If the function f : ( - ? , - 1 ] ? ( a , b ] defined by f ( x ) = e x 3 - 3 x + 1 is one-one and onto, then the distance of the point P ( 2 b + 4 , a + 2 ) from the line x + e - 3 y = 4 is : [2024]

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

More

Solution

Q.
If the function $f : (- \infty, - 1] \to (a, b]$ defined by $f (x) = e^{x^{3} - 3 x + 1}$ is one-one and onto, then the distance of the point $P (2 b + 4, a + 2)$ from the line $x + e^{- 3} y = 4$ is : [2024]

1 $2 \sqrt{1 + e^{6}}$

2 $4 \sqrt{1 + e^{6}}$

3 $3 \sqrt{1 + e^{6}}$

4 $\sqrt{1 + e^{6}}$

Ans.
(1)

Given $f (x) = e^{x^{3} - 3 x + 1}$

$∴$    $f' (x) = e^{x^{3} - 3 x + 1} [3 (x - 1) (x + 1)] \geq 0$

As, $e^{x^{3} - 3 x + 1}$ is always positive.

$∴$    $3 (x - 1) (x + 1) \geq 0$

$\Rightarrow x \in (- \infty, - 1] \cup [1, \infty)$

For onto function, Range = Co-domain

$∴$    $a = f (- \infty) = e^{- \infty} = 0$

and $b = f (- 1) = e^{- 1 + 3 + 1} = e^{3}$

$∴$    $P (2 b + 4, a + 2) = P (2 e^{3} + 4, 2)$

Now, distance of the point $P (2 e^{3} + 4, 2)$ from the line $x + e^{- 3} y - 4 = 0$

$= \frac{2 e^{3} + 4 + 2 e^{- 3} - 4}{\sqrt{1 + e^{- 6}}} = \frac{2 (e^{3} + e^{- 3})}{\sqrt{1 + e^{- 6}}} = 2 (\frac{e^{6} + 1}{e^{3}}) \times \frac{1}{\sqrt{1 + e^{6}}} \times e^{3}$

$= 2 \sqrt{1 + e^{6}}$

Want Us to Email you About Special Offers & Updates?