Question

The difference between maximum and minimum value of the expression $y=1+2sinx+3co{s}^{2}x$ is

• A. $\frac{16}{3}$
• B. $\frac{13}{3}$
• C. $7$
• D. $8$

Answer 1

The correct option is A $\frac{16}{3}$

$y=1+2sin\left(x\right)+3co{s}^2\left(x\right)$

$y=4+2sin\left(x\right)-3si{n}^2\left(x\right)$

$y=4-3\left[si{n}^2\right(x)-\frac{2}{3}sin(x)+\frac{1}{9}]+\frac{1}{3}$

$y=\frac{13}{3}-3{\left[sin\right(x)-\frac{1}{3}]}^2$

Max value for y is $\frac{13}{3}$

$y=\frac{13}{3}-3{\left[sin\right(x)-\frac{1}{3}]}^2$ has max value when ${\left[sin\right(x)-\frac{1}{3}]}^2$ is zero

$y=\frac{13}{3}-3{\left[sin\right(x)-\frac{1}{3}]}^2$ has min value when ${\left[sin\right(x)-\frac{1}{3}]}^2$ has max value

Because of ${}^2$ expression $\left[sin\right(x)-\frac{1}{3}]$ has a ABSOLUTE value $-\frac{4}{3}$when $sin\left(x\right)=-1$ then

min value is $y=\frac{13}{3}-3{[-\frac{4}{3}]}^2=\frac{13}{3}-\frac{16}{3}=-1$

So the difference is $\frac{13}{3}+1=\frac{16}{3}$