UPSC CDS 1 2022 Math Paper

Solution Given:
y=cos2x+sec2x;0≤x<‌

π

2

Concept Used:
- A.M ≥ G.M
- If a and b are two numbers then (a+b)∕2≥√‌ab
- cos2x×sec2x=1
Calculation:
We have y=cos2x+sec2x‌‌−⋯ (i)
Let the two numbers be cos2x and sec2x
According to the concept used A. M≥G.M
⇒[cos2x+sec2x]∕2≥√‌[cos2x×sec2x]
⇒[cos2x+sec2x]∕2≥1
⇒[cos2x+sec2x]≥2
Now, from (i)
⇒y≥2
∴ The correct relation is y≥2.