Explanation
To find the value of p, we use the conditions for continuity and the continuity of the derivative f′(x).
Step 1: Continuity at x=3
For f(x) to be continuous at x=3, the left-hand limit must equal the right-hand limit:
x→3−limf(x)=x→3+limf(x)
x→3−lim(x−1)=x→3+lim(px2+qx+2)
3−1=p(3)2+q(3)+2
2=9p+3q+2
9p+3q=0⟹q=−3p…(Equation 1)
Step 2: Continuity of f′(x) at x=3
The derivative f′(x) is continuous at x=3 if the left-hand derivative equals the right-hand derivative:
x→3−limf′(x)=x→3+limf′(x)
For 1 < x < 3, f′(x)=dxd(x−1)=1.
For x > 3, f′(x)=dxd(px2+qx+2)=2px+q.
1=2p(3)+q
6p+q=1…(Equation 2)
Step 3: Solving for p
Substitute Equation 1 into Equation 2:
6p+(−3p)=1
3p=1
p=31
Thus, the value of p is 31, and the correct option is (c).
