Why is the least value of $\alpha$ equal to the maximum value of the function?

Think of $\alpha$ as a ceiling. The condition $f(x) \le \alpha$ means that $\alpha$ must be greater than or equal to every single value the function takes on the interval. To make $\alpha$ as small as possible (the 'least value'), we must bring the ceiling down until it just touches the highest point of the function. Thus, $\alpha$ must equal the absolute maximum of $f(x)$.

JEE Main 2021 (17 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: 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 $f^{''} (x)$ is $1/2$ . If $f (x) \leq α$ for $x \in [- 1, 1]$ , then the least value of $α$ is equal to ____.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Look closely at the second derivative

f^{''} (x) = 2 a

. Since it is a constant, the condition that the maximum value of

f^{''} (x)

1/2

implies that

2 a

must be exactly

1/2

. This immediately gives you the value of

a

Hint 2

Hint 2: The Arsenal. You now have

a = 1/4

. Use the given conditions

f (- 1) = 2

and

f^{'} (- 1) = 1

to set up a system of linear equations. Remember that

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

. Substitute your known

a

and the value of

x

to solve for

b

, then use

f (- 1)

to find

c

Hint 3

Hint 3: The Execution. Once you have the full expression for

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

, calculate the first derivative

f^{'} (x)

. Check its sign on the interval

[- 1, 1]

. If

f^{'} (x) > 0

throughout the interval, the function is strictly increasing, meaning the maximum value must occur at the right endpoint,

x = 1

Visualized Solution

The Quadratic Function

Given function: $f (x) = a x^{2} + b x + c$
Domain: $x \in [- 1, 1]$
Goal: Find the least value $α$ such that $f (x) \leq α$

Using the Second Derivative

First derivative: $f^{'} (x) = 2 a x + b$
Second derivative: $f^{''} (x) = 2 a$
Given max $f^{''} (x) = \frac{1}{2} ⟹ 2 a = \frac{1}{2}$
Therefore, $a = \frac{1}{4}$

The Starting Point

Given condition: $f (- 1) = 2$
This means the curve passes exactly through the point $(- 1, 2)$ .

The Tangent Slope

Given condition: $f^{'} (- 1) = 1$
The slope of the tangent line at $x = - 1$ is exactly $1$ .

Determining Coefficient $b$

Substitute $a = \frac{1}{4}$ and $x = - 1$ into $f^{'} (x)$ :
$2 (\frac{1}{4}) (- 1) + b = 1$
$- \frac{1}{2} + b = 1 ⟹ b = \frac{3}{2}$

Determining Coefficient $c$

Substitute $a = \frac{1}{4}, b = \frac{3}{2}$ into $f (- 1) = 2$ :
$\frac{1}{4} (- 1)^{2} + \frac{3}{2} (- 1) + c = 2$
$\frac{1}{4} - \frac{3}{2} + c = 2 ⟹ c = \frac{13}{4}$

The Complete Function

The fully determined function is:
$f (x) = \frac{1}{4} x^{2} + \frac{3}{2} x + \frac{13}{4}$

Analyzing Monotonicity

Check $f^{'} (x)$ on the interval $[- 1, 1]$ :
$f^{'} (x) = \frac{1}{2} x + \frac{3}{2}$
For $x \geq - 1$ , $f^{'} (x) \geq \frac{1}{2} (- 1) + \frac{3}{2} = 1 > 0$
Conclusion: $f (x)$ is strictly increasing on $[- 1, 1]$ .

Finding the Maximum Value

Since $f (x)$ is increasing, the maximum value occurs at $x = 1$ .
$f (1) = \frac{1}{4} (1)^{2} + \frac{3}{2} (1) + \frac{13}{4}$
$f (1) = \frac{1}{4} + \frac{6}{4} + \frac{13}{4} = \frac{20}{4} = 5$

The Least Value of $α$

We need $f (x) \leq α$ for all $x \in [- 1, 1]$ .
The absolute maximum value of $f (x)$ is $5$ .
Therefore, the least possible value for $α$ is $5$ .

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We begin with the quadratic function defined as

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

. Our first task is to understand its curvature through the second derivative.

The second derivative is given by:

f^{''} (x) = 2 a

The problem states that the maximum value of

f^{''} (x)

on the interval

[- 1, 1]

1/2

. Since

2 a

is a constant, it does not change with

x

. Therefore, we must have:

2 a = \frac{1}{2} \Rightarrow a = \frac{1}{4}

The Detective Work

Now that we have determined

a

, we utilize the given anchors

f (- 1) = 2

and

f^{'} (- 1) = 1

to find

b

and

c

. We start with the slope condition using the derivative

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

Substituting

a = 1/4

and

x = - 1

2 (\frac{1}{4}) (- 1) + b = 1

- \frac{1}{2} + b = 1 \Rightarrow b = \frac{3}{2}

Next, we apply the point condition

f (- 1) = 2

to the original function:

