# 120 - Mathematics Questions Answers

**Asked By: AARSHI**

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**Joshi sir comment**

I dont know what the question is but if the question is to solve this eq. then the method will be as given below

tan^{2}x tan^{2}3x tan4x = tan^{2}x - tan^{2}3x + tan4x

so tan4x = [tan^{2}x - tan^{2}3x] / [tan^{2}x tan^{2}3x-1]

now split RHS terms by formula x^{2}-y^{2}= (x+y)(x-y)

so RHS = tan4xtan2x

now solve

A(3,4) and B is a variable point on the line |X| = 6. Also, AB ≤ 4. Then the number of positions of B with integral co-ordinates is:

(a)5 (b) 6 (c) 10 (d) 12

**Asked By: GAUTHAM GANESH**

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**Joshi sir comment**

by diagram (6,4),(6,3),(6,2) are points having AB<4 besides it two upside points are also there by symmetry not shown in diagram. These points are (6,5) and (6,6)

Total no. of points = 5

help me understand lamis theorem

**Asked By: TANYE COLLINS KOFI**

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**Joshi sir comment**

It is the theorem for the equilibrium of a body under 3 forces

According to the theorem if three forces are acting on a body and body is in equilibrium then

P/sinA = Q/sinB = R/sinC

**Asked By: NEHA GUPTA**

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**Joshi sir comment**

let us consider the case of ellipse with x and y axes as their axes

eq. is x^{2}/a^{2} + y^{2}/b^{2}= 1

on differentiating we get 2x/a^{2} + 2yy^{'}/b^{2 }= 0 y'is first differential

so yy^{'}/x = -b^{2}/a^{2}

again differentiate and get answer.

You should remember that you should differentiate as many times as the number of constants.

for ex. in the case of parabola only first diffrentiation is sufficient.

now complete it for all conics

prove using property of determinant

1. a+b+c -c -b

-c a+b+c -a = 2(a+b)( b+c ) (c+a)

-b -a a+b+c

**Asked By: SONAM SHARMA**

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**Joshi sir comment**

first use C1 = C1+C2+C3

then R1 = R1+R2 and R2 = R2+R3

after it solve mathematically

using the transformation x=r cosθ and y=r sinθ,find the singular solutionof the differential equation x+py=(x-y)(p^{2}+1)½ where p=dy/dx

**Asked By: DEBANJAN GHOSHAL**

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**Joshi sir comment**

x=rcosθ and y=rsinθ

so dx=-rsinθdθ and dy = rcosθdθ

so p = -cotθ

so given eq. will become x-cotθy = (x-y)cosecθ

on putting values of x and y we get

0 = r(cosθ-sinθ)/sinθ

so tanθ = 1 so θ = nπ+π/4

so x = rcos(nπ+π/4) and y = rsin(nπ+π/4)

show that the family of parabolas y^{2}=4a(x+a) is self orthogonal.

**Asked By: DEBANJAN GHOSHAL**

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**Joshi sir comment**

y^{2}= 4a(x+a)

so 2yy^{'} = 4a so a = yy^{'}/2

on putting a we get y^{2}= 4yy^{'}/2(x+yy^{'}/2)

so y^{2} = yy^{'} (2x+yy^{'}) or y = 2xy^{'} + yy^{'2} (1)

now on putting -1/y^{'} in the place of y^{'}

we get y^{2} = -y/y^{'}[2x-y/y^{'}]

so -yy^{'2} = 2xy^{'} - y (2)

similarity of (1) and (2) shows that the given curve is self orthogonal

rank of the number 3241?(digits cannot be repeated)

**Asked By: HARSHITH**

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**Joshi sir comment**

total numbers = 4*3*2*1 = 24

on arranging these numbers in order we get 1234

so rank = [24/4]*2+[6/3]*1+[2/2]*1+1 = 16

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**which is the 4 digit number whose second digit is thrice the first digit and 3'rd digit is sum of 1'st and 2'nd and last digit is twice the second digit.**

**Asked By: SAURABH TIWARI**

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**Joshi sir comment**

let the 4 digit number is 1000a+100b+10c+d

then according to the given condition b=3a, c=a+b, d=2b

so c=4a, d=6a and b=3a

on putting all the values we get number = 1000a+100(3a)+10(4a)+6a = 1346a

on putting a =1, number = 1346

Sir , i have solved limx-->0 (sin^{-1}x - tan^{-1}x )/x^{3 }. plz have a look on others.

**Asked By: VISHAL PHOGAT**

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**Joshi sir comment**

limx-->0 (sin^{-1}x - tan^{-1}x )/x^{3}

use D L Hospital rule or expansion of sin^{-1}x and tan^{-1}x to solve it