Let for a differentiable functionf:(0,infty)rightarrow R$f:(0,\infty)\rightarrow R$, f(x)-f(y)ge log_eleft(fracxyright)+x-y, forall x, y in (0,infty)$f(x)-f(y)\ge \log_{e}\left(\frac{x}{y}\right)+x-y, \forall x, y \in (0,\infty)$. Then sum_n=1^20f'left(frac1n^2right)$\sum_{n=1}^{20}f'\left(\frac{1}{n^{2}}\right)$ is equal to:
Numerical Answer Type:
Enter a numerical valueAnswer: 2890 to 2890+4 marks
Solution & Explanation
### Related Formula
f'(x) = lim_y to x fracf(x)-f(y)x-y$f'(x) = \lim_{y \to x} \frac{f(x)-f(y)}{x-y}$sum_n=1^k n^2 = frack(k+1)(2k+1)6$\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}$
### Core Logic
Expand the given logarithmic inequality:
f(x) - f(y) ge ln x - ln y + x - y$f(x) - f(y) \ge \ln x - \ln y + x - y$
Divide both sides by (x - y)$(x - y)$. The inequality sign will behave differently based on whether (x - y)$(x - y)$ is positive or negative.
Case 1: Let x > y$x > y$ (so x - y > 0$x - y > 0$).
fracf(x)-f(y)x-y ge fracln x - ln yx-y + 1$\frac{f(x)-f(y)}{x-y} \ge \frac{\ln x - \ln y}{x-y} + 1$
Taking the limit as y to x^-$y \to x^-$:
f'(x) ge fracddx(ln x) + 1 Rightarrow f'(x) ge frac1x + 1$f'(x) \ge \frac{d}{dx}(\ln x) + 1 \Rightarrow f'(x) \ge \frac{1}{x} + 1$
### Step 1: Analyzing the second case
Case 2: Let x < y$x < y$ (so x - y < 0$x - y < 0$).
Dividing flips the inequality sign:
fracf(x)-f(y)x-y le fracln x - ln yx-y + 1$\frac{f(x)-f(y)}{x-y} \le \frac{\ln x - \ln y}{x-y} + 1$
Taking the limit as y to x^+$y \to x^+$:
f'(x) le fracddx(ln x) + 1 Rightarrow f'(x) le frac1x + 1$f'(x) \le \frac{d}{dx}(\ln x) + 1 \Rightarrow f'(x) \le \frac{1}{x} + 1$
### Step 2: Squeeze Theorem deduction
Since f$f$ is given as a differentiable function, both the left-hand and right-hand limits must yield the exact same derivative.
Thus, it is sandwiched between the bounds:
f'(x) = frac1x + 1$f'(x) = \frac{1}{x} + 1$
### Step 3: Evaluating the Summation
We need to compute sum_n=1^20 f'left(frac1n^2right)$\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right)$.
