JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 21)
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
5
Explanation
$$f(x) = a{x^2} + bx + c$$
$$f'(x) = 2ax + b,$$
$$f''(x) = 2a$$
Given, $$f''( - 1) = {1 \over 2}$$
$$ \Rightarrow a = {1 \over 4}$$
$$f'( - 1) = 1 \Rightarrow b - 2a = 1$$
$$ \Rightarrow b = {3 \over 2}$$
$$f( - 1) = a - b + c = 2$$
$$ \Rightarrow c = {{13} \over 4}$$
Now, $$f(x) = {1 \over 4}({x^2} + 6x + 13),x \in [ - 1,1]$$
$$f'(x) = {1 \over 4}(2x + 6) = 0$$
$$ \Rightarrow x = - 3 \notin [ - 1,1]$$
$$f(1) = 5,f( - 1) = 2$$
$$f(x) \le 5$$
So, $$\alpha$$minimum = 5
$$f'(x) = 2ax + b,$$
$$f''(x) = 2a$$
Given, $$f''( - 1) = {1 \over 2}$$
$$ \Rightarrow a = {1 \over 4}$$
$$f'( - 1) = 1 \Rightarrow b - 2a = 1$$
$$ \Rightarrow b = {3 \over 2}$$
$$f( - 1) = a - b + c = 2$$
$$ \Rightarrow c = {{13} \over 4}$$
Now, $$f(x) = {1 \over 4}({x^2} + 6x + 13),x \in [ - 1,1]$$
$$f'(x) = {1 \over 4}(2x + 6) = 0$$
$$ \Rightarrow x = - 3 \notin [ - 1,1]$$
$$f(1) = 5,f( - 1) = 2$$
$$f(x) \le 5$$
So, $$\alpha$$minimum = 5
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