166 - Mathematics Questions Answers

first use C1 = C1+C2+C3

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

after it solve mathematically

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)

y²= 4a(x+a)

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

on putting a we get y²= 4yy^'/2(x+yy^'/2)      

so y² = yy^' (2x+yy^')  or y = 2xy^' + yy^'2    (1)

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

we get y² = -y/y^'[2x-y/y^']

so -yy^'2 = 2xy^' - y (2)

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

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

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

limx-->0    (sin^-1x  - tan^-1x )/x³

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

two intersecting points means two solutions

multiply with the conjugate in numerator again and again you will get 

lim_x->∞ ((x+1)(x+2)(x+3)(x+4) - x⁴)/((x+1)(x+2)(x+3)(x+4))^1/4 + x )*((x+1)(x+2)(x+3)(x+4))^1/2 + x² )

after solving the numerator get a 3 power expression of x then take 3 power of x common, similarly take x and 2 power of x from the first and second denominator terms

its I. E. Maron 

There should be any variabe in the question