Substitute x = frac1n^2$x = \frac{1}{n^2}$ into the derivative:
f'left(frac1n^2right) = frac11/n^2 + 1 = n^2 + 1$f'\left(\frac{1}{n^2}\right) = \frac{1}{1/n^2} + 1 = n^2 + 1$
Apply the summation:
sum_n=1^20 (n^2 + 1) = sum_n=1^20 n^2 + sum_n=1^20 1$\sum_{n=1}^{20} (n^2 + 1) = \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1$= frac20 times 21 times 416 + 20$= \frac{20 \times 21 \times 41}{6} + 20$= 2870 + 20 = 2890$= 2870 + 20 = 2890$
### Pattern Recognition
Symmetric functional inequalities bounded by identical structural forms always compress down to an equality via the Sandwich/Squeeze theorem. Create the difference quotient limit to extract the derivative directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Derivatives
Class 11 Maths: Sequences and Series
Class 11 Maths: Limits and Derivatives
Keywords:#differentiable function#JEE Main 2024 Morning Q22#Application of Derivatives JEE Main 2024#Monotonicity JEE Main 2024
More Application of Derivatives Previous-Year Questions — Page 4
Q5jee_main_2024_30_january_eveningMaxima and Minima
Let f(x) = (x + 3)^2 (x - 2)^3, x in [-4, 4]$f(x) = (x + 3)^2 (x - 2)^3, x \in [-4, 4]$ . If M$M$ and m$m$ are the maximum and minimum values of f$f$ , respectively in [-4, 4]$[-4, 4]$ , then the value of M - m$M - m$ is:
A.600$600$
B.392$392$
C.608$608$
D.108$108$
Solution
### Related Formula
textTo find absolute extrema, evaluate f(x) text at critical points and domain boundaries.$\text{To find absolute extrema, evaluate } f(x) \text{ at critical points and domain boundaries.}$
### Core Logic
Calculate the derivative of f(x)$f(x)$ using the product rule:
f'(x) = fracddx [(x + 3)^2] cdot (x - 2)^3 + (x + 3)^2 cdot fracddx [(x - 2)^3]$f'(x) = \frac{d}{dx} [(x + 3)^2] \cdot (x - 2)^3 + (x + 3)^2 \cdot \frac{d}{dx} [(x - 2)^3]$f'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2$f'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2$
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)]$f'(x) = (x + 3)(x - 2)^2 [2(x - 2) + 3(x + 3)]$f'(x) = (x + 3)(x - 2)^2 [2x - 4 + 3x + 9]$f'(x) = (x + 3)(x - 2)^2 [2x - 4 + 3x + 9]$f'(x) = 5(x + 3)(x - 2)^2 (x + 1)$f'(x) = 5(x + 3)(x - 2)^2 (x + 1)$
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$x = -4$: f(-4) = (-4 + 3)^2(-4 - 2)^3 = (1)(-216) = -216$f(-4) = (-4 + 3)^2(-4 - 2)^3 = (1)(-216) = -216$
For x = -3$x = -3$: f(-3) = 0$f(-3) = 0$
For x = -1$x = -1$: f(-1) = (-1 + 3)^2(-1 - 2)^3 = (4)(-27) = -108$f(-1) = (-1 + 3)^2(-1 - 2)^3 = (4)(-27) = -108$
For x = 2$x = 2$: f(2) = 0$f(2) = 0$
For x = 4$x = 4$: f(4) = (4 + 3)^2(4 - 2)^3 = 49 times 8 = 392$f(4) = (4 + 3)^2(4 - 2)^3 = 49 \times 8 = 392$
### 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$M - m = 392 - (-216) = 392 + 216 = 608$
### Pattern Recognition
For polynomials on a closed interval, critical points combined with boundary values reliably expose the absolute max and min.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Derivatives
Q5jee_main_2024_30_jan_morningMaxima and Minima
The maximum area of a triangle whose one vertex is at (0, 0)$(0, 0)$ and the other two vertices lie on the curve y = -2x^2 + 54$y = -2x^2 + 54$ at points (x, y)$(x, y)$ and (-x, y)$(-x, y)$ where y > 0$y > 0$ is:
A.88$88$
B.122$122$
C.92$92$
D.108$108$
Solution
### Related Formula
textArea of Delta = frac12 times textbase times textheight$\text{Area of } \Delta = \frac{1}{2} \times \text{base} \times \text{height}$
### 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 Delta = frac12 (2x) (y) = xy$\Delta = \frac{1}{2} (2x) (y) = xy$ (Assuming x > 0$x > 0$).
