Question

ques for others

Answer 1

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)

Answer 2

@amistre64 @iGreen @TheSmartOne @Nnesha @tkhunny @Luigi0210 @sammixboo

Answer 3

@ganeshie8 @rational @IrishBoy123 @Loser66 @mathmath333

Answer 4

Nice questions :)

I've solved Q1, Q2, Q4, Q6, Q7 thus far...

Answer 5

wow!

Answer 6

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 \(\large |z_2|=1\)

Answer 7

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 \[{G_1}^4+2{G_2}^4+{G_3}^4=nl(n^2+2nl+l^2)=nl(n+l)^2=4nlm^2\]

Answer 8

Q 3)

\(x^2+2xy−3y^2=0\)
\((x-y)(x+3y) = 0\)
\(y = x, \ y = -\frac{1}{3}x\)


b)meets the curve again in the fourth quadrant

Answer 9

Nice! any quadratic equation of form \(ax^2+bxy+cy^2=0\) represents a pair of straight lines passing thru origin

Answer 10

@rational that did come as a surprise :p

Answer 11

Q4 is easy

new mean = (15*6 + 3+4+5)/(15+3) = 14

Answer 12

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
\[\frac{1}{2}\int\limits_2^4 1~dx = 1\]

Answer 13

bravo!!!!!!

Answer 14

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

Answer 15

great job rational .....

Answer 16

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)

Answer 17

wow I feel so stupid now!! NICE

Answer 18

\(\vec n = \nabla [\frac{x^2}{9} + \frac{y^2}{5} - 1] = <\frac{2x}{9}, \frac{2y}{5}>\)
\( \vec n(2, \frac{5}{3}) = <\frac{4}{9},\frac{2}{3}> = \frac{1}{18}<2,3>\)
\(\vec t = <3, -2>\)
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: \((0,3) \ and \ (\frac{9}{2}, 0)\)
\(A = 4 \times [\frac{1}{2} \times 3 \times \frac{9}{2} ] = 27\)