if `x=2+2^(2//3)+2^(1//3)`
, then the value of `x^3-6x^2+6x`
is
(a)`3`
b. `2`
c. `1`
d. `-2`
step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 let x denote the number of sixes. p(getting a six when a die is thrown) = p = `(1)/(6)` ∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)` given, n = 6∴ x ∼ b`(6, 1/6)`the p.m.f. of x is given by p(x = x) = `""^6"c"\_x (1/6)^x (5/6)^(6 - x), x` = 0, 1,...,6p(getting at most 2 sixes) = p(x ≤ 2) = p(x = 0 or x = 1 or x = 2) = p(x = 0) + p(x = 1) + p(x = 2) = `""^6"c"\_0(1/6)^0 (5/6)^6 + ""^6"c"\_1(1/6)(5/6)5 + ""^6"c"\_2(1/6)^2 (5/4)^4` = `(5/6)^4 + 6(1/6)(5/6)^5 + (6!)/(2! xx 4!)(1/6)^2 (5/6)^4` = `(5/6)^4 [(5/6)^2 + (5/6) + (6 xx 5 xx 4!)/(2 xx 1 xx 4) xx (1/6)^2]` = `(5/6)^4 [25/6^2 + 30/6^ + 15/6^2]` = `(5/6)^4 (70/36)` = `(5/6)^4 (35/18)` = `(5/4)^4 ((5 xx 7)/(6 xx 3))` = `(7)/(3)(5/6)^5`. given, x3 – 1, x – 1 as pair of polynomialsneed to find the quotient and remainder⇒ to find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b⇒ since, the polynomial is of third degree we can write the q(x) as x2 + ax + b∴ p(x) = (x – a)( x2 + ax + b) + c⇒ x3 + 1 = (x + 1)(x2 + ax + b) + c⇒ x3 + 1 = (x3 + x2 + ax2 + ax + bx + b) + c⇒ x3 + 1 = x3 + (a + 1)x2 + (b + a)x + (c + b)∴ a + 1 = 0, b + a = 0, c + b = 1⇒ a = – 1, b = 1⇒ c + b = 1⇒ c = 0quotient = x2 + ax + b = x2 – x + 1remainder = 0 we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). 4x3 – 12x2 + 11x – 5 is divided by 2x – 1remainder theorem states that if p(x) is any polynomial and a is any real number and if p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).let p(x) = 4x3 – 12x2 + 11x – 5 and we have (2x – 1)the zero of (2x – 1) is now using remainder theorem,p(x) = 4x3 – 12x2 + 11x – 5 is divided by 2x – 1 then, is the remainderremainder = –2 y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6) formula :- therefore ,put -2x dx = dt