## Category Archives: ISI Kolkatta Entrance Exam

### Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

Question 1.

If the point on $y = x \tan {\alpha} - \frac{ax^{2}}{2u^{2}\cos^{2}{\alpha}}$, where $\alpha>0$, where the tangent is parallel to $y=x$ has an ordinate $\frac{u^{2}}{4a}$, then what is the value of $\alpha$?

Question 2:

Prove that the segment of the tangent to the curve $y=c/x$, which is contained between the coordinate axes is bisected at the point of tangency.

Question 3:

Find all the tangents to the curve $y = \cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$.

Question 4:

Prove that the curves $y=f(x)$, where $f(x)>0$, and $y=f(x)\sin{x}$, where $f(x)$ is a differentiable function have common tangents at common points.

Question 5:

Find the condition that the lines $x \cos{\alpha} + y \sin{\alpha} = p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$.

Question 6:

Find the equation of a straight line which is tangent to one point and normal to the point on the curve $y=8t^{3}-1$, and $x=4t^{2}+3$.

Question 7:

Three normals are drawn from the point $(c,0)$ to the curve $y^{2}=x$. Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the two other normals are perpendicular to each other.

Question 8:

If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2} + p_{2}^{2}=a^{2}$.

Question 9:

Show that the curve $x=1-3t^{2}$, and $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$. If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t= \tan {\psi} +\sec {\psi}$. If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles.

Question 10:

Find the condition that the curves $ax^{2}+by^{2}=1$ and $a^{'}x^{2} + b^{'}y^{2}=1$ intersect orthogonality and hence show that the curves $\frac{x^{2}}{(a^{2}+b_{1})} + \frac{y^{2}}{(b^{2}+b_{1})} = 1$ and $\frac{x^{2}}{a^{2}+b_{2}} + \frac{y^{2}}{(b^{2}+b_{2})} =1$ also intersect orthogonally.

More later,

Nalin Pithwa.

### Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

“Easy” questions:

Question 1:

Find the slope of the tangent to the curve represented by the curve $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$ at the point $(2,-1)$.

Question 2:

Find the co-ordinates of the point P on the curve $y^{2}=2x^{3}$, the tangent at which is perpendicular to the line $4x-3y+2=0$.

Question 3:

Find the co-ordinates of the point $P(x,y)$ lying in the first quadrant on the ellipse $x^{2}/8 + y^{2}/18=1$ so that the area of the triangle formed by the tangent at P and the co-ordinate axes is the smallest.

Question 4:

The function $f(x) = \frac{\log (\pi+x)}{\log (e+x)}$, where $x \geq 0$ is

(a) increasing on $(-\infty, \infty)$

(b) decreasing on $[0, \infty)$

(c) increasing on $[0, \pi/e)$ and decreasing on $[\pi/e, \infty)$

(d) decreasing on $[0, \pi/e)$ and increasing on $[\pi/e, \infty)$.

Fill in the correct multiple choice. Only one of the choices is correct.

Question 5:

Find the length of a longest interval in which the function $3\sin(x) -4\sin^{3}(x)$ is increasing.

Question 6:

Let $f(x)=x e^{x(1-x)}$, then $f(x)$ is

(a) increasing on $[-1/2, 1]$

(b) decreasing on $\Re$

(c) increasing on $\Re$

(d) decreasing on $[-1/2, 1]$.

Fill in the correct choice above. Only one choice holds true.

Question 7:

Consider the following statements S and R:

S: Both $\sin(x)$ and $\cos (x)$ are decreasing functions in the interval $(\pi/2, \pi)$.

R: If a differentiable function decreases in the interval $(a,b)$, then its derivative also decreases in $(a,b)$.

Which of the following is true?

(i) Both S and R are wrong.

(ii) Both S and R are correct, but R is not the correct explanation for S.

(iii) S is correct and R is the correct explanation for S.

(iv) S is correct and R is wrong.

Indicate the correct choice. Only one choice is correct.

