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%)

Let f, g: (0, infty) to mathbbR be two functions defined by f(x) = int_-x^x left( |t| - t^2 right) e^-t^2 dt and g(x) = int_0^x^2 t^frac12 e^-t dt. Then the value of left( fleft( sqrtlog_e 9 right) + gleft( sqrtlog_e 9 right) right) is equal to

Solution & Explanation

### Related Formula textLeibnitz Rule: fracddxint_alpha(x)^beta(x) h(t)dt = h(beta(x))beta'(x) - h(alpha(x))alpha'(x) ### Core Logic For x>0, f(x) = int_-x^x left( |t| - t^2 right) e^-t^2 dt. f'(x) = left(|x|-x^2right)e^-x^2 cdot 1 - left(|-x|-(-x)^2right)e^-(-x)^2 cdot (-1) = 2(x-x^2)e^-x^2 For g(x) = int_0^x^2 t^frac12 e^-t dt: g'(x) = (x^2)^1/2 e^-x^2 cdot 2x = 2x^2 e^-x^2 Sum of derivatives: f'(x) + g'(x) = 2xe^-x^2 - 2x^2e^-x^2 + 2x^2e^-x^2 = 2xe^-x^2 Integrate both sides with respect to x: f(x) + g(x) = int_0^x 2t e^-t^2 dt Let t^2 = u implies 2t dt = du: f(x) + g(x) = int_0^x^2 e^-u du = [-e^-u]_0^x^2 = 1 - e^-x^2 Substitute x = sqrtlog_e 9: fleft(sqrtlog_e 9right) + gleft(sqrtlog_e 9right) = 1 - e^-log_e 9 = 1 - frac19 = frac89 The question asks for 9(f+g) based on typical JEE formatting of this problem (the PDF states "9(f(x)+g(x))" in the solution line). So 9 times frac89 = 8. ### Pattern Recognition When dealing with definite integrals with variable limits, apply Leibnitz rule to convert to differential form, combine terms, and integrate back. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals

Reference Study Guides

Integrals

More Integrals Previous-Year Questions — Page 6

Q21 jee_main_2024_31_jan_evening Properties of Definite Integrals
left|frac120pi^3int_0^pi fracx^2sin xcos xsin^4x + cos^4x dxright| is equal to
Numerical Answer. Answer: 15 to 15

Solution

### Related Formula int_0^a x f(x) dx = fraca2 int_0^a f(x) dx quad textif f(a-x) = f(x) ### Core Logic Let I = int_0^pi fracx^2sin xcos xsin^4x + cos^4x dx. Split the integral into int_0^pi/2 + int_pi/2^pi. For the second integral, substitute x = pi - t: I = int_0^pi/2 fracsin xcos xsin^4x + cos^4x (x^2 - (pi - x)^2) dx I = int_0^pi/2 fracsin xcos xsin^4x + cos^4x (2pi x - pi^2) dx = 2pi int_0^pi/2 x f(x) dx - pi^2 int_0^pi/2 f(x) dx where f(x) = fracsin x cos xsin^4 x + cos^4 x. Since f(pi/2 - x) = f(x), we have int_0^pi/2 x f(x) dx = fracpi4 int_0^pi/2 f(x) dx. I = 2pi left(fracpi4right) int_0^pi/2 f(x) dx - pi^2 int_0^pi/2 f(x) dx = -fracpi^22 int_0^pi/2 fracsin xcos xsin^4x + cos^4x dx To evaluate this simpler integral: I = -fracpi^22 int_0^pi/2 fracsin xcos x1 - 2sin^2 xcos^2 x dx = -fracpi^22 int_0^pi/2 fracsin 2x2 - sin^2 2x dx I = -fracpi^22 int_0^pi/2 fracsin 2x1 + cos^2 2x dx Let cos 2x = t implies -2sin 2x dx = dt. Limits: 1 to -1. I = -fracpi^22 int_1^-1 frac-dt/21+t^2 = -fracpi^24 int_-1^1 fracdt1+t^2 I = -fracpi^24 [arctan t]_-1^1 = -fracpi^24 left(fracpi4 - left(-fracpi4right)right) = -fracpi^38 Finally: left| frac120pi^3 left(-fracpi^38right) right| = 15 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals
Q21 jee_main_2024_31_jan_morning Definite Integration by Substitution
If the integral 525 int_0^fracpi2 sin 2x cos^frac112x (1 + cos^frac52x)^frac12 dx is equal to (nsqrt2 - 64), then n is equal to
Numerical Answer. Answer: 176 to 176

