1-lim-x-3-3x-x-2-4x-1-2-0-1-arctan-2x-1-1-x-x-2-dx-3-how-many-integer-solution-sets-exist-for-the-equation-x-2-y-2-2-

Question Number 110307 by bemath last updated on 28/Aug/20

(1) \lim_{x \to - \infty} \frac{3 - 3 x}{\sqrt{x^{2} - 4 x + 1}} ?

(2) \underset{0}{\int^{1}} \arctan (\frac{2 x - 1}{1 + x - x^{2}}) d x

(3) h o w m a n y i n t e g e r s o l u t i o n s e t s

e x i s t f o r t h e e q u a t i o n x^{2} + y^{2} = 2

Answered by john santu last updated on 28/Aug/20

★ \frac{J S}{\approx} ★

l e t x = - t, t h e n t \to \infty

\lim_{t \to \infty} \frac{3 + 3 t}{\sqrt{t^{2} + 4 t + 1}} = \lim_{t \to \infty} \frac{t (3 + \frac{3}{t})}{t \sqrt{1 + \frac{4}{t} + \frac{1}{t^{2}}}} = 3

Answered by Dwaipayan Shikari last updated on 28/Aug/20

\int_{0}^{1} {t a n}^{- 1} (\frac{x - (1 - x)}{1 + x (1 - x)})

\int_{0}^{1} {t a n}^{- 1} x - {t a n}^{- 1} (1 - x) = \int_{0}^{1} {t a n}^{- 1} (1 - x) - {t a n}^{- 1} x = I

2 I = 0

I = 0

Commented by Dwaipayan Shikari last updated on 28/Aug/20

I s i t r i g h t ?

Answered by Aziztisffola last updated on 28/Aug/20

(3) (x; y) \in {(1; 1); (1; - 1); (- 1; 1); (- 1; - 1)}

Answered by Rio Michael last updated on 28/Aug/20

\lim_{x \to - \infty} \frac{3 - 3 x}{\sqrt{x^{2} - 4 x + 1}} = \lim_{x \to - \infty} \frac{3 - 3 x}{x \sqrt{1 - \frac{4}{x} + \frac{1}{x^{2}}}}

= \lim_{x \to - \infty} \frac{\frac{3}{x} - 3}{\sqrt{1 - \frac{4}{x} + \frac{1}{x^{2}}}} = - 3

Commented by bemath last updated on 28/Aug/20

w r o n g s i r . x \to - \infty i t s h o u l d b e

\sqrt{x^{2}} = - x n o t x

Commented by Rio Michael last updated on 28/Aug/20

correct . thanks for the correction

Answered by 1549442205PVT last updated on 28/Aug/20

(1) \lim_{x \to - \infty} \frac{3 - 3 x}{\sqrt{x^{2} - 4 x + 1}} = \lim_{x \to - \infty} \frac{3 - 3 x}{(- x) \sqrt{1 - \frac{4}{x} + \frac{1}{x^{2}}}}

= \lim_{x \to - \infty} \frac{\frac{- 3}{x} + 3}{\sqrt{1 - \frac{4}{x} + \frac{1}{x^{2}}}} = \frac{0 + 3}{\sqrt{1 - 0 + 0}} = 3

3) Solve x^{2} + y^{2} = 2 in Z .

x^{2} = 2 - y^{2} ⩽ 2 but since x \in Z, ∣ x ∣⩽ 1

\Rightarrow x \in - 1, 1 \Rightarrow y \in {\pm 1, \pm 1}

The solution set of the given eq .

has four elements

S = {(- 1, - 1), (- 1, 1), (1, - 1), (1, 1)}

Answered by john santu last updated on 28/Aug/20

(2) \int \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x = I

s e t t i n g x = u - \frac{1}{2}

I = \underset{- 1 / 2}{\int^{1 / 2}} \tan^{- 1} (\frac{2 u}{\frac{5}{4} - u^{2}}) d u

s i n c e t h e r e s u l t i n g i n t e g r a n d i s

a c o n t i n o u s o d d f u n c t i o n, s o w e g o t

I = 0 .