Question 8:

For which of the following functions on $[0,1]$, the Lagrange’s Mean Value theorem is not applicable:

(i) $f(x) = 1/2 -x$, when $x<1/2$; and $f(x) = (1/2-x)^{2}$, when $x \geq 1/2$.

(ii) $f(x) = \frac{\sin(x)}{x}$, when $x \neq 0$; and $f(x)=1$, when $x=0$.

(iii) $f(x)=x |x|$

(iv) $f(x)=|x|$.

Only one choice is correct. Which one?

Question 9:

How many real roots does the equation $e^{x-1}+x-2=0$ have?

Question 10:

What is the difference between the greatest and least values of the function $f(x) = \cos(x) + \frac{1}{2}\cos(2x) -\frac{1}{3}\cos(3x)$?

More later,

Nalin Pithwa.

### Circles and System of Circles: IITJEE Mains: some solved problems I

Part I: Multiple Choice Questions:

Example 1:

Locus of the mid-points of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the centre is (a) $x+y=2$ (b) $x^{2}+y^{2}=1$ (c) $x^{2}+y^{2}=2$ (d) $x-y=0$

Solution 1:

Let O be the centre of the circle $x^{2}+y^{2}=4$, and let AB be any chord of this circle, so that $\angle AOB=\pi /2$. Let $M(h,x)$ be the mid-point of AB. Then, OM is perpendicular to AB. Hence, $(AB)^{2}=(OA)^{2}+(AM)^{2}=4-2=2 \Longrightarrow h^{2}+k^{2}=2$. Therefore, the locus of $(h,k)$ is $x^{2}+y^{2}=2$.

Example 2:

If the equation of one tangent to the circle with centre at $(2,-1)$ from the origin is $3x+y=0$, then the equation of the other tangent through the origin is (a) $3x-y=0$ (b) $x+3y=0$ (c) $x-3y=0$ (d) $x+2y=0$.

Solution 2:

Since $3x+y=0$ touches the given circle, its radius equals the length of the perpendicular from the centre $(2,-1)$ to the line $3x+y=0$. That is,

$r= |\frac{6-1}{\sqrt{9+1}}|=\frac{5}{\sqrt{10}}$.

Let $y=mx$ be the equation of the other tangent to the circle from the origin. Then,

$|\frac{2m+1}{\sqrt{1+m^{2}}}|=\frac{5}{\sqrt{10}}=25(1+m^{2})=10(2m+1)^{2} \Longrightarrow 3m^{2}+8m-3=0$, which gives two values of m and hence, the slopes of two tangents from the origin, with the product of the slopes being -1. Since the slope of the given tangent is -3, that of the required tangent is 1/3, and hence, its equation is $x-3y=0$.

Example 3.

A variable chord is drawn through the origin to the circle $x^{2}+y^{2}-2ax=0$. The locus of the centre of the circle drawn on this chord as diameter is (a) $x^{2}+y^{2}+ax=0$ (b) $x^{2}+y^{2}+ay=0$ (c) $x^{2}+y^{2}-ax=0$ (d) $x^{2}+ y^{2}-ay=0$.

Solution 3:

Let $(h,k)$ be the centre of the required circle. Then, $(h,k)$ being the mid-point of the chord of the given circle, its equation is $hx+ky-a(x+h)=h^{2}+k^{2}-2ah$.

Since it passes through the origin, we have $-ah=h^{2}+k^{2}-2ah \Longrightarrow h^{2}+k^{2}-ah=0$.

Hence, locus of $(h,k)$ is $x^{2}+y^{2}-ax=0$.

Quiz problem:

A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) $m(m+n)$ (b) $m+n$ (c) $n(m+n)$ (d) $(1/2)(m+n)$.

To be continued,

Nalin Pithwa.

### Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II

I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.

Problem 1:

Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points $R_{1}$, $R_{2}$, $R_{3}$, $\ldots$, $R_{n}$ and on it a point R is taken such that $\frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}$

Show that the locus of R is a straight line.

