Question and Answers Forum

The product of the four terms of an increasing arithmetic progression is a, their sum is b, and the sum of their reciprocal is c. Suppose that a,b,c form a geometric progression whose product is 8000, find the sum of the first and fourth term.

$The product of the four terms of an$ $increasing arithmetic progression is a,$ $their sum is b, and the sum of their$ $reciprocal is c . Suppose that a, b, c form$ $a geometric progression whose$ $product is 8000, find the sum of the$ $first and fourth term .$

Question Number 110425 Answers: 2 Comments: 3

Which of the following is not a factor of x^6 −56x+55 A. x−1 B. x^2 −x+5 C. x^3 +2x^2 −2x−11 D. x^4 +x^3 +4x^2 −9x+11 E. x^5 +x^4 +x^3 +x^2 +x−55 Please show all workings clearly. Thanks.

$Which of the following is not a factor$ $of x^{6} - 56 x + 55$ $A . x - 1 B . x^{2} - x + 5 C .$ $x^{3} + {2 x}^{2} - 2 x - 11 D .$ $x^{4} + x^{3} + {4 x}^{2} - 9 x + 11 E .$ $x^{5} + x^{4} + x^{3} + x^{2} + x - 55$ $Please show all workings clearly .$ $Thanks .$

Question Number 110420 Answers: 1 Comments: 0

Prove that x^5 −3x^4 −17x^3 −x^2 −3x+17 cannot be factorized completely over the set of polynomials with integral coefficients.

$Prove that$ $x^{5} - {3 x}^{4} - {17 x}^{3} - x^{2} - 3 x + 17 cannot be$ $factorized completely over the set of$ $polynomials with integral coefficients .$

Question Number 110419 Answers: 1 Comments: 2

Let t be a root of x^3 −3x+1=0, if ((t^2 +pt+1)/(t^2 −t+1)) can be written as t+c for some p,c ∈ Z, then p−c equals?

$Let t be a root of x^{3} - 3 x + 1 = 0, if$ $\frac{t^{2} + pt + 1}{t^{2} - t + 1} can be written as t + c for$ $some p, c \in Z, then p - c equals ?$

Question Number 110410 Answers: 0 Comments: 0

Question Number 110403 Answers: 0 Comments: 1

if W represents the Runesky determinant of the two independent solutions linearly (y_1 ,y_2 )of the equation y^(′′) +p(x)y^′ +Q(x)=0 then demonstrate that W satisfies the differential equation (W^( ′) +p(x)W=0) and solve this equation to qet W ? help me sir please

$i f W r e p r e s e n t s t h e R u n e s k y d e t e r m i n a n t o f t h e t w o$ $i n d e p e n d e n t s o l u t i o n s l i n e a r l y (y_{1}, y_{2}) o f t h e e q u a t i o n y^{^{″}} + p (x) y^{^{'}} + Q (x) = 0 t h e n d e m o n s t r a t e t h a t W s a t i s f i e s t h e d i f f e r e n t i a l e q u a t i o n (W^{^{'}} + p (x) W = 0) a n d s o l v e t h i s e q u a t i o n t o q e t W ?$ $h e l p m e s i r p l e a s e$

Question Number 110399 Answers: 0 Comments: 5

The identity 2[16a^4 +81b^4 +c^4 ]=[4a^2 +9b^2 +c^2 ]^2 cannot result from which of the following equations? A. 6b=4a+2c B. 6a=9b+3c C. 6b=−4a+2c D.c= −2a−3b E. 6c=2b+3a

$The identity$ $2 [{16 a}^{4} + {81 b}^{4} + c^{4}] = {[{4 a}^{2} + {9 b}^{2} + c^{2}]}^{2}$ $cannot result from which of the$ $following equations ?$ $A . 6 b = 4 a + 2 c B . 6 a = 9 b + 3 c C .$ $6 b = - 4 a + 2 c D . c = - 2 a - 3 b E .$ $6 c = 2 b + 3 a$

