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JEE Main 202224 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2022 · 24 June · Evening Shift · Mathematics · Definite Integration

_ n → ∞ ( n ^ 2 ( n ^ 2 + 1)(n + 1) + n ^ 2 ( n ^ 2 + 4)(n + 2) + n...

lim⁡n→∞(n2(n2+1)(n+1)+n2(n2+4)(n+2)+n2(n2+9)(n+3)+…+n2(n2+n2)(n+n))\lim_{n \to \infty}\left(\dfrac{{n}^{2}}{\left({n}^{2} + 1\right)\left(n + 1\right)} + \dfrac{{n}^{2}}{\left({n}^{2} + 4\right)\left(n + 2\right)} + \dfrac{{n}^{2}}{\left({n}^{2} + 9\right)\left(n + 3\right)} + \ldots + \dfrac{{n}^{2}}{\left({n}^{2} + {n}^{2}\right)\left(n + n\right)}\right)
is equal to

Options

  1. Aπ8+14ln⁡2\dfrac{\pi}{8} + \dfrac{1}{4}\ln 2
  2. Bπ4+18ln⁡2\dfrac{\pi}{4} + \dfrac{1}{8}\ln 2
  3. Cπ4−18ln⁡2\dfrac{\pi}{4} - \dfrac{1}{8}\ln 2
  4. Dπ8+ln2\dfrac{\pi}{8} + ln\sqrt{2}

Correct answer

A. π8+14ln⁡2\dfrac{\pi}{8} + \dfrac{1}{4}\ln 2

Step-by-step solution

lim⁡n→∞∑r=1n1n(1+r2n2)(1+rn)=∫01dx(1+x2)(1+x)\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\dfrac{\dfrac{1}{n}}{\left(1 + \dfrac{{r}^{2}}{{n}^{2}}\right)\left(1 + \dfrac{r}{n}\right)} = \mathop{∫}_{0}^{1}\dfrac{dx}{\left(1 + {x}^{2}\right)\left(1 + x\right)}

Let I=∫01dx(1+x2)(1+x)I = \mathop{∫}_{0}^{1}\dfrac{dx}{\left(1 + {x}^{2}\right)\left(1 + x\right)}

Now x=tanθ,dx=sec⁡2\thetadθx = tan\theta, dx = \sec^{2}\thetad\theta

I=∫0π4dθ1+tanθ=12∫0π42cos\thetadθsinθ+cosθI = \mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{d\theta}{1 + tan\theta} = \dfrac{1}{2}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{2cos\thetad\theta}{sin\theta + cos\theta}

=12∫0π4(sinθ+cosθ)dθsinθ+cosθ+12∫0π4(cosθ−sinθ)dθsinθ+cosθ= \dfrac{1}{2}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{\left(sin\theta + cos\theta\right)d\theta}{sin\theta + cos\theta} + \dfrac{1}{2}\mathop{∫}_{0}^{\dfrac{\pi}{4}}\dfrac{\left(cos\theta - sin\theta\right)d\theta}{sin\theta + cos\theta}

=12×π4+12[ln⁡∣sin⁡θ+cosθ∣]0π4= \dfrac{1}{2} \times \dfrac{\pi}{4} + \dfrac{1}{2}{\left[\ln \left|\sin\theta + cos\theta\right|\right]}_{0}^{\dfrac{\pi}{4}} =π8+14ln⁡2= \dfrac{\pi}{8} + \dfrac{1}{4}\ln 2

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