Question

If $3{\sin }^{2}x+13\cos x-7=0$ has $5$ solutions in $x\in [0,\frac{n\pi }{2}],n\in N$ then the sum of maximum & minimum value of $n$ is divisible by

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A), (C), (D)

The correct options are
A 2
C 4
D 5

$3{\sin }^{2}x+13\cos x-7=0$

$\Rightarrow 3(1-{\cos }^{2}x)+13\cos x-7=0$

$\Rightarrow 3{\cos }^{2}x-13\cos x+4=0$

$\Rightarrow \cos x=\frac{13\pm \sqrt{169-48}}{6}=\frac{13\pm \sqrt{121}}{6}=\frac{13\pm 11}{6}$

$\cos x=\frac{13-11}{6}<1$

$\cos x=\frac{1}{3}\to 5$ solution in $[0,\frac{4\pi }{2}]$

$\cos x=\frac{1}{3}$ has $5$ solution in

$[0,\frac{9\pi }{2}];[0,\frac{11\pi }{2}]$

$\Rightarrow$ $n_{min}=9$

$n_{max}=11$

$n_{min}+n_{max}=20$

divisible by $2,4,5$