Question

20) Find the range of the function 1/2cos∅-1​

Answer 1

EXPLANATION :

$\sf \frac{1}{2} \cos \phi \: - 1$

$= \sf \: \frac{1}{2} [ - 1 ,\: 1] - 1$

For pt -1

$\sf - \frac{1}{2} - 1$

$= \sf - \frac{3}{2}$

2nd pt

$\sf \frac{1}{2} - 1$

$= - \frac{1}{2}$

hence range of function,

$[ - \frac{3}{2} , - \frac{1}{2}]$