Let f: R to R be a twice differentiable function such that the quadratic equation f(x)m^2 - 2f'(x)m + f''(x) = 0 in m , has two equal roots for every x in R . If f(0) = 1 , f'(0) = 2 and (alpha, beta) is the largest interval in which the function f(log_e x - x) is increasing, then alpha + beta is equal to

Numerical Answer Type:
Enter a numerical value Answer: 1 to 1 +4 marks

Solution & Explanation

### Related Formula For a quadratic equation Am^2 + Bm + C = 0 having equal roots, Discriminant D = 0 Rightarrow B^2 - 4AC = 0. ### Core Logic Given quadratic equation in m: f(x)m^2 - 2f'(x)m + f''(x) = 0 has equal roots. D = 0 Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0 4(f'(x))^2 = 4f(x)f''(x) Rightarrow (f'(x))^2 = f(x)f''(x) ### Step 1: Solve the Differential Equation Rewrite the DE: fracf''(x)f'(x) = fracf'(x)f(x) Integrate both sides: int fracf''(x)f'(x) dx = int fracf'(x)f(x) dx ln|f'(x)| = ln|f(x)| + ln|c| Rightarrow f'(x) = c cdot f(x) Using given f(0) = 1 and f'(0) = 2: f'(0) = c cdot f(0) Rightarrow 2 = c(1) Rightarrow c = 2 Now we have f'(x) = 2f(x) Rightarrow fracf'(x)f(x) = 2. Integrate again: ln|f(x)| = 2x + d Use f(0) = 1 Rightarrow ln(1) = 0 + d Rightarrow d = 0. So, ln f(x) = 2x Rightarrow f(x) = e^2x. ### Step 2: Investigate increasing interval Let g(x) = f(ln x - x) = e^2(ln x - x). For g(x) to be increasing, g'(x) geq 0. g'(x) = 2e^2(ln x - x) cdot fracddx(ln x - x) g'(x) = 2e^2(ln x - x) left(frac1x - 1right) Since exponential is always positive, g'(x) geq 0 Rightarrow frac1x - 1 geq 0. frac1 - xx geq 0 The critical points are x=0, x=1. Based on domain of ln x, x > 0. Sign scheme yields positive derivative in x in (0, 1]. Therefore, (alpha, beta) = (0, 1) Rightarrow alpha = 0, beta = 1. ### Step 3: Final Output alpha + beta = 0 + 1 = 1 ### Pattern Recognition The relation (f')^2 = f cdot f'' is a classical indicator of exponential functions (f = Ce^kx). Solving via double logarithmic integration collapses the differential equation almost instantaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Applications of Derivatives

Reference Study Guides

Differential Equations

More Differential Equations Previous-Year Questions — Page 7

Q27 jee_main_2024_30_jan_morning Linear Differential Equations
Let y = y(x) be the solution of the differential equation (1 - x^2) dy = left[ xy + (x^3 + 2)sqrt3(1 - x^2) right] dx, quad -1 < x < 1, y(0) = 0. If yleft(frac12right) = fracmn, m and n are co-prime numbers, then m + n is equal to
Numerical Answer. Answer: 97 to 97

