1nishant sir from the above thing i have done t lies in the II or Iv quadrant , t is negative,
the number of solutions is true
the no. of quation of roots satisfying the equation
2(sinA +cosA)=csin2A
c2<8 in the interval [0,2∩]
a)0
b)1
c)2
d)3
-
UP 0 DOWN 0 0 11
11 Answers
62Draw the graph!
You will solve this one in a flash
draw for LHS and for RHS!
1i solved it like this
4(sinA+cosA)2=4c2sin2Acos2A
4(1+sin2A)=c2sin22A
let t=tanA
then
4(1+t2+2t) = 4c2t2/1+t2
( sin2A=2t/1+t2)
(1+t2)(1+t2+2t)=c2t2
(1+t2)2+2t(1+t2)+t2=(c2+1)t2<9t2 ( c2<8)
(1+t+t2)(1+4t+t2)<0
(1-t)2(t2+4t+1)<0
t2+4t+1<0
hence
-2-√3<t<-2+√3
nishant sir what can be concluded from this
62Amit this method of yours is unfortunately not correct....
Just try to do the graphs of both the sides...
162Amit, what you have to do afterthis is that take the value of tan -1 of the values you have got!!!
and then try to find the number of solutions in the interval 0-2pi!
341Let sin A + cos A = t. Then sin 2A = t2-1
The equation is therefore 2t = c(t2-1) or ct2-2t-c = 0
We have to remember here that -\sqrt 2 \le t \le \sqrt 2
Solving the quadratic gives t = \frac{1}{c} \pm \sqrt{\frac{1}{c^2}+1}
We are given that c2<8
Now, if c>0, we have \frac{1}{c} > \frac{1}{2 \sqrt 2} and so \sqrt{\frac{1}{c^2}+1} > \frac{3}{2 \sqrt 2}
So \frac{1}{c} + \sqrt{\frac{1}{c^2}+1} > \frac{1}{2 \sqrt 2} + \frac{3}{2 \sqrt 2}>\sqrt 2
So this root is not admissible.
Note that the product of the roots is -1. That means the other roots is admissible since it will be in \left(-\frac{1}{\sqrt 2},0 \right)).
Again, if c<0, \frac{1}{c} - \sqrt{1+\frac{1}{c^2}} < -\sqrt 2. But by the same argument as above, the other root will be admissible.
For c = 0 we must have sin A + cos A = 0
In all cases we have to solve sin A + cos A = b for some constant b.
This obviously has two solutions in [0, 2 \pi]
