Class 12 Continuity And Differentiability · Case Study4 Markscase_based_application
Class 12 Maths · Continuity And Differentiability · Case Study #34
For h>0 approaching 0, the left and right derivatives at a are limits of [f(a−h)−f(a)]/(−h) and [f(a+h)−f(a)]/h. A derivative exists when both limits exist, are finite and agree. Let f(x)=|x−3| for x≥1 and f(x)=x²/4−3x/2+13/4 for x<1.
Part (1) 1 Marks
Find the right derivative at x=1.
Answer: −1
- For 0<h<2, f(1+h)=|h−2|=2−h and f(1)=2. The quotient is [(2−h)−2]/h=−1.
- Thus the right derivative at 1 is −1.
Part (2) 1 Marks
Find the left derivative at x=1.
Answer: −1
- f(1−h)=2+h+h²/4. The left quotient is [h+h²/4]/(−h)=−1−h/4.
- Its limit is −1, so the left derivative is −1.
Part (3) 2 Marks
Determine whether f is differentiable at x=1.
Answer: Yes; f′(1)=−1.
- Both branches approach 2=f(1), so continuity holds.
- The left and right difference-quotient limits are each −1; they exist and agree.
- Therefore f is differentiable at 1 and f′(1)=−1.
Part (4) 2 Marks
Alternatively, find f′(2) and f′(−1).
Answer: f′(2)=−1; f′(−1)=−2.
- Near x=2, |x−3|=3−x, whose derivative is −1.
- For x<1, the derivative of the quadratic branch is x/2−3/2, which equals −2 at −1.
- Thus f′(2)=−1 and f′(−1)=−2.