Q.

The period of sec(x0+x04+x016+...∞) is

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a

90

b

180

c

270

d

360

answer is C.

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Detailed Solution

a1−r=11−14=43sec(x0+x04+....∞)=sec(π180x.43)period =2ππ180(43)=270
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Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
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