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Question

Let f:R(0,π) be defined as f(x)=cot1(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 limx23f(x)=5π12.
f(x) is an even function.
Hence, f is not injective.
We have f(x)=cot1(2|x|2+|x|)
=π2tan1(2|x|2+|x|)=π2tan1⎜ ⎜ ⎜1|x|21+|x|2⎟ ⎟ ⎟
=π2(tan11tan1|x|2)f(x)=π4+tan1|x|2
Since tan1|x|2[0,π2),
f(x)[π4,3π4)
Hence, f(x) is not surjective.

f(x)=π4tan1x2,x0π4+tan1x2,x>0f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪11+(x/2)212,x011+(x/2)212,x>0

f(0+)=12 and f(0)=12
Clearly, f(x) is non-differentiable at x=0

Also, f is non-periodic.

limx23f(x)=limx23(π4tan1x2)=5π12

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