Mathematics: Post your doubts here!

[Hamnah Zahoor]

hellodjfos;s'ff Thank you for your help

 

[Hamnah Zahoor]

You must find the combinations when only the batsman is being included in the squad(1st), only the bowler is being included in the squad(2nd) and when none of them are being included(3rd).
1st: 1C1(for the batsman having a confirmed place)*6C4*4C4(bowler has been excluded here)*2C1*2C1=60
2nd: 6C5*1C1(bowler having confirmed place)*4C3*2C1*2C1=96
3rd: 6C5(batsman excluded)*4C4(bowlers excluded)*2C1*2C1=24
24+96+60=180

In the third why are we excluding both the batsman and the bowler.

 

[hellodjfos;s'ff]

In the third why are we excluding both the batsman and the bowler.

Because when we exclude both the batsmen and bowler, the batsman is still refusing to be in a team with a bowler but if they are not picked, they don't end up being in a team together anyways so the condition is met.

 

[Hamnah Zahoor]

Because when we exclude both the batsmen and bowler, the batsman is still refusing to be in a team with a bowler but if they are not picked, they don't end up being in a team together anyways so the condition is met.

Thank you

 

[A*****]

|z-10i|=2|z-4i|
(z-10i)(z+10i)=4(z-4i)(z+41) - using the concept from first equation in (i)
(z-10i)(z-10i)*=4(z-4i)(z-4i)* - now use concept from 2nd equation in (ii)
(z-10i)(z*+10i)=4(z-4i)(z*+4i)
zz*+10iz-10iz*+100=4(zz*+4iz-4iz*+16)
3zz*+6iz-6iz*-36=0
3(zz*-2iz*+2iz-12)=0 - first part proven
|z-2i|=(z-2i)(z*+2i)=zz*-2iz*+2iz+4
Now compare this with the earlier equation we have proven. If we subtract 16 from this new equation, both the equations would be equal right? So...
zz*-2iz*+2iz+4=4^2 - notice i out 4^2 not 16 because we were squaring both sides of the equation as the question had told us to do earlier.
Hence radius=4 making the equation |z-2i|=4

Can I work the 2nd part like this....
zz*-2iz*+2iz-12=0
zz*-2iz*+2iz+4-16=0
zz*-2iz*+2iz-4i²-16=0
z*(z-2i)+2i(z-2i)=4²
(z-2i)(z*+2i)=4²
(z-2i)(z-2i)*=4²
|z-2i|²=4²
|z-2i|=4
P.s can we take z-2i as whole in place of z because the statements in (i) are for z?

 

[hellodjfos;s'ff]

Can I work the 2nd part like this....
zz*-2iz*+2iz-12=0
zz*-2iz*+2iz+4-16=0
zz*-2iz*+2iz-4i²-16=0
z*(z-2i)+2i(z-2i)=4²
(z-2i)(z*+2i)=4²
(z-2i)(z-2i)*=4²
|z-2i|²=4²
|z-2i|=4
P.s can we take z-2i as whole in place of z because the statements in (i) are for z?

That works well too.

 

[A*****]

hellodjfos;s'ff in these questions how do we know that we have to take x in radians or degrees?

 

[A*****]

(iii) part plz...a bit confused about the minus sign in the argumemt of u

 

[hellodjfos;s'ff]

hellodjfos;s'ff in these questions how do we know that we have to take x in radians or degrees?

Usually they tell us whether to use radians or degrees. For complex number questions, try to use radians always (since that's the conventional rule) but they do accept answers in degrees. For a question like this where it's unclear whether to use radians or degrees, just look at how reasonable the trigonometric term is... so try out different values of x in sin(x/3) in degrees and radians mode and see which is more reasonable. Since most values of x produce a value too small in degree mode and most values of x produce a value large enough in radians, we do this question in radians.

 

[hellodjfos;s'ff]

(iii) part plz...a bit confused about the minus sign in the argumemt of u

arg(u*/u) means arg(u*)-arg(u)
so arg(u*/u)=arg(u*)-arg(u)
Find these arguments and you will be able to prove the equation.

 

[A*****]

arg(u*/u) means arg(u*)-arg(u)
so arg(u*/u)=arg(u*)-arg(u)
Find these arguments and you will be able to prove the equation.

u is 3-i and u* 3+i
The problem is when it would be written as arg(u*)-arg(u), which would be tan-¹(1/3)-tan-¹(-1/3), what would be done to the minus sign in arg(u)?

 

[hellodjfos;s'ff]

u is 3-i and u* 3+i
The problem is when it would be written as arg(u*)-arg(u), which would be tan-¹(1/3)-tan-¹(-1/3), what would be done to the minus sign in arg(u)?

-tan-¹(1/3) and tan-¹(-1/3) are the same so tan-¹(1/3)-tan-¹(-1/3) becomes tan-¹(1/3)+tan-¹(1/3) hence 2tan-¹(1/3)

 

[A*****]

-tan-¹(1/3) and tan-¹(-1/3) are the same so tan-¹(1/3)-tan-¹(-1/3) becomes tan-¹(1/3)+tan-¹(1/3) hence 2tan-¹(1/3)

Thank u sooo much!!!

 

[A*****]

-tan-¹(1/3) and tan-¹(-1/3) are the same so tan-¹(1/3)-tan-¹(-1/3) becomes tan-¹(1/3)+tan-¹(1/3) hence 2tan-¹(1/3)

Will tan-¹(1/-3) also equal -tan-¹(1/3)?

 

[hellodjfos;s'ff]

Will tan-¹(1/-3) also equal -tan-¹(1/3)?

Yes they're the same. You can even put the values in your calculator and clarify that.

 

[A*****]

Yes they're the same. You can even put the values in your calculator and clarify that.

So if a complex number is -5+2i so how will its argument be written in a presentable way? And it won't be equal to -tan-¹(2/5) right?

 

[hellodjfos;s'ff]

-5+2i would be in the 2nd quadrant so the argument would be pi/2+tan-¹(5/2)

 

[Hamnah Zahoor]

Can someone help me with the second part why are we subtracting 50 from 72?
Nov 2010 62

 

[hellodjfos;s'ff]

Can someone help me with the second part why are we subtracting 50 from 72?
Nov 2010 62View attachment 64597

To find standard deviation, you need the value of either Σx and Σx^2 or Σ(x-50) and Σ(x-50)^2. Since we only have the latter, we have to solve the question using them.
We found Σ(x-50) and Σ(x-50)^2 in the first part... these coded total represents the value of Σx-50 which means that every data added to the set must have 50 subtracted from its value(this is the basis of coding - if you didn't know this, you should probably learn coding from scratch). Which is why when we add a data with value 72, it effectively has a value of 22 in the Σ(x-50) coding.

Edit: The alternative method would be to find Σx^2 using earlier standard deviation and value of Σx. In this case, you would add 72 to the coding since these are the original values. But this method is more time-consuming since you got to spend time finding the value of Σx^2.

 

[PLAyer2002002]

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