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Relation and FunctionsQuestion and Answers: Page 4

Question Number 185211 Answers: 3 Comments: 0

Question Number 183712 Answers: 1 Comments: 0

solve: W(In(4x))=(√((x−1)))

$s o l v e :$ $W (I n (4 x)) = \sqrt{(x - 1)}$

Question Number 183158 Answers: 1 Comments: 0

Given f(x)= (([(1/3)x]∣2x∣+Ax)/(∣4−x^2 ∣)) if f ′(−1)= 5 then A=? [ ] = floor function

$G i v e n f (x) = \frac{[\frac{1}{3} x] ∣ 2 x ∣ + A x}{∣ 4 - x^{2} ∣}$ $i f f^{'} (- 1) = 5 t h e n A = ?$ $[] = f l o o r f u n c t i o n$

Question Number 183119 Answers: 0 Comments: 0

Question Number 180901 Answers: 1 Comments: 0

∫_1 ^( n) ((⌊x⌋)/x^2 )dx =

$\int_{1}^{n} \frac{⌊ x ⌋}{x^{2}} d x =$

Question Number 178635 Answers: 1 Comments: 2

solution set of log_x^(2 ) ((x/(∣x∣))−x)≥0

$solution set of \log_{x^{2}} (\frac{x}{∣ x ∣} - x) ⩾ 0$

Question Number 178624 Answers: 1 Comments: 0

let f:[0,1]→ R be given by f(x) = (((1+x^(1/3) )^3 +(1−x^(1/3) )^3 )/(8(1+x))) then max{f(x): x∈[0,1]}−min{f(x):x∈[0,1]} is

$let f : [0, 1] \to R be given by$ $f (x) = \frac{{(1 + x^{\frac{1}{3}})}^{3} + {(1 - x^{\frac{1}{3}})}^{3}}{8 (1 + x)} then$ $\max {f (x) : x \in [0, 1]} - \min {f (x) : x \in [0, 1]}$ $is$

Question Number 178032 Answers: 0 Comments: 3

• D={z : ∣z∣<1} • H (A→B) denotes the set of holomorfic functions from A to B • We define: W={f∈H (D→R) : ∣∣f∣∣_W <∞ } where ∣∣ ∙ ∣∣_W : { (W,→,R_+ ),(f, ,(Σ_(n=0) ^∞ ((∣f^((n)) (0)∣)/(n!)))) :} Let f∈W Show that ∀g∈H ( f(D^ )), g○f∈W tip: show that ∣∣h∣∣_W ≤cste × Sup_(z∈D) {∣h(z)∣+∣h′′(z)∣} and that W is an algebra then, re−wright f=f_1 +f_2 with f_2 : z Σ_(n=N) ^∞ ((f^((n)) (0))/(n!))z^n with N great enough to make sure that Σ_(n=0) ^∞ ((g^((n)) (0))/(n!))f_2 ^( n) is well defined and converges over W. ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙

Question Number 177626 Answers: 3 Comments: 0

Question Number 177146 Answers: 2 Comments: 0

f(x)=f(x−1)+x^2 +2x f(6)=33 faind volue of f(50)=?

$f (x) = f (x - 1) + x^{2} + 2 x$ $f (6) = 33 f a i n d v o l u e o f f (50) = ?$

Question Number 176501 Answers: 1 Comments: 1

find the range of x+y such that (x−2)^2 + (y−4)^2 = 49

$find the range of x + y such that$ ${(x - 2)}^{2} + {(y - 4)}^{2} = 49$

Question Number 176336 Answers: 2 Comments: 0

in AB_ ^Δ C : ((b−c)/(h_a )) =k , and A^ is given. B^ , C^ =?

$i n A {\overset{Δ}{B}}_{} C : \frac{b - c}{h_{a}} = k,$ $a n d \hat{A} i s g i v e n .$ $\hat{B}, \hat{C} = ?$

Question Number 175863 Answers: 0 Comments: 0

Question Number 175801 Answers: 2 Comments: 5

solve the follwing equation x(√(x )) + y(√(y )) = 3 and x(√(y )) + y(√(x )) = 2 someone solve the above equations in the following way x^3 + y^3 + 2xy(√(xy )) = 9.....(1) and x^2 y + y^2 x + 2xy(√(xy )) = 4......(2) (1) − (2) ⇒ (x − y)(x^2 − y^2 ) = 5 hence x = 3 and y = 2 which is obiviusly does not satisfy the original equations. where is the fallacy in the above solution? Please explain.

$s o l v e t h e f o l l w i n g e q u a t i o n$ $x \sqrt{x} + y \sqrt{y} = 3 a n d x \sqrt{y} + y \sqrt{x} = 2$ $s o m e o n e s o l v e t h e a b o v e e q u a t i o n s i n t h e f o l l o w i n g w a y$ $x^{3} + y^{3} + 2 x y \sqrt{x y} = 9 . . . . . (1) a n d x^{2} y + y^{2} x + 2 x y \sqrt{x y} = 4 . . . . . . (2)$ $(1) - (2) \Rightarrow (x - y) (x^{2} - y^{2}) = 5$ $h e n c e x = 3 a n d y = 2 w h i c h i s o b i v i u s l y d o e s n o t s a t i s f y t h e$ $o r i g i n a l e q u a t i o n s .$ $w h e r e i s t h e f a l l a c y i n t h e a b o v e s o l u t i o n ? Please explain .$