Definite IntegrationmediumFree

Definite Integration: Question

JEE Maths question with a full step-by-step solution.

Question
If f(n)=∑k=1k=n∫kk+1(k+1)dxx(x+1)f(n) = \displaystyle\sum_{k=1}^{k=n}\int_{k}^{k+1}\frac{\left(k+1\right)dx}{x\left(x+1\right)} and g(x)=(x+1)ln⁡(x+2x+1)+f(x)g(x) = \left(x+1\right)\ln\left(\dfrac{x+2}{x+1}\right)+f(x), then
Ag′(0)=2g'(0) = 2
Bg′′(0)=−1g''(0) = -1correct
Cg′(1)=12g'(1) = \dfrac12correct
Dlim⁡n→∞∑r=1r=n(12eg(r)−1+n)=log⁡e2\displaystyle\lim_{n\to\infty}\sum_{r=1}^{r=n}\left(\frac{1}{2e^{g(r)}-1+n}\right) = \log_e 2
Solution
Step 1:
1x(x+1)=1x−1x+1⟹∫kk+1dxx(x+1)=[ln⁡xx+1]kk+1=ln⁡k+1k+2−ln⁡kk+1\frac{1}{x\left(x+1\right)} = \frac1x-\frac{1}{x+1} \quad\Longrightarrow\quad \int_{k}^{k+1}\frac{dx}{x\left(x+1\right)} = \left[\ln\frac{x}{x+1}\right]_{k}^{k+1} = \ln\frac{k+1}{k+2}-\ln\frac{k}{k+1}
 Writing ak=ln⁡kk+1a_k = \ln\dfrac{k}{k+1}, the kkth term of ff is (k+1)(ak+1−ak)\left(k+1\right)\left(a_{k+1}-a_k\right). Step 2: Summing by parts,
f(n)=∑k=1n(k+1)(ak+1−ak)=(n+1)an+1−2a1−∑j=2najf(n) = \sum_{k=1}^{n}\left(k+1\right)\left(a_{k+1}-a_k\right) = \left(n+1\right)a_{n+1}-2a_1-\sum_{j=2}^{n}a_j
 Step 3:
a1=ln⁡12=−ln⁡2,an+1=ln⁡n+1n+2a_1 = \ln\tfrac12 = -\ln2 ,\qquad a_{n+1} = \ln\frac{n+1}{n+2}
∑j=2naj=ln⁡(23⋅34⋯nn+1)=ln⁡2n+1\sum_{j=2}^{n}a_j = \ln\left(\frac23\cdot\frac34\cdots\frac{n}{n+1}\right) = \ln\frac{2}{n+1}
f(n)=(n+1)ln⁡n+1n+2+2ln⁡2−ln⁡2n+1=(n+1)ln⁡n+1n+2+ln⁡(2(n+1))f(n) = \left(n+1\right)\ln\frac{n+1}{n+2}+2\ln2-\ln\frac{2}{n+1} = \left(n+1\right)\ln\frac{n+1}{n+2}+\ln\left(2\left(n+1\right)\right)
 Step 4: The two (n+1)ln⁡\left(n+1\right)\ln terms are negatives of one another, so
g(n)=(n+1)ln⁡n+2n+1+f(n)=ln⁡(2(n+1))g(n) = \left(n+1\right)\ln\frac{n+2}{n+1}+f(n) = \ln\left(2\left(n+1\right)\right)
 and, as a function of a continuous variable,
g(x)=ln⁡(2(x+1))=ln⁡2+ln⁡(x+1)g(x) = \ln\left(2\left(x+1\right)\right) = \ln 2+\ln\left(x+1\right)
 Check: f(1)=2ln⁡43=0.5754f(1) = 2\ln\tfrac43 = 0.5754 and g(1)=2ln⁡32+0.5754=1.3863=ln⁡4=ln⁡(2⋅2)g(1) = 2\ln\tfrac32+0.5754 = 1.3863 = \ln4 = \ln\left(2\cdot2\right) Step 5:
g′(x)=1x+1,g′′(x)=−1(x+1)2g'(x) = \frac{1}{x+1},\qquad g''(x) = -\frac{1}{\left(x+1\right)^2}
g′(0)=1≠2 - (1) false;g′′(0)=−1 - (2) true;g′(1)=12 - (3) trueg'(0) = 1 \ne 2 \ \textbf{- (1) false};\qquad g''(0) = -1 \ \textbf{- (2) true};\qquad g'(1) = \tfrac12 \ \textbf{- (3) true}
 Step 6:
eg(r)=2(r+1)⟹2eg(r)−1+n=4r+3+ne^{g(r)} = 2\left(r+1\right) \quad\Longrightarrow\quad 2e^{g(r)}-1+n = 4r+3+n
∑r=1n14r+3+n=1n∑r=1n14(rn)+3n+1 ⟶ ∫01dx4x+1=14ln⁡5≈0.4024\sum_{r=1}^{n}\frac{1}{4r+3+n} = \frac1n\sum_{r=1}^{n}\frac{1}{4\left(\tfrac rn\right)+\tfrac{3}{n}+1} \ \longrightarrow\ \int_{0}^{1}\frac{dx}{4x+1} = \frac14\ln5 \approx 0.4024
 which is not ln⁡2≈0.6931\ln2 \approx 0.6931. (4) is false. Answer: (2) and (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Definite Integration · easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Definite Integration · easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
Definite Integration · easy
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx, n∈Nn\in\mathbb{N}, then n=n=
Definite Integration · easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath