Question

if the maximum and minimum value of f(x)=a+bsinx are 7and 1 respectively then find the value of f(π/6) .

Answer 1

$f(x) = a + b \sin( \alpha )$

Maximum value of f(x) is when sin a = 1

And minimum value of f(x) is when sin a = 0

Therefore ,
Max.(f(x)) = a + b(1) = a + b = 7
Min.(f(x)) = a + b(0) = a + 0 = 1

Hereby , we get -

a = 1 and b = 6

$f(x) = 1 + 6 \sin( \alpha )$

At
$\alpha = \frac{\pi}{6}$

$f(x) = 1 + 6 \sin( \frac{\pi}{6} ) \\ \\ f(x) = 1 + 6( \frac{1}{2} ) = 4$

Hence , 4 is the correct answer.

Answer 2

Answer:

Step-by-step explanation:

min.value of f(x) ,when sinx=-1

therefore a-b=1............(i)

max value of f(x),when sinx=1

therefore a+b=7...........(ii)

Adding (i)+(ii)

2a=8

Therefore a=4

Putting the value of a in eqn. 1

therefore b=3

Therefore

f(x)=4+3sinx...... therefore, f(π/6)=4+3sin(π/6)

=4+(3×1/2)=11/2