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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Mathematics
Chapter
Limits

JEE Advanced 2016 · Paper 2 · Mathematics · Limits

Let f(x) = _ n → ∞ ( n ^ n (x + n)(x + (n)/(2)) . . (x + (n)/(n)) n...

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

Options

  1. Af(12)≥f(1)f\left(\dfrac{1}{2}\right) \ge f\left(1\right)
  2. Bf(13)≤f(23)f\left(\dfrac{1}{3}\right) \le f\left(\dfrac{2}{3}\right)
  3. Cf′(2)≤0{f}^{′}\left(2\right) \le 0
  4. D\dfrac{{f}^{′}\left(3\right)}{f\left(3\right)} \ge \dfrac{{f}^{′}\left(2\right)}{f\left(2\right)}

Correct answer

B. f(13)≤f(23)f\left(\dfrac{1}{3}\right) \le f\left(\dfrac{2}{3}\right)

Step-by-step solution

Taking log both side
lnf(x)=lim⁡n→∞xnln⁡[∏r=1n(x+1rn)∏r=1n(x2+1(rn)2)1∏r=1n(rn)]lnf\left(x\right) = \lim_{n \to \infty} \dfrac{x}{n}\ln \left[\dfrac{\mathop{∏}_{r = 1}^{n}\left(x + \dfrac{1}{\dfrac{r}{n}}\right)}{\mathop{∏}_{r = 1}^{n}\left({x}^{2} + \dfrac{1}{{\left(\dfrac{r}{n}\right)}^{2}}\right)}\dfrac{1}{\mathop{∏}_{r = 1}^{n}\left(\dfrac{r}{n}\right)}\right]
=xlim⁡n→∞1n∑r=1nln⁡(xrn+1(xrn)2+1)= x\lim_{n \to \infty} \dfrac{1}{n}\mathop{∑}_{r = 1}^{n}\ln \left(\dfrac{x\dfrac{r}{n} + 1}{{\left(x\dfrac{r}{n}\right)}^{2} + 1}\right)
Express summation in definite integration using rn=t and 1n=dt\dfrac{r}{n} = t \text{ and } \dfrac{1}{n} = dt
=x∫01ln⁡(1+tx1+t2x2)dt= x\mathop{∫}_{0}^{1}\ln \left(\dfrac{1 + tx}{1 + {t}^{2}{x}^{2}}\right)dt
Put tx=ztx = z
lnf(x)=∫0xln⁡(1+z1+z2)dzlnf\left(x\right) = \mathop{∫}_{0}^{x}\ln \left(\dfrac{1 + z}{1 + {z}^{2}}\right)dz
⇒⇒ f′(x)f(x)=ln⁡(1+x1+x2)\dfrac{f′\left(x\right)}{f\left(x\right)} = \ln \left(\dfrac{1 + x}{1 + {x}^{2}}\right)( ∵ f( x )isalwayspositive )<br/><br />\forall x \in ( 0, 1 )<br/><br />ln\dfrac{1 + x}{1 + {x}^{2}} > 0<br/><br />⇒ \dfrac{f ' ( x )}{f( x )} > 0 \forall x \in ( 0, 1 )<br/><br />asf( x ) > 0 ⇒ f ' ( x ) > 0 \forall x \in ( 0, 1 )<br/>So<br />Sof( x )isincreasingis increasing( 0, 1 )andand{f}^{′}( 1 ) = 0 ( x > 1 ⇒ {f}^{′}( x ) < 0 )<br/><br />⇒ f\left(\dfrac{1}{2}\right) < f\left(1\right), f\left(\dfrac{1}{3}\right) < f\left(\dfrac{2}{3}\right), {f}^{′}\left(2\right) < 0<br/>Also<br />Also\dfrac{{f}^{′}\left(3\right)}{f\left(3\right)} - \dfrac{{f}^{′}\left(2\right)}{f\left(2\right)} = \ln \left(\dfrac{4}{10}\right) - \ln \left(\dfrac{3}{5}\right)
=ln⁡(46)<0⇒f′(3)f(3)<f′(2)f(2)= \ln \left(\dfrac{4}{6}\right) < 0 ⇒ \dfrac{{f}^{′}\left(3\right)}{f\left(3\right)} < \dfrac{{f}^{′}\left(2\right)}{f\left(2\right)}
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