Solution

### Related Formula For a linear differential equation fracdydx + P(x)y = Q(x): IF = e^int P(x) dx Solution: y cdot IF = int Q(x) cdot IF \, dx + C ### Core Logic Rearrange the given differential equation into standard linear form: (1 - x^2) fracdydx = xy + (x^3 + 2)sqrt3(1 - x^2) fracdydx - fracx1 - x^2 y = frac(x^3 + 2)sqrt3(1 - x^2)1 - x^2 fracdydx - fracx1 - x^2 y = fracsqrt3(x^3 + 2)sqrt1 - x^2 Identify P(x) = -fracx1 - x^2. ### Step 1: Finding Integrating Factor (IF) IF = e^int -fracx1 - x^2 dx Let 1 - x^2 = t Rightarrow -2x dx = dt Rightarrow -x dx = fracdt2. IF = e^frac12 int frac1t dt = e^frac12 ln t = e^ln sqrtt = sqrt1 - x^2 ### Step 2: General Solution The solution is given by: y sqrt1 - x^2 = int left( fracsqrt3(x^3 + 2)sqrt1 - x^2 right) sqrt1 - x^2 \, dx + C y sqrt1 - x^2 = sqrt3 int (x^3 + 2) \, dx + C y sqrt1 - x^2 = sqrt3 left( fracx^44 + 2x right) + C ### Step 3: Finding Constant C Using y(0) = 0: 0 cdot 1 = sqrt3(0 + 0) + C Rightarrow C = 0 So, y(x) = fracsqrt3sqrt1 - x^2 left( fracx^44 + 2x right). ### Step 4: Evaluating required point Substitute x = 1/2: yleft(frac12right) = fracsqrt3sqrt1 - 1/4 left( frac1/164 + 2left(frac12right) right) Wait, frac(1/2)^44 = frac1/164 = frac164. Let's recheck the expression: yleft(frac12right) = fracsqrt3sqrt3/4 left( frac164 + 1 right) = fracsqrt3sqrt3/2 left( frac6564 right) = 2 times frac6564 = frac6532 Here, m = 65 and n = 32. They are co-prime. Thus, m + n = 65 + 32 = 97. ### Pattern Recognition Whenever roots matching the integration denominator appear on the RHS of a linear differential setup, it is a high-confidence signal that the integrating factor cleanly annihilates the fractional root component during the solution stage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q8 jee_main_2024_31_jan_evening Newton's Law of Cooling
The temperature T(t) of a body at time t = 0 is 160^circ mathrmF and it decreases continuously as per the differential equation fracmathrmdTmathrmdt = -mathrmK(T - 80), where K is positive constant. If T(15) = 120^circmathrmF, then T(45) is equal to
  • A. 85^circmathrmF
  • B. 95^circmathrmF
  • C. 90^mathrmomathrmF
  • D. 80^circmathrmF

Solution

### Related Formula int fracdTT-T_s = -K int dt implies ln|T-T_s| = -Kt + C ### Core Logic Given fracdTdt = -K(T-80). Integrating from t=0 to t: int_160^T fracdTT-80 = -K int_0^t dt [ln|T-80|]_160^T = -Kt lnleft(fracT-8080right) = -Kt implies T(t) = 80 + 80e^-Kt Given T(15) = 120: 120 = 80 + 80e^-15K implies 40 = 80e^-15K implies e^-15K = frac12 To find T(45): T(45) = 80 + 80e^-45K = 80 + 80(e^-15K)^3 = 80 + 80left(frac12right)^3 = 80 + 80left(frac18right) = 90 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q29 jee_main_2024_31_jan_evening Linear Differential Equations
Let y = y(x) be the solution of the differential equation sec^2 x \, dx + left( e^2y tan^2 x + tan x right) dy = 0, 0 < x < fracpi2, yleft(fracpi4right) = 0. If yleft(fracpi6right) = alpha. Then e^8alpha is equal to
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula textIntegrating factor for fracdudy + P(y)u = Q(y) text is I.F. = e^int P(y)dy ### Core Logic Given DE: sec^2 x fracdxdy + e^2y tan^2 x + tan x = 0 Substitute t = tan x implies fracdtdy = sec^2 x fracdxdy. The DE becomes: fracdtdy + t = -t^2 e^2y This is a Bernoulli equation in t. Divide by t^2: frac1t^2fracdtdy + frac1t = -e^2y Substitute u = frac1t implies fracdudy = -frac1t^2fracdtdy. -fracdudy + u = -e^2y implies fracdudy - u = e^2y This is a linear DE in u with respect to y. P(y) = -1, Q(y) = e^2y. Integrating Factor: I.F. = e^int -1 dy = e^-y. Solution: u e^-y = int e^2y e^-y dy = int e^y dy = e^y + C Substitute u = frac1tan x: frace^-ytan x = e^y + C Use given condition y(pi/4) = 0: frace^0tan(pi/4) = e^0 + C implies 1 = 1 + C implies C = 0 Therefore, frace^-ytan x = e^y implies tan x = e^-2y. Evaluate at x = pi/6, y = alpha: tan(pi/6) = e^-2alpha implies frac1sqrt3 = e^-2alpha e^2alpha = sqrt3 implies (e^2alpha)^4 = (sqrt3)^4 = 9 Thus, e^8alpha = 9. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q9 jee_main_2024_31_jan_morning Homogeneous Differential Equations
The solution curve of the differential equation yfracdxdy = x(log_e x - log_e y + 1), x > 0, y > 0 passing through the point (e, 1) is
  • A. |log_e fracyx| = x
  • B. |log_e fracyx| = y^2
  • C. |log_e fracxy| = y
  • D. 2|log_e fracxy| = y + 1

