Question

The roots of equation 3 = + 1 lies between

Answer 1

Answer:

f(x)=3(x−a)(x−b)

f(1)=3(1−a)(1−b)= -ve as a<1<b

on f(1)=3−3sinα−2(1−sin

2

α)<0

or 2sin

2

α−3sinα+1<0

(2sinα−1)(sinα−1)<0

2(sinα−

2

1

)(sinα−1)<0

∴

2

1

<sinα<1. Now sin30

∘

=sin150

∘

=

2

1

∴α lies between

6

π

to

6

5π

excluding

2

π

as sinα is not equal to 1 and hence (d) is correct.

Solve any