what will be Ymax and Ymin
11th- trigno
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y = (27)^cos2x .(81)^sin2x
= (3)^3cos2x.(3)^4sin2x
= (3)^(3cos2x + 4sin2x)
y will be maximum when ( 3cos2x + 4sin2x ) will be maximum and vice - versa .
we know,
-√(3² + 4²) ≤ ( 3cos2x + 4sin2x ) ≤ √(3² +4²)
-5 ≤ ( 3cos2x + 4sin2x ) ≤ 5
hence,
Ymax = (3)^5 = 243
Ymin = (3)^-5 = 1/243
= (3)^3cos2x.(3)^4sin2x
= (3)^(3cos2x + 4sin2x)
y will be maximum when ( 3cos2x + 4sin2x ) will be maximum and vice - versa .
we know,
-√(3² + 4²) ≤ ( 3cos2x + 4sin2x ) ≤ √(3² +4²)
-5 ≤ ( 3cos2x + 4sin2x ) ≤ 5
hence,
Ymax = (3)^5 = 243
Ymin = (3)^-5 = 1/243
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