arthham

JEE Advanced 2016 · Integrals

Let f(x)=lim⁡n→∞(nn(x+n)(x+n2)⋯(x+nn)n!(x2+n2)(x2+n24)⋯(x2+n2n2))xnf(x)=\lim_{n\to\infty}\left(\frac{n^{n}(x+n)\left(x+\frac{n}{2}\right)\cdots\left(x+\frac{n}{n}\right)}{n!\left(x^{2}+n^{2}\right)\left(x^{2}+\frac{n^{2}}{4}\right)\cdots\left(x^{2}+\frac{n^{2}}{n^{2}}\right)}\right)^{\frac{x}{n}}, for all x>0x>0. Then

Answers: B. f(13)≤f(23)f\left(\frac{1}{3}\right)\leq f\left(\frac{2}{3}\right) · C. f′(2)≤0f'(2)\leq 0

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Written solution

Let f(x)=lim_(n→∞)((nⁿ(x+n)(x+n/2)·s(x+n/n))/(n!(x²+n²)(x²+n²/4)·s(x²+n²/n²)))^(x/n), for all x>0. Then

f(x)=lim⁡n→∞(product of (x+n/k)n!nn product of (x2+n2/k2))xnf(x)=\lim_{n\to\infty}\left(\dfrac{\text{product of }(x+n/k)}{\dfrac{n!}{n^{n}}\ \text{product of }(x^{2}+n^{2}/k^{2})}\right)^{\frac{x}{n}}

To find: which of (1)–(4) are correct\text{To find: which of } (1)\text{--}(4) \text{ are correct}

Ln=nn∏k=1n(x+nk)n!∏k=1n(x2+n2k2)L_n=\dfrac{n^{n}\prod_{k=1}^{n}(x+\frac{n}{k})}{n!\prod_{k=1}^{n}(x^{2}+\frac{n^{2}}{k^{2}})}

ln⁡f=lim⁡n→∞xnln⁡Ln\ln f=\lim_{n\to\infty}\dfrac{x}{n}\ln L_n

∏k=1n(x+nk)=nnn!∏k=1n(1+xkn)\prod_{k=1}^{n}\left(x+\dfrac{n}{k}\right)=\dfrac{n^{n}}{n!}\prod_{k=1}^{n}\left(1+\dfrac{xk}{n}\right)

∏k=1n(x2+n2k2)=n2n(n!)2∏k=1n(1+x2k2n2)\prod_{k=1}^{n}\left(x^{2}+\dfrac{n^{2}}{k^{2}}\right)=\dfrac{n^{2n}}{(n!)^{2}}\prod_{k=1}^{n}\left(1+\dfrac{x^{2}k^{2}}{n^{2}}\right)

Ln=∏k=1n(1+xkn)∏k=1n(1+x2k2n2)L_n=\dfrac{\prod_{k=1}^{n}\left(1+\dfrac{xk}{n}\right)}{\prod_{k=1}^{n}\left(1+\dfrac{x^{2}k^{2}}{n^{2}}\right)}

ln⁡f=x[∫01ln⁡(1+xt) dt−∫01ln⁡(1+x2t2) dt]\ln f=x\left[\int_0^1\ln(1+xt)\,dt-\int_0^1\ln(1+x^{2}t^{2})\,dt\right]

∫01ln⁡(1+xt) dt=1+xxln⁡(1+x)−1\int_0^1\ln(1+xt)\,dt=\dfrac{1+x}{x}\ln(1+x)-1

∫01ln⁡(1+x2t2) dt=ln⁡(1+x2)−2+2xarctan⁡x\int_0^1\ln(1+x^{2}t^{2})\,dt=\ln(1+x^{2})-2+\dfrac{2}{x}\arctan x

ln⁡f(x)=(1+x)ln⁡(1+x)−xln⁡(1+x2)+x−2arctan⁡x\ln f(x)=(1+x)\ln(1+x)-x\ln(1+x^{2})+x-2\arctan x

f′(x)f(x)=ddxln⁡f(x)=ln⁡1+x1+x2\dfrac{f'(x)}{f(x)}=\dfrac{d}{dx}\ln f(x)=\ln\dfrac{1+x}{1+x^{2}}

1+x1+x2≥1  ⟺  0≤x≤1  ⇒  f′>0 on (0,1), f′<0 on (1,∞)\dfrac{1+x}{1+x^{2}}\ge 1 \iff 0\le x\le 1 \;\Rightarrow\; f'>0 \text{ on }(0,1),\ f'<0 \text{ on }(1,\infty)

f(13)≤f(23)  ⇒  (2) TRUEf\left(\dfrac{1}{3}\right)\le f\left(\dfrac{2}{3}\right) \;\Rightarrow\; (2)\ \text{TRUE}

f(12)<f(1)  ⇒  (1) FALSEf\left(\dfrac{1}{2}\right)<f(1) \;\Rightarrow\; (1)\ \text{FALSE}

f′(2)f(2)=ln⁡35<0  ⇒  f′(2)<0  ⇒  (3) TRUE\dfrac{f'(2)}{f(2)}=\ln\dfrac{3}{5}<0 \;\Rightarrow\; f'(2)<0 \;\Rightarrow\; (3)\ \text{TRUE}

ln⁡410=ln⁡0.4<ln⁡0.6=ln⁡35  ⇒  (4) FALSE\ln\dfrac{4}{10}=\ln 0.4<\ln 0.6=\ln\dfrac{3}{5} \;\Rightarrow\; (4)\ \text{FALSE}

Correct: (2) and (3)\text{Correct: } (2)\ \text{and}\ (3)

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