Solution 1:

Let equations of the given lines be $a_{i}x+b_{i}y+c_{i}=0$, $i=1,2,\ldots, n$, and the point O be the origin $(0,0)$.

Then, the equation of the line through O can be written as $\frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r$ where $\theta$ is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O.

Let $r, r_{1}, r_{2}, \ldots, r_{n}$ be the distances of the points $R, R_{1}, R_{2}, \ldots, R_{n}$ from O which in turn $\Longrightarrow OR=r$ and $OR_{i}=r_{i}$, where $i=1,2,3 \ldots n$.

Then, coordinates of R are $(r\cos{\theta}, r\sin{\theta})$ and of $R_{i}$ are $(r_{i}\cos{\theta},r_{i}\sin{\theta})$ where $i=1,2,3, \ldots, n$.

Since $R_{i}$ lies on $a_{i}x+b_{i}y+c_{i}=0$, we can say $a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0$ for $i=1,2,3, \ldots, n$

$\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}$, for $i=1,2,3, \ldots, n$

$\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$

$\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$ …as given…

$\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0$

Hence, the locus of R is $(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0$ which is a straight line.

Problem 2:

Determine all values of $\alpha$ for which the point $(\alpha,\alpha^{2})$ lies inside the triangle formed by the lines $2x+3y-1=0$, $x+2y-3=0$, $5x-6y-1=0$.

Solution 2:

Solving equations of the lines two at a time, we get the vertices of the given triangle as: $A(-7,5)$, $B(1/3,1/9)$ and $C(5/4, 7/8)$.

So, AB is the line $2x+3y-1=0$, AC is the line $x+2y-3=0$ and BC is the line $5x-6y-1=0$

Let $P(\alpha,\alpha^{2})$ be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line $5x-6y-1=0$, both $5(-7)-6(5)-1$ and $5\alpha-6\alpha^{2}-1$ must have the same sign.

$\Longrightarrow 5\alpha-6\alpha^{2}-1<0$ or $6\alpha^{2}-5\alpha+1>0$ which in turn $\Longrightarrow (3\alpha-1)(2\alpha-1)>0$ which in turn $\Longrightarrow$ either $\alpha<1/3$ or $\alpha>1/2$….call this relation I.

Again, since B and P lie on the same side of the line $x+2y-3=0$, $(1/3)+(2/9)-3$ and $\alpha+2\alpha^{2}-3$ have the same sign.

$\Longrightarrow 2\alpha^{2}+\alpha-3<0$ and $\Longrightarrow (2\alpha+3)(\alpha-1)<0$, that is, $-3/2 <\alpha <1$…call this relation II.

Lastly, since C and P lie on the same side of the line $2x+3y-1=0$, we have $2 \times (5/4) + 3 \times (7/8) -1$ and $2\alpha+3\alpha^{2}-1$ have the same sign.

$\Longrightarrow 3\alpha^{2}+2\alpha-1>0$ that is $(3\alpha-1)(\alpha+1)>0$

$\alpha<-1$ or $\alpha>1/3$….call this relation III.

Now, relations I, II and III hold simultaneously if $-3/2 < \alpha <-1$ or $1/2<\alpha<1$.

Problem 3:

A variable straight line of slope 4 intersects the hyperbola $xy=1$ at two points. Find the locus of the point which divides the line segment between these two points in the ratio $1:2$.

Solution 3:

Let equation of the line be $y=4x+c$ where c is a parameter. It intersects the hyperbola $xy=1$ at two points, for which $x(4x+c)=1$, that is, $\Longrightarrow 4x^{2}+cx-1=0$.

Let $x_{1}$ and $x_{2}$ be the roots of the equation. Then, $x_{1}+x_{2}=-c/4$ and $x_{1}x_{2}=-1/4$. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are $(x_{1}, \frac{1}{x_{1}})$ and that of B are $(x_{2}, \frac{1}{x_{2}})$.

