Question : If $a^2+b^2=2$ and $c^2+d^2=1$, then the value of $(ad-bc)^2+(ac+bd)^2$ is:
Option 1: $\frac{4}{9}$
Option 2: $\frac{1}{2}$
Option 3: $1$
Option 4: $2$
Answer (1)
25th Jan, 2024
Correct Answer: $2$
Solution :
Given: $a^2+b^2=2$ and $c^2+d^2=1$
$(ad-bc)^2+(ac+bd)^2$
$= (ad)^2+(bc)^2-2abcd+(ac)^2+(bd)^2+2abcd$
$=d^2(a^2+b^2)+c^2(a^2+b^2)$
$=(a^2+b^2)(c^2+d^2)$
$=2 \times 1$
$=2$
Hence, the correct answer is 2.
Comments (0)
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Get Updates Brochure
Thank You!
Your Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Application Date:
1 Dec, 2024
- 30 Dec, 2024
Admit Card Date:
9 Dec, 2024
- 16 Dec, 2024
Mains Exam
Exam Date:
14 Dec, 2024
- 14 Dec, 2024
