JEE Main 2025 (January)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ If the range of f is $[α, β]$ , then $4 (α + β)$ equals

Q: Why do we check $f(1)$ and $f(5)$? Shouldn't we just look at the critical points?

The critical points only tell us about local extrema. On a closed interval, the absolute maximum or minimum can occur at the boundaries. Always check the endpoints to ensure you have captured the global range!

Q: Why is $f(4)$ a local maximum?

By checking the sign of $f'(x) = x(x-4)(x-5)$ around $x=4$, we see it changes from positive to negative. A positive-to-negative transition in the derivative signifies a peak in the function's graph.

Q: Is it possible for the range to be wider than the values at the critical points?

No, for a continuous function on a closed interval, the range is strictly bounded by the absolute minimum and maximum, which must occur at either a critical point or an endpoint. You have everything you need!

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Given function: $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$
Domain of interest: $x \in [1, 5]$
Goal: Find the range $[α, β]$ and evaluate $4 (α + β)$

Applying the Leibniz Rule

To find the range, we first find $f^{'} (x)$ using the Leibniz Rule.
$\frac{d}{d x} \int_{0}^{x} g (t) d t = g (x) \cdot \frac{d}{d x} (x) - g (0) \cdot \frac{d}{d x} (0)$
$f^{'} (x) = x (x^{2} - 9 x + 20)$

Factorizing the Derivative

Factorize the quadratic term: $x^{2} - 9 x + 20 = (x - 4) (x - 5)$
So, $f^{'} (x) = x (x - 4) (x - 5)$
Critical points are at $x = 0, 4, 5$ .
Within our domain $[1, 5]$ , the critical points are $x = 4$ and $x = 5$ .

Analyzing Monotonicity

For $x \in (1, 4)$ : $f^{'} (x) = (+) (-) (-) > 0 ⟹ f (x)$ is strictly increasing.
For $x \in (4, 5)$ : $f^{'} (x) = (+) (+) (-) < 0 ⟹ f (x)$ is strictly decreasing.
Therefore, a local maximum occurs at $x = 4$ .

Integrating to find $f (x)$

Let's evaluate the integral to find the explicit function.
$f (x) = \int_{0}^{x} (t^{3} - 9 t^{2} + 20 t) d t$
$f (x) = [\frac{t ^{4}}{4} - 3 t^{3} + 10 t^{2}]_{0}^{x}$
$f (x) = \frac{x ^{4}}{4} - 3 x^{3} + 10 x^{2}$

Evaluating $f (1)$ (Left Boundary)

Substitute $x = 1$ into $f (x)$ :
$f (1) = \frac{1 ^{4}}{4} - 3 (1)^{3} + 10 (1)^{2}$
$f (1) = \frac{1}{4} - 3 + 10 = 7 + \frac{1}{4}$
$f (1) = \frac{29}{4}$

Evaluating $f (4)$ (Local Maximum)

Substitute $x = 4$ into $f (x)$ :
$f (4) = \frac{4 ^{4}}{4} - 3 (4)^{3} + 10 (4)^{2}$
$f (4) = 64 - 192 + 160$
$f (4) = 32$

Evaluating $f (5)$ (Right Boundary)

Substitute $x = 5$ into $f (x)$ :
$f (5) = \frac{5 ^{4}}{4} - 3 (5)^{3} + 10 (5)^{2}$
$f (5) = \frac{625}{4} - 375 + 250 = \frac{625}{4} - 125$
$f (5) = \frac{625 - 500}{4} = \frac{125}{4}$

Determining Range $[α, β]$

Comparing the values: $f (1) = \frac{29}{4} = 7.25$ , $f (4) = 32$ , $f (5) = \frac{125}{4} = 31.25$
Absolute minimum $α = \frac{29}{4}$
Absolute maximum $β = 32$
Range is $[\frac{29}{4}, 32]$

Final Calculation

We need to calculate $4 (α + β)$ :
$4 (\frac{29}{4} + 32) = 4 \cdot \frac{29}{4} + 4 \cdot 32$
$= 29 + 128$
$= 157$

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Landscape

The function is defined as:

f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t

We are tasked with finding the range of this function over the interval

x \in [1, 5]

. This requires identifying the absolute minimum

α

and the absolute maximum

β

within this domain.

The X-Ray Vision

Leibniz Rule

To understand the behavior of the function, we apply the Leibniz Rule to find the derivative

f^{'} (x)

f^{'} (x) = x (x^{2} - 9 x + 20)

Factoring the quadratic expression, we obtain:

f^{'} (x) = x (x - 4) (x - 5)

This derivative acts as our compass. Within the interval

[1, 5]

, the critical points occur where

f^{'} (x) = 0

, specifically at

x = 4

and

x = 5

Tracing the Curve

We analyze the monotonicity of the function. For

x \in (1, 4)

, the derivative

f^{'} (x)

is positive, indicating that the function is strictly increasing.

x = 4

, the function reaches a local maximum. For

x \in (4, 5)

, the derivative

f^{'} (x)

becomes negative, indicating that the function is strictly decreasing.

