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 3
Q16jee_main_2024_29_january_eveningIncreasing and Decreasing Functions
The function f(x) = fracxx^2 - 6x - 16, x in mathbbR - \-2, 8\$f(x) = \frac{x}{x^2 - 6x - 16}, x \in \mathbb{R} - \{-2, 8\}$,
A. decreases in (-2, 8)$(-2, 8)$ and increases in (-infty, -2) cup (8, infty)$(-\infty, -2) \cup (8, \infty)$
B. decreases in (-infty, -2) cup (-2, 8) cup (8, infty)$(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$
C. decreases in (-infty, -2)$(-\infty, -2)$ and increases in (8, infty)$(8, \infty)$
D. increases in (-infty, -2) cup (-2, 8) cup (8, infty)$(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$
Solution
### Related Formula
Using the quotient rule:
f'x = fracu'v - uv'v^2$f'x = \frac{u'v - uv'}{v^2}$
If f'(x) < 0$f'(x) < 0$, the function decreases.
### Core Logic
Let us compute f'(x)$f'(x)$:
f'(x) = frac1(x^2 - 6x - 16) - x(2x - 6)(x^2 - 6x - 16)^2$f'(x) = \frac{1(x^2 - 6x - 16) - x(2x - 6)}{(x^2 - 6x - 16)^2}$f'(x) = fracx^2 - 6x - 16 - 2x^2 + 6x(x^2 - 6x - 16)^2 = frac-x^2 - 16(x^2 - 6x - 16)^2 = -fracx^2 + 16(x^2 - 6x - 16)^2$f'(x) = \frac{x^2 - 6x - 16 - 2x^2 + 6x}{(x^2 - 6x - 16)^2} = \frac{-x^2 - 16}{(x^2 - 6x - 16)^2} = -\frac{x^2 + 16}{(x^2 - 6x - 16)^2}$
### Step 1: Sign Assessment
Observe that x^2 + 16 > 0$x^2 + 16 > 0$ for all real x$x$ and the denominator (x^2 - 6x - 16)^2 > 0$(x^2 - 6x - 16)^2 > 0$ everywhere except at the excluded boundary asymptotes x = -2$x = -2$ and x = 8$x = 8$.
Therefore:
f'(x) < 0 quad forall x in mathbbR - \-2, 8\$f'(x) < 0 \quad \forall x \in \mathbb{R} - \{-2, 8\}$
Thus, the function decreases throughout its domain intervals:
(-infty, -2) cup (-2, 8) cup (8, infty)$(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$
### Pattern Recognition
When the numerator of the derivative simplifies to a strictly negative or positive constant/expression (like -x^2 - 16$-x^2 - 16$), the function exhibits monotonic behavior over all continuous subsets.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Derivatives
Q28jee_main_2024_27_jan_morningHigher Order Derivatives
Let f(x)=x^3+x^2f'(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:
Numerical Answer.Answer: 202 to 202
Solution
### Related Formula
fracddx(x^n) = nx^n-1$\frac{d}{dx}(x^n) = nx^{n-1}$
### Core Logic
Since f'(1)$f'(1)$, f''(2)$f''(2)$, and f'''(3)$f'''(3)$ are evaluated at specific points, they act strictly as numerical constants.
Let's differentiate the function sequentially:
f'(x) = 3x^2 + 2x f'(1) + f''(2)$f'(x) = 3x^2 + 2x f'(1) + f''(2)$f''(x) = 6x + 2f'(1)$f''(x) = 6x + 2f'(1)$f'''(x) = 6$f'''(x) = 6$
### Step 1: Finding Unknown Constants
From the third derivative, evaluate it at x=3$x=3$:
f'''(3) = 6$f'''(3) = 6$
Next, evaluate the second derivative at x=2$x=2$:
f''(2) = 6(2) + 2f'(1) = 12 + 2f'(1)$f''(2) = 6(2) + 2f'(1) = 12 + 2f'(1)$
Finally, evaluate the first derivative at x=1$x=1$:
f'(1) = 3(1)^2 + 2(1)f'(1) + f''(2)$f'(1) = 3(1)^2 + 2(1)f'(1) + f''(2)$f'(1) = 3 + 2f'(1) + f''(2)$f'(1) = 3 + 2f'(1) + f''(2)$
Rearranging gives:
f'(1) + f''(2) = -3$f'(1) + f''(2) = -3$
### Step 2: Solving the System
Substitute f''(2) = 12 + 2f'(1)$f''(2) = 12 + 2f'(1)$ into the rearranged equation:
