- Prove that \(x>\ln x\), for \(x>0\). (2 marks)
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- Using part (i), or otherwise, prove that for all positive integers \(n\),
\( e^{n^2+n}>(n !)^2 .\) (3 marks)
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\( e^{n^2+n}>(n !)^2 .\) (3 marks)
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i. \(\text{See Worked Solutions}\)
ii. \(\text{See Worked Solutions}\)
i. \(\text{Prove}\ \ x > \ln x\ \ \text{for} \ \ x>0: \)
\(\Rightarrow \ \text{Show}\ \ f(x) = x-\ln x > 0 \)
\(\text{SP’s occur when}\ \ f^{′}(x) = 1-\dfrac{1}{x} = 0\)
\(\text{SP at}\ (1,1) \)
\(f^{″} = x^{-2}>0,\ \ \forall x>0 \)
\(\text{SP at (1, 1) is a global minimum for}\ x>0 \)
\(\Rightarrow f(x) \geq 1 > 0 \)
\(\therefore x > \ln x\ \ \text{for} \ \ x>0 \)
ii. \(x > \ln x\ \ \text{for} \ \ x>0 \ \ \Rightarrow \ \ e^x > x\ \ \text{(by definition)} \)
\(\text{Choose any positive integer}\ n: \)
\(e^n\) | \(>n \) | |
\(e^{n-1}\) | \(>n-1 \) | |
\(\ \ \vdots \) | ||
\(e^2\) | \(>2\) | |
\(e^1\) | \(>1\) |
\(\text{Multiply each side of the equations above:}\)
\(e^n \times e^{n-1} \times \cdots \times e^{1} \) | \(>n(n-1)(n-2) \cdots (2)(1) \) | |
\(e^{n+(n-1)+(n-2)+ \cdots + 2+1}\) | \(>n!\) | |
\(e^{\frac{n(n+1)}{2}} \) | \(>n!\ \ (\text{using AP formula}\ \ S_n=\frac{n}{2}(a+l) ) \) | |
\(e^{\frac{n(n+1)}{2} \times 2}\) | \(>(n!)^2\) | |
\(e^{n^2+n}\) | \(>(n!)^2\ \ …\ \text{as required}\) |
It is given that for positive numbers `x_(1),x_(2),x_(3),dots,x_(n)` with arithmetic mean `A`,
`(x_(1)xxx_(2)xxx_(3)xx cdots xxx_(n))/(A^(n)) <= 1` (Do NOT prove this.)
Suppose a rectangular prism has dimensions `a,b,c` and surface area `S`.
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i. `text{Show}\ \ abc <= ((S)/(6))^((3)/(2))`
`S=2(ab+bc+ac)`
`A=(ab+bc+ac)/3`
`text{Using given relationship, where}\ x_1=ab, x_2=bc, …`
`(ab xx bc xx ca)/((ab+bc+ac)/3)^3` | `<=1` | |
`(ab xx bc xx ca)` | `<=((ab+bc+ac)/3)^3` | |
`(abc)^2` | `<=((2(ab+bc+ac))/6)^3` | |
`abc` | `<=(S/6)^(3/2)` |
ii. `text{If prism is a cube}\ \ a=b=c`
`=> V=a^3, \ \ S=6a^2`
`(S/6)^(3/2)=((6a^2)/6)^(3/2)=a^3`
`text{In the case of a cube:}`
`V=(S/6)^(3/2)`
`text{Also,}\ \ V<=(S/6)^(3/2)\ \ \ text{(using part (i))}`
`:.\ text{For rectangular prisms with a given surface area, a cube}`
`text{has the maximum volume.}`
For all non-negative real numbers `x` and `y, \ sqrt(xy) <= (x + y)/2`. (Do NOT prove this.)
