Let f(x)=x3+x2f'(1)+xf''(2)+f'''(3), x R. Then f'(10) is equal to:

Year	2026	2025	2024	Total
Questions	5	8	12	25

Q5 jee_main_2024_30_january_evening Maxima and Minima

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:

A. $600$ $600$
B. $392$ $392$
C. $608$ $608$
D. $108$ $108$

Solution

Related Formula

To find absolute extrema, evaluate f(x) at critical points and domain boundaries.

Core Logic

Calculate the derivative of $f(x)$ $f(x)$ using the product rule:

f'(x) = (d)/(dx) [(x + 3)²] \cdot (x - 2)³ + (x + 3)² \cdot (d)/(dx) [(x - 2)³]

f'(x) = 2(x + 3)(x - 2)³ + 3(x + 3)²(x - 2)²

Factor out the common terms $(x+3)(x-2)²$ $(x+3)(x-2)^2$:

f'(x) = (x + 3)(x - 2)² [2(x - 2) + 3(x + 3)]

f'(x) = (x + 3)(x - 2)² [2x - 4 + 3x + 9]

f'(x) = 5(x + 3)(x - 2)² (x + 1)

Setting $f'(x) = 0$ $f'(x) = 0$ gives the critical points: $x = -3, -1, 2$ $x = -3, -1, 2$

Step 1: Evaluating at Critical Points and Boundaries

Evaluate $f(x)$ $f(x)$ at $x = -4, -3, -1, 2, 4$ $x = -4, -3, -1, 2, 4$:

Maxima and Minima diagram for Q5 - JEE Main 2024 Evening

For

x = -4

:

f(-4) = (-4 + 3)²(-4 - 2)³ = (1)(-216) = -216

For

x = -3

:

f(-3) = 0

For

x = -1

:

f(-1) = (-1 + 3)²(-1 - 2)³ = (4)(-27) = -108

For

x = 2

:

f(2) = 0

For

x = 4

:

f(4) = (4 + 3)²(4 - 2)³ = 49 \times 8 = 392

Step 2: Finding M and m

From the evaluated values, the maximum $M = 392$ $M = 392$ and the minimum $m = -216$ $m = -216$.

M - m = 392 - (-216) = 392 + 216 = 608

Pattern Recognition

For polynomials on a closed interval, critical points combined with boundary values reliably expose the absolute max and min.

Chapter Mix

Class 12 Maths: Application of Derivatives

Q5 jee_main_2024_30_jan_morning Maxima and Minima

The maximum area of a triangle whose one vertex is at

(0, 0)

and the other two vertices lie on the curve

y = -2x² + 54

at points

(x, y)

and

(-x, y)

where

y > 0

is:

A. $88$ $88$
B. $122$ $122$
C. $92$ $92$
D. $108$ $108$

Solution

Related Formula

Area of Δ = (1)/(2) \times base \times height

Core Logic

Maxima and Minima diagram for Q5 - JEE Main 2024 Morning

The vertices of the triangle are $(0, 0)$ $(0, 0)$, $(x, y)$ $(x, y)$, and $(-x, y)$ $(-x, y)$. The base of the triangle lies along the horizontal line segment joining $(-x, y)$ $(-x, y)$ and $(x, y)$ $(x, y)$. Length of base $= 2x$ $= 2x$. The height of the triangle from $(0,0)$ $(0,0)$ to the line segment is $y$ $y$. So, Area $Δ = (1)/(2) (2x) (y) = xy$ $\Delta = \frac{1}{2} (2x) (y) = xy$ (Assuming $x > 0$ $x > 0$).

Step 1: Setting up the area function

Substitute $y = -2x² + 54$ $y = -2x^2 + 54$ into the area function:

Area(Δ) = A(x) = x(-2x² + 54) = -2x³ + 54x

Step 2: Differentiating to find maximum area

To find the maximum area, take the derivative with respect to $x$ $x$ and set it to zero:

A'(x) = (dA)/(dx) = -6x² + 54

Setting $A'(x) = 0$ $A'(x) = 0$:

-6x² + 54 = 0 \Rightarrow 6x² = 54 \Rightarrow x² = 9

Since $x$ $x$ represents half the base length, $x = 3$ $x = 3$.

Step 3: Calculating maximum area

Substitute $x = 3$ $x = 3$ back into the area function:

Amax = 3(-2(3)² + 54) = 3(-18 + 54) = 3(36) = 108

Pattern Recognition

For symmetric figures inscribed under a parabolic arch, the area function $A(x) = x \cdot y(x)$ $A(x) = x \cdot y(x)$ is standard. Differentiate and find the critical point directly.

