Indian Statistical Institute (ISI) B.Math & B.Stat : Algebra

If $p,q,r$ are positive real numbers such that $pqr=1$, then find the value of $\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}$. Throught the simplification we will use $1=pqr,q^{-1}=pr,r^{-1}=pq \quad and \quad p^{-1} = qr$ Given expression is $$\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}$$ $$= \frac{pqr}{pqr+p+pr}+\frac{pqr}{pqr+q+pq}+\frac{1}{1+r+p^{-1}}$$ $$= \frac{qr}{qr+1+r}+\frac{pr}{pr+1+p}+\frac{1}{1+r+p^{-1}}$$ $$= \frac{qr}{p^{-1}+1+r}+\frac{pr}{pr+pqr+p}+\frac{1}{1+r+p^{-1}}$$ $$= \frac{qr}{p^{-1}+1+r}+\frac{r}{r+qr+1}+\frac{1}{1+r+p^{-1}}$$ $$= \frac{qr}{p^{-1}+1+r}+\frac{r}{r+p^{-1}+1}+\frac{1}{1+r+p^{-1}}$$ $$= \frac{qr+r+1}{p^{-1}+1+r}$$ $$= \frac{p^{-1}+r+1}{p^{-1}+1+r}$$ $$= 1$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Number Theory

Consider the equation $x^2 + y^2 = 2007$. How many solutions $(x, y)$ exist such that $x$ and $y$ are positive integers? $2007 = 2000 + 7 \equiv 0 + 3 \equiv 3 (mod \quad 4)$. Now we know that square on an integer is either divisible by $4$ or leaves a remainder $1$ when divided by $4$, said otherwise $x \in \mathbb{Z} \implies x^2 \equiv 0 \quad or \quad 1 (mod \quad 4)$. Thus for integers $x$ and $y$, $x^2+y^2 \equiv 0 \quad or \quad 1 \quad or \quad 2 (mod \quad 4)$. Since we have different remainders $mod \quad 4$ on the two sides, it follows there cannot be any solution in $\mathbb{Z}$ hence no solution in $\mathbb{Z^+}$

Indian Statistical Institute ( ISI ) B.Math & B.Stat :Complex Numbers

Let $z$ be a non-zero complex number such that $|z −\frac{1}{z}| = 2.$ What is the maximum value of $|z|$? Given $2 = |z −\frac{1}{z}| \geq \big||z|- |\frac{1}{z}|\big|$ Let $t = |z|$ $\implies \big|t- \frac{1}{t}\big| \leq 2$ $\implies -2 \leq t- \frac{1}{t} \leq 2$ $\implies -2t \leq t^2- 1 \leq 2t$ $\implies t^2 +2t- 1 \geq 0 \quad and \quad t^2 -2t- 1 \leq 0$ The first inequality gives $t \in ( - \infty, -1-\sqrt{2}] \cup [\sqrt{2}-1, \infty)$. Since $t \geq 0$ $\implies t \in [\sqrt{2}-1, \infty)$. The second inequality gives $t \in [1-\sqrt{2} , 1+\sqrt{2}]$. Again since $t \geq 0$ $\implies t \in [0,1+\sqrt{2}]$ Combining the two case we see $t \in [\sqrt{2}-1,\sqrt{2}+1] \implies |z| \in [\sqrt{2}-1,\sqrt{2}+1]$. Thus the maximum value of $|z|$ is $\sqrt{2}+1$. $Practice-Problem$ Let $z$ be a non-zero complex number such that $|z +\frac{1}{z}| = a, a \in \mathbb{R^+}.$ What is the maximum and minimum value of $|z|$?

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Co-ordinate Geometry

Let $A$ be the set of all points $(h, k)$ such that the area of the triangle formed by $(h, k), (5, 6)$ and $(3, 2)$ is $12$ square units. What is the least possible length of a line segment joining $(0, 0)$ to a point in $A$? Take the base of the triangle to be the line segment obtained by joining the points $(5,6)$ and $(3,2)$. Equation of the base is $2x-y-4 =0$. Length of the base is $\sqrt{(5-3)^2+(6-2)^2} = 2 \sqrt{5}.$ Let $p$ be the length of the perpendicular from the point $(h,k)$ onto the base. ( Note that the point $(h,k)$ cannot lie on the base. Why?) Since the area is given to be $12$, $12 = \frac{1}{2} \times p \times 2 \sqrt{5} \implies p = \frac{12}{\sqrt{5}}$. Therefore the point $(h,k)$ lies at a distance of $\frac{12}{\sqrt{5}}$ units from the base on both sides. Thus $A$ is the set of all points on the line $parallel$ to the base and at a distance $\frac{12}{\sqrt{5}}$ units away from the base. In the diagram, the lines colored green represents the set $A$. Clearly the least possible length of a line segment joining $(0, 0)$ to a point in $A$? is the distance between the point $(0,0)$ and the line drawn parallel to the base and to the left side of the base. Let $XY$ be the line segment perpendicular to the base and the line and passing through the orgin as shown in the diagram. Required distance is $OX$ and $OX = XY - OY = \frac{12}{\sqrt{5}} - \big|\frac{-4}{\sqrt{2^2+(-1)^2}}\big| = \frac{12}{\sqrt{5}} - \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$

