Aditya Gupta
Last Activity: 6 Years ago
at k=-2, f(x)= 8x+3cosx
so f’(x)=8 – 3sinx which is always greater than zero since sinx lies in – 1 to 1.
so f is an increasing function.
theerfore f(πx-x2) greater than f(sinx) implies that πx-x2 greater than sinx.
or x^2 – pix+sinx less than 0
let g(x)= x^2 – pix+sinx
g’(x)= 2x+cosx – pi
g”(x)= 2 – sinx which is always greater than 0.
so g is a concave up function.
now g(0)=0
also, g(3) is less than zero while g(4) is greater than zero.
hence all integral values of x in between 0 to 3.something will satisfy the inequality x^2 – pix+sinx less than 0.
these values correspond to 1, 2, 3.
hence 1+2+3=6.