Solution

### Core Logic Given DE: fracdxdy = fracxy left(lnleft(fracxyright) + 1right) Let fracxy = t implies x = ty. Differentiating w.r.t y: fracdxdy = t + yfracdtdy ### Step 1: Substitution and Integration t + yfracdtdy = t(ln(t) + 1) = tln t + t yfracdtdy = tln t implies fracdttln t = fracdyy Integrate both sides. Let ln t = p implies frac1t dt = dp. int fracdpp = int fracdyy ln|p| = ln y + C implies ln|ln t| = ln y + C lnleft|lnleft(fracxyright)right| = ln y + C ### Step 2: Applying Boundary Conditions Given curve passes through (e, 1): lnleft|lnleft(frace1right)right| = ln(1) + C implies C = 0 lnleft|lnleft(fracxyright)right| = ln y left|lnleft(fracxyright)right| = e^ln y = y ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q11 jee_main_2024_31_jan_morning Linear Differential Equations
Let y = y(x) be the solution of the differential equation fracdydx = frac(tan x) + ysin x(sec x - sin x tan x), x in left(0, fracpi2right) satisfying the condition yleft(fracpi4right) = 2. Then, yleft(fracpi3right) is
  • A. sqrt3(2 + log_esqrt3)
  • B. fracsqrt32(2 + log_e 3)
  • C. sqrt3(1 + 2log_e 3)
  • D. sqrt3(2 + log_e 3)

Solution

### Core Logic fracdydx = fracfracsin xcos x + ysin x left(frac1cos x - fracsin^2 xcos xright) = fracsin x + ycos xsin x (1 - sin^2 x) fracdydx = fracsin x + ycos xsin x cos^2 x = sec^2 x + frac2ysin 2x fracdydx - 2csc(2x)y = sec^2 x ### Step 1: Integrating Factor This is an LDE of form fracdydx + Py = Q. I.F. = e^int -2csc(2x) dx Let 2x = t implies 2dx = dt. I.F. = e^-int csc t dt = e^-ln|tan(t/2)| = e^-ln|tan x| = frac1|tan x| ### Step 2: Solution of LDE y(I.F.) = int Q(I.F.) dx + C yfrac1tan x = int sec^2 x frac1tan x dx + C Let tan x = t implies sec^2 x dx = dt. yfrac1tan x = int fracdtt + C = ln|tan x| + C y = tan x(ln|tan x| + C) ### Step 3: Boundary Value Given y(pi/4) = 2: 2 = 1(ln 1 + C) implies C = 2 Thus, y = tan x (ln|tan x| + 2). At x = pi/3: y(pi/3) = sqrt3(lnsqrt3 + 2) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

More Differential Equations Questions — jee_main_2026_21_jan_morning

Practice all Differential Equations previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

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.

Now Live Open
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.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...