JEE 2023 Mathematics PYQ — Consider the triangles with vertices , and , . If the maximum and… | Mathem Solvex | Mathem Solvex
Tip:A–D to answerE for explanationV for videoS to reveal answer
JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Consider the triangles with vertices A(2,1), B(0,0) and C(t,4), t∈[0,4]. If the maximum and the minimum perimeters of such triangles are obtained at t=α and t=β, respectively, then 6α+21β is equal to ________.
Choose the correct answer:
A.
48
(Correct Answer)
B.
49
C.
50
D.
51
Correct Answer:
48
Explanation
Solution
Triangle ABC ka perimeter P(t) teeno sides ki length ka sum hota hai:
P(t)=AB+BC+CA
1. Side Lengths nikalna
Side AB:AB=(2−0)2+(1−0)2=4+1=5 (Yeh constant hai).
Side BC:BC=(t−0)2+(4−0)2=t2+16.
Side CA:CA=(t−2)2+(4−1)2=(t−2)2+9.
Ab perimeter function banega:
P(t)=5+t2+16+(t−2)2+9
2. Minimum Perimeter (β) nikalna
Minima find karne ke liye hum derivative P′(t)=0 set karenge:
P′(t)=t2+16t+(t−2)2+9t−2=0
t2+16t=(t−2)2+92−t
Dono side square karne par aur cross-multiply karne par:
t2((t−2)2+9)=(2−t)2(t2+16)
t2(t−2)2+9t2=(t−2)2t2+16(t−2)2
9t2=16(t−2)2⟹3t=4(2−t) or 3t=−4(2−t)
3t=8−4t⟹7t=8⟹t=78
Toh, β=78.
3. Maximum Perimeter (α) nikalna
Interval t∈[0,4] mein maximum value endpoints par check ki jati hai:
At t=0: P(0)=5+16+4+9=5+4+13
At t=4: P(4)=5+16+16+4+9=5+42+13
Clearly, P(4) > P(0), isliye maximum t=4 par hoga.
Toh, α=4.
4. Final Value Calculate karna
Hume 6α+21β ki value nikalni hai:
6(4)+21(78)
24+3(8)
24+24=48
Answer:48
Explanation
Solution
Triangle ABC ka perimeter P(t) teeno sides ki length ka sum hota hai:
P(t)=AB+BC+CA
1. Side Lengths nikalna
Side AB:AB=(2−0)2+(1−0)2=4+1=5 (Yeh constant hai).
Side BC:BC=(t−0)2+(4−0)2=t2+16.
Side CA:CA=(t−2)2+(4−1)2=(t−2)2+9.
Ab perimeter function banega:
P(t)=5+t2+16+(t−2)2+9
2. Minimum Perimeter (β) nikalna
Minima find karne ke liye hum derivative P′(t)=0 set karenge:
P′(t)=t2+16t+(t−2)2+9t−2=0
t2+16t=(t−2)2+92−t
Dono side square karne par aur cross-multiply karne par:
t2((t−2)2+9)=(2−t)2(t2+16)
t2(t−2)2+9t2=(t−2)2t2+16(t−2)2
9t2=16(t−2)2⟹3t=4(2−t) or 3t=−4(2−t)
3t=8−4t⟹7t=8⟹t=78
Toh, β=78.
3. Maximum Perimeter (α) nikalna
Interval t∈[0,4] mein maximum value endpoints par check ki jati hai:
At t=0: P(0)=5+16+4+9=5+4+13
At t=4: P(4)=5+16+16+4+9=5+42+13
Clearly, P(4) > P(0), isliye maximum t=4 par hoga.