Q.
The equation ksinθ+cos2θ=2k-7 possesses a solution if:
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a
2≤k≤6
b
k>2
c
k > 6
d
k < 2
answer is A.
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Detailed Solution
Given, ksinθ+cos2θ=2k−7⇒ksinθ+1−2sin2θ=2k−7⇒2sin2θ−ksinθ+(2k−8)=0 For the existence of real roots, discriminant ≥0.⇒k2−4×2(2k−8)≥0⇒k-82≥0, which is always true. Roots of the quadratic equation are k−42,2, but sinθ≠2∴sinθ=k−42 but −1≤sinθ≤1⇒−1≤k−42≤1⇒−2≤k−4≤2⇒2≤k≤6
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