JEE Advanced 1988

LEVELJEE Main

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

Q: Wait, why can't I just integrate the function first and then differentiate it? Isn't that safer?

That is a classic trap! While mathematically valid, it is a massive waste of time. In JEE Advanced, time is your most precious resource. By using the Newton-Leibniz rule, you bypass the integration entirely. Always look for the 'shortcut' that the problem designer has hidden in plain sight.

Q: Why is $x = 2$ not a local maximum or minimum? The derivative is zero there, so shouldn't it be an extremum?

Great observation! A zero derivative is a necessary condition for an extremum, but not a sufficient one. At $x = 2$, the factor $(x - 2)^2$ has an even power. This means the sign of $f'(x)$ does not change as you cross $x = 2$. Since the function doesn't switch from increasing to decreasing (or vice versa), it is just a momentary pause—a point of inflection, not an extremum.

Q: How do I know if it's a local maximum or a local minimum without checking the second derivative?

You do not need the second derivative! The First Derivative Test is your best friend. If the sign of $f'(x)$ changes from positive to negative, you have climbed a hill (Maximum). If it changes from negative to positive, you have dipped into a valley (Minimum). It is all about the 'flow' of the function.

Visualized Solution

Analyzing the Function $f (x)$

Given: $f (x) = \int_{1}^{x} [2 (t - 1) (t - 2)^{3} + 3 (t - 1)^{2} (t - 2)^{2}] d t$
Goal: Find points of local maxima and minima.
Strategy: Differentiate $f (x)$ to find critical points and analyze the sign of $f^{'} (x)$ .

Applying Newton-Leibniz Rule

Use the Newton-Leibniz Rule for differentiating under the integral sign:
$\frac{d}{d x} \int_{a (x)}^{b (x)} g (t) d t = g (b (x)) \cdot b^{'} (x) - g (a (x)) \cdot a^{'} (x)$
Here, the upper limit is $x$ and the lower limit is $1$ .
$f^{'} (x) = [2 (x - 1) (x - 2)^{3} + 3 (x - 1)^{2} (x - 2)^{2}] \cdot \frac{d}{d x} (x) - 0$

Factoring the Derivative

$f^{'} (x) = 2 (x - 1) (x - 2)^{3} + 3 (x - 1)^{2} (x - 2)^{2}$
Extract the common factors: $(x - 1)$ and $(x - 2)^{2}$ .
$f^{'} (x) = (x - 1) (x - 2)^{2} [2 (x - 2) + 3 (x - 1)]$

Simplifying the Expression

Simplify the terms inside the square bracket:
$2 (x - 2) + 3 (x - 1) = 2 x - 4 + 3 x - 3$
Combine like terms: $5 x - 7$
Final factored form: $f^{'} (x) = (x - 1) (x - 2)^{2} (5 x - 7)$

Identifying Critical Points

Set $f^{'} (x) = 0$ to find the critical points.
$(x - 1) (x - 2)^{2} (5 x - 7) = 0$
Equating each factor to zero gives:
$x = 1$
$x = \frac{7}{5}$
$x = 2$

Sign Scheme: $x > 2$

We use the Wavy Curve Method to find the sign of $f^{'} (x)$ .
For $x > 2$ , choose a test value, say $x = 3$ .
$f^{'} (3) = (3 - 1) (3 - 2)^{2} (5 (3) - 7) = (+) (+) (+) > 0$
Thus, $f^{'} (x)$ is positive in $(2, \infty)$ .

Sign Scheme: $\frac{7}{5} < x < 2$

Moving left across $x = 2$ .
The factor $(x - 2)$ has an even power of $2$ .
Rule: The sign of the expression does not change when crossing a root with an even power.
Thus, $f^{'} (x)$ remains positive in $(\frac{7}{5}, 2)$ .

Sign Scheme: $1 < x < \frac{7}{5}$

Moving left across $x = \frac{7}{5}$ .
The factor $(5 x - 7)$ has an odd power of $1$ .
Rule: The sign of the expression changes when crossing a root with an odd power.
Thus, $f^{'} (x)$ becomes negative in $(1, \frac{7}{5})$ .

Sign Scheme: $x < 1$

Moving left across $x = 1$ .
The factor $(x - 1)$ has an odd power of $1$ .
The sign changes again.
Thus, $f^{'} (x)$ becomes positive in $(- \infty, 1)$ .

Local Maxima at $x = 1$

Apply the First Derivative Test at $x = 1$ .
As $x$ increases through $1$ , the sign of $f^{'} (x)$ changes from positive ( $+$ ) to negative ( $-$ ).
This means the function changes from increasing to decreasing.
Therefore, $f (x)$ has a Local Maximum at $x = 1$ .

Local Minima at $x = \frac{7}{5}$

Apply the First Derivative Test at $x = \frac{7}{5}$ .
As $x$ increases through $\frac{7}{5}$ , the sign of $f^{'} (x)$ changes from negative ( $-$ ) to positive ( $+$ ).
This means the function changes from decreasing to increasing.
Therefore, $f (x)$ has a Local Minimum at $x = \frac{7}{5}$ .

Point of Inflection at $x = 2$

Apply the First Derivative Test at $x = 2$ .
As $x$ increases through $2$ , the sign of $f^{'} (x)$ remains positive ( $+$ to $+$ ).
Since there is no sign change, it is neither a maximum nor a minimum.
Therefore, $x = 2$ is a Point of Inflection.