Let $R(h,k)$ be the point which divides AB in the ratio $1:2$, then $h=\frac{2x_{1}+x_{2}}{3}$ and $k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}$, that is, $\Longrightarrow 2x_{1}+x_{2}=3h$…call this equation I.

and $x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k$….call this equation II.

Adding I and II, we get $3(x_{1}+x_{2})=3(h-\frac{k}{4})$, that is,

$3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}$….call this equation III.

Subtracting II from I, we get $x_{1}-x_{2}=3(h+\frac{k}{4})$

$\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}$

$\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}$

$\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}$

$\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})$

$\Longrightarrow 16h^{2}+10hk+k^{2}-2=0$

so that the locus of $R(h,k)$ is $16x^{2}+10xy+y^{2}-2=0$

More later,

Nalin Pithwa.

### Cartesian system and straight lines: IITJEE Mains: Problem solving skills

Problem 1:

The line joining $A(b\cos{\alpha},b\sin{\alpha})$ and $B(a\cos{\beta},a\sin{\beta})$ is produced to the point $M(x,y)$ so that $AM:MB=b:a$, then find the value of $x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}$.

Solution 1:

As M divides AB externally in the ratio $b:a$, we have $x=\frac{b(a\cos{\beta})-a(b\cos{\alpha})}{b-a}$ and $y=\frac{b(a\sin{\beta})-a(b\sin{\alpha})}{b-a}$ which in turn

$\Longrightarrow \frac{x}{y} = \frac{\cos{\beta}-cos{\alpha}}{\sin{\beta}-\sin{\alpha}}$

$= \frac{2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}}{2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\beta-\alpha}{2}}}$

$\Longrightarrow x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}=0$

Problem 2:

If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points $(a^{2}+1,a^{2}+1)$ and $(2a,-2a)$, then where does the orthocentre lie?

Solution 2:

From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre $(0,0)$ and the centroid $(\frac{(a+1)^{2}}{2},\frac{(a-1)^{2}}{2})$, that is, $y.\frac{(a+1)^{2}}{2} = x.\frac{(a-1)^{2}}{2}$, or $(a-1)^{2}x-(a+1)^{2}y=0$. That is, the orthocentre lies on this line.

Problem 3:

If a, b, c are unequal and different from 1 such that the points $(\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1})$, $(\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1})$ and $(\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1})$ are collinear, then which of the following option is true?

a: $bc+ca+ab+abc=0$

b: $a+b+c=abc$

c: $bc+ca+ab=abc$

d: $bc+ca+ab-abc=3(a+b+c)$

Solution 3:

Suppose the given points lie on the line $lx+my+n=0$ then a, b, c are the roots of the equation :

$lt^{3}+m(t^{2}-3)+n(t-1)=0$, or

$lt^{3}+mt^{2}+nt-(3m+n)=0$

$\Longrightarrow a+b+c=-\frac{m}{l}$ and $ab+bc+ca=\frac{n}{l}$, that is, $abc=(3m+n)/l$

Eliminating l, m, n, we get $abc=-3(a+b+c)+bc+ca+ab$

$\Longrightarrow bc+ca+ab-abc=3(a+b+c)$, that is, option (d) is the answer.

Problem 4:

If $p, x_{1}, x_{2}, \ldots, x_{i}, \ldots$ and $q, y_{1}, y_{2}, \ldots, y_{i}, \ldots$ are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points $A_{i}(x_{i},y_{i})$ with $i=1,2,3 \ldots, n$ lie?

Solution 4:

Note: Centre of Mean Position is $(\frac{\sum{xi}}{n},\frac{\sum {yi}}{n})$.

