JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023The value of 12∫03∣x2−3x+2∣dx is
Choose the correct answer:
- A.
22
(Correct Answer) - B.
23
- C.
24
- D.
25
22
Explanation
Step 1: Critical points nikalna
Modulus ke andar ka function f(x)=x2−3x+2 hai. Ise factorise karte hain:
Function ke signs:
-
x∈[0,1] par: Positive (+)
-
x∈[1,2] par: Negative (−)
-
x∈[2,3] par: Positive (+)
Step 2: Integral ko break karein
Modulus hatane ke liye hum limit ko teen parts mein divide karenge:
Step 3: Integration solve karein
∫(x2−3x+2)dx=3x3−23x2+2x
-
First Part [0,1]: (31−23+2)−(0)=62−9+12=65
-
Second Part [1,2]: (38−6+4)−(65)=32−65=64−5=−61 (Modulus ki wajah se ye +1/6 ho jayega)
-
Third Part [2,3]: (9−227+6)−(32)=23−32=69−4=65
Step 4: Total sum nikalna
Sum =65−(−61)+65=65+61+65=611
Step 5: Question ki value
Hume 12×Integral nikalna hai:
Final Answer:
Iska sahi jawab 22 hai.
Explanation
Step 1: Critical points nikalna
Modulus ke andar ka function f(x)=x2−3x+2 hai. Ise factorise karte hain:
Function ke signs:
-
x∈[0,1] par: Positive (+)
-
x∈[1,2] par: Negative (−)
-
x∈[2,3] par: Positive (+)
Step 2: Integral ko break karein
Modulus hatane ke liye hum limit ko teen parts mein divide karenge:
Step 3: Integration solve karein
∫(x2−3x+2)dx=3x3−23x2+2x
-
First Part [0,1]: (31−23+2)−(0)=62−9+12=65
-
Second Part [1,2]: (38−6+4)−(65)=32−65=64−5=−61 (Modulus ki wajah se ye +1/6 ho jayega)
-
Third Part [2,3]: (9−227+6)−(32)=23−32=69−4=65
Step 4: Total sum nikalna
Sum =65−(−61)+65=65+61+65=611
Step 5: Question ki value
Hume 12×Integral nikalna hai:
Final Answer:
Iska sahi jawab 22 hai.

