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 , for all . Then
Options
- A
- B
- C
- D\dfrac{{f}^{′}\left(3\right)}{f\left(3\right)} \ge \dfrac{{f}^{′}\left(2\right)}{f\left(2\right)}
Correct answer
B.
Step-by-step solution
Taking log both side
Express summation in definite integration using
Put
( ∵ f( x )isalwayspositive )\forall x \in ( 0, 1 )ln\dfrac{1 + x}{1 + {x}^{2}} > 0⇒ \dfrac{f ' ( x )}{f( x )} > 0 \forall x \in ( 0, 1 )asf( x ) > 0 ⇒ f ' ( x ) > 0 \forall x \in ( 0, 1 )f( x )( 0, 1 ){f}^{′}( 1 ) = 0 ( x > 1 ⇒ {f}^{′}( x ) < 0 )⇒ 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\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)
Express summation in definite integration using
Put
( ∵ f( x )isalwayspositive )\forall x \in ( 0, 1 )ln\dfrac{1 + x}{1 + {x}^{2}} > 0⇒ \dfrac{f ' ( x )}{f( x )} > 0 \forall x \in ( 0, 1 )asf( x ) > 0 ⇒ f ' ( x ) > 0 \forall x \in ( 0, 1 )f( x )( 0, 1 ){f}^{′}( 1 ) = 0 ( x > 1 ⇒ {f}^{′}( x ) < 0 )⇒ 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\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)