Chapter Mix

Class 12 Maths: Application of Derivatives

Q7 jee_main_2024_30_jan_morning Mean Value Theorems

Let

g : R \to R

be a non-constant twice differentiable such that

g'((1)/(2)) = g'((3)/(2))

. If a real valued function

f

is defined as

f(x) = (1)/(2)[g(x) + g(2 - x)]

, then

A. $f'(x) = 0$ $f'(x) = 0$ for at least two $x$ $x$ in $(0,2)$ $(0,2)$
B. $f'(x) = 0$ $f'(x) = 0$ for exactly one $x$ $x$ in $(0,1)$ $(0,1)$
C. $f'(x) = 0$ $f'(x) = 0$ for no $x$ $x$ in $(0,1)$ $(0,1)$
D. $f((3)/(2)) + f((1)/(2)) = 1$

Solution

Related Formula

Rolle's Theorem: If $f(x)$ $f(x)$ is continuous on $[a, b]$ $[a, b]$, differentiable on $(a, b)$ $(a, b)$, and $f(a) = f(b)$ $f(a) = f(b)$, then there exists at least one $c in (a, b)$ $c \in (a, b)$ such that $f'(c) = 0$ $f'(c) = 0$.

Core Logic

Given $f(x) = (1)/(2)[g(x) + g(2 - x)]$ . Differentiating $f(x)$ $f(x)$ with respect to $x$ $x$:

f'(x) = (g'(x) - g'(2 - x))/(2)

Step 1: Evaluating the derivative at given points

Evaluate $f'((3)/(2))$ $f'\left(\frac{3}{2}\right)$:

f'((3)/(2)) = (g'((3)/(2)) - g'((1)/(2)))/(2)

Since $g'((1)/(2)) = g'((3)/(2))$ , we get:

f'((3)/(2)) = 0

Also evaluate $f'((1)/(2))$ $f'\left(\frac{1}{2}\right)$:

f'((1)/(2)) = (g'((1)/(2)) - g'((3)/(2)))/(2) = 0

Additionally, observe that $f'(1) = (g'(1) - g'(1))/(2) = 0$ $f'(1) = \frac{g'(1) - g'(1)}{2} = 0$.

Step 2: Applying Rolle's Theorem

We have $f'((1)/(2)) = 0$ $f'\left(\frac{1}{2}\right) = 0$, $f'(1) = 0$ $f'(1) = 0$, and $f'((3)/(2)) = 0$ $f'\left(\frac{3}{2}\right) = 0$. By Rolle's Theorem on $f'(x)$ $f'(x)$ (though we are asked about $f'(x)=0$ $f'(x)=0$ directly), we already found three distinct roots for $f'(x) = 0$ $f'(x) = 0$: $x = (1)/(2), 1, (3)/(2)$ $x = \frac{1}{2}, 1, \frac{3}{2}$. Notice that $x = 1/2$ $x = 1/2$ and $x=1$ $x=1$ are in $(0, 2)$ $(0, 2)$, and $x=3/2$ $x=3/2$ is in $(0, 2)$ $(0, 2)$. Thus $f'(x) = 0$ $f'(x) = 0$ for at least three values in $(0, 2)$ $(0, 2)$, which satisfies the condition "at least two $x$ $x$ in $(0, 2)$ $(0, 2)$".

Pattern Recognition

Symmetry in function definition $g(x) + g(a-x)$ $g(x) + g(a-x)$ guarantees a critical point at $x = a/2$ $x = a/2$. Additional critical points arise due to the given condition on $g'(x)$ $g'(x)$, creating a scenario perfectly suited for Rolle's implications.

Chapter Mix

Class 12 Maths: Application of Derivatives

Q13 jee_main_2024_31_jan_evening Monotonicity and Distances

If the function

f: (-\infty, -1] \to (a, b]

defined by

f(x) = ex³ - 3x + 1

is one-one and onto, then the distance of the point

P(2b + 4, a + 2)

from the line

x + e⁻³y = 4

is :

A. $2 1 + e⁶$ $2\sqrt{1 + \mathrm{e}^6}$
B. $4 1 + e⁶$ $4\sqrt{1 + \mathrm{e}^6}$
C. $3 1 + e⁶$ $3\sqrt{1 + \mathrm{e}^{6}}$
D. $1 + e⁶$ $\sqrt{1 + \mathrm{e}^{6}}$

Solution

Related Formula

Distance of point (x₁, y₁) from line Ax+By+C=0 is d = |Ax₁ + By₁ + C|\sqrt(A² + B²)

