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Author Topic: math doubts [NEW]  (Read 15871 times)

Offline astarmathsandphysics

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Re: functions
« Reply #75 on: October 14, 2009, 07:20:58 pm »
its paper 1 ...pure 1

I will look now
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Offline astarmathsandphysics

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Re: functions
« Reply #76 on: October 14, 2009, 07:31:40 pm »
i)a_4=a+4d
a_14=a+14d
ii)\frac {a_5}{a}=\frac{a_{15}}{a_5}
\frac {a+4d}{a}= \frac {a+14d}{a+4d}
(a+4d)^2 = a(a+14d)
a^2 +8ad +16d^2 =a^2 +14ad
16d^2 =6ad
8d=3a
iii)r=\frac {a+4d}{a}=\frac {a+1.5a}{a}=\frac {2.5a}{a}=2.5
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Offline sameer210394

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Re: functions
« Reply #77 on: October 19, 2009, 09:05:47 am »
Quote from: ~Am~ on September 26, 2009, 09:05:46 pm
shudnt -3>x>5


abbey Adi, u and the guy wrote the same answer and arguing?  ???
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Offline Ghost Of Highbury

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Re: functions
« Reply #78 on: October 19, 2009, 09:33:24 am »
Quote from: sameer210394 on October 19, 2009, 09:05:47 am
abbey Adi, u and the guy wrote the same answer and arguing?  ???

yea lol...v realized it later.. :P :P
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Offline helllife

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Re: functions
« Reply #79 on: October 29, 2009, 05:06:48 pm »
i have a questions....
1) Find the sets of values of k for which the roots of the equation x^2+kx-k+3=0 are real and of the same sign.
2)Given that y=px^2+2qs+r, shows that y will have same sign for all real values of x if and only if q^2<pr.

can anyone explain this?
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Offline astarmathsandphysics

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Re: functions
« Reply #80 on: October 29, 2009, 05:19:41 pm »
1)b^2-4ac=k^2-4*1*(-k+3)>0 and -k>sqrt(k^2-4*1*(-k+3))>0
k^2+4k-12=(k+6)(k-2)>0 so k>6 or k<-2 and k^2>k^2-4*1*(-k+3) so-k+3>0 so k<3

so k>6 or k<-2 anf k<3ie k<-2

")b^2-4ac<0 since y=0 has no roots
(3q)^2-4pr<0 so 4q^2<4pr so q^2<pr.
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zara

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Re: math doubts [NEW]
« Reply #81 on: October 31, 2009, 11:42:13 am »
NEW QUESTIONS! PLZ HELP!


Q1)Given that x=sin-1(2/5), find the exact value of
(i) cos2x
(ii)tan2x
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zara

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Re: math doubts [NEW]
« Reply #82 on: October 31, 2009, 11:45:10 am »
Q2) Prove this identity

1/cos(theta) + tan(theta) = cos(theta)/1-sin(theta)
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #83 on: October 31, 2009, 11:49:56 am »
first find x

sin-1 2/5 = 23.5

cos2 x = 0.84

and

tan2x = 0.19

just substitute and find it
---
next

(1/cos x + sinx/cosx) = (1+sinx/ cos x) * (1-sinx/1-sinx) = 1-sin2x/cosx - sinxcosx = cos2x/cosx-sinxcosx

= cos2x / cosx(1-sinx) = cosx/1-sinx

x = theta

LHS = RHS

hence proved
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zara

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Re: math doubts [NEW]
« Reply #84 on: October 31, 2009, 11:55:29 am »
Q3)Given that the equation


[sin(theta) - cos(theta)]/[sin(theta)+cos(theta)] =6sin(theta)/cos(theta),

find the exact values of tan(theta).Hence find (theta) for 0<=(theta)<=360
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zara

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Re: math doubts [NEW]
« Reply #85 on: October 31, 2009, 12:02:08 pm »
Quote from: A@di on October 31, 2009, 11:49:56 am
first find x

sin-1 2/5 = 23.5

cos2 x = 0.84

and

tan2x = 0.19

just substitute and find it
---
next

(1/cos x + sinx/cosx) = (1+sinx/ cos x) * (1-sinx/1-sinx) = 1-sin2x/cosx - sinxcosx = cos2x/cosx-sinxcosx

= cos2x / cosx(1-sinx) = cosx/1-sinx

x = theta

LHS = RHS

hence proved
adi how u got cos2=0.84? n same for tan?
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #86 on: October 31, 2009, 12:04:33 pm »
find cos x --> square the answer
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zara

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Re: math doubts [NEW]
« Reply #87 on: October 31, 2009, 12:16:21 pm »
Quote from: A@di on October 31, 2009, 12:04:33 pm
find cos x --> square the answer
how?
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Offline Ghost Of Highbury

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Re: math doubts [NEW]
« Reply #88 on: October 31, 2009, 12:23:14 pm »
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zara

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Re: math doubts [NEW]
« Reply #89 on: October 31, 2009, 12:52:32 pm »
oh shoot!
lol thts wht hppns ven im doin many things at a time...
now i get the whole thing....Thanks...
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