f'(1) + (12 + 2f'(1)) = -3$f'(1) + (12 + 2f'(1)) = -3$3f'(1) = -15 Rightarrow f'(1) = -5$3f'(1) = -15 \Rightarrow f'(1) = -5$
Now find f''(2)$f''(2)$:
f''(2) = 12 + 2(-5) = 2$f''(2) = 12 + 2(-5) = 2$
### Step 3: Final Output Calculation
We now have the complete explicitly defined first derivative equation:
f'(x) = 3x^2 + 2x(-5) + 2 = 3x^2 - 10x + 2$f'(x) = 3x^2 + 2x(-5) + 2 = 3x^2 - 10x + 2$
Evaluate this at x=10$x=10$:
f'(10) = 3(10)^2 - 10(10) + 2$f'(10) = 3(10)^2 - 10(10) + 2$f'(10) = 300 - 100 + 2 = 202$f'(10) = 300 - 100 + 2 = 202$
### Pattern Recognition
Functional equations that contain derivatives evaluated at specific points are just polynomials with unknown coefficients. Treat those evaluations as scalar constants (A, B, C$A, B, C$), differentiate systematically, and plug the anchors back in to build a basic system of linear equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Derivatives
Class 11 Maths: Limits and Derivatives
Q14jee_main_2024_29_jan_morningRoots of Polynomial Equations
Consider the function f:[frac12,1]rightarrow R$f:[\frac{1}{2},1]\rightarrow R$ defined by f(x)=4sqrt2x^3-3sqrt2x-1$f(x)=4\sqrt{2}x^3-3\sqrt{2}x-1$. Consider the statements
(I) The curve y=f(x)$y=f(x)$ intersects the x-axis exactly at one point
(II) The curve y=f(x)$y=f(x)$ intersects the x-axis at x=cosfracpi12$x=\cos\frac{\pi}{12}$
Then
A.textOnly (II) is correct$\text{Only (II) is correct}$
B.textBoth (I) and (II) are incorrect$\text{Both (I) and (II) are incorrect}$
C.textOnly (I) is correct$\text{Only (I) is correct}$
D.textBoth (I) and (II) are correct$\text{Both (I) and (II) are correct}$
Solution
### Related Formula
cos 3theta = 4cos^3theta - 3costheta$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$
Intermediate Value Theorem (IVT): If a function is continuous and monotonic on [a,b]$[a,b]$, and f(a)$f(a)$ and f(b)$f(b)$ have opposite signs, there is exactly one root in (a,b)$(a,b)$.
### Core Logic
Analyze the derivative of f(x)$f(x)$ on the interval [frac12, 1]$[\frac{1}{2}, 1]$:
f'(x) = 12sqrt2x^2 - 3sqrt2$f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2}$
For x in [frac12, 1]$x \in [\frac{1}{2}, 1]$, the minimum value of x^2$x^2$ is frac14$\frac{1}{4}$.
f'(x) ge 12sqrt2left(frac14right) - 3sqrt2 = 3sqrt2 - 3sqrt2 = 0$f'(x) \ge 12\sqrt{2}\left(\frac{1}{4}\right) - 3\sqrt{2} = 3\sqrt{2} - 3\sqrt{2} = 0$
Thus, f'(x) ge 0$f'(x) \ge 0$ for all x in [frac12, 1]$x \in [\frac{1}{2}, 1]$, meaning the function is strictly increasing on this interval.
Now, check the boundary values:
fleft(frac12right) = 4sqrt2left(frac18right) - 3sqrt2left(frac12right) - 1 = fracsqrt22 - frac3sqrt22 - 1 = -sqrt2 - 1 lt 0$f\left(\frac{1}{2}\right) = 4\sqrt{2}\left(\frac{1}{8}\right) - 3\sqrt{2}\left(\frac{1}{2}\right) - 1 = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} - 1 = -\sqrt{2} - 1 \lt 0$f(1) = 4sqrt2(1) - 3sqrt2(1) - 1 = sqrt2 - 1 gt 0$f(1) = 4\sqrt{2}(1) - 3\sqrt{2}(1) - 1 = \sqrt{2} - 1 \gt 0$
Since f(frac12) lt 0$f(\frac{1}{2}) \lt 0$ and f(1) gt 0$f(1) \gt 0$ and the function is strictly increasing, it must cross the x-axis exactly once. Statement (I) is correct.
