JEE Mains 29-Jan-2023 Shift 2 Solved Paper

‌f′′(x)=g′′(x)+6x . . . (1)
‌f′(1)=4g′(1)−3=9 . . . (2)
‌f(2)=3g(2)=12 . . . (3)
By integrating (1)
f′(x)=g′(x)+6‌

x2

2

+C
‌ At ‌‌x=1
‌f′(1)=g′(1)+3+C
⇒9=4+3+C⇒C=3
∴f′(x)=g′(x)+3x2+3
Again by integrating,
‌f(x)=g(x)+‌

3x3

3

+3x+D
‌‌ At ‌x=2
‌f(2)=g(2)+8+3(2)+D
‌⇒12=4+8+6+D⇒D=−6
So, f(x)=g(x)+x3+3x−6
‌⇒f(x)−g(x)=x3+3x−6
‌‌ At ‌x=−2
‌⇒g(−2)−f(−2)=20‌‌‌ (Option (1) is true) ‌
Now, for −1<x,2
‌h(x)=f(x)−g(x)=x3+3x−6
‌⇒h′(x)=3x2+3
‌⇒h(x)↑
‌‌ So, ‌h(−1)<h(x)<h(2)
‌⇒−10<h(x)<8
‌⇒|h(x)|<10‌‌‌ (option (2) is NOT true) ‌
Now, h′(x)=f′(x)−g′(x)=3x2+3
If |h′(x)|<6⇒|3x2+3|<6
⇒3x2+3<6
⇒x2<1
⇒−1<x<1‌‌ (option (3) is True)
If x∈(−1,1)|f′(x)−g′(x)|<6
option (3) is true and now to solve
‌f(x)−g(x)=0
‌⇒x3+3x−6=0
‌h(x)=x3+3x−6
here, h(1)=−ve and h(‌

3

2

)=+ve
So there exists x0∈(1,‌

3

2

) such that f(x0)=g(x0)
(option (4) is true)