If the domain of the function f ( x ) = sec - 1 ( 2 x 5 x + 3 ) is then | 3+ 10 + 21| is equal to ______ . [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

More

Solution

Q.
If the domain of the function $f (x) = \sec^{- 1} (\frac{2 x}{5 x + 3})$ is $[α, β) \cup (γ, δ],$ then $| 3 α + 10 (β + γ) + 21 δ |$ is equal to ______ . [2023]

Ans.
(24)

$Given f (x) = \sec^{- 1} (\frac{2 x}{5 x + 3})$

$So, \frac{2 x}{5 x + 3} \geq 1$

$\Rightarrow \frac{2 x}{5 x + 3} - 1 \geq 0 \Rightarrow \frac{- 3 x - 3}{5 x + 3} \geq 0$

$\Rightarrow \frac{x + 1}{5 x + 3} \leq 0 \Rightarrow x \in [- 1, \frac{- 3}{5})$

$Now, \frac{2 x}{5 x + 3} \leq - 1 \Rightarrow \frac{2 x}{5 x + 3} + 1 \leq 0$

$\Rightarrow \frac{7 x + 3}{5 x + 3} \leq 0 \Rightarrow x \in (\frac{- 3}{5}, \frac{- 3}{7}]$

$∴$ $x \in [- 1, \frac{- 3}{5}) \cup (\frac{- 3}{5}, \frac{- 3}{7}]$

$α = - 1,$ $β = \frac{- 3}{5},$ $γ = \frac{- 3}{5},$ $δ = \frac{- 3}{7}$

$So,$ $| 3 α + 10 (β + γ) + 21 δ |$

$=$ $| 3 \times (- 1) + 10 (\frac{- 3}{5} - \frac{3}{5}) + 21 \times (\frac{- 3}{7}) |$ $=$ $| - 3 - 12 - 9 |$ $= 24$

Want Us to Email you About Special Offers & Updates?