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The equation $2 si n^{2} x - (p + 3) sinx + 2 p - 2 = 0$ posses a real solution, if

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2 ⩽ p ⩽ 6

0 ⩽ p ⩽ 1

4 ⩽ p ⩽ 6

p ⩾ 6

answer is B.

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Detailed Solution

It is given that

2 si n^{2} x - (p + 3) sinx + 2 p - 2 = 0

.
We know that solutions or the roots of the equation

a x^{2} + bx + c = 0

are given by

\frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

.
Comparing the given equation with the equation

a x^{2} + bx + c = 0

and substituting the values of a, b and c in the quadratic formula,

\frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

we get,

sinx = \frac{p + 3 \pm \sqrt{(p + 3)^{2} - 4 \times 8 \times 2 (p - 1)}}{4}

= \frac{p + 3 \pm \sqrt{(p + 3)^{2} - 64 (p - 1)}}{4}

for this to be real

∴ (p + 3) 2 − 64 (p − 1) ≥ 0 ⇒ p 2 − 58 p − 73 ≥ 0 ⇒ (p − 29 − 163) (p − 29 + 163) ≥ 0

Hence, either

p \leq 29 - 16 \sqrt{3}

p \geq 29 + 16 \sqrt{3}

0 \leq p \leq 1

.
Hence, the correct option is 2.

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