### Step 1: Evaluate Statement II
To find where f(x) = 0$f(x) = 0$:
4sqrt2x^3 - 3sqrt2x - 1 = 0$4\sqrt{2}x^3 - 3\sqrt{2}x - 1 = 0$sqrt2(4x^3 - 3x) = 1$\sqrt{2}(4x^3 - 3x) = 1$4x^3 - 3x = frac1sqrt2$4x^3 - 3x = \frac{1}{\sqrt{2}}$
Recognizing the trigonometric identity 4cos^3theta - 3costheta = cos 3theta$4\cos^3\theta - 3\cos\theta = \cos 3\theta$, substitute x = costheta$x = \cos\theta$ (which is valid since x in [1/2, 1]$x \in [1/2, 1]$ implies theta in [0, pi/3]$\theta \in [0, \pi/3]$):
cos 3theta = frac1sqrt2$\cos 3\theta = \frac{1}{\sqrt{2}}$3theta = fracpi4$3\theta = \frac{\pi}{4}$theta = fracpi12$\theta = \frac{\pi}{12}$
Hence, the root is x = cosfracpi12$x = \cos\frac{\pi}{12}$. Statement (II) is correct.
### Pattern Recognition
The structural polynomial form 4x^3 - 3x$4x^3 - 3x$ is an immediate trigger for the cos 3theta$\cos 3\theta$ substitution. Proving monotonicity first guarantees that the single trig root found is the *only* root in the constrained domain.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Derivatives
Class 11 Mathematics: Trigonometric Functions
Q24jee_main_2024_29_jan_morningGraphing and Intersections
Let f(x)=2^x-x^2,xin R$f(x)=2^x-x^2,x\in R$. If m$m$ and n$n$ are respectively the number of points at which the curves y=f(x)$y=f(x)$ and y=f'(x)$y=f'(x)$ intersects the x-axis, then the value of m+n$m+n$ is
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
To find points where a curve intersects the x-axis, set y = 0$y = 0$.
### Core Logic
First, evaluate m$m$, the number of roots for f(x) = 0$f(x) = 0$:
2^x - x^2 = 0 Rightarrow 2^x = x^2$2^x - x^2 = 0 \Rightarrow 2^x = x^2$
Plotting the exponential curve y = 2^x$y = 2^x$ against the parabola y = x^2$y = x^2$:
- At x = 2$x = 2$, 2^2 = 4$2^2 = 4$ and 2^2 = 4$2^2 = 4$ (Intersection 1).
- At x = 4$x = 4$, 2^4 = 16$2^4 = 16$ and 4^2 = 16$4^2 = 16$ (Intersection 2).
- For x lt 0$x \lt 0$, y=2^x$y=2^x$ decays to 0 while y=x^2$y=x^2$ grows to infinity. They strictly cross exactly once at x = alpha$x = \alpha$ (where alpha approx -0.76$\alpha \approx -0.76$).
Hence, the total number of intersections is 3.
Therefore, m = 3$m = 3$.
Graphing and Intersections
### Step 1: Evaluate n for derivative roots
Now, evaluate n$n$, the number of roots for f'(x) = 0$f'(x) = 0$:
f'(x) = 2^x ln 2 - 2x$f'(x) = 2^x \ln 2 - 2x$
Set f'(x) = 0 Rightarrow 2^x ln 2 = 2x$f'(x) = 0 \Rightarrow 2^x \ln 2 = 2x$
Plotting the scaled exponential curve $
Plotting the scaled exponential curve $y = 2^x \ln 2 against the line $ against the line $y = 2x:
- The line $:
- The line $y=2x passes through origin with slope 2.
- The curve $ passes through origin with slope 2.
- The curve $y = 2^x \ln 2 has a y-intercept of $ has a y-intercept of $\ln 2 (approx 0.693).
Because exponential growth eventually overtakes any linear function, it crosses the line twice for $ (approx 0.693).
Because exponential growth eventually overtakes any linear function, it crosses the line twice for $x > 0. Since the line goes negative for $. Since the line goes negative for $x < 0 and the exponential stays positive, they do not cross on the negative axis.
Hence, the total number of intersections is 2.
Therefore, $ and the exponential stays positive, they do not cross on the negative axis.
Hence, the total number of intersections is 2.
Therefore, $n = 2.
Graphing and Intersections
### Step 2: Final Computation
Sum the variables:
$.
Graphing and Intersections
### Step 2: Final Computation
Sum the variables:
$m + n = 3 + 2 = 5$m + n = 3 + 2 = 5$
### Pattern Recognition
The transcendental equation $
### Pattern Recognition
The transcendental equation $a^x = x^p typically produces exactly 3 intersections if $ typically produces exactly 3 intersections if $a \gt 1, p \gt 1 (two positive, one negative). When differentiating to $ (two positive, one negative). When differentiating to $a^x \ln a = px$, you strip away the left-sided negative quadratic wall, reducing it to a linear intersection which yields only 2 roots.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Derivatives
Class 11 Mathematics: Sets, Relations and Functions
More Application of Derivatives Questions — jee_main_2024_27_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Under Development
PATH B — TARGETED PRACTICE
Topic-wise Practice Hub
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
