The maximum value of 1−x−3x2,x∈R  is

The given equation is 1−x−3x2 Now,  1−x−3x2=1  −  (x+3x2)    =  1−3  (x2+13x)    =  1−3  (x2+13x+136)+336    =  (1+112)−3  (x+16)2≤  1312∀x∈R. Alternatively.   maximum value of ax2+bx+c,  a<0  and x∈R  is 4ac−b24a. In this case The given equation is 1−x−3x2 ⇒a=−3,  b=−1  and c=1. ∴   Required maximum value =  4×(−3)×1−(−1)24(−3)  =  −12−1−12  =  1312.