Let the coordinates of the centre of mean position of the points $A_{i}$, $i=1,2,3, \ldots,n$ be $(x,y)$ then

$x=\frac{x_{1}+x_{2}+x_{3}+\ldots + x_{n}}{n}$ and $y=\frac{y_{1}+y_{2}+\ldots + y_{n}}{n}$

$\Longrightarrow x = \frac{np+a(1+2+\ldots+n)}{n}$, $y=\frac{nq+b(1+2+\ldots+n)}{n}$

$\Longrightarrow x=p+ \frac{n(n+1)}{2n}a$ and $y=q+ \frac{n(n+1)}{2n}b$

$\Longrightarrow x=p+\frac{n+1}{2}a$, and $y=q+\frac{n+1}{2}b$

$\Longrightarrow 2\frac{(x-p)}{a}=2\frac{(y-q)}{b} \Longrightarrow bx-ay=bp-aq$, that is, the CM lies on this line.

Problem 5:

The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}$?

Solution 5:

Equation of the line L in the two coordinate systems is $\frac{x}{a} + \frac{y}{b}=1$, and $\frac{X}{p} + \frac{Y}{q}=1$ where $(X,Y)$ are the new coordinate of a point $(x,y)$ when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.

$\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}}$

$\Longrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}$

or $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}=0$. So, the value is zero.

Problem 6:

Let O be the origin, $A(1,0)$ and $B(0,1)$ and $P(x,y)$ are points such that $xy>0$ and $x+y<1$, then which of the following options is true:

a: P lies either inside the triangle OAB or in the third quadrant

b: P cannot lie inside the triangle OAB

c: P lies inside the triangle OAB

d: P lies in the first quadrant only.

Solution 6:

Since $xy>0$, P either lies in the first quadrant or in the third quadrant. The inequality $x+y<1$ represents all points below the line $x+y=1$. So that $xy>0$ and $x+y<1$ imply that either P lies inside the triangle OAB or in the third quadrant.

Problem 7:

An equation of a line through the point $(1,2)$ whose distance from the point $A(3,1)$ has the greatest value is :

option i: $y=2x$

option ii: $y=x+1$

option iii: $x+2y=5$

option iv: $y=3x-1$

Solution 7:

Let the equation of the line through $(1,2)$ be $y-2=m(x-1)$. If p denotes the length of the perpendicular from $(3,1)$ on this line, then $p=|\frac{2m+1}{\sqrt{m^{2}+1}}|$

$\Longrightarrow p^{2}=\sqrt{\frac{4m^{2}+4m+1}{m^{2}+1}}=4+ \frac{4m-3}{m^{2}+1}=s$, say

then $p^{2}$ is greatest if and only if s is greatest.

Now, $\frac{ds}{dm} = \frac{(m^{2}+1)(4)-2m(4m-3)}{(m^{2}+1)^{2}} = \frac{-2(2m-1)(m-2)}{(m^{2}+1)^{2}}$

$\frac{ds}{dm} = 0$ so that $\Longrightarrow m = \frac{1}{2}, 2$. Also, $\frac{ds}{dm}<0$, if $m<\frac{1}{2}$, and

$\frac{ds}{dm} >0$, if $1/2

and $\frac{ds}{dm} <0$, if $m>2$. So s is greatest for $m=2$. And, thus, the equation of the required line is $y=2x$.

Problem 8:

The points $A(-4,-1)$, $B(-2,-4)$, Slatex C(4,0)\$ and $D(2,3)$ are the vertices of a :

option a: parallelogram

option b: rectangle

option c: rhombus

option d: square.

Note: more than one option may be right. Please mark all that are right.

Solution 8:

Mid-point of AC = $(\frac{-4+4}{2},\frac{-1+0}{2})=(0, \frac{-1}{2})$

Mid-point of BD = $(\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})$

$\Longrightarrow$ the diagonals AC and BD bisect each other.

$\Longrightarrow$ ABCD is a parallelogram.

Next, $AC= \sqrt{(-4-4)^{2}+(-1+0)^{2}}=\sqrt{64+1}=\sqrt{65}$ and $BD=\sqrt{(-2-2)^{2}+(-4+3)^{2}}=\sqrt{16+49}=\sqrt{65}$ and since the diagonals are also equal, it is a rectangle.