Solution

### Core Logic I = int_0^fracpi2 sin 2x cdot (cos x)^frac112 (1 + (cos x)^frac52)^frac12 dx Substitute cos x = t^2 implies sin x dx = -2t dt. Since sin 2x = 2sin x cos x, we have 2(t^2)(-2t dt) = -4t^3 dt. Limits: x=0 to t=1, x=fracpi2 to t=0. I = 4 int_0^1 t^2 cdot (t^2)^frac112 sqrt1 + t^5 t dt = 4 int_0^1 t^14 sqrt1 + t^5 dt ### Step 1: Second Substitution Put 1 + t^5 = k^2 implies 5t^4 dt = 2k dk. t^5 = k^2 - 1. I = 4 int_1^sqrt2 (k^2 - 1)^2 cdot k cdot frac2k5 dk I = frac85 int_1^sqrt2 (k^6 - 2k^4 + k^2) dk ### Step 2: Evaluate the Integral I = frac85 left[ frack^77 - frac2k^55 + frack^33 right]_1^sqrt2 I = frac85 left[ frac8sqrt27 - frac8sqrt25 + frac2sqrt23 - frac17 + frac25 - frac13 right] I = frac85 left[ frac120sqrt2 - 168sqrt2 + 70sqrt2105 - frac15 - 42 + 35105 right] I = frac85 left[ frac22sqrt2105 - frac8105 right] ### Step 3: Equate with Given Form Given 525 I = nsqrt2 - 64. 525 times frac85 left( frac22sqrt2 - 8105 right) = 8 times (22sqrt2 - 8) = 176sqrt2 - 64 Comparing with (nsqrt2 - 64), we get n = 176. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals
Q30 jee_main_2024_31_jan_morning Properties of Definite Integrals
Let f : mathbbR to mathbbR be a function defined by f(x) = frac4^x4^x + 2 and M = int_f(a)^f(1 - a) x sin^4(x(1 - x)) dx, N = int_f(a)^f(1 - a) sin^4(x(1 - x)) dx; a neq frac12. If alpha M = beta N, alpha, beta in mathbbN, then the least value of alpha^2 + beta^2 is equal to
Numerical Answer. Answer: 5 to 5

Solution

### Related Formula f(x) + f(1-x) = frac4^x4^x + 2 + frac4^1-x4^1-x + 2 = 1 ### Core Logic Using the property f(x) + f(1-x) = 1, we have f(a) + f(1-a) = 1. Let limits be A = f(a) and B = f(1-a). Then A + B = 1. M = int_A^B x sin^4(x(1 - x)) dx ### Step 1: Apply King's Property Apply the property int_A^B g(x) dx = int_A^B g(A + B - x) dx. M = int_A^B (1 - x) sin^4((1 - x)(1 - (1 - x))) dx M = int_A^B (1 - x) sin^4(x(1 - x)) dx M = int_A^B sin^4(x(1 - x)) dx - int_A^B x sin^4(x(1 - x)) dx M = N - M ### Step 2: Conclusion 2M = N Given alpha M = beta N, we get alpha = 2 and beta = 1 (for least integral values). Thus, alpha^2 + beta^2 = 2^2 + 1^2 = 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals Class 12 Maths: Relations and Functions

More Integrals Questions — jee_main_2024_31_jan_evening

Practice all Integrals previous-year questions →

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.

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.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...