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i. `text(Show)\ \ sqrt(abc) <= (a^2 + b^2 + 2c)/4`
`sqrt(abc) = sqrt((ab)c) <= (ab + c)/2\ …\ (1)`
`text(Let)\ \ x = a^2\ \ text(and)\ \ y = b^2:`
`sqrt(xy) = sqrt(a^2b^2) ` | `<= (a^2 + b^2)/2` |
`ab` | `<= (a^2 + b^2)/2\ …\ (2)` |
`text{Substitute (2) into (1):}`
`sqrt(abc) <= ((a^2 + b^2)/2 + c)/2`
`sqrt(abc) <= (a^2 + b^2 + 2c)/4`
ii. `text(Show)\ sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`
`text{Similarly (to part a):}`
`text(If)\ \ x = b^2\ \ text(and)\ \ y = c^2`
`sqrt(abc) <= (b^2 + c^2 + 2a)/4`
`text(If)\ \ x = c^2\ \ text(and)\ \ y = a^2`
`sqrt(abc) <= (c^2 + a^2 + 2b)/4`
`:.3sqrt(abc)` | `<= (a^2 + b^2 + 2c + b^2 + c^2 + 2a + c^2 + a^2 + 2b)/4` |
`3sqrt(abc)` | `<= (2(a^2 + b^2 + c^2 + a + b + c))/4` |
`sqrt(abc)` | `<= (a^2 + b^2 + c^2 + a + b + c)/6` |
If `x, y, z ∈ R` and `x ≠ y ≠ z`, then
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i. `x^2 + y^2 + z^2 – yz – zx – xy > 0`
`text(Multiply) × 2`
`2x^2 + 2y^2 + 2z^2 – 2yz – 2zx – 2xy > 0`
`(x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 -2xz + x^2) > 0`
`(x – y)^2 + (y – z)^2 + (z – x)^2 > 0`
`text(Square of any rational number) > 0`
`:.\ text(Statement is true.)`
ii. | `(x + y + z)^2` | `= x^2 + xy + xz + yx + y^2 + yz + zx + zy + z^2` |
`1` | `= x^2 + y^2 + z^2 + 2xy + 2yz + 2xz` |
`text{Consider statement in part (i):}`
`x^2 + y^2 + z^2 – yz – zx – xy > 0`
`=> (x + y + z)^2 – (x^2 + y^2 + z^2 – yz – zx – xy)<1`
`3xy + 3yz + 3zx` | `< 1` |
`:. \ yz + zx + xy` | `< (1)/(3)` |
Suppose `0 <= t <= 1/sqrt 2.`
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i. | `0` | `<= t <= 1/sqrt 2` |
`0` | `<= t^2 <= 1/2` | |
`0` | `>= -t^2 >= -1/2` | |
`1` | `>= 1 – t^2 >= 1/2` | |
`1` | `<= 1/(1 – t^2) <= 2` | |
`2t^2` | `<= (2t^2)/(1 – t^2) <= 4t^2,\ \ \ \ (2t^2>0)` | |
`:. 0` | `<= (2t^2)/(1 – t^2) <= 4t^2` |
ii. `1/(1 + t) + 1/(1 – t) – 2`
`=((1-t)+(1+t)-2(1-t^2))/(1-t^2)`
`=(2t^2)/(1-t^2)`
`text{Substituting into part (i)}`
`:. 0 <= 1/(1 + t) + 1/(1 – t) – 2 <= 4t^2`
iii. | `int_0^x 0\ dt` | `<= int_0^x (1/(1 + t) + 1/(1 – t) – 2)\ dt <= int_0^x 4t^2\ dt` |
`0` | `<= [log_e (1 + t) – log_e (1 – t) – 2t]_0^x <= [(4t^3)/3]_0^x` | |
`0` | `<= [log_e (1 + x) – log_e (1 – x) – 2x]<= [(4x^3)/3]` | |
`0` | `<= log_e ((1 + x)/(1 – x)) – 2x <= (4x^3)/3` |
iv. `text(S)text(ince)\ \ e^a>e^b\ \ text(when)\ \ a>b`
`e^0` | `<= e^([ln ((1 + x)/(1 – x)) – 2x]) <= e^((4x^3)/3)` |
`1` | `<= e^(ln ((1 + x)/(1 – x))) xx e^(-2x) <= e^((4x^3)/3)` |
`1` | `<= ((1 + x)/(1 – x)) e^(-2x) <= e^((4x^3)/3)` |
If `p, q` and `r` are positive real numbers and `p + q >= r`, prove that
`p/(1 + p) + q/(1 + q) - r/(1 + r) >= 0.` (3 marks)
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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(If)\ \ p + q >= r,\ \ text(then)\ \ p + q – r >= 0`
`text(LHS)` | `= p/(1 + p) + q/(1 + q) – r/(1 + r)` |
`= (p(1 + q)(1 + r) + q(1 + p)(1 + r) – r(1 + p)(1 + q))/((1 + p)(1 + q)(1 + r))` | |
`= (p(1 + q + r + qr) + q(1 + p + r + pr) – r(1 + p + q + pq))/((1 + p)(1 + q)(1 + r))` | |
`= (p + pq + pr + pqr + q + pq + qr + pqr – r – pr – qr – pqr)/((1 + p)(1 + q)(1 + r))` | |
`= ((p + q – r) + pq(2 + r))/((1 + p)(1 + q)(1 + r))` | |
`>=(pq(2 + r))/((1 + p)(1 + q)(1 + r))\ \ \ \ text{(S}text{ince}\ \ p + q – r >= 0 text{)}` | |
`>= 0\ \ \ \ \ \ \ text{(S}text{ince}\ \ p > 0,\ q > 0,\ r > 0text{)}` |
Three positive real numbers `a`, `b` and `c` are such that `a + b + c = 1` and `a ≤ b ≤ c`.
By considering the expansion of `(a + b + c)^2`, or otherwise, show that
`qquad 5a^2 + 3b^2 +c^2 ≤ 1`. (2 marks)
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`text{Proof (See Worked Solutions)}`
`a + b+ c = 1,\ \ a ≤ b ≤ c`
`(a + b+ c)^2 = 1`
`a^2 + b^2 + c^2 + 2ab+ 2bc + 2ac = 1`
`text(S)text(ince)\ a ≤ b\ \ =>a^2 ≤ ab`
`text(S)text(ince)\ b ≤ c\ \ =>b^2 ≤ bc`
`text(S)text(ince)\ a ≤ c\ \ =>a^2 ≤ ac`
`=> a^2 + b^2 + c^2 + 2a^2 + 2b^2 + 2a^2` | `≤ 1` |
`:.5a^2 + 3b^2 + c^2` | `≤ 1\ \ \ text(… as required)` |