Math, asked by swapnil8149, 11 months ago

Show that if x is real, the expression x^2-bc÷2x-b-c

has no value between b and C.

Answers

Answered by akbar1993malda
3

Answer:

Step-by-step explanation:

Answered by arindam999
0

Answer:

lue between [two values]”, it’s unclear whether you mean the domain or range of the expression. However, the rational expression has at most two vertical asymptotes (as the denominator has at most 2 zeros), so the domain lacks at most two values of x, and you must be referring to the range.

Looking back at the four solutions offered, we can immediately rule two of them out; we know the expression has a horizontal asymptote (due to matching degree of numerator and denominator) at  y=1  (quotient of equal leading coefficients of numerator and denominator), and two vertical asymptotes, both of which exclude the two “lies between” options.

Thus, if we can find one or more local maximum or minimum values of the expression, we should be able to narrow down to one solution. Knowing that  (FG)′=F′G−FG′G2 , we want to solve:

0=(2x+34)(x2+2x−7)−(x2+34x−71)(2x+2)(x2+2x−7)2  

Since we only need zeroes, we can ignore the denominator (unless it has zeros at any of the same x-values as the numerator) and solve for numerator=0; the numerator simplifies to:

0=−33x2+128x−96  

Solving via your method of choice yields two zeros,  x={1,3} ; substituting those values into your initial expression yields output values of 9 and 5, respectively, as extrema. Since we already established that all of the values are not constrained between any two values, there must be no values between 5 and 9.

Graphical check confirms:

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Pratyush Kumar, Student

Answered August 13, 2016

Originally Answered: If x is real, then the expression x^2 + 34x - 71 / x^2 + 2x - 7 lies between 4 and 7 or between 5 and 9 or has no value between 4 and 7 or between 5 and 9?

Take x^2+34x-71/x^2+2x-7=y

then by transposing denominator to other side and rearrànging

(y-1)x^2+(2y-34)x-7y+71=0

As x is real

D must be positive>=0

Applying b^2-4ac>=0

4y^2+1156–136x-4(y-1)(-7y+71)>=0

32y^2–448y+114y>=0

Y^2-14y+45>=0

(y-9)(y-5)>=0

Therefore y belongs to (-infinity, 5] U [ 9,infinity)

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As x is real, so D > 0.

This is possible only when either both terms are (+)ve or both (–)ve.

(y – c) (y – b) = (+)ve, when y > c and y > b

(y – c) (y – b) = (–)ve, when y < c and y < b

Hence, the given expression has no real value between b and c.

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