Skip to main contentSkip to solution

If f(x)={ax2+b,x<−1bx2+ax+4,x≥−1f(x) = \begin{cases} ax^2 + b, & x < -1 \\ bx^2 + ax + 4, & x \geq -1 \end{cases} is everywhere differentiable, then :

Solution

✅ Correct Option: 1

For continuity at x=−1x=-1: a+b=b−a+4⇒2a=4⇒a=2a+b=b-a+4 \Rightarrow 2a=4 \Rightarrow a=2.

For differentiability at x=−1x=-1: left derivative =2ax∣x=−1=−2a=−4=2ax|_{x=-1}=-2a=-4; right derivative =2bx+a∣x=−1=−2b+a=2bx+a|_{x=-1}=-2b+a.

Setting equal: −4=−2b+2⇒b=3-4=-2b+2 \Rightarrow b=3. So a=2,b=3a=2,b=3.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question