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Question

If dθcos2θtan2θ+sec2θ=λtanθ+2logefθ+C , where C is the constant of integration, then the ordered pair λ,fθ is equal to:


A

-1,1-tanθ

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B

-1,1+tanθ

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C

1,1+tanθ

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D

1,1-tanθ

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Solution

The correct option is B

-1,1+tanθ


Explanation for the correct option:

Step 1: Simplify the left side of the given equation

The given integration equation is,

dθcos2θtan2θ+sec2θ=λtanθ+2logefθ+C

Let's consider, the left-hand side of the above equation, we get

I=dθcos2θtan2θ+sec2θ

I=sec2θdθ2tanθ1-tan2θ+1+tan2θ1-tan2θ sec2θ=1cos2θ,tan2θ=2tanθ1-tan2θ,sec2θ=1+tan2θ1-tan2θ

I=sec2θdθtan2θ+2tanθ+11-tan2θ

I=1-tan2θsec2θdθ1+tanθ2

Step 2: Apply the substitute method

Substitute tanθ=u

sec2θdθ=du

We get,

I=1-u2du1+u2

I=1-udu1+u

I=21+u-1du

I=211+udu-1du

I=2loge1+u-u+C

I=2loge1+tanθ-tanθ+C

Comparing the given equation, we get

λ=-1 and fθ=1+tanθ

Thus, the value of the ordered pair λ,fθ is -1,1+tanθ.

Hence, the correct option is B)-1,1+tanθ.


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