166 - Questions Answers

I think n is printed by mistake, so considering it as x

the question will be limit x tends to ∞

x{ [tan^-1 (x+1/x+4)] - (π/4)}

consider [tan^-1 (x+1/x+4)] - (π/4) = θ

so [tan^-1 (x+1/x+4)] = (π/4)+θ

so (x+1/x+4) = tan(π/4+θ)

or x = (-3-5tanθ)/2tanθ

so converted question will be limθ->0  (-3-5tanθ)θ/2tanθ

solve

I think n is printed by mistake, so considering it as x

the question will be limit x tends to ∞

x{ [tan^-1 (x+1/x+4)] - (π/4)}

consider [tan^-1 (x+1/x+4)] - (π/4) = θ

so [tan^-1 (x+1/x+4)] = (π/4)+θ

so (x+1/x+4) = tan(π/4+θ)

or x = (-3-5tanθ)/2tanθ

so converted question will be limθ->0  (-3-5tanθ)θ/2tanθ

now we know that limθ->0 tanθ/θ = limθ->0 θ/tanθ = 1

so next line will be limθ->0 (-3-5tanθ)/2 = -3/2 

We have to calculate limit n -> ∞ [{³√(n²-n³) }+ n ]

Use the formula a³+b³ = (a+b)(a²+b²-ab) in the format 

(a+b) = (a³+b³)/(a²+b²-ab)

here a = ³√(n²-n³)  and b = n

on solving we get 1/(1+1+1) = 1/3

f(x) = x|x|  => f(x) = -x²  for negative real values of x  and f(x) = x² for positive real values of x

so f^-1(x) = -√|x|  for negative real values and = +√|x| for positive real values of x

Total number of terms in the expansion = 4n-2+1 = 4n-1

for a negative real number i² is compulsory so associated terms are 3rd, 7th, 11th ..........

Total number of terms = n

sin(|x|)+|x| is differentiable at x = 0

LHD of first = {cos|0+h|+|0+h|-cos|0|-|0|}/0+h-0  = (cosh+h-1)/h = (1-2sin²h+h-1)/h = 1 (on taking limits)

now check RHD, its value will be -1

similarly in second both values are 0 so it is differentiable

^{x²-2y²-2√2x-4√2y-6=0} 

^{Arrange this equation in form of standard hyperbola as} 

(x-√2)²/4 - (y+√2)²/2 = 1

so X = x-√2, Y = y+√2

vertex coordinate :  X = 0 and Y = 0 so x = √2, y = -√2

similarly focus : X = ae, Y = 0,     here a = 2, b = √2 and b² = a²(1-e²)

and end point of latus rectum : X = ae, Y = b²/a

solve the area and get the answer