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Question

The minimum value of sec2x+cosec2x equals maximum value of asin2x+bcos2x, where a>b>0, the value of a is


A

a=1

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B

a=2

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C

a=3

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D

a=4

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Solution

The correct option is D

a=4


Explanation for the correct option:

Step 1: Find the minimum value of first equation

Given functions are, sec2x+cosec2x and asin2x+bcos2x

We have,
sec2x+cosec2x=2+tan2x+cot2x[sec2x=1+tan2x,cosec2x=1+cot2x]

We know that,
atan2x+bcot2x2ab

Comparing with the given equation,
a=1 and b=1

Thus, the minimum value of 2+tan2x+cot2x is,
2+21×1=4

Step 2: Find the maximum value of second equation

We know that,
sin2x+cos2x=1

So,
asin2x+bcos2x=asin2x+b1-sin2x=asin2x+b-bsin2x=a-bsin2x+b

We see that the function is maximum when sin2x=1 which happens at x=π2, since a>b
i.e., a-b+b=a

So the maximum value of asin2x+bcos2x is a

Step 3: Set equal the maximum and minimum value of the equations

From given,
minsec2x+cosec2x=maxasin2x+bcos2x4=aa=4

Therefore, the value of a is 4

Hence, option A is correct.


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