JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023The sum of the absolute maximum and minimum values of the function f(x)=∣x2−5x+6∣−3x+2 in the interval [−1,3] is equal to:
Choose the correct answer:
- A.
12
- B.
13
- C.
10
(Correct Answer) - D.
24
10
Explanation
Solution
Step 1: Redefine the function.
x2−5x+6=(x−2)(x−3). In [−1,3], this is positive in [−1,2] and negative in [2,3].
-
For x∈[−1,2], f(x)=x2−5x+6−3x+2=x2−8x+8
-
For x∈[2,3], f(x)=−(x2−5x+6)−3x+2=−x2+2x−4
Step 2: Check critical points and endpoints.
-
f(−1)=1+8+8=17 (Absolute Max)
-
f(2)=4−16+8=−4
-
f(3)=−9+6−4=−7 (Absolute Min)
-
Critical point in [−1,2]: f′(x)=2x−8=0⟹x=4 (not in interval).
-
Critical point in [2,3]: f′(x)=−2x+2=0⟹x=1 (not in interval).
Step 3: Sum of values.
Sum =17+(−7)=10.
Correct Option: (3)
Explanation
Solution
Step 1: Redefine the function.
x2−5x+6=(x−2)(x−3). In [−1,3], this is positive in [−1,2] and negative in [2,3].
-
For x∈[−1,2], f(x)=x2−5x+6−3x+2=x2−8x+8
-
For x∈[2,3], f(x)=−(x2−5x+6)−3x+2=−x2+2x−4
Step 2: Check critical points and endpoints.
-
f(−1)=1+8+8=17 (Absolute Max)
-
f(2)=4−16+8=−4
-
f(3)=−9+6−4=−7 (Absolute Min)
-
Critical point in [−1,2]: f′(x)=2x−8=0⟹x=4 (not in interval).
-
Critical point in [2,3]: f′(x)=−2x+2=0⟹x=1 (not in interval).
Step 3: Sum of values.
Sum =17+(−7)=10.
Correct Option: (3)