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CUET
-- Mathematics - Section B1
Applications of Derivatives
The minimum value of the function f(x)=a2x+b2a−x, a > 0, b > 0, in (0, a), is |
a+b 1a+b (a+b)2a a+ba2 |
(a+b)2a |
We have, f(x)=a2x+b2a−x ⇒f′(x)=−a2x2+b2(a−x)2 ∴ f′(x)=0 ⇒a2x2=b2(a−x)2 ⇒a2(a−x)2=b2x2 ⇒ a(a−x)=±bx ⇒ x=a2a±b⇒x=a2a+b \left[∵ \frac{a^2}{a-b} \notin(0, a)\right] Now, f''(x)=\frac{2 a^2}{x^3}+\frac{2 b^2}{(a-x)^2}>0 for all x \in(0, a) Hence, f(x) is minimum at x=\frac{a^2}{a+b} with the minimum value given by f\left(\frac{a^2}{a+b}\right)=(a+b)+\frac{b^2}{a-\frac{a^2}{a+b}}=(a+b)+\frac{b^2(a+b)}{a b}=\frac{(a+b)^2}{a} |