JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions

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 differentiatble 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.