Let $f:[-1,1] \rightarrow R$ be defined as $f(x)=a x^{2}+b x+c$ for all $x \in[-1,1]$, where $a, b, c \in R$ such that $f(-1)=2$, $f^{\prime}(-1)=1$ and for $x \in(-1,1)$ the maximum value of $f^{\prime \prime}(x)$ is $\frac{1}{2}$. If $f(x) \leq \alpha$, $x \in[-1,1]$, then the least value of $\alpha$ is equal to ____.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Application of Derivatives

Let

f : [- 1, 1] \to R

be defined as

f (x) = a x^{2} + b x + c

for all

x \in [- 1, 1]

, where

a, b, c \in R

such that

f (- 1) = 2

f^{'} (- 1) = 1

and for

x \in (- 1, 1)

the maximum value of

f^{''} (x)

\frac{1}{2}

. If

f (x) \leq α

x \in [- 1, 1]

, then the least value of

α

is equal to ____.

Answer: 5

For

f (x) = a x^{2} + b x + c

the second derivative is the constant

2 a

, so its maximum value on

(- 1, 1)

2 a

, and

2 a = \frac{1}{2}

gives

a = \frac{1}{4}

.
From

f^{'} (- 1) = - 2 a + b = 1

comes

b = \frac{3}{2}

, and from

f (- 1) = a - b + c = 2

comes

c = \frac{13}{4}

, so

f (x) = \frac{x ^{2}}{4} + \frac{3 x}{2} + \frac{13}{4}

[- 1, 1]

f^{'} (x) = \frac{x}{2} + \frac{3}{2} > 0

, so

f

increases and attains its greatest value at the right endpoint:

f (1) = \frac{1}{4} + \frac{3}{2} + \frac{13}{4} = 5

The smallest

α

for which

f (x) \leq α

[- 1, 1]

5

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer