# 315 - Mathematics Questions Answers

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**Submit By: MANISH SIR**6 year ago

limit n tends to ∞

then

[³√(n²-n³) + n ] equals

**Asked By: AMIT DAS**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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

Use the formula a^{3}+b^{3} = (a+b)(a^{2}+b^{2}-ab) in the format

(a+b) = (a^{3}+b^{3})/(a^{2}+b^{2}-ab)

here a = ³√(n²-n³) and b = n

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

if f(X)=xlxl then find f^{-1}(x) can you please explain me the meaning and use of sgn

**Asked By: NIKHIL VARSHNEY**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

f(x) = x|x| => f(x) = -x^{2} for negative real values of x and f(x) = x^{2} for positive real values of x

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

**The number of real negative terms in the binomial expansion of (1+ix) ^{(4n-2)}, n **

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**N and x > 0 is?**

**Asked By: KAMAL**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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

for a negative real number i^{2} is compulsory so associated terms are 3rd, 7th, 11th ..........

Total number of terms = n

which of the following is differntiable at x=0 ?

cos(lxl)+lxl

sin(lxl)-lxl

**Asked By: NIKHIL VARSHNEY**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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 = (1-2sin^{2}h+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

##
^{Consider a branch of the hyperbola x²-2y²-2√2x-4√2y-6=0, with vertex at the point A. Let B be one of the end points of its latus rectum .If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is ??????}

**Asked By: AMIT DAS**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

##
^{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 b^{2} = a^{2}(1-e^{2})

and end point of latus rectum : X = ae, Y = b^{2}/a

solve the area and get the answer