interval [a,b] then a+b+c is (where c is point such that f′(c)=0 )

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Q. Consider the function $f (x) = max {x^{2}, (1 - x)^{2}, 2 x (1 - x)}$ where $0 \leq x \leq 1$ . Let Rolle's Theorem is applicable for $f (x)$ on greatest interval $[a, b]$ then $a + b + c$ is (where $c$ is point such that $f^{'} (c) = 0$ )

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{2}$