As $AB=\sqrt{(-4+2)^{2}+(-1+4)^{2}}=\sqrt{13}$ and $BC=\sqrt{(-2-4)^{2}+(-4)^{2}}=\sqrt{36+16}=sqrt{52}$, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.

Problem 9:

Equations $(b-c)x+(c-a)y+(a-b)=0$ and $(b^{3}-c^{3})x+(c^{3}-a^{3})y+a^{3}-b^{3}=0$ will represent the same line if

option i: $b=c$

option ii: $c=a$

option iii: $a=b$

option iv: $a+b+c=0$

Solution 9:

The two lines will be identical if there exists some real number k, such that

$b^{3}-c^{3}=k(b-c)$, and $c^{3}-a^{3}=k(c-a)$, and $a^{3}-b^{3}=k(a-b)$.

$\Longrightarrow b-c=0$ or $b^{2}+c^{2}+bc=k$

$\Longrightarrow c-a=0$ or $c^{2}+a^{2}+ac=k$, and

$\Longrightarrow a-b=0$ or $a^{2}+b^{2}+ab=k$

That is, $b=c$ or $c=a$, or $a=b$.

Next, $b^{2}+c^{2}+bc=c^{2}+a^{2}+ca \Longrightarrow b^{2}-a^{2}=c(a-b)$. Hence, $a=b$, or $a+b+c=0$.

Problem 10:

The circumcentre of a triangle with vertices $A(a,a\tan{\alpha})$, $B(b, b\tan{\beta})$ and $C(c, c\tan{\gamma})$ lies at the origin, where $0<\alpha, \beta, \gamma < \frac{\pi}{2}$ and $\alpha + \beta + \gamma = \pi$. Show that it’s orthocentre lies on the line $4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}x-4\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}y=y$

Solution 10:

As the circumcentre of the triangle is at the origin O, we have $OA=OB=OC=r$, where r is the radius of the circumcircle.

Hence, $OA^{2}=r^{2} \Longrightarrow a^{2}+a^{2}\tan^{2}{\alpha}=r^{2} \Longrightarrow a = r\cos{\alpha}$

Therefore, the coordinates of A are $(r\cos{\alpha},r\sin{\alpha})$. Similarly, the coordinates of B are $(r\cos{\beta},r\sin{\beta})$ and those of C are $(r\cos{\gamma},r\sin{\gamma})$. Thus, the coordinates of the centroid G of $\triangle ABC$ are

$(\frac{1}{3}r(\cos{\alpha}+\cos{\beta}+\cos{\gamma}),\frac{1}{3}r(\sin{\alpha}+\sin{\beta}+\sin{\gamma}))$.

Now, if $P(h,k)$ is the orthocentre of $\triangle ABC$, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.

Hence, $\frac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{\cos{\alpha}+\cos{\beta}+\cos{\gamma}}=\frac{k}{h}$

$\Longrightarrow \frac{4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}}{1+4\sin{(\frac{\alpha}{2})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}}= \frac{k}{h}$

because $\alpha+\beta+\gamma=\pi$.

Hence, the orthocentre $P(h,k)$ lies on the line

$4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}x-4\sin{(\frac{}{})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}y=y$.

Hope this gives an assorted flavour. More stuff later,

Nalin Pithwa.

