Let f ( x ) = 2 x + tan - 1 x and g ( x ) = log e ( 1 + x 2 + x ) , x [ 0 , 3 ] . Then [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q.
Let $f (x) = 2 x + \tan^{- 1} x$ and $g (x) = \log_{e} (\sqrt{1 + x^{2}} + x), x \in [0, 3] .$ Then [2023]

1 $\min f' (x) = 1 + \max g' (x)$

2 $there exist 0 < x_{1} < x_{2} < 3 such that f (x) < g (x), \forall x \in (x_{1}, x_{2})$

3 $\max f (x) > \max g (x)$

4 $there exists \hat{x} \in [0, 3] such that f' (\hat{x}) < g' (\hat{x})$

Ans.
(3)

We have, $f (x) = 2 x + \tan^{- 1} x and g (x) = \log_{e} (\sqrt{1 + x^{2}} + x)$

$g' (x) = \frac{1}{\sqrt{1 + x^{2}} + x} [\frac{1}{2 \sqrt{1 + x^{2}}} \times 2 x + 1]$

$= \frac{x + \sqrt{1 + x^{2}}}{(\sqrt{1 + x^{2}} + x) (\sqrt{1 + x^{2}})} = \frac{1}{\sqrt{1 + x^{2}}}; f' (x) = 2 + \frac{1}{1 + x^{2}}$

Both $f (x)$ and $g (x)$ are strictly increasing functions in [0, 3].

$Max f (x) = f (3) = 6 + \tan^{- 1} 3$

$Max g (x) = g (3) = \ln (\sqrt{10} + 3)$

$Max f (x) > Max g (x), x \in [0, 3]$

Solution

Q.
Let $f (x) = 2 x + \tan^{- 1} x$ and $g (x) = \log_{e} (\sqrt{1 + x^{2}} + x), x \in [0, 3] .$ Then [2023]

1 $\min f' (x) = 1 + \max g' (x)$

2 $there exist 0 < x_{1} < x_{2} < 3 such that f (x) < g (x), \forall x \in (x_{1}, x_{2})$

3 $\max f (x) > \max g (x)$

4 $there exists \hat{x} \in [0, 3] such that f' (\hat{x}) < g' (\hat{x})$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let f(x)=2x+tan-1x and g(x)=loge(1+x2+x), x∈[0,3]. Then [2023]

1 minf'(x)=1+maxg'(x)

2 there exist 0<x1<x2<3 such that f(x)<g(x),∀x∈(x1,x2)

3 maxf(x)>maxg(x)

4 there exists x^∈[0,3] such that f'(x^)<g'(x^)

Ans. (3) We have, f(x)=2x+tan-1x and g(x)=loge(1+x2+x) g'(x)=11+x2+x[121+x2×2x+1] =x+1+x2(1+x2+x)(1+x2)=11+x2; f'(x)=2+11+x2 Both f(x) and g(x) are strictly increasing functions in [0, 3]. Max f(x)=f(3)=6+tan-13 Max g(x)=g(3)=ln(10+3) Max f(x)>Max g(x), x∈[0,3]

Want Us to Email you About Special Offers & Updates?

Q.
Let $f (x) = 2 x + \tan^{- 1} x$ and $g (x) = \log_{e} (\sqrt{1 + x^{2}} + x), x \in [0, 3] .$ Then [2023]

1 $\min f' (x) = 1 + \max g' (x)$

2 $there exist 0 < x_{1} < x_{2} < 3 such that f (x) < g (x), \forall x \in (x_{1}, x_{2})$

3 $\max f (x) > \max g (x)$

4 $there exists \hat{x} \in [0, 3] such that f' (\hat{x}) < g' (\hat{x})$