Let f(x)=a(1−xsin⁡x)+bcos⁡x+5x2,    x<03,                                         x=01+P(x)x21/x                 ,    x>0 where P(x) is a cubic function and f is continuous at x = 0.The range of function g(x) = 3a sin x - b cos x isThe value of P''(0) isIf the leading coefficient of P(x) is positive, then the equation P(x) = b has

f(x)=a(1−xsin⁡x)+bcos⁡x+5x2,    x<03,                                           x=01+P(x)x1/x                   ,    x>0where P(x)=a0+a1x+a2x2+a3x3f (0) = 3R.H.L =limx→0+ f(x)=limh→0 f(0+h)=limh→0 f(h)=limh→0 1+P(h)h1/hSince f is continuous at x = 0, R.H.L. exists.For the existence of R.H.L., a0,a1=0. Thus,R.H.L. = limh→0 1+a2h+a3h21/h 1∞ form =elimh→0 1+a2h+a3h2−1(1/h)=eo2L.H.L = limx→0− f(x)=limh→0 f(0−h)=limh→0 a(1−(−h)sin⁡(−h))+bcos⁡(−h)+5(−h)2=limh→0 a(1−h(h))+b1−h22!+5h2For finite value of L.H.L., a + b +5 = 0 and −a−b2=3Solving, we get a = -l , b = -4.Now, S(x) = 3a sin x - b cos x = - 3 sin x + 4 cos x which has range [- 5, 5].Also, P(x)=a3x3+loge⁡3x2P′′(x)=6a3x+2loge⁡3∴P′′(0)=2loge⁡3Further, P(x)=b or a3x3+loge⁡3x2=−4 has only one real root, as the graph of P(x)=a3x3+loge3x2 meets y = - 4only once for negative value of x.