(\frac{1}{4}) (- 1)^{2} + (\frac{3}{2}) (- 1) + c = 2

\frac{1}{4} - \frac{3}{2} + c = 2

Simplifying the fractions:

\frac{1}{4} - \frac{6}{4} + c = 2 \Rightarrow - \frac{5}{4} + c = 2

c = 2 + \frac{5}{4} = \frac{13}{4}

The fully determined function is:

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

The Monotonicity Reveal

To find the maximum value of

f (x)

on the interval

[- 1, 1]

, we examine the derivative:

f^{'} (x) = \frac{1}{2} x + \frac{3}{2}

On the interval

[- 1, 1]

, the minimum value of the derivative occurs at

x = - 1

, where

f^{'} (- 1) = 1

. Since

f^{'} (x) > 0

for all

x

in the interval, the function is strictly increasing.

Because the function is strictly increasing, the maximum value must occur at the rightmost boundary of the interval,

x = 1

The Final Victory

We calculate the value of the function at the boundary

x = 1

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

f (1) = \frac{1}{4} + \frac{6}{4} + \frac{13}{4} = \frac{20}{4} = 5

The problem asks for the least value of

α

such that

f (x) \leq α

for all

x \in [- 1, 1]

. Since the maximum value of the function is

5

, the least such value is:

α = 5

Common Traps & FAQs

Wait, why is the maximum value of

f^{''} (x)

equal to

1/2

f^{''} (x)

is just a constant?

That is a fantastic observation! In the case of a quadratic function

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

, the second derivative

f^{''} (x) = 2 a

is indeed a constant. If a constant function has a maximum value, that constant *is* the maximum value. Therefore,

2 a = 1/2

is the only logical conclusion.

Why can't I just find the vertex of the parabola to determine the maximum value?

You absolutely could, but you must be careful! The vertex of a parabola

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

occurs at

x = - b / (2 a)

. You should calculate this value and check if it lies within the interval

[- 1, 1]

. In this specific problem, you will find that the vertex lies outside the interval, which is why the function is monotonic (always increasing) on the given domain.

Why is the least value of

α

equal to the maximum value of the function?

Think of

α

as a ceiling. The condition

f (x) \leq α

means that

α

must be greater than or equal to every single value the function takes on the interval. To make

α

as small as possible (the 'least value'), we must bring the ceiling down until it just touches the highest point of the function. Thus,

α

must equal the absolute maximum of

f (x)

Enter Numerical Value:

Visualized Solution

The Quadratic Function

Using the Second Derivative

The Starting Point

The Tangent Slope

Determining Coefficient $b$

Determining Coefficient $c$

The Complete Function

Analyzing Monotonicity

Finding the Maximum Value

The Least Value of $α$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Detective Work

The Monotonicity Reveal

The Final Victory

Similar Questions

The least value of $a \in R$ for which $4 α x^{2} + \frac{1}{x} \geq 1$ , for all $x > 0$ , is

Comprehension Passage

Which of the following is true?

Which of the following is true?

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

The minimum value of the twice differentiable function $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ , is :

Comprehension Passage

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

Let $g (x) = f (x) + f (1 - x)$ and $f^{''} (x) > 0, x \in (0, 1)$ . If $g$ is decreasing in the interval $(0, α)$ and increasing in the interval $(α, 1)$ , then $tan^{- 1} 2 α + tan^{- 1} (\frac{1}{α}) + tan^{- 1} (\frac{α + 1}{α})$ is equal to

If $f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ , for every real number $x$ , then the minimum value of $f$

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{∣ t + 1∣}{t ^{2}}, t < 0$ , is strictly decreasing. Then the local maximum value of the function $g (x) = 2 lo g_{e} (x - 2) + α x^{2} + 4 x - α, x > 2$ , is .........

Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x ^{2} + x + 2}{x ^{2} + 5 x + 6} < 0$ , be $(α, β)$ . Then $α^{2} + β^{2}$ is equal to _________

Enter Numerical Value:

Visualized Solution

The Quadratic Function

Using the Second Derivative

The Starting Point

The Tangent Slope

Determining Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Determining Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

The Complete Function

Analyzing Monotonicity

Finding the Maximum Value

The Least Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Detective Work

The Monotonicity Reveal

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least value of a∈R for which 4αx2+x1​≥1, for all x>0, is

Comprehension Passage

Which of the following is true?

Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0ex​1+t2f′(t)​dt. Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The minimum value of the twice differentiable function f(x)=∫0x​ex−tf′(t)dt−(x2−x+1)ex,x∈R, is :

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is true for 0<x<1?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function e−xf(x) assumes its minimum in the interval [0,1] at x=1/4, which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x2+1x2−1​, for every real number x, then the minimum value of f

Determining Coefficient $b$

Determining Coefficient $c$

The Least Value of $α$

The least value of $a \in R$ for which $4 α x^{2} + \frac{1}{x} \geq 1$ , for all $x > 0$ , is

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

The minimum value of the twice differentiable function $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ , is :

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

If $f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ , for every real number $x$ , then the minimum value of $f$