If the domain of the function f ( x ) = cos - 1 ( 2 x - 5 11 - 3 x ) + sin - 1 ( 2 x 2 - 3 x + 1 ) is the interval then the value of i s : [2026]

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Solution

Q.
If the domain of the function $f (x)$ $= \cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + \sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ then $α + 2 β$ is equal to: [2026]

1 5

2 2

3 3

4 1

Ans.
(3)

$f (x)$ $= \cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + \sin^{- 1} (2 x^{2} - 3 x + 1)$

$- 1 \leq \frac{2 x - 5}{11 - 3 x} \leq 1$

$- 1 \leq 2 x^{2} - 3 x + 1 \leq 1$

$2 x^{2} - 3 x + 2 \geq 0, 2 x^{2} - 3 x \leq 0$

$x \in [0, \frac{3}{2}] . . . . . . . . . . (i)$

$\frac{2 x - 5}{11 - 3 x} + 1 \geq 0 \frac{2 x - 5}{11 - 3 x} - 1 \leq 0$

$\frac{2 x - 5 + 11 - 3 x}{11 - 3 x} \geq 0 \frac{5 x - 16}{11 - 3 x} \leq 0$

$\frac{6 - x}{11 - 3 x} \geq 0$

$x \in (- \infty, \frac{16}{5}] \cup (\frac{11}{3}, \infty)$

$x \in (- \infty, \frac{11}{3}) \cup (6, \infty)$

Intersection:

$x \in (- \infty, \frac{16}{5}] \cup [6, \infty) . . . . (ii)$

Intersection of (i) & (ii) $x \in [0, \frac{3}{2}]$

$α = 0, β = \frac{3}{2} \Rightarrow α + 2 β = 3$

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