Question

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

• A. $0⩽p⩽1$
• B. $-1⩽p⩽3$
• C. $4⩽p⩽6$
• D. $p⩾6$

Answer 1

Answer: (A)

$0⩽p⩽1$

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

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

$=\frac{p+3\pm \sqrt{(p+3{)}^2-64(p-1)}}{4}$ for this to be real

$\therefore (p+3{)}^2-64(p-1)\ge 0$

$\Rightarrow p^2-58p-73\ge 0$

$\Rightarrow (p-29-16\sqrt{3})(p-29+16\sqrt{3})\ge 0$

Hence, either $p\le 29-16\sqrt{3}$ or $p\ge 29+16\sqrt{3}$

$0\le p\le 1$