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Question

# Let f:R→(0,π) be defined as f(x)=cot−1(2−|x|2+|x|). Then which of the following statements is(are) TRUE?

A
f(x) is neither injective nor surjective.
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B
f(x) is continuous and differentiable on R.
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C
f(x) is both even function and non-periodic function.
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D
limx23f(x)=5π12.
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Solution

## The correct option is D limx→−2√3f(x)=5π12.f(x) is an even function. Hence, f is not injective. We have f(x)=cot−1(2−|x|2+|x|) =π2−tan−1(2−|x|2+|x|)=π2−tan−1⎛⎜ ⎜ ⎜⎝1−|x|21+|x|2⎞⎟ ⎟ ⎟⎠ =π2−(tan−11−tan−1|x|2)f(x)=π4+tan−1|x|2 Since tan−1|x|2∈[0,π2), ∴f(x)∈[π4,3π4) Hence, f(x) is not surjective. f(x)=⎧⎪⎨⎪⎩π4−tan−1x2,x≤0π4+tan−1x2,x>0f′(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩−11+(x/2)2⋅12,x≤011+(x/2)2⋅12,x>0 f′(0+)=12 and f′(0−)=−12 Clearly, f(x) is non-differentiable at x=0 Also, f is non-periodic. limx→−2√3f(x)=limx→−2√3(π4−tan−1x2)=5π12

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