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+1 vote
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in Mathematics by (80 points)

find the no. of values belonging to the range of f(x) = sin-1x - cos-1 (x+2) + tan-1 ​​​​​​(x+1) .

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1 Answer

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Given function is f(x) = sin-1 x - cos-1(x + 2) + tan-1(x + 1)

∵ Domain of sin-1x and cos-1 x is [-1, 1].

∴ -1 ≤ x ≤ 1 ------(1) [∵ Domain of sin-1 x is (-1, 1).]

And -1 ≤ x+2 ≤ 1 [∵ Domain of cos-1 x is (-1, 1).]

⇒ -1 -2 ≤ x ≤ 1 - 2

⇒ -3 ≤ x ≤ -1 -------(2)

From equation (1) and (2),

we obtain x = -1, (∵ x > -1, x ≤ -1 ⇒ x = -1)

Hence, the domain of f(x) is {-1}.

Now the range of f(x) is

f(-1) = sin-1(-1) - cos-1(-1+2) + tan-1(-1+1)

= -sin-11 - cos-11+tan-1(0)

\(\frac{-\pi}{2}\) - 0+0 = \(\frac{-\pi}{2}\)

Hence, range of f(x) is \(\{\frac{-\pi}{2}\}\)

Therefore, the number of elements in range of given function f(x) is 1.

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