Question of The Day26-04-2021

Find the maximum and minimum value of 4 tan2 θ + 16 cot2 θ.


Correct Answer : a ) ∞, 16

Explanation :

According to the question

We know that,

Minimum value of atan2θ+ bcot2θ = 2√a√b

And maximum value of atan2θ + bcot2θ = ∞

⇒Minimum value of 4tan2θ+16cot2θ=2*√4*√16   

⇒ 2 * 2 * 4

⇒ 16

And maximum value of 4tan2θ + 16cot2θ = ∞

Thus, minimum and maximum value of 4tan2θ+16cot2θ is (16 and ∞). 

Hence, (a) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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