Core Logic

Monotonicity and Distances diagram for Q13 - JEE Main 2024 Evening

Given $f(x) = ex³ - 3x + 1$ $f(x) = e^{x^3 - 3x + 1}$. Check monotonicity on $(-\infty, -1]$ $(-\infty, -1]$:

f'(x) = ex³ - 3x + 1 \cdot (3x² - 3) = 3(x-1)(x+1)ex³ - 3x + 1

For $x \leq -1$ $x \le -1$, $(x+1) \leq 0$ $(x+1) \le 0$ and $(x-1) < 0$ $(x-1) < 0$, making $f'(x) \geq 0$ $f'(x) \ge 0$. Hence $f(x)$ $f(x)$ is strictly increasing. Since $f$ $f$ is onto $(a,b]$ $(a,b]$, the range is dictated by the domain boundaries:

a = x\to-\infty f(x) = e-\infty = 0

b = f(-1) = e-1 + 3 + 1 = e³

Point $P(2b + 4, a + 2) \equiv P(2e³ + 4, 2)$ . Find distance from line $x + e⁻³y - 4 = 0$ $x + e^{-3}y - 4 = 0$:

d = |2e³ + 4 + e⁻³(2) - 4| 1² + (e⁻³)² = 2e³ + 2e⁻³ 1 + e⁻⁶

d = 2e³(1 + e⁻⁶) 1 + e⁻⁶ = 2e³ 1 + e⁻⁶ = 2\sqrt(e⁶ + 1)

Chapter Mix

Class 12 Maths: Applications of Derivatives Class 11 Maths: Straight Lines

Q22 jee_main_2024_31_jan_morning Maxima and Minima

Let

S = (-1, \infty)

and

f : S \to R

be defined as

f(x) = \int₋₁x (e^t - 1)¹¹ (2t - 1)⁵ (t - 2)⁷ (t - 3)¹² (2t - 10)⁶¹ dt

. Let

p =

Sum of square of the values of

x

, where

f(x)

attains local maxima on

S

and

q =

Sum of the values of

x

, where

f(x)

attains local minima on

S

. Then, the value of

p² + 2q

is

Numerical Answer. Answer: 27 to 27

Solution

Related Formula

f'(x) = 0 for critical points. Change from + \to - means local maxima, - \to + means local minima.

Core Logic

Using Newton-Leibniz formula:

f'(x) = (e^x - 1)¹¹ (2x - 1)⁵ (x - 2)⁷ (x - 3)¹² (2x - 10)⁶¹

The critical points are roots of $f'(x) = 0$ $f'(x) = 0$: $x = 0, (1)/(2), 2, 3, 5$ $x = 0, \frac{1}{2}, 2, 3, 5$.

Maxima and Minima diagram for Q22 - JEE Main 2024 Morning

Step 1: Sign Change Analysis

Perform the Wavy Curve Method for $x in (-1, \infty)$ $x \in (-1, \infty)$: At $x=5$ $x=5$ (odd power 61): sign changes $- \to +$ $- \to +$ (Local Minima) At $x=3$ $x=3$ (even power 12): sign doesn't change (Neither) At $x=2$ $x=2$ (odd power 7): sign changes $+ \to -$ $+ \to -$ (Local Maxima) At $x=1/2$ $x=1/2$ (odd power 5): sign changes $- \to +$ $- \to +$ (Local Minima) At $x=0$ $x=0$ (from $e^x - 1$ $e^x - 1$, odd power 11): sign changes $+ \to -$ $+ \to -$ (Local Maxima)

Step 2: Compute p and q

Local minima at $x = 5, (1)/(2)$ $x = 5, \frac{1}{2}$. Local maxima at $x = 2, 0$ $x = 2, 0$.

p = 0² + 2² = 4

q = 5 + (1)/(2) = (11)/(2)

Step 3: Final Computation

p² + 2q = 16 + 2((11)/(2)) = 16 + 11 = 27

Chapter Mix

Class 12 Maths: Application of Derivatives Class 12 Maths: Integrals

Let f(x)=x³+x²f'(1)+xf''(2)+f'''(3)$f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)$, xin R$x\in R$. Then f'(10)$f'(10)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Unknown Constants

Step 2: Solving the System

Step 3: Final Output Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Derivatives Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Evaluating at Critical Points and Boundaries

Step 2: Finding M and m

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the area function

Step 2: Differentiating to find maximum area

Step 3: Calculating maximum area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating the derivative at given points

Step 2: Applying Rolle's Theorem

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Sign Change Analysis

Step 2: Compute p and q

Step 3: Final Computation

Chapter Mix

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Let $f(x)=x³+x²f'(1)+xf''(2)+f'''(3)$ $f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)$, $xin R$ $x\in R$. Then $f'(10)$ $f'(10)$ is equal to: