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The other two; are these answers correct?: $1) $ y = tan^{-1}\sqrt{x^2 - 1} \\ \\ \text { Domain: } x \leq -1 $ or $ x \geq 1 \text { and Range: } 0 \leq y < \frac {\pi}{2} \\ \\ \\ $2) $ y = cos^{-1}(e^x) \\ \\ \text{ Domain: } -\infty < x \leq 0 \text { and Range: } 0...

Is last one: sin^{-1} \frac {\sqrt {15} + \sqrt 8}{12}??

Nice diagram. But the 3 red lines must be equal in length, being radii of the unit circle.

z_1 = cis \theta \cdots (0 \leq \theta \leq \frac {\pi}{2}) You can argue geometrically that the sum of the squares of the distances remains unchanged (is invariant) when we rotate the figure anticlockwise until theta = 0 (your special case) or any other angle.

z_1 \text{ not necessarily } = 1 but answer still = 6. z_1 = cis\theta; $ $ z_2 = cis(\theta + \frac{2\pi}{3}); $ $ z_3 = cis(\theta + \frac {4\pi}{3})\\ \\ |z_1 -1|^2 + |z_2 - 1|^2 + |z_3-1|^2 = (z_1-1)\overline{(z_1 - 1)} + (z_2 - 1)\overline{(z_2 - 1)} + (z_3 - 1)\overline{(z_3 - 1)}\\...

For Q5: \text {Assume true for }n=k \geq 6 \implies \frac{1}{k!} < \frac {1}{e^k}\\ \\ \therefore \frac {1}{(k+1)!} = \frac {1}{k!} \times \frac {1}{k+1} < \frac {1}{e^k} \times \frac {1}{k+1} < \frac {1}{e^k} + \frac {1}{e} = \frac {1}{e^{k+1}} \\ \\ \text {since, for k greater than 5: }...

See Note-2 above.

Q8(e) The 2 values are related to the roots of: 4x^2+2x-1 =0, $ viz $ \frac {-1 \pm \sqrt 5}{4}\\ \\ $So: $ sin\frac {\pi}{10} = \frac {-1+\sqrt 5}{4} $ and $ sin \frac {3\pi}{10} = - \frac{-1-\sqrt 5 }{4} = \frac {1+\sqrt 5}{4} Note: 1) (4x^2+bx+c)^2 =16x^4 + b^2x^2 + c^2 +2(4bx^3 +...

a) \int xlnxdx = \frac{1}{2} \int lnxdx^2 = \frac{1}{2} x^2 lnx - \frac {1}{2} \int x^2 dlnx = \cdots -\frac{1}{2} \int x^2 \times \frac {1}{x}dx \\ \\ = \frac {1}{2}x^2lnx - \frac {1}{4} x^2 + C b) \int x(ln x)^2dx = \frac {1}{2} \int (ln x)^2dx^2 = \frac {1}{2}x^2(ln x)^2 - \frac {1}{2}...

I think the process of writing down whatever you want to memorise is far more effective than repeated recitation. You should write it down, try to recall, try writing it down again as far as you can by recall, referring to the original the parts you cannot recall. Repeat this process until you...

Looks like it. 5x^4-11x^3+16x^2-11x+5 = x^2(5(x^2+\frac{1}{x^2}) -11(x+\frac{1}{x}) + 16) = 0

I think \overrightarrow {OC} = \sqrt 3 i\omega OBC is an equilateral triangle \angle AOC \text { is right-angled and }\overrightarrow {OC} \text { is therefore a multiple of } i\omega; \text { indeed: } \sqrt 3 i\omega \\ \\ \text {since: }OC = \sqrt 3 \times OA $ as $ tan\frac {\pi}{3} =...

What's the question?

Thank you. You know I struggle with LaTeX. Now you know why my proof comes out in dribs and drabs.

Let OAEB be a //gram. \overrightarrow {OE} = z_1 + z_2 \\ \\ \overrightarrow {DB} =\overrightarrow {DO} + \overrightarrow {OB} = z_1 + z_2 \\ \\ \overrightarrow {DB} = \overrightarrow {OE} \implies DOEB \text { is a //gram } \\ \\ \overrightarrow {AB} = z_2 - z_1 \\ \\ \overrightarrow {AC} =...

Let M be intersection of AC & BD and O the Origin \overrightarrow {OC} = \sqrt 3 + (1+\sqrt 3)i\\ \\ \overrightarrow {OA} = 1 \\ \\ \overrightarrow {OM} = \frac{1}{2} (\overrightarrow {OA} + \overrightarrow {OC}) = \frac {1}{2} (1 + \sqrt 3 + (1 + \sqrt 3)i) = \frac {1+\sqrt 3}{2} + \frac...

Can I come?

No. You're not stupid. It took me a while to do it. Not that straightforward. You can derive, if necessary, that: 2sinAsinB = cos(A-B) - cos(A+B) from: cos(A+B) = cosAcosB - sinAsinB and cos(A-B) = cosAcosB + sinAsinB $Now: $ cos2A = 2cos^2A - 1 \\ \\ 2sin4Asin2A = cos(4A-2A) -...

Can show to be true for n=3 \text {Now assume true for } 3 \leq n \leq k , $ and $ k \geq 3 \text {(Strong Law)}\\ \\ \therefore U_k = 2^k + 5^k \\ \\ \therefore U_{k+1} = 7U_k - 10 U_{k-1} = 7(2^k + 5^k) - 10(2^{k-1} + 5^{k-1}) \\ \\ = 7\times2^k + 7\times 5^k - 10 \times 2^{k-1} - 10 \times...

"worth doing"? "safe doing"?