Question Number 110395 Answers: 0 Comments: 5

prove to ealier problem of ∫_0 ^1 ∫_0 ^1 ((tanh^(−1) ((x)^(1/4) )tanh^(−1) ((y)^(1/4) ))/(x(√y)))dxdy=π^2 solution let I=∫_0 ^1 ((tanh^(−1) ((x)^(1/4) ))/x)dx∫_0 ^1 ((tanh^(−1) ((y)^(1/4) ))/(√y))dy=A.B let m=x^(1/4) and dx=4m^3 A=∫_0 ^1 ((tanh^(−1) (m))/m^4 )×4m^3 dm=4∫_0 ^1 ((tanh^(−1) (m))/m)dm but series of tanh^(−1) (m)=Σ_(k=0) ^∞ (m^(2k+1) /(2k+1)) A=4Σ_(k=0) ^∞ (1/(2k+1))∫_0 ^1 (m^(2k+1) /m)dm=4Σ_(k=o) ^∞ (1/(2k+1))∫_0 ^1 m^(2k) dm A=4Σ_(k=0) ^∞ (1/((2k+1)^2 ))=4×(1/2)Σ_(k=−∞) ^∞ (1/((2k+1)^2 )) recall that Σ_(n=−∞) ^∞ (1/((an+1)^m ))=−(π/((a)^m ))lim_(z→−(1/a) ) (1/((m−1)!))(d^((m−1)) /dz^((m−1)) )[cot(πz)] ∵A=2[−(π/((2)^2 ))lim_(z→−(1/2)) (1/((2−1)!))(d/dz)[cot(πz)] A=−(π/2)lim_(z→−(1/2)) [−πcosec^2 (−(π/2))]=(π^2 /2).....(1) then B=∫_0 ^1 ((tanh^(−1) ((y)^(1/4) ))/(√y))dy let n=y^(1/4) and dy=4n^3 B=∫_0 ^1 ((tanh^(−1) (n))/n^2 )×4n^3 dn=4∫_0 ^1 ntanh^(−1) (n)dn B=4Σ_(k=0) ^∞ (1/(2k+1))∫_0 ^1 n^(2k+2) dn=4Σ_(k=0) ^∞ (1/((2k+1)(2k+3))) B=4×(1/2)Σ_(k=0) ^∞ ((1/(2k+1))−(1/(2k+3))) B=2lim_(k→∞) (1−(1/3)+(1/3)−(1/5)+(1/5).......+(1/(2k+1)))=2.....(2) the series is telescoping but I=A×B=(π^2 /2)×2=π^2 ..............(3) ∫_0 ^1 ∫_0 ^1 ((tanh^(−1) ((x)^(1/4) )tanh^(−1) ((y)^(1/4) ))/(x(√y)))dxdy=π^2 Q.E.D by mathdave(28/08/2020)

