A) 0
B) 1
C) 2
D) infinite
Correct Answer: C
Solution :
Put \[x=0\] \[{{b}^{2}}=\cos \] \[{{b}^{2}}-1\] or \[\cos {{b}^{2}}=1+{{b}^{2}}\]\[\Rightarrow \]\[b=0\] |
Now the equation becomes \[-\,2\,\,a{{\sin }^{2}}\frac{x}{2}=\cos a\,x-1=-\,2{{\sin }^{2}}\frac{ax}{2}\] |
or \[a{{\sin }^{2}}\frac{x}{2}={{\sin }^{2}}\frac{ax}{2}\] must be true \[\forall \,\,x\in R\]\[\Rightarrow \]\[a=0\] or \[a=1\] |
You need to login to perform this action.
You will be redirected in
3 sec