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are given the function:

f (x) = \int_{1}^{x} [2 (t - 1) (t - 2)^{3} + 3 (t - 1)^{2} (t - 2)^{2}] d t

When encountering an integral function of this form, avoid the temptation to integrate the polynomial directly. In JEE Advanced, the objective is to identify the underlying structure of the expression rather than performing brute-force calculation.

The Newton-Leibniz Insight

To determine the maxima and minima, we must find the derivative

f^{'} (x)

. By applying the Newton-Leibniz rule, the derivative of an integral with a variable upper limit is simply the integrand evaluated at that limit.

The derivative is:

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

The integral sign is eliminated, revealing the core polynomial structure of the derivative.

The Art of Factoring

Do not expand the polynomial terms, as this will lead to unnecessary complexity. Instead, identify the common factors, which are

(x - 1)

and

(x - 2)^{2}

Factoring these out yields:

f^{'} (x) = (x - 1) (x - 2)^{2} [2 (x - 2) + 3 (x - 1)]

Simplifying the expression inside the brackets:

2 x - 4 + 3 x - 3 = 5 x - 7

Thus, the final factored form of the derivative is:

f^{'} (x) = (x - 1) (x - 2)^{2} (5 x - 7)

The Wavy Curve Method

To analyze the sign of

f^{'} (x)

, we plot the critical points on the number line:

x = 1

x = 7/5

, and

x = 2

Starting from the right (

x > 2

), the expression is positive. As we cross

x = 2

, note that the factor

(x - 2)^{2}

has an even power. Consequently, the sign does not change, and the curve touches the axis before remaining positive in the interval

(7/5, 2)

Next, crossing

x = 7/5

(which corresponds to the factor

(5 x - 7)

with an odd power), the sign changes from positive to negative. Finally, crossing

x = 1

(which corresponds to

(x - 1)

with an odd power), the sign changes again from negative to positive.

The Conclusion

We interpret the sign changes of

f^{'} (x)

to identify the extrema:

1. At $x = 1$ , the derivative changes from positive to negative, indicating a local maximum. 2. At $x = 7/5$ , the derivative changes from negative to positive, indicating a local minimum. 3. At $x = 2$ , the derivative remains positive on both sides, indicating a point of inflection.

By utilizing these structural insights, you can efficiently solve complex integral function problems without falling into common algebraic traps.

Common Traps & FAQs

Wait, why can't I just integrate the function first and then differentiate it? Isn't that safer?

That is a classic trap! While mathematically valid, it is a massive waste of time. In JEE Advanced, time is your most precious resource. By using the Newton-Leibniz rule, you bypass the integration entirely. Always look for the 'shortcut' that the problem designer has hidden in plain sight.

Why is

x = 2

not a local maximum or minimum? The derivative is zero there, so shouldn't it be an extremum?

Great observation! A zero derivative is a necessary condition for an extremum, but not a sufficient one. At

x = 2

, the factor

(x - 2)^{2}

has an even power. This means the sign of

f^{'} (x)

does not change as you cross

x = 2

. Since the function doesn't switch from increasing to decreasing (or vice versa), it is just a momentary pause—a point of inflection, not an extremum.

How do I know if it's a local maximum or a local minimum without checking the second derivative?

You do not need the second derivative! The First Derivative Test is your best friend. If the sign of

f^{'} (x)

changes from positive to negative, you have climbed a hill (Maximum). If it changes from negative to positive, you have dipped into a valley (Minimum). It is all about the 'flow' of the function.

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

Visualized Solution

Analyzing the Function $f (x)$

Applying Newton-Leibniz Rule

Factoring the Derivative

Simplifying the Expression

Identifying Critical Points

Sign Scheme: $x > 2$

Sign Scheme: $\frac{7}{5} < x < 2$

Sign Scheme: $1 < x < \frac{7}{5}$

Sign Scheme: $x < 1$

Local Maxima at $x = 1$

Local Minima at $x = \frac{7}{5}$

Point of Inflection at $x = 2$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Newton-Leibniz Insight

The Art of Factoring

The Wavy Curve Method

The Conclusion

Similar Questions

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Investigate for maxima and minima the function f(x)=∫1x​[2(t−1)(t−2)3+3(t−1)2(t−2)2]dt

Visualized Solution

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

Applying Newton-Leibniz Rule

Factoring the Derivative

Simplifying the Expression

Identifying Critical Points

Sign Scheme: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>2

Sign Scheme: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 57​<x<2

Sign Scheme: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1<x<57​

Sign Scheme: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<1

Local Maxima at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Local Minima at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=57​

Point of Inflection at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Newton-Leibniz Insight

The Art of Factoring

The Wavy Curve Method

The Conclusion

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

Analyzing the Function $f (x)$

Sign Scheme: $x > 2$

Sign Scheme: $\frac{7}{5} < x < 2$

Sign Scheme: $1 < x < \frac{7}{5}$

Sign Scheme: $x < 1$

Local Maxima at $x = 1$

Local Minima at $x = \frac{7}{5}$

Point of Inflection at $x = 2$

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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:

The total number of local maxima and local minima of the function $f (x) = {(2 + x)^{3}, x^{2/3}, - 3 < x \leq - 1 - 1 < x < 2$ is

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

The function $f (x) = 2∣ x ∣ + ∣ x + 2∣ - ∣∣ x + 2∣ - 2∣ x ∣∣$ has a local minimum or a local maximum at $x =$

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

The sum of absolute maximum and absolute minimum values of the function $f (x) = ∣2 x^{2} + 3 x - 2∣ + sin x cos x$ in the interval $[0, 1]$ is :