$p r o v e t o e a l i e r p r o b l e m o f$ $\int_{0}^{1} \int_{0}^{1} \frac{\tanh^{- 1} (\sqrt[4]{x}) \tanh^{- 1} (\sqrt[4]{y})}{x \sqrt{y}} d x d y = π^{2}$ $s o l u t i o n$ $l e t$ $I = \int_{0}^{1} \frac{\tanh^{- 1} (\sqrt[4]{x})}{x} d x \int_{0}^{1} \frac{\tanh^{- 1} (\sqrt[4]{y})}{\sqrt{y}} d y = A . B$ $l e t m = x^{\frac{1}{4}} a n d d x = 4 m^{3}$ $A = \int_{0}^{1} \frac{\tanh^{- 1} (m)}{m^{4}} \times 4 m^{3} d m = 4 \int_{0}^{1} \frac{\tanh^{- 1} (m)}{m} d m$ $b u t s e r i e s o f \tanh^{- 1} (m) = \underset{k = 0}{\sum^{\infty}} \frac{m^{2 k + 1}}{2 k + 1}$ $A = 4 \underset{k = 0}{\sum^{\infty}} \frac{1}{2 k + 1} \int_{0}^{1} \frac{m^{2 k + 1}}{m} d m = 4 \underset{k = o}{\sum^{\infty}} \frac{1}{2 k + 1} \int_{0}^{1} m^{2 k} d m$ $A = 4 \underset{k = 0}{\sum^{\infty}} \frac{1}{{(2 k + 1)}^{2}} = 4 \times \frac{1}{2} \underset{k = - \infty}{\sum^{\infty}} \frac{1}{{(2 k + 1)}^{2}}$ $r e c a l l t h a t$ $\underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(a n + 1)}^{m}} = - \frac{π}{{(a)}^{m}} \lim_{z \to - \frac{1}{a}}$ $\frac{1}{(m - 1)!} \frac{d^{(m - 1)}}{{d z}^{(m - 1)}} [\cot (π z)]$ $∵ A = 2 [- \frac{π}{{(2)}^{2}} \lim_{z \to - \frac{1}{2}} \frac{1}{(2 - 1)!} \frac{d}{d z} [\cot (π z)]$ $A = - \frac{π}{2} \lim_{z \to - \frac{1}{2}} [- π {cosec}^{2} (- \frac{π}{2})] = \frac{π^{2}}{2} . . . . . (1)$ $t h e n B = \int_{0}^{1} \frac{\tanh^{- 1} (\sqrt[4]{y})}{\sqrt{y}} d y$ $l e t n = y^{\frac{1}{4}} a n d d y = 4 n^{3}$ $B = \int_{0}^{1} \frac{\tanh^{- 1} (n)}{n^{2}} \times 4 n^{3} d n = 4 \int_{0}^{1} n \tanh^{- 1} (n) d n$ $B = 4 \underset{k = 0}{\sum^{\infty}} \frac{1}{2 k + 1} \int_{0}^{1} n^{2 k + 2} d n = 4 \underset{k = 0}{\sum^{\infty}} \frac{1}{(2 k + 1) (2 k + 3)}$ $B = 4 \times \frac{1}{2} \underset{k = 0}{\sum^{\infty}} (\frac{1}{2 k + 1} - \frac{1}{2 k + 3})$ $B = 2 \underset{k \to \infty}{li m} (1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} . . . . . . . + \frac{1}{2 k + 1}) = 2 . . . . . (2)$ $t h e s e r i e s i s t e l e s c o p i n g$ $b u t I = A \times B = \frac{π^{2}}{2} \times 2 = π^{2} . . . . . . . . . . . . . . (3)$ $\int_{0}^{1} \int_{0}^{1} \frac{\tanh^{- 1} (\sqrt[4]{x}) \tanh^{- 1} (\sqrt[4]{y})}{x \sqrt{y}} d x d y = π^{2} Q . E . D$ $b y m a t h d a v e (28 / 08 / 2020)$

Question Number 110385 Answers: 0 Comments: 3

Question Number 110374 Answers: 2 Comments: 0

If P(x) is a polynomial whose sum of coefficients is 3 and P(x) can be factorised into two polynomials Q(x),R(x) with integer coefficients, the sum of the coefficients Q(x)^2 +R(x)^2 is

$If P (x) is a polynomial whose sum of$ $coefficients is 3 and P (x) can be$ $factorised into two polynomials$ $Q (x), R (x) with integer coefficients,$ $the sum of the coefficients$ $Q {(x)}^{2} + R {(x)}^{2} is$

Question Number 110367 Answers: 2 Comments: 0

Question Number 110365 Answers: 3 Comments: 0

Question Number 110359 Answers: 1 Comments: 0

Let f(x)=∣x−2∣+∣x−4∣−∣2x−6∣, for 2≤x≤8. The sum of the largest and smallest values of f(x) is

$Let$ $f (x) =∣ x - 2 ∣ + ∣ x - 4 ∣ - ∣ 2 x - 6 ∣,$ $for 2 ⩽ x ⩽ 8 . The sum of$ $the largest and$ $smallest values of f (x) is$

Pg 1020 Pg 1021 Pg 1022 Pg 1023 Pg 1024 Pg 1025 Pg 1026 Pg 1027 Pg 1028 Pg 1029

Contact: info@tinkutara.com