If the domain of the function $f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^{2}-3 x+1\right)$ is the interval $[\alpha, \beta]$, then $\alpha+2 \beta$ is equal to :

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Inverse Trigonometric Functions

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 :

Answer: (B)

For

sin^{- 1} (2 x^{2} - 3 x + 1)

the argument must lie in

[- 1, 1]

: the lower bound

2 x^{2} - 3 x + 2 \geq 0

holds for all

x

, while the upper bound

2 x^{2} - 3 x \leq 0

gives

0 \leq x \leq \frac{3}{2}

. On this interval

11 - 3 x > 0

, and

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

reduces to

x \leq \frac{16}{5}

and

x \leq 6

, both already satisfied. Hence the domain is

[0, \frac{3}{2}]

, so

α = 0, β = \frac{3}{2}

and

α + 2 β = 3

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