13. AM >=GM. Thus, true.
Pls Solve And EXPLAIN these to me
1)The equation secθ =-0.7 has a solution in the second and third quadrant.(True/False)
2)Find the maximum value of |sinα|-cosβ
3)The minimum value of sec2 θ +cos2 θ is 2 (True/false)
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10 Answers
11. cos x can be neg. in 3rd quadrant but it can never be <1 . cos x=1/sec x = 1/(-0.7) i.e sumthing less than -1 for sure. Thus i think sec x cant have value of -0.7
12. Im not sure what the 1st part of the q is.I think it is |sin a|. Then the q. is |sin a|-cos B max. value. Let Sin A take its max. value of 1. Cos B take its min. value of -1. Then it becomes 1- -1,i.e 1+1=2.
Therefore max. value of expression is 2.
1sec2x=1/cos2x.
(1/cos2x +cos2x)/2 >= √(1/cos2x * cos2x) Since AM >= GM
Therefore it comes--sec2x + cos2x >= 2.
1Another way of solving question no 3 ....which is a bit long but easier to understand...
cos2θ + sec2θ
=cos2θ + sec2θ +2secθcosθ - 2secθcosθ
=(cosθ - secθ)2+2secθcosθ
=(cosθ - secθ)2+2
since,
(cosθ - secθ)2≥ 0
therefore (cosθ - secθ)2 +2 ≥ 2
i,e cos2θ + sec2θ≥2
hence proved....
1ronald's answer is much more logical for question no. 3. the concept of am>=gm is very impt from iit point of view......
1definitely it was logical...but as arpan was not getting it...thats y i made it more easier...
1Thanks for ur answers.I had also found my technique for q no 3.
Since we are required 2 find minimum value of the expression and we have a '+' sign in between hence the minimum value of the expression has to b the minimum values of the individual functions.thus minimum value of secθ is -1 and that of cosθ is -1.Thus squaring and adding the minimum value of the expression must be =2
11)secθ=-0.7 .cosθ=-10/7.therefore no quadrant False
3)sec2 θ +cos2 θ= 1+cos^4 θ=cos has a maximum value of 1 therefore it has a maximum value of 2 true