UPSC NDA I 2015 Mathematics Solved Paper

For a function to have a maximum value at x, the first derivative of the function at x should be 0, and the second differential of the function at x should be less than 0.
Hence,
f(x)=0.75x4−x3−9x2+7
f′(x)=3x3−3x2−18x=0
∴3x(x2−x−6)=0
3x(x−3)(x+2)=0
x=0,3,−2
f′′(x)=9x2−6x−18
=3(3x2−2x−6)
f′′(0)=−18
f′′(3)=45
f′′(−2)=30
The second derivative at x = 0 is negative. Hence, it gives maximum value.
Maximum value of the function= f(0) = 7