### IITJEE Foundation Math and PRMO (preRMO) practice: another random collection of questions

Problem 1: Find the value of $\frac{x+2a}{2b--x} + \frac{x-2a}{2a+x} + \frac{4ab}{x^{2}-4b^{2}}$ when $x=\frac{ab}{a+b}$

Problem 2: Reduce the following fraction to its lowest terms:

$(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \div (\frac{x+y+z}{x^{2}+y^{2}+z^{2}-xy-yz-zx} - \frac{1}{x+y+z})+1$

Problem 3: Simplify: $\sqrt[4]{97-56\sqrt{3}}$

Problem 4: If $a+b+c+d=2s$, prove that $4(ab+cd)^{2}-(a^{2}+b^{2}-c^{2}-d^{2})^{2}=16(s-a)(s-b)(s-c)(s-d)$

Problem 5: If a, b, c are in HP, show that $(\frac{3}{a} + \frac{3}{b} - \frac{2}{c})(\frac{3}{c} + \frac{3}{b} - \frac{2}{a})+\frac{9}{b^{2}}=\frac{25}{ac}$.

May u discover the joy of Math! 🙂 🙂 🙂

Nalin Pithwa.

### Solutions to Birthday Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

Problem 1:

What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?

Solution 1:

The probability of the second person having a different birthday from the first person is $\frac{364}{365}$. The probability of the first three persons having different birthdays is $\frac{364}{365} \times \frac{363}{365}$. In this way, the probability of all n persons in a room having different birthdays is $P(n) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+1}{365}$. For the value of n, when P(n) falls just below 1/2 is the least number of people in a room when the probability of at least two people having the same birthday becomes greater than one half (that is, more likely than not). Now, one can make the following table:

$\begin{tabular}{|c|c|}\hline N & P(n) \\ \hline 2 & 364/365 \\ \hline 3 & 0.9918 \\ \hline 4 & 0.9836 \\ \hline 5 & 0.9729 \\ \hline 6 & 0.9595 \\ \hline 7 & 0.9438 \\ \hline 8 & 0.9257 \\ \hline 9 & 0.9054 \\ \hline 10 & 0.8830 \\ \hline 11 & 0.8589 \\ \hline 12 & 0.8330 \\ \hline 13 & 0.8056 \\ \hline 14 & 0.7769 \\ \hline 15 & 0.7471 \\ \hline 16 & 0.7164 \\ \hline 17 & 0.6850 \\ \hline 18 &0.6531 \\ \hline 19 & 0.6209 \\ \hline 20 & 0.5886 \\ \hline 21 & 0.5563 \\ \hline 22 & 0.5258 \\ \hline 23 & 0.4956 \\ \hline \end{tabular}$

Thus, the answer is 23. One may say that during a football match with one referee, it is more likely than not that at least two people on the field have the same birthday! 🙂 🙂 🙂

Problem 2:

You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?

Solution 2:

The probability of the first person having a different birthday from yours is $\frac{364}{365}$. Similarly, the probability of the first two persons not having the same birthday as yours is $\frac{(364)^{2}}{(365)^{2}}$. Thus, the probability of n persons not  having the same birthday as yours is $\frac{(364)^{n}}{(365)^{n}}$. When this value falls below 0.5, then it becomes more likely than not that at least another person has the same birthday as yours. So, the least value of n is obtained from $(\frac{364}{365})^{n}<\frac{1}{2}$. Taking log of both sides, we solve to get $n>252.65$. So, the least number of people required is 253.

Problem 3:

A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

Solution 3:

For the nth person to earn a free entry, first $(n-1)$ persons must have different birthdays and the nth person must have the same birthday as that of one of these previous $(n-1)$ persons. The probability of such an event can we written as

$P(n) = [\frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+2}{365}] \times \frac{n-1}{365}$

For a maximum, we need $P(n) > P(n+1)$. Alternatively, $\frac{P(n)}{P(n+1)} >1$. Using this expression for P(n), we get $\frac{365}{365-n} \times \frac{n-1}{n} >1$. Or, $n^{2}-n-365>0$. For positive n, this inequality is satisfied first for some n between 19 and 20. So, the best place in the queue to get a free entry is the 20th position.

More later,

Nalin Pithwa.

### Can anyone have fun with infinite series?

Below is list of finitely many puzzles on infinite series to keep you a bit busy !! 🙂 Note that these puzzles do have an academic flavour, especially concepts of convergence and divergence of an infinite series.

