If x1 and x2 are the solutions of ex2(sinx+cos x)=1/2 , where x1 ,x2≠0 , then the maximum number of the solutions of the equation ex2(sin x-cos x)=x lying between x1 and x2 can be
a.2
b.0
c.1
d.none of these
please give a solution
341consider f(x) \equiv \frac{1}{2} e^{-x^2} - (\sin x + \cos x)
f(x_1) = f(x_2) = 0 \Rightarrow \exists x_3 \in (x_1,x_2) such that f'(x3) = 0.
\Rightarrow \exists x_3 \in (x_1,x_2)\rightarrow \sin x - \cos x = x e^{-x^2}
which implies that the equation e^{x^2}(\sin x - \cos x) = x has at least one root in (x1, x3).
1basically 'rolle's theoram question', u can call it..
x1 and x2 are also the solutions of
(sinx + cosx)-1/2 e^(-x2)=0
by rolle's theoram ,der exists atleast 1 'x' in (x1,x2) such dat ,
f'(x)=0
or (cosx-sinx)+xe^(-x2)=0
or (sinx-cosx)e^(-x2)=x
cheers
