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Question

. lf y=1+sec2θ+cos2θ,θ0 then

A
y=0
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B
y2
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C
y3
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D
y>2
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Solution

The correct option is C y3

f(θ)=y=1+sec2θ+cos2θ,θ0
we have to find range of y,
y=1+1cos2θ+cos2θ
as cosθ is a continuous function,
Let g(θ)=1cos2θ+cos2θ
Put x=cosθ
g(x)=1x2+x2
for minima, diff. g(x) with respect to x and equal to 0.
g(x)=2x3+2x=0
x4=1
cos2θ=1
g(x)=2
minimum value of y=1+min(g(x))=1+2=3
yϵ[3,]


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