Puzzle 1: A grandmother’s vrat (fast) requires her to keep odd number of lamps of finite capacity lit in a temple at any time during 6pm to 6am the next morning. Each oil-filled lamp lasts 1 hour and it burns oil at a constant rate. She is not allowed to light any lamp after 6pm but she can light any number of lamps before 6pm and transfer oil from some to the others throughout the night while keeping odd number of lamps lit all the time. How many fully-filled oil lamps does she need to complete her vrat?

Puzzle 2: Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 liters of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that $p+2$ is also a prime number. Then, the first number theorist would pipette out $\frac{1}{p}$ litres of chemical and the second $\frac{1}{(p+2)}$ litres. How many times do they have to play this game to empty the flask completely?

Puzzle 3: How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Puzzle 4: Imagine a tank that can be filled with infinite taps and can be emptied with infinite drains. The taps, turned on alone, can fill the empty tank to its full capacity in 1 hour, 3 hours, 5 hours, 7 hours and so on. Likewise, the drains opened alone, can drain a full tank in 2 hours, 4 hours, 6 hours, and so on. Assume that the taps and drains are sequentially arranged in the ascending order of their filling and emptying durations.

Now, starting with an empty tank, plumber A alternately turns on a tap for 1 hour and opens the drain for 1 hour, all operations done one at a time in a sequence. His sequence, by using $t_{i}$ for $i^{th}$ tap and $d_{j}$ for $j^{th}$ drain, can be written as follows: $\{ t_{1}, d_{1}, t_{2}, d_{2}, \ldots\}_{A}$.

When he finishes his operation, mathematically, after using all the infinite taps and drains, he notes that the tank is filled to a certain fraction, say, $n_{A}<1$.

Then, plumber B turns one tap on for 1 hour and then opens two drains for 1 hour each and repeats his sequence: $\{ (t_{1},d_{1},d_{2}), (t_{2},d_{3},d_{4}), (t_{3},d_{4},d_{5}) \ldots \}_{B}$.

At the end of his (B’s) operation, he finds that the tank is filled to a fraction that is exactly half of what plumber A had filled, that is, $0.5n_{A}$.

How is this possible even though both have turned on all taps for 1 hour and opened all drains for 1 hour, although in different sequences?

I hope u do have fun!!

-Nalin Pithwa.

### Lagrange’s Mean Value Theorem and Cauchy’s Generalized Mean Value Theorem

Lagrange’s Mean Value Theorem:

If a function $f(x)$ is continuous on the interval $[a,b]$ and differentiable at all interior points of the interval, there will be, within $[a,b]$, at least one point c, $a, such that $f(b)-f(a)=f^{'}(c)(b-a)$.

Cauchy’s Generalized Mean Value Theorem:

If $f(x)$ and $phi(x)$ are two functions continuous on an interval $[a,b]$ and differentiable within it, and $phi(x)$ does not vanish anywhere inside the interval, there will be, in $[a,b]$, a point $x=c$, $a, such that $\frac{f(b)-f(a)}{phi(b)-phi(a)} = \frac{f^{'}(c)}{phi^{'}(c)}$.

Some questions based on the above:

Problem 1:

Form Lagrange’s formula for the function $y=\sin(x)$ on the interval $[x_{1},x_{2}]$.

Problem 2:

Verify the truth of Lagrange’s formula for the function $y=2x-x^{2}$ on the interval $[0,1]$.

Problem 3:

Applying Lagrange’s theorem, prove the inequalities: (i) $e^{x} \geq 1+x$ (ii) $\ln (1+x) , for $x>0$. (iii) $b^{n}-a^{n} for $b>a$. (iv) $\arctan(x) .

Problem 4:

Write the Cauchy formula for the functions $f(x)=x^{2}$, $phi(x)=x^{3}$ on the interval $[1,2]$ and find c.

More churnings with calculus later!

Nalin Pithwa.