To find the explicit form of the function, we integrate the integrand

t^{3} - 9 t^{2} + 20 t

f (x) = \int (t^{3} - 9 t^{2} + 20 t) d t = \frac{x ^{4}}{4} - 3 x^{3} + 10 x^{2}

The Grand Finale

Evaluating the Heights

We now evaluate the function at the boundaries and the critical point to determine the range:

At the start,

x = 1

f (1) = \frac{1}{4} - 3 + 10 = 7.25 = \frac{29}{4}

At the peak,

x = 4

f (4) = \frac{256}{4} - 3 (64) + 10 (16) = 64 - 192 + 160 = 32

At the end of the interval,

x = 5

f (5) = \frac{625}{4} - 3 (375) + 10 (25) = \frac{625}{4} - 375 + 250 = \frac{125}{4} = 31.25

Comparing these values, the absolute minimum is

α = \frac{29}{4}

and the absolute maximum is

β = 32

. The range of the function is

[\frac{29}{4}, 32]

Finally, we compute the requested value:

4 (α + β) = 4 (\frac{29}{4} + 32) = 29 + 128 = * * 157 * *

Common Traps & FAQs

Why do we check

f (1)

and

f (5)

? Shouldn't we just look at the critical points?

The critical points only tell us about local extrema. On a closed interval, the absolute maximum or minimum can occur at the boundaries. Always check the endpoints to ensure you have captured the global range!

Why is

f (4)

a local maximum?

By checking the sign of

f^{'} (x) = x (x - 4) (x - 5)

around

x = 4

, we see it changes from positive to negative. A positive-to-negative transition in the derivative signifies a peak in the function's graph.

Is it possible for the range to be wider than the values at the critical points?

No, for a continuous function on a closed interval, the range is strictly bounded by the absolute minimum and maximum, which must occur at either a critical point or an endpoint. You have everything you need!

Animated Solution for Mathematics - Differentiation: Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ If the range of f is $[α, β]$ , then $4 (α + β)$ equals

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Applying the Leibniz Rule

Factorizing the Derivative

Analyzing Monotonicity

Integrating to find $f (x)$

Evaluating $f (1)$ (Left Boundary)

Evaluating $f (4)$ (Local Maximum)

Evaluating $f (5)$ (Right Boundary)

Determining Range $[α, β]$

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Landscape

The X-Ray Vision

Tracing the Curve

The Grand Finale

Similar Questions

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Select Answer:

Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Applying the Leibniz Rule

Factorizing the Derivative

Analyzing Monotonicity

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Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1) (Left Boundary)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(4) (Local Maximum)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(5) (Right Boundary)

Determining Range .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [α,β]

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Landscape

The X-Ray Vision

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The Grand Finale

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Animated Solution for Mathematics - Differentiation: Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ If the range of f is $[α, β]$ , then $4 (α + β)$ equals

Understanding the Function $f (x)$

Integrating to find $f (x)$

Evaluating $f (1)$ (Left Boundary)

Evaluating $f (4)$ (Local Maximum)

Evaluating $f (5)$ (Right Boundary)

Determining Range $[α, β]$

Investigate for maxima and minima the function $f (x) = \int_{1}^{x} [2 (t - 1) (t - 2)^{3} + 3 (t - 1)^{2} (t - 2)^{2}] d t$

The function $f (x) = \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t$ has a local minimum at $x =$

Let $f (x) = (x + 3)^{2} (x - 2)^{3}, x \in [- 4, 4]$ . If $M$ and $m$ are the maximum and minimum values of $f$ , respectively in $[- 4, 4]$ , then the value of $M - m$ is :

Let $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 8 t + 15}{e ^{t}}$ dt, $x \in R$ Then the numbers of local maximum and local minimum points of f, respectively, are:

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If m and n respectively are the number of local maximum and local minimum points of the function $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 5 t + 4}{2 + e ^{t}} d t$ , then the ordered pair (m, n) is equal to

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The sum of the absolute maximum and minimum values of the function $f (x) = ∣ x^{2} - 5 x + 6∣ - 3 x + 2$ in the interval $[- 1, 3]$ is equal to :

The sum of all local minimum values of the function $f (x) = ⎩ ⎨ ⎧ 1 - 2 x, \frac{1}{3} (7 + 2∣ x ∣), \frac{1}{18} (x - 4) (x - 5), x < - 1 - 1 \leq x \leq 2 x > 2$ is