some dbts from mtg.

1)No of ordered triplets (x,y,z) satisfying
(1+sin4x)(2+cot2y)(4+sin4z)12sin2x
2)Let P be a point on the hyperbola x2-y2=a2,wher a is parameter,such that P is nearest to the line y=2a.The locus of P is _____

3)suppose a sequence {bn} is defined as b1=1 and bn=cosbn-1

and another sequence {pn} p1=1 and pn=pn-1+cospn-1

then prove that the two sequences were non monotonic.

7 Answers

1
Unicorn--- Extinct!! ·

Q1)
Is there no 'y' in the inequality?

3
msp ·

lolz sry,edited

11
Devil ·

3)
Monotonicity of {Pn} & {Bn} both depend on the inequality cosθ<θ or the reverse....now from the graph itself we can cocnclude that the line y=x and y=cosθ have no specific inequality-relation....i mean we can have both cosθ>θ as well as the reverse....From which we can cocnclude that the sequences are not monotonic.

11
Devil ·

I don't get the 2nd qsn...there's a point of intersection of that line and the given hyperbola....i.e. (a5,2a)....what is the locus for?

3
msp ·

soumi i thot of the same for the second and the third,and i am clueless in the first.tx for the confirmation.

1
b_k_dubey ·

(1 + sin4xsin2x)(1 + (1 + cot2y))(4 + sin4z) ≤ 12

(cosec2x + sin2x)(1 + cosec2y)(4 + sin4z) ≤ 12 ...... (1)

now, cosec2x + sin2x2 ≥ (cosec2x sin2x)1/2 = 1

cosec2x + sin2x ≥ 2 .... (a)

1 + cosec2y ≥ 2 (since min of cosec2y is 1) .... (b)

4 + sin4z ≥ 3 (since min of sin4z is -1) ......... (c)

multiplying (a), (b), (c) :
(cosec2x + sin2x)(1 + cosec2y)(4 + sin4z) ≥ 12 ...... (2)

from (1) and (2) :

(cosec2x + sin2x)(1 + cosec2y)(4 + sin4z) = 12

for which : sin2x = 1, cosec2y = 1, sin4z = -1

3
msp ·

great work sir,thanq.

Your Answer

Close [X]