Indian Statistical Institute (ISI) B.Math & B.Stat : Trigonometry

Find the ratio of the areas of the regular pentagons inscribed and circumscribed around a given circle. Let $a$ be the side of the circumscribed pentagon and $b$ be that of the inscribed pentagon. First note that for the circle is inscribed for the exterior pentagon and circumscribed for the interior pentagon. Therefore the $in-radius$ of the exterior polygon, say $r$ is equal to the $circum-radius$, say $R$ of the interior pentagon, i.e., $R=r$. See the figure below. Using standard formula, $$a = 2 r \tan \frac{\pi}{5}, \quad b = 2 R \sin \frac{\pi}{5}$$ . Area of a regular polygon having $n$ sides is $n \times \frac {(side)^2}{4} \cot \frac{\pi}{n}$. Therefore the required ratio is $\bigg( \frac{5 \times \frac {b^2}{4} \cot \frac{\pi}{5}}{5 \times \frac {a^2}{4} \cot \frac{\pi}{5}} \bigg) = \frac{b^2}{a^2} = \frac{(2 R \sin \frac{\pi}{5})^2} {(2 r \tan \frac{\pi}{5})^2} = \cos^2 \frac{\pi}{5}$

Indian Statistical Institute B.Math & B.Stat : Trigonometry

Let $\theta_1 = \frac{2 \pi}{3}, \theta_2 = \frac{4 \pi}{7}, \theta_3 = \frac{7 \pi}{3}$. Then show that $( \sin \theta_1)^{ \sin \theta_1} < ( \sin \theta_3)^{ \sin \theta_3} < ( \sin \theta_2)^{ \sin \theta_2}$. First note that $\pi > \theta_1 > \theta_3 > \theta_2 > 0$ and all of them belong to the $second$ quadrant. $Sine$ function strictly decreases from $1$ to $0$ in the $second$ quadrant. Also $\sin \theta_1 \neq \sin \theta_2 \neq \sin \theta_3 \neq 0$ and each of them are posititve. Using the strictly decreasing property of $Sine$ in the second quadrant we have $\sin \theta_1 < \sin \theta_3 < \sin \theta_2$. Now the result follows the standard inequality $x^c < y^d$ for $x,y,c,d > 0 \quad where \quad x < y, \quad c < d$.

Indian Statistical Institute B.Math & B.Stat : Trigonometry

Find the value of the sum $\cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000}$. Let $z = \cos \frac{\pi}{1000} + i \sin \frac{\pi}{1000} = \cos \theta + i \sin \theta$ where $\theta = \frac{\pi}{1000}$ . It is easy to see that $z \neq 1,-1$. Consider the sum $1 +z^2+z^4+ \dots + z^{1998}$, $z \neq 1,-1$. Putting $w = z^2$ the sum reduces to $1 +w+w^2+ \dots + w^{999}$, $w \neq 1$. Now, $1 +w+w^2+ \dots + w^{999} = \frac{w^{1000}-1}{w-1}$ Substituting back $w$ we have the following identity $1 +z^2+z^4+ \dots + z^{1998} = \frac{z^{2000}-1}{z^2-1}$,$z \neq 1,-1$. Using $De-Moivre's$ theorem we have $z^n = \cos n \theta + i \sin n \theta$ for $n \in \mathbb{N}$. Substituting back in the above identity we have, $\big(1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta \big) + i \big(1+ \sin 2 \theta + \sin 4 \theta + \dots + \sin 1998 \theta \big) = \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1}$ Equating the real part from both side we have. $1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re \bigg( \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1} \bigg) = Re \bigg( \frac{\cos 2 \pi + i \sin 2 \pi -1 }{\cos 2 \theta + i \sin 2 \theta -1}\bigg)$, since $\theta = \frac{\pi}{1000}$. Therefore $1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re (0) = 0 \implies \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = -1$. $\implies \cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000} = -1$

