JEE Main 18 March 2021 Shift 1 Solved Paper

Given, f(x)={

‌ 1 |x| ,	|x|≥1
ax2+b,	|x|<1.

⇒‌‌f(x)={

‌ 1 |x| ,	x≤−1‌ or ‌x≥1
ax2+b,	−1<x<1.

⇒‌‌f(x)={

‌ −1 x ,	x≤−1
ax2+b,	−1<x<1
‌ 1 x ,	x≥1.

Given, f(x) is differentiable at every point of domain.
∴‌‌f′(x)={

‌ 1 x2 ,	x<−1
2ax,	−1<x<1
‌ −1 x2 ,	x>1.

∵f(x) is differentiable at x=1
∴( LHD at x=1)=( RHD at x=1)
⇒‌‌f′(1)=f′(1+)
⇒‌‌2a=−1⇒a=−‌

1

2

As, we know that, a function is differentiable at x=a, if it is continuous at x=a.
Hence, f(x) is also continuous at x=1.
i.e., ( LHL at x=1)=( RHL at x=1)=f(1)
⇒‌‌a+b=1
⇒‌‌(−‌

1

2

)+b=1⇒b=‌

3

2

Hence, a=−‌

1

2

,b=‌

3

2

Note You can also (or apply) continuity and differentiability at x=−1.