CUET PG 2025 — Mathematics PYQ
CUET PG | Mathematics | 2025The points (K,2−2K), (−K+1,2K) and (−4−K,6−2K) are collinear if:
(A) K=21
(B) K=−21
(C) K=23
(D) K=−1
(E) K=1
Choose the correct answer from the options given below:
Choose the correct answer:
- A.
(A) and (D) only
(Correct Answer) - B.
(A) and (E) only
- C.
(B) and (D) only
- D.
(D) only
(A) and (D) only
Explanation
1. Values rakhne par:
K−K+1−4−Kamp;2−2Kamp;2Kamp;6−2Kamp;1amp;1amp;1=0
2. Operations R2→R2−R1 aur R3→R1 apply karne par:
K−2K+1−4−2Kamp;2−2Kamp;4K−2amp;4amp;1amp;0amp;0=0
3. Column 3 (C3) ke along expand karne par:
1⋅[(−2K+1)(4)−(−4−2K)(4K−2)]=0
4. Equation solve karne par:
−8K+4−[−16K+8−8K2+4K]=0
−8K+4−[−12K+8−8K2]=0
−8K+4+12K−8+8K2=0
8K2+4K−4=0
5. 4 se divide karne par:
2K2+K−1=0
6. Factorization:
2K2+2K−K−1=0
2K(K+1)−1(K+1)=0
(2K−1)(K+1)=0
7. Final Values:
K=21,K=−1
8. Final Answer
(A) and (D) only
Explanation
1. Values rakhne par:
K−K+1−4−Kamp;2−2Kamp;2Kamp;6−2Kamp;1amp;1amp;1=0
2. Operations R2→R2−R1 aur R3→R1 apply karne par:
K−2K+1−4−2Kamp;2−2Kamp;4K−2amp;4amp;1amp;0amp;0=0
3. Column 3 (C3) ke along expand karne par:
1⋅[(−2K+1)(4)−(−4−2K)(4K−2)]=0
4. Equation solve karne par:
−8K+4−[−16K+8−8K2+4K]=0
−8K+4−[−12K+8−8K2]=0
−8K+4+12K−8+8K2=0
8K2+4K−4=0
5. 4 se divide karne par:
2K2+K−1=0
6. Factorization:
2K2+2K−K−1=0
2K(K+1)−1(K+1)=0
(2K−1)(K+1)=0
7. Final Values:
K=21,K=−1
8. Final Answer
(A) and (D) only