# 161 - Mathematics Questions Answers

### Tangents are drawn to the circle x^2 + y^2=12 at the points where it is met by the circle x^2 +y^2 - 5x + 3y - 2=0.Find the point of intersection of these tangents.

**Asked By: SUSHOVAN HALDAR**

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**Joshi sir comment**

sovle the two eqs for x and y,

method :

### x^2 + y^2=12

on putting this in 2nd eq. we get -5x+3y = -10 so x = (3y+10)/5

so [(3y+10)/5]^{2 }+ y^{2}= 12

now get y then x, then get the tangents in the first circle and then intersection point of the tangents.

If x+y = 1 ,then ∑ r ^{n}C_{r}x^{r}y^{n-r}equals nx....How??

**Asked By: RESHU SINGH**

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**Joshi sir comment**

∑ r ^{n}C_{r}x^{r}y^{n-r}

= ∑ r ^{n}C_{r}x^{r}(1-x)^{n-r}

= ∑ r n!/r!(n-r)! x^{r}(1-x)^{n-r}

= n (n-1)!/(r-1)!(n-r)! x^{r}(1-x)^{n-r }

= n ^{(}^{n-1)}C_{(}_{r-1) }x^{r}(1-x)^{n-r}

= n ^{(}^{n-1)}C_{r }x^{r+1}(1-x)^{n-r-1 }let r = r+1

= nx ^{(}^{n-1)}C_{r }x^{r}(1-x)^{n-r-1 }

^{now solve}

How to inegrate ( In t) dt /(t-1)

**Asked By: PARTHASARATHY**

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**Joshi sir comment**

let lnt = x so dt/t = dx so dt = tdx = e^{x}dx

now ∫(ln t)dt/(t-1) = ∫xe^{x}dx/(e^{x}-1) = ∫x(e^{x}-1+1)dx/(e^{x}-1)

now solve

ƒ(x)=|x-2| and g=ƒoƒ so g'(x)=.......? ; 2<x<4

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

for 2<x<4, f(x) = x-2

so g(x) = f(x-2) = x-2-2 = x-4

so g'(x) = 1

ƒ(x)=3|2+x| so ƒ'(-3)=? ,if its -3 , how?

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

ƒ(x)=3|2+x|

for x>-2, f(x) = 3(2+x) so f^{'}(x) = 3

for x<-2, f(x) = -3(2+x) so f^{'}(x) = -3

∫√tanx

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

let √tanx = t

so sec^{2}xdx/2√tanx = dt

so (1+t^{4})dx/2t = dt

so dx = 2tdt/(1+t^{4})

now solve

What is the value of 0⁄0? if its otherthan 1, then how this can be true: lim (e^{x}-1)/x = 1

^{ x −−>0}

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

exact 0 and lim tends to 0 are different

what is the angular difference between the +j and -j ?

OR

+4j and -4j ?

**Asked By: ADARSH KUMAR**

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**Joshi sir comment**

180 degree

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)

= (π-2) + (π-4) + (6-2π)

= 0

if ƒ:R−{0} -> R, 2ƒ(x) − 3ƒ(1⁄x) = x² then ƒ(3)= ?

**Asked By: BONEY HAVELIWALA**

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**Joshi sir comment**

2ƒ(x) − 3ƒ(1⁄x) = x² (1)

so 2ƒ(1/x) − 3ƒ(x) = 1/x² (2)

multiply (1) to 2 and (2) to 3, then add the two equations, you will get f(x) then calculate f(3)