Let x = -1 and x = 2 be the critical points of the function f(x) = x3 + ax2 + b |x| + 1, x neq 0 . Let m and M…

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More Application of Derivatives Previous-Year Questions — Page 2

Q jee_main_2025_02_april_morning Maxima and Minima

If the function

f(x) = 2x³ - 9ax² + 12a²x + 1

, where

a > 0

, attains its local maximum and local minimum values at

p

and

q

, respectively, such that

p² = q

, then

f(3)

is equal to:

A. $55$ $55$
B. $10$ $10$
C. $23$ $23$
D. $37$ $37$

Solution

Related Formula

For local extrema of a differentiable function, set the first derivative to zero:

$f'(x) = 0$ $f'(x) = 0$

Core Logic

Differentiate $f(x)$ $f(x)$ to obtain its critical points $p$ $p$ and $q$ $q$ in terms of $a$ $a$, use the constraint $p² = q$ $p^2 = q$ to fix $a$ $a$, and evaluate $f(3)$ $f(3)$.

Step 1: Differentiate and Find Critical Points

$f'(x) = 6x² - 18ax + 12a²$ $ f'(x) = 6x^2 - 18ax + 12a^2 $

Set$

Set $f'(x) = 0 $:$

6(x² - 3ax + 2a²) = 0 6(x-a)(x-2a) = 0

The critical points are $ The critical points are $x = a and $ and $ x = 2a . Since $. Since $ a > 0, checking the sign change of $, checking the sign change of $ f'(x) shows that the local maximum occurs at the smaller root ($ shows that the local maximum occurs at the smaller root ($ p = a) and the local minimum at the larger root ($) and the local minimum at the larger root ($ q = 2a). Step 2: Apply Root Constraint Given $). Step 2: Apply Root Constraint Given $p^2 = q :

a² = 2a a(a-2) = 0

Since $ Since $a > 0, we get $, we get $ a = 2 . Step 3: Evaluate function at x = 3 Substitute $. Step 3: Evaluate function at x = 3 Substitute $a = 2 back into the definition of $ back into the definition of $ f(x) :

f(x) = 2x³ - 18x² + 48x + 1

Now compute $ Now compute $f(3) :

f(3) = 2(3)³ - 18(3)² + 48(3) + 1 = 54 - 162 + 144 + 1 = 37

Pattern Recognition For cubic equations with two distinct real critical roots, the smaller root is always the local maximum if the leading coefficient is positive ($ Pattern Recognition For cubic equations with two distinct real critical roots, the smaller root is always the local maximum if the leading coefficient is positive ($2 > 0). This ensures $). This ensures $ p=a and $ and $ q=2a$ immediately without relying on secondary derivative checks. Chapter Mix Class 12 Mathematics: Application of Derivatives

Q53 jee_main_2025_08_april_evening Monotonicity

Let the function

f(x) = (x)/(3) + (3)/(x) + 3, x \neq 0

be strictly increasing in

(-\infty, α₁) (α₂, \infty)

and strictly decreasing in

(α₃, α₄) (α₄, α₅)

. Then

Σi=1⁵ αᵢ²

is equal to:

A. $48$ $48$
B. $28$ $28$
C. $40$ $40$
D. $36$ $36$

Solution

Related Formula

f'(x) > 0 Increasing

f'(x) < 0 Decreasing

Core Logic

Differentiate the rational function and determine the critical intervals by assessing where the derivative flips signs around critical points and domain boundaries.

Step 1: Derivative Assessment

Given $f(x) = (x)/(3) + (3)/(x) + 3$ $f(x) = \frac{x}{3} + \frac{3}{x} + 3$

f'(x) = (1)/(3) - (3)/(x²) = (x² - 9)/(3x²)

Critical points occur at $x = \pm 3$ $x = \pm 3$, and a domain discontinuity sits at $x = 0$ $x = 0$.

Step 2: Sign Scheme Mapping

Analyzing interval signs:

Strictly Increasing ( $f'(x) > 0$ $f'(x) > 0$): $x in (-\infty, -3) (3, \infty)$ $x \in (-\infty, -3) \cup (3, \infty)$
Strictly Decreasing ( $f'(x) < 0$ $f'(x) < 0$): $x in (-3, 0) (0, 3)$ $x \in (-3, 0) \cup (0, 3)$

Step 3: Interval Summation

Comparing bounds with assigned symbols:

α₁ = -3, α₂ = 3, α₃ = -3, α₄ = 0, α₅ = 3

Σi=1⁵ αᵢ² = (-3)² + (3)² + (-3)² + (0)² + (3)² = 9 + 9 + 9 + 0 + 9 = 36

Pattern Recognition

Functions of the form $x + (k)/(x)$ $x + \frac{k}{x}$ always present localized extrema turning symmetric zones at $\pm\sqrt(k)$ $\pm\sqrt{k}$. Always include the asymptotes ( $x=0$ $x=0$) when stating precise disjoint decreasing intervals.

