Math, asked by bhumiputramobile, 5 months ago

IF AVERAGE OF x AND 1/x IS 1, THEN 8x^10+4/x^5​

Answers

Answered by sushamabhopale
1

Answer:

Don't get into the algebra for this answer as this is a simple one and can be understood by assuming values of x… to make it more simple let us assume the value of x be 1(one).

Now, if mean of x and 1/x is M

=> Mean of 1 and 1/1 is 1 ( because we assumed value of x to be 1)

=> Mean of 1 and 1 is 1.

Now mean of x^2 and 1/x^2 will also be equal to 1 ( again because we assumed x= 1)

Conclusion is ( x and 1/x) & ( x^2 and 1/x^2 ) both will have same means….

:-)

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Mean of x and 1/x is M. So,

(x+1/x)/2 = M

Squaring on both sides,

(x+1/x)²/2² = M²

(x² + 1/x² + 2×x×1/x)/4 = M²

x² + 1/x² + 2 = 4M²

x² + 1/x² = 4M² - 2

Dividing by 2 on both sides,

(x² + 1/x²)/2 = (4M² - 2)/2

=> (x² + 1/x²)/2 = 2×(2M² - 1)/2 = 2M² - 1

=> Mean of x² and 1/x² = (2M²-1)

If mean of x+1/x is M then the mean of x^2 +1/x^2 will always be 4(M^2) - 2 for any value of x.

(A+B)^2= A^2 +B^2 +2AB

(A+B)/2 = M or A+B= 2M on squaring both side, we get A^2+B^2+2AB =4M^2 or A2 +B2 =4M^2–2AB and mean of x^2+1/x^2=(4M^2–2)/2

Here A is x and B is 1/x

I hope you got it

Best of luck

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Mean of x and 1/x=(x+1/x)/2=M given

x+1/x=2M

Squaring both sides we have,

(x+1/x)^2=(2M)^2

=> x^2. + 2 x.1/x + 1/x^2 = 4M^2

=> x^2 + 1/x^2. = 4M^2 -2=2(M^2–1). Ans.

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Mean of x and 1/x is M implies that:

x + 1/x = 2M

squaring both side

x^2 + 1/ x^2 + 2 = 4 M^2

so x^2 + 1/ x^2 = 4 M^2 -2

x^2 + 1/ x^2 = 2(2M^2 -1)

Mean of x and 1/x is M.

(x+1/x)/2==M

x+1/x=2M………….(1)

Mean of x^2 and 1/x^2 =(x^2+1/x^2)/2

= [(x+1/x)^2–2]/2

= (4M^2-2)/2.

= 2M^2 -1. Answer.

m square minus 2

2(M^2)-1

2

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