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Hello these r some tough ques I'm posting for people to practice, I don't know the solutions but answers will be given at the end :) Q1.)Suppose \[z_{1}\] and \[z_{2}\] are complex numbers such that \[\frac{z_{1}-2z_{2}}{2-z_{1} z^*_{2} }\] is unimodular. Then the point z1 lies on a Here the star denotes conjugate denotes a)circle of radius 2 b)circle of radius root 2 c)straight line parallel to x-axis d) straight line parallel to y axis Q2.)If m is the A.M. of two distinct real numbers l and n(l,n>1) and G1, G2 and G3 are three geometric means between l and n, then \[G_{1}^4+2G_{2}^4+G_{3}^4\] equals a) 4lmn^2 b)(2lmn)^2 c)4l^2mn d)4lm^2n Q3) The normal to the curve, \[x^2+2xy-3y^2=0\] at (1,1) a)meets the curve again in the third quadrant b)meets the curve again in the fourth quadrant c)does not meet the curve again d)meets the curve again in second quadrant Q4) The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3,4, and 5 are added to the data, then the mean of the resultant data is, a)15.8 b)14.0 c)16.8 d)16.0 Q5)The Integral \[\int\limits_{2}^{4}\frac{\log(x^2)}{\log(x^2)+\log(36-12x+x^2)}dx\] is equal to a)1 b)6 c)2 d)4 Q6)Let alpha and beta be the roots of the equation \[x^2-6x-2=0\] If \[a_{n}=\alpha^n-\beta^n\] for n>=1 then the value of\[\frac{a_{10}-2a_{8}}{2a_{9}}\] is equal to a)3 b)-3 c)6 d)-6 Q7)Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2 if \[\lim_{x \rightarrow 0}[1+\frac{f(x)}{x^2}]=3\] then f(2) is equal to a)0 b)4 c)-8 d)-4 Q8) The area(in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse \[\frac{x^2}{9}+\frac{y^2}{5}=1\] is a)27/2 b)27 c)27/4 d)18 Q9) If 12 identical balls are placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is: a)\[220(\frac{1}{3})^{12}\] b)\[22(\frac{1}{3})^{11}\] c)\[\frac{55}{3}(\frac{2}{3})^{11}\] d)\[55(\frac{2}{3})^{10}\] ANSWERS Q1a) Q2d) Q3b) Q4b) Q5a) Q6a) Q7a) Q8b) Q9c)

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Nice questions :) I've solved Q1, Q2, Q4, Q6, Q7 thus far...
wow!
Q1 : we use the properties \(\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}\) and \(|z|^2 = z\overline{z}\) \(\large \left|\dfrac{z_{1}-2z_{2}}{2-z_{1} \overline{z_{2}} }\right|=1 \implies \left|z_{1}-2z_{2}\right|=\left|2-z_{1} \overline{z_{2}} \right|\) squaring both sides \(\large \left|z_{1}-2z_{2}\right|^2=\left|2-z_{1} \overline{z_{2} }\right|^2 \) \(\large (z_{1}-2z_{2})(\overline{z_{1}-2z_{2}})=(2-z_{1} \overline{z_{2}}) (\overline{2-z_{1} \overline{z_{2}}}) \) \(\large (z_{1}-2z_{2})(\overline{z_{1}}-2\overline{z_{2}})=(2-z_{1} \overline{z_{2}}) (2- \overline{z_{1}}z_2) \) distributing and factoring gives \(\large (z_1\overline{z_1}-4)(z_2\overline{z_2}-1)=0\) \(\large |z_1|=2\) or \(\large |z_2|=1\)
Q2 : \(l,~G_1, ~G_2, ~G_3,~n\) are in GP \(G_2\) is the geometric mean of \(n\) and \(l\) : \[ G_2=\sqrt{nl}\tag{1}\] \(G_1\) is the geometric mean of \(l\) and \(G_2\) : \[G_1 = \sqrt{lG_2}\tag{2}\] \(G_3\) is the geometric mean of \(G_2\) and \(n\) : \[G_3 = \sqrt{nG_2}\tag{3}\] so \[{G_1}^4+2{G_2}^4+{G_3}^4=nl(n^2+2nl+l^2)=nl(n+l)^2=4nlm^2\]
Q 3) \(x^2+2xy−3y^2=0\) \((x-y)(x+3y) = 0\) \(y = x, \ y = -\frac{1}{3}x\) |dw:1433497346633:dw| b)meets the curve again in the fourth quadrant
Nice! any quadratic equation of form \(ax^2+bxy+cy^2=0\) represents a pair of straight lines passing thru origin
@rational that did come as a surprise :p
Q4 is easy new mean = (15*6 + 3+4+5)/(15+3) = 14
Q5 \[\begin{align} \int\limits_{2}^{4}\frac{\log(x^2)}{\log(x^2)+\log(36-12x+x^2)}dx &=\int\limits_{2}^{4}\frac{\log(x^2)}{\log(x^2)+\log((6-x)^2)}dx\\~\\ &=\int\limits_{2}^{4}\frac{\log(x)}{\log(x(6-x))}dx~~~\color{red}{\star}\\~\\ &=\int\limits_{2}^{4}\frac{\log(\color{blue}{2+4-x})}{\log((\color{blue }{2+4-x})(6-(\color{blue}{2+4-x})))}dx\\~\\ &=\int\limits_{2}^{4}\frac{\log(6-x)}{\log(x(6-x))}dx~~~\color{blue}{\star}\\~\\ \end{align}\] Averaging \(\color{red}{\star},~\color{blue}{\star} \) the integral becomes \[\frac{1}{2}\int\limits_2^4 1~dx = 1\]
bravo!!!!!!
