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Take quadratic equation x^2-6x+14=0 and its solutions a and b. What is a/b+b/a?

In this kind of exersise it is easy to use Viète's formula for quadratic equations: Let: a+b = S a b=P then x 2 -Sx + P=0 In our case, S=6 and P=14 by multiplying the fractions we get: (a a)/(a b)+(b b)/(a b...
AM
3407 Views

a) Given f(x) = ln(x), use the Mean Value Theorem to show that for 0 <a <b, (b-a)/b < ln(b/a) < (b-a)/a. b)Hence show that ln(1.2) lies between 1/m and 1/n, with m, n consecutive integers to be determined.

a) f(x)=ln(x), so f'(x)=1/x. By MVT, f'(c) = (f(b)-f(a))/(b-a) = (ln b - ln a)/(b-a) = ln(b/a)/(b-a), where c lies between a and b. Now, since 1/x is a decreasing function, and a &lt; c &lt; b, we get 1/b &l...
CR
9450 Views

In statistics, what is the benefit of taking a sample survey rather than a census?

Taking a sample survey is quicker, cheaper and easier to carry out than a census.
EG
4840 Views

write showing all working the following algebraic expression as a single fraction in its simplest form: 4-[(x+3)/ ((x^2 +5x +6)/(x-2))]

4-[(x+3)*((x-2)/(x^2 +5x +6))] 4-[(x+3)(x-2)/(x^2 +5x +6))] factorise denominator 4-[(x+3)(x-2)/(x+3)(x+2)] cancel down (x+3) 4-[(x-2)/(x+2)] expand 4(x+2)/(x+2) - (x-2)/(x+2) now share a denominator so make...
ST
12820 Views

Given y=x^3-x^2+6x-1, use diffferentiation to find the gradient of the normal at (1,5).

dy/dx = 3x^2-2x+6 At (1,5), dy/dx = 7. 7 is the gradient of the tangent, therefore the gradient of the normal at (1,5) is -1/7.
KR
2906 Views