Let f : [$-$1, 1] $\to$ R be defined as f(x) = ax2 + bx + c for all x$\in$[$-$1, 1], where a, b, c$\in$R such that f($-$1) = 2, f'($-$1) = 1 for x$\in$($-$1, 1) the maximum value of f ''(x) is ${{1 \over 2}}$. If f(x) $\le$ $\alpha$, x$\in$[$-$1, 1], then the least value of $\alpha$ is equal to _________.
Answer (integer)
5
Solution
$f(x) = a{x^2} + bx + c$<br><br>$f'(x) = 2ax + b,$<br><br>$f''(x) = 2a$<br><br>Given, $f''( - 1) = {1 \over 2}$<br><br>$\Rightarrow a = {1 \over 4}$<br><br>$f'( - 1) = 1 \Rightarrow b - 2a = 1$<br><br>$\Rightarrow b = {3 \over 2}$<br><br>$f( - 1) = a - b + c = 2$<br><br>$\Rightarrow c = {{13} \over 4}$<br><br>Now, $f(x) = {1 \over 4}({x^2} + 6x + 13),x \in [ - 1,1]$<br><br>$f'(x) = {1 \over 4}(2x + 6) = 0$<br><br>$\Rightarrow x = - 3 \notin [ - 1,1]$<br><br>$f(1) = 5,f( - 1) = 2$<br><br>$f(x) \le 5$<br><br>So, $\alpha$<sub>minimum</sub> = 5
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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