JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, f(x)‌‌={

−55x	‌	‌‌x<−5
2x3−3x2−120x	‌	−5≤x<4
2x3−3x2−36x+10	‌	x≥4.

∴‌‌f′(x)‌‌={

−55	x<−5
6(x2−x−20)	‌	−5≤x<4
6(x2−x−6)	‌	x≥4.

f′(x)‌‌={

−55	x<−5
6(x−5)(x+4)	‌	−5≤x<4
6(x−3)(x+2)	‌	x≥4.

For f to be increasing, f′(x)>0
Now, f′(x)=−55 is always less than zero.
f′(x)=6(x−5)(x+4)<0,−5≤x<4
Critical points =5,−4‌‌[x∈(−5,−4)
and f′(x)=6(x−3)(x+2)<0,x≥4
Critical point, =3,−2x∈(4,∞) ... (ii)
From Eqs. (i) and (ii), f(x) is increasing in x∈(−5,−4)∪(4,∞)