JEE Main 24 June 2022 Shift 1 Solved Paper

f(x)=|(2x−1)(x+2)|+‌

sin‌2‌x

2

0≤x<‌

1

2

‌‌f(x)=(1−2x)(x+2)+‌

sin‌2‌x

2

f′(x)=−4x−3+cos‌2‌x<0
For x≥‌

1

2

:‌‌f′(x)=4x+3+cos‌2‌x>0
So, minima occurs at x=‌

1

2

.f(x)|min‌‌=|2(‌

1

2

)2+‌

3

2

−2|+sin(‌

1

2

)⋅cos(‌

1

2

)
‌‌=‌

1

2

‌sin‌1
So, maxima is possible at x=0 or x=1
Now checking for x=0 and x=1, we can see it attains its maximum value at x=1
c.f(x)|max=|2+3−2|+‌

sin‌2

2

=3+‌

1

2

‌sin‌2
Sum of absolute maximum and minimum value =3+‌

1

2

(sin‌1+sin‌2)