Indian Statistical Institute B.Math & B.Stat : Quadratic Equations

Consider the function $f(x) = ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are real numbers with $a > 0$. If $f$ is strictly increasing, then show that the function $g(x) =f′(x) − f′′(x) + f′′′(x)$ is positive for all $x \in \mathbb{R}$. First we calculate the derivatives up to the third order. $f'(x) = 3ax^2+2bx+c, \quad f''(x) = 6ax+2b \quad and \quad f'''(x) = 6a$. It is given that $f$ is strictly increasing which implies $f' > 0$ which in turn implies $3ax^2+2bx+c > 0$. Let $y = 3ax^2+2bx+c$ It is easy to see that $y = 3a \big( x + \frac{b}{3a} \big)^2 + \frac{3ac-b^2}{3a}$. Since $y > 0$ and $a$ is given to be positive $3ac$ must be strictly greater than $b^2$. Note $\big( x + \frac{b}{3a} \big)^2$ is always non-negative. Now $g(x) = f′(x) − f′′(x) + f′′′(x) = 3ax^2+2bx+c -(6ax+2b) + 6a = 3ax^2 + 2x(b-3a)+(c-2b+6a)$ $= 3a\big( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2}+ \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2} \big)$ = $3a\bigg( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2} \bigg) + 3a \bigg( \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2}\bigg)$ $= 3a \big( x + \frac{b-3a}{3a} \big)^2 + \frac{9a^2+3ac-b^2}{3a}$ $3a \big( x + \frac{b-3a}{3a} \big)^2 \geq 0$ for all $x \in \mathbb{R}$. (since $a$ is given to be positive) We have already shown that $3ac > b^2$ therefore $\frac{9a^2+3ac-b^2}{3a} > 0$. Thus $g(x) > 0$ for all $x \in \mathbb{R}$.

Indian Statistical Institute B.Math & B.Stat : Real Analysis

Suppose $f$ is a differentiable and increasing function on $[0, 1]$ such that $f(0) < 0 < f(1)$. Let $F(t) = \int_{0}^{t} f(x) dx$. Then show that $F$ has a unique minimum in the open interval $(0, 1)$. Since $f$ is differentiable, $f$ is continuous. Therefore $F(t)$ is differentiable and $F'(t) = f(t) , t \in [0,1]$ ( students interested in the proof are advised to read INTRODUCTION TO REAL ANALYSIS, Author : ROBERT G. BARTLE, DONALD R. SHERBERT ). Given that $f$ is a differentiable and increasing function on $[0, 1]$ such that $f(0) < 0 < f(1)$ it follows that $F'(t)$ satisfies the same conditions of $f$ in $[0,1]$ since $F'(t) = f(t)$. Now see that $F'(0)F'(1) = f(0)f(1) < 0$, continuity of $F'$ implies $\exists$ $c \quad \in (0,1)$ such that $F'(c)=0$. Again since $F'$ is increasing, the points at which $F'$ vanishes forms a sub-interval $(c,d)$ of $[0,1]$ (i.e, the points at which the function $F'$ vanishes forms a connected set !) where $0 < c < d < 1$. Thus $F'(t) < 0$ for $t \in [0,c)$ and $F'(t) > 0$ for $t \in (d,1]$ which implies $F$ is strictly decreasing in $[0,c)$, strictly increasing in $(d,1]$ and $F$ is constant on $[c,d]$ which in turn implies $F$ has a unique minimum on $[c,d]$. Note that if it was given $f$ to be strictly increasing then $c=d$ and the unique minimum will be attained at a unique point.

Indian Statistical Institute B.Math & B.Stat : Complex Numbers

Show that the set of complex numbers $z$ satisfying the equation $(3+7i)z+(10-2i)\overline{z}+100 = 0$ represents, in the Argand plane, a point. Let $z=x+iy$, taking the conjugate of the given equation we have $(3-7i)\overline{z}+(10+2i)z+100 = 0$ Adding the two equations we get, $26x-18y+200 = 0$ (do the calculations yourself!), this shows that $z$ lies on the line $26x-18y+200 = 0$ Subtracting the two equations we get, $10x-4y = 0$, this again shows that that $z$ lies on the line $10x-4y = 0$ Thus $z$ satisfies both the equations $26x-18y+200 = 0$ and $10x-4y = 0$, thus $z$ represents a point in the Argand Plane.