### Step 1: Setting up the area function
Substitute y = -2x^2 + 54$y = -2x^2 + 54$ into the area function:
textArea(Delta) = A(x) = x(-2x^2 + 54) = -2x^3 + 54x$\text{Area}(\Delta) = A(x) = x(-2x^2 + 54) = -2x^3 + 54x$
### 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) = fracdAdx = -6x^2 + 54$A'(x) = \frac{dA}{dx} = -6x^2 + 54$
Setting A'(x) = 0$A'(x) = 0$:
-6x^2 + 54 = 0 Rightarrow 6x^2 = 54 Rightarrow x^2 = 9$-6x^2 + 54 = 0 \Rightarrow 6x^2 = 54 \Rightarrow x^2 = 9$
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:
A_textmax = 3(-2(3)^2 + 54) = 3(-18 + 54) = 3(36) = 108$A_{\text{max}} = 3(-2(3)^2 + 54) = 3(-18 + 54) = 3(36) = 108$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Derivatives
Q7jee_main_2024_30_jan_morningMean Value Theorems
Let g : mathbbR to mathbbR$g : \mathbb{R} \to \mathbb{R}$ be a non-constant twice differentiable such that g'left(frac12right) = g'left(frac32right)$g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$. If a real valued function f$f$ is defined as f(x) = frac12left[g(x) + g(2 - x)right]$f(x) = \frac{1}{2}\left[g(x) + g(2 - x)\right]$, 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)$(0,1)$
### 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) = frac12left[g(x) + g(2 - x)right]$f(x) = \frac{1}{2}\left[g(x) + g(2 - x)\right]$.
Differentiating f(x)$f(x)$ with respect to x$x$:
f'(x) = fracg'(x) - g'(2 - x)2$f'(x) = \frac{g'(x) - g'(2 - x)}{2}$
### Step 1: Evaluating the derivative at given points
Evaluate f'left(frac32right)$f'\left(\frac{3}{2}\right)$:
f'left(frac32right) = fracg'left(frac32right) - g'left(frac12right)2$f'\left(\frac{3}{2}\right) = \frac{g'\left(\frac{3}{2}\right) - g'\left(\frac{1}{2}\right)}{2}$
Since g'left(frac12right) = g'left(frac32right)$g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$, we get:
f'left(frac32right) = 0$f'\left(\frac{3}{2}\right) = 0$
Also evaluate f'left(frac12right)$f'\left(\frac{1}{2}\right)$:
f'left(frac12right) = fracg'left(frac12right) - g'left(frac32right)2 = 0$f'\left(\frac{1}{2}\right) = \frac{g'\left(\frac{1}{2}\right) - g'\left(\frac{3}{2}\right)}{2} = 0$
Additionally, observe that f'(1) = fracg'(1) - g'(1)2 = 0$f'(1) = \frac{g'(1) - g'(1)}{2} = 0$.
### Step 2: Applying Rolle's Theorem
We have f'left(frac12right) = 0$f'\left(\frac{1}{2}\right) = 0$, f'(1) = 0$f'(1) = 0$, and f'left(frac32right) = 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 = frac12, 1, frac32$x = \frac{1}{2}, 1, \frac{3}{2}$.
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Derivatives
Q13jee_main_2024_31_jan_eveningMonotonicity and Distances
If the function f: (-infty, -1] to (a, b]$f: (-\infty, -1] \to (a, b]$ defined by f(x) = e^x^3 - 3x + 1$f(x) = e^{x^3 - 3x + 1}$ is one-one and onto, then the distance of the point P(2b + 4, a + 2)$P(2b + 4, a + 2)$ from the line x + e^-3y = 4$x + e^{-3}y = 4$ is :
A.2sqrt1 + mathrme^6$2\sqrt{1 + \mathrm{e}^6}$
B.4sqrt1 + mathrme^6$4\sqrt{1 + \mathrm{e}^6}$
C.3sqrt1 + mathrme^6$3\sqrt{1 + \mathrm{e}^{6}}$
D.sqrt1 + mathrme^6$\sqrt{1 + \mathrm{e}^{6}}$
Solution
### Related Formula
textDistance of point (x_1, y_1) text from line Ax+By+C=0 text is d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$\text{Distance of point } (x_1, y_1) \text{ from line } Ax+By+C=0 \text{ is } d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
### Core Logic
Monotonicity and Distances diagram for Q13 - JEE Main 2024 Evening
Given f(x) = e^x^3 - 3x + 1$f(x) = e^{x^3 - 3x + 1}$.
