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Question

# If f(x)=tan−1x−cot−1x, g(x)=sec−1x−cosec−1x, h(x)=sin−1x+cos−1x+tan−1x, i(x)=sin−1x−cos−1x, j(x)=√x, then

A
Domain of jof(x)+jog(x) is [2,)
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B
Domain of jof(x)+joi(x) is [0,1]
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C
Domain of joi(x)+joh(x) is
[12,1]
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D
Domain of jog(x)+joh(x) is {1}
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Solution

## The correct option is C Domain of joi(x)+joh(x) is [1√2,1] jof(x)=√tan−1x−cot−1x 2tan−1x≥π2 ⇒ Domain of (jof(x))=[1,∞) ⋯(i) jog(x)=√sec−1x−cosec−1x 2sec−1x≥π2 ⇒ Domain of (jog(x))=(−∞,−1]∪[√2,∞) ⋯(ii) joh(x)=√sin−1x+cos−1x+tan−1x Domain of sin−1x,cos−1x is [−1,1] and sin−1x+cos−1x+tan−1x≥0⇒tan−1x≥−π2 ⇒ Domain of (joh(x))=[−1,1] ⋯(iii) joi(x)=√sin−1x−cos−1x 2sin−1x≥π2 ⇒ Domain of (joi(x))=[1√2,1] ⋯(iv) ∴ From (i),(ii),(iii),(iv) a) Domain of jof(x)+jog(x) is [√2,∞) b) Domain of jof(x)+joi(x) is {1} c) Domain of joi(x)+joh(x) is [1√2,1] d) Domain of jog(x)+joh(x) is {−1}

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