If limx→0 cos⁡4x+acos⁡2x+bx4 is finite, then(a,b)=

limx→0 cos⁡4x+acos⁡2x+bx4 is finite.Therefore cos⁡4x+acos⁡2x+bx4 should be of the form 00atx=0.Consequently, the value cos 4x+acos⁡2x+b must be zero at x=0.i.e.,1+a+b=0                  …(i)Using L' Hospital's rule the given limit is limx→0 −4sin⁡4x−2asin⁡2x4x3 00 form =limx→0 −16cos⁡4x−4acos⁡2x12x2 [Using L Hospitals rule] This should be of the form 00.∴ −16−4a=0                                …(ii)Solving (i) and (ii), we get a=−4 and b=3.