Chapter Mix

Class 12 Mathematics: Application of Derivatives

Q68 jee_main_2025_29_jan_evening Maxi-Mini of Functions

Let

f(x) = \int₀x t(t² - 9t + 20) dt

1 \leq x \leq 5

. If the range of

f

[α, β]

, then

4(α + β)

equals:

A. $157$ $157$
B. $253$ $253$
C. $125$ $125$
D. $154$ $154$

Solution

Related Formula

Leibniz Rule for differentiation under integral sign:

(d)/(dx) ( \int₀x g(t) dt ) = g(x)

Core Logic

Find critical points inside the interval by finding $f'(x) = 0$ $f'(x) = 0$:

f'(x) = x(x² - 9x + 20) = x(x - 4)(x - 5) = 0

Maxi-Mini of Functions diagram for Q68 - JEE Main 2025 Evening

Critical points inside the interval $[1, 5]$ $[1, 5]$ are $x = 4$ $x = 4$ and $x = 5$ $x = 5$.

Step 1: Perform Integration

Integrate the function to evaluate boundary metrics:

f(x) = \int₀x (t³ - 9t² + 20t) dt = [ (t⁴)/(4) - 3t³ + 10t² ]₀x

f(x) = (x⁴)/(4) - 3x³ + 10x²

Step 2: Evaluate Points and Sum up Range Boundaries

Evaluate at endpoints and critical points:

f(1) = (1)/(4) - 3 + 10 = (29)/(4) = 7.25

f(4) = (256)/(4) - 3(64) + 10(16) = 64 - 192 + 160 = 32

f(5) = (625)/(4) - 3(125) + 10(25) = 156.25 - 375 + 250 = 31.25

Thus, the global minimum is $α = f(1) = (29)/(4)$ $\alpha = f(1) = \frac{29}{4}$, and the global maximum is \beta = f(4) = 32.

4(α + β) = 4 ( (29)/(4) + 32 ) = 29 + 128 = 157

Pattern Recognition

Always check boundary points along with interior critical values when finding the absolute range on a closed interval.

Chapter Mix

Class 12 Mathematics: Application of Derivatives Class 12 Mathematics: Integral Calculus

Q jee_main_2025_28_jan_morning Local Maxima and Minima

The sum of all local minimum values of the function

f(x) = cases 1 - 2x, & x < -1 (1)/(3)(7 + 2|x|), & -1 \leq x \leq 2 (11)/(18)(x - 4)(x - 5), & x > 2 cases

is:

A. $(171)/(72)$ $\frac{171}{72}$
B. $(131)/(72)$ $\frac{131}{72}$
C. $(157)/(72)$ $\frac{157}{72}$
D. $(167)/(72)$ $\frac{167}{72}$

Solution

Related Formula

Local minima occur at points where the derivative changes sign from negative to positive, or at sharp corners where the function reaches a local low point.

Core Logic

Analyze the piecewise sections across all transition domains:

Local Maxima and Minima diagram for Q64 - JEE Main 2025 Morning

For

x in [-1, 2]

f(x) = (1)/(3)(7 + 2|x|)

. This curve has a sharp structural minimum at

x = 0

since

|x|

decreases then increases. Value at

x = 0 f(0) = (7)/(3)

Step 1: Analyzing the Quadratic Domain Boundaries

For $x > 2$ $x > 2$, $f(x) = (11)/(18)(x² - 9x + 20)$ $f(x) = \frac{11}{18}(x^2 - 9x + 20)$. Finding the vertex point by setting the derivative to zero:

f^ (x) = (11)/(18)(2x - 9) = 0 x = (9)/(2) = 4.5

Since $4.5 > 2$ $4.5 > 2$, this vertex is within the valid domain and represents a local minimum. Value at $x = 4.5 f((9)/(2)) = (11)/(18)((1)/(2))(-(1)/(2)) = -(11)/(72)$ .

Step 2: Calculating Final Combined Summation

Summing both local minima:

Total Minima Sum = (7)/(3) + (-(11)/(72)) = (168 - 11)/(72) = (157)/(72)

Pattern Recognition

For piecewise function systems, always verify if the quadratic vertices fall inside their restricted intervals before adding them to your solution path.

Chapter Mix

Class 12 Maths: Application of Derivatives

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve for Coefficients

Step 2: Check Critical Points in Target Interval

Step 3: Calculate Absolute Sum

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Derivatives Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Differentiate and Find Critical Points

Step 2: Apply Root Constraint

Step 2: Apply Root Constraint

Step 3: Evaluate function at x = 3

Step 3: Evaluate function at x = 3

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Derivative Assessment

Step 2: Sign Scheme Mapping

Step 3: Interval Summation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Perform Integration

Step 2: Evaluate Points and Sum up Range Boundaries

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyzing the Quadratic Domain Boundaries

Step 2: Calculating Final Combined Summation

Pattern Recognition

Chapter Mix

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