Q6 using the identity \[x^n-y^n=(x+y)(x^{n-1}-y^{n-1})-xy(x^{n-2}-y^{n-2})\] \[a_{n}=\alpha^n-\beta^n=(\alpha+\beta)(\alpha^{n-1}-\beta^{n-1})-\alpha\beta(\alpha^{n-2}-\beta^{n-2}) = 6a_{n-1}+2a_{n-2}\] plugging in \(n=10\) and rearranging gives the desired result
great job rational .....
Q7) \(f(x) = ax^4 + bx^3 + cx^2 + dx + e\) \(1 + \frac{f(x)}{x^2} = 1 + ax^2 + bx + c +\frac{ d}{x} + \frac{e}{x^2}\) \( \lim_{x-> 0} [1 + \frac{f(x)}{x^2}] = 3 :=> d,e = 0, c = 2 \) \(f'(x) = 4ax^3 + 3bx^2 + 4x\) \(f'(1) = f'(2) = 0\) \(4a + 3b + 4 = 0\) \(32a + 12b + 8 = 0\) \(a = \frac{1}{2}, b = -2\) \(f(2) = \frac{1}{2}(16) -2(8) + 2(4) = 0\) a)
wow I feel so stupid now!! NICE
|dw:1433502854097:dw| \(\vec n = \nabla [\frac{x^2}{9} + \frac{y^2}{5} - 1] = <\frac{2x}{9}, \frac{2y}{5}>\) \( \vec n(2, \frac{5}{3}) = <\frac{4}{9},\frac{2}{3}> = \frac{1}{18}<2,3>\) \(\vec t = <3, -2>\) tangent line : \(y = -\frac{2}{3}x + c, \ y(2) = \frac{5}{3}, \ c = 3\) by symmetry, Area = \(4 \times \) Area under \(y = -\frac{2}{3} x + 3\) in first quadrant tangent line intercepts: \((0,3) \ and \ (\frac{9}{2}, 0)\) \(A = 4 \times [\frac{1}{2} \times 3 \times \frac{9}{2} ] = 27\)
Q9) binomial distribution, with p (success) = 1/3, q (fail) = 2/3: for 1 box only, the odds of getting exactly 3 balls (getting a ball is a "success") from 12 trials is : \(P = \ ^{12}C_{3}.(\frac{1}{3})^3.(\frac{2}{3})^9\) but there are 3 boxes and so this can happen in 2 other ways. meaning [and this is a problem with the wording of the question - see below] \(P_{T} = \ ^{12}C_{3}.(\frac{1}{3})^3.(\frac{2}{3})^9.3\) = \(220.(\frac{1}{3})^2.(\frac{2}{3})^9\) = \(55.(\frac{2 \times2}{3 \times3}).(\frac{2}{3})^9\) = \(55.(\frac{2}{3})^{11}\) but that's \(1/3 \times\) the "answer" meaning that the answer only makes sense if you are looking at a single box, ie "given" you have selected a particular box, what are the odds that that particular box ends up with 3 balls. in which case we go back from \(P_T\) to \(P\), address the simpler problem, and we get : = \(\frac{55}{3}.(\frac{2}{3})^{11}\) [ie c)]

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