Check monotonicity on (-infty, -1]$(-\infty, -1]$:
f'(x) = e^x^3 - 3x + 1 cdot (3x^2 - 3) = 3(x-1)(x+1)e^x^3 - 3x + 1$f'(x) = e^{x^3 - 3x + 1} \cdot (3x^2 - 3) = 3(x-1)(x+1)e^{x^3 - 3x + 1}$
For x le -1$x \le -1$, (x+1) le 0$(x+1) \le 0$ and (x-1) < 0$(x-1) < 0$, making f'(x) ge 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 = lim_xto-infty f(x) = e^-infty = 0$a = \lim_{x\to-\infty} f(x) = e^{-\infty} = 0$b = f(-1) = e^-1 + 3 + 1 = e^3$b = f(-1) = e^{-1 + 3 + 1} = e^3$
Point P(2b + 4, a + 2) equiv P(2e^3 + 4, 2)$P(2b + 4, a + 2) \equiv P(2e^3 + 4, 2)$.
Find distance from line x + e^-3y - 4 = 0$x + e^{-3}y - 4 = 0$:
d = frac|2e^3 + 4 + e^-3(2) - 4|sqrt1^2 + (e^-3)^2 = frac2e^3 + 2e^-3sqrt1 + e^-6$d = \frac{|2e^3 + 4 + e^{-3}(2) - 4|}{\sqrt{1^2 + (e^{-3})^2}} = \frac{2e^3 + 2e^{-3}}{\sqrt{1 + e^{-6}}}$d = frac2e^3(1 + e^-6)sqrt1 + e^-6 = 2e^3 sqrt1 + e^-6 = 2sqrte^6 + 1$d = \frac{2e^3(1 + e^{-6})}{\sqrt{1 + e^{-6}}} = 2e^3 \sqrt{1 + e^{-6}} = 2\sqrt{e^6 + 1}$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Applications of Derivatives
Class 11 Maths: Straight Lines
Q22jee_main_2024_31_jan_morningMaxima and Minima
Let S = (-1, infty)$S = (-1, \infty)$ and f : S to mathbbR$f : S \to \mathbb{R}$ be defined as f(x) = int_-1^x (e^t - 1)^11 (2t - 1)^5 (t - 2)^7 (t - 3)^12 (2t - 10)^61 dt$f(x) = \int_{-1}^{x} (e^t - 1)^{11} (2t - 1)^5 (t - 2)^7 (t - 3)^{12} (2t - 10)^{61} dt$. Let p =$p =$ Sum of square of the values of x$x$, where f(x)$f(x)$ attains local maxima on S$S$ and q =$q =$ Sum of the values of x$x$, where f(x)$f(x)$ attains local minima on S$S$. Then, the value of p^2 + 2q$p^2 + 2q$ is
Numerical Answer.Answer: 27 to 27
Solution
### Related Formula
f'(x) = 0 text for critical points. Change from + to - text means local maxima, - to + text means local minima.$f'(x) = 0 \text{ for critical points. Change from } + \to - \text{ means local maxima, } - \to + \text{ means local minima.}$
### Core Logic
Using Newton-Leibniz formula:
f'(x) = (e^x - 1)^11 (2x - 1)^5 (x - 2)^7 (x - 3)^12 (2x - 10)^61$f'(x) = (e^x - 1)^{11} (2x - 1)^5 (x - 2)^7 (x - 3)^{12} (2x - 10)^{61}$
The critical points are roots of f'(x) = 0$f'(x) = 0$: x = 0, frac12, 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, frac12$x = 5, \frac{1}{2}$.
Local maxima at x = 2, 0$x = 2, 0$.
p = 0^2 + 2^2 = 4$p = 0^2 + 2^2 = 4$q = 5 + frac12 = frac112$q = 5 + \frac{1}{2} = \frac{11}{2}$
### Step 3: Final Computation
p^2 + 2q = 16 + 2left(frac112right) = 16 + 11 = 27$p^2 + 2q = 16 + 2\left(\frac{11}{2}\right) = 16 + 11 = 27$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Derivatives
Class 12 Maths: Integrals
More Application of Derivatives Questions — jee_main_2024_27_jan_morning
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