Indian Statistical Institute, ISI BStat & BMath 2023 UGB Solutions & Discussions

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ISI BMath & BSTAT 2023 Subjective Questions UGB: solutions and discussions

Past Year’s BMATH/BSTAT Papers & Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2023 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Answer Keys: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Question Paper: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGB Question Paper: Click Here

Problem 1.

Determine all integers $n>1$ such that every power of $n$ has an odd number of digits.

Topic: Number Theory
Difficulty level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 1 UGB Solution

Solution:

We will prove that all powers of and integer $n>1$ have odd number of digits if and only if $n=10^{2k}$ for some $k\geq 1.$ We note that the number of digits of an integer $n>1$ is exactly

$1+\lfloor \log_{10}(n)\rfloor.$

Indeed, if $k=1+\lfloor \log_{10}(n)\rfloor,$ then $10^{k-1}\leq n<10^k$ and $n$ has exactly $k$ digits. If $n=10^{2k},$ then for any $m\geq 1$ the number of digits of $n^m$ is odd:

$1+\lfloor \log_{10}(n^m)\rfloor =1+\lfloor 2km\rfloor =1+2km.$

Assume that all powers of $n$ have odd number of digits, but $n$ is not of the form $10^{2k}.$ If $n$ is a power of $10,$ then $n$ is an odd power of $10,$ hence $n^1=n$ has even number of digits. This is impossible.
So, $n$ is not a power of $10$ and $\log_{10}(n)$ is not an integer number. Let us write $\log_{10}(n)=l+r,$ where $l\geq 0$ is an integer and $r\in (0,1).$ Since $n^1=n$ has odd number of digits, we deduce that

$1+\lfloor \log_{10}(n)\rfloor=1+l$

is odd. In particular, $l$ is even.

If $r=\frac{1}{m}$ for some integer $m\geq 2,$ then

$1+\lfloor \log_{10}(n^m)\rfloor =1+\lfloor m(l+r)\rfloor=1+ml+1=ml+2$

is even (since $l$ is even). So, $r\ne \frac{1}{m}$ for any $m\geq 2,$ and $r\in \left(\frac{1}{m+1},\frac{1}{m}\right)$ for some integer $m\geq 1.$ But then $n^{m+1}$ has even number of digits:

$1+\lfloor \log_{10}(n^{m+1})\rfloor=1+\lfloor (m+1)l+(m+1)r \rfloor=1+(m+1)l+1=(m+1)l+2.$

The latter relation follows from the fact that $(m+1)r\in\left(1,\frac{m+1}{m}\right)\subset (1,2).$

Video Solution:

Problem 2.

Let $a_0=\frac{1}{2}$ and $a_n$ be defined inductively by

$a_n=\sqrt{\frac{1+a_{n-1}}{2}}, n \geq 1 .$

(a) Show that for $n=0,1,2, \ldots$ .

$a_n=\cos \theta_n \text { for some } 0<\theta_n<\frac{\pi}{2} \text {. }$

and determine $\theta_n$ .

(b) Using (a) or otherwise, calculate

$\lim _{n \rightarrow \infty} 4^n\left(1-a_n\right)$

Topic: Trigonometric Sequences
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 2 UGB Solution.

Solution:

(a) We note that $a_0=\frac{1}{2}=\cos \frac{\pi}{3}.$ If $a_{n-1}=\cos \theta_{n-1},$ $0<\theta_{n-1}<\frac{\pi}{2},$ then

$a_n=\sqrt{\frac{1+a_{n-1}}{2}}=\sqrt{\frac{1+\cos\theta_{n-1}}{2}}=\sqrt{\cos^2\left(\frac{\theta_{n-1}}{2}\right)}=\cos\left(\frac{\theta_{n-1}}{2}\right).$

So, $\theta_n=\frac{\theta_{n-1}}{2}.$ Since $\theta_0=\frac{\pi}{3},$ it follows that

$\theta_n=\frac{\pi}{3\times 2^n}.$

(b)

$\lim_{n\to\infty} 4^n (1-a_n)=\lim_{n\to\infty}\frac{1-\cos \left(\frac{\pi}{3\times 2^{n}}\right)}{\frac{1}{4^{n}}}=$

since $1-\cos x\sim \frac{x^2}{2},$ $x\to 0,$

$=\lim_{n\to\infty}\frac{\frac{\pi^2}{18\times 4^n}}{\frac{1}{4^n}}=\frac{\pi^2}{18}$

Video Solution:

Problem 3.

In a triangle $A B C$ , consider points $D$ and $E$ on $A C$ and $A B$ . respectively, and assume that they do not coincide with any of the vertices $A, B, C$ . If the segments $B D$ and $C E$ intersect at $F$ , consider the areas $w, x, y, z$ of the quadrilateral $A E F D$ and the triangles $B E F, B F C, C D F$ , respectively.

(a) Prove that $y^2>xz$ .

(b) Determine $w$ in terms of $x, y, z$ .

Topic: Geometry
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 3 UGB Solution.

Solution:

(a) Consider altitudes $EG,$ $FH,$ $DI$ of length $\alpha,\beta,\gamma$ respectively:

Denote length of $BG,$ $GH,$ $HI,$ $IC$ by $a,b,c,d$ respectively, and let $l$ be the length of $BC.$ Then

$x+y=\frac{1}{2}\alpha l, \ z+y=\frac{1}{2}\gamma l, \ y=\frac{1}{2}\beta l.$

The inequality $y^2>xz$ is equivalent to

$y^2>\left(\frac{1}{2}\alpha l-y\right)\left(\frac{1}{2}\gamma l-y\right),$

or to

$\frac{1}{4}\alpha \gamma l^2<\frac{1}{2} y l \left(\alpha+\gamma\right).$

After simplifications we obtain, equivalently,

$y>\frac{\alpha \gamma l}{2(\alpha+\gamma)}.$

Since $y=\frac{1}{2}\beta l,$ the latter is equivalent to

$\beta>\frac{\alpha \gamma}{\alpha+\gamma},$

or to

$\frac{\beta}{\alpha}+\frac{\beta}{\gamma}>1.$

From triangles $EGC$ and $FHI$ we find that

$\frac{\beta}{\alpha}=\frac{c+d}{b+c+d}.$

From triangles $BDI$ and $BFH$ we find that

$\frac{\beta}{\gamma}=\frac{a+b}{a+b+c}.$

It remains to prove that

$\frac{a+b}{a+b+c}+\frac{c+d}{b+c+d}>1.$

But it follows from simple inequalities $\frac{a+b}{a+b+c}>\frac{b}{b+c},$ $\frac{c+d}{b+c+d}>\frac{c}{b+c}.$

(b) Consider a segment $AF$ and let $u$ be the area of $AEF,$ $v$ be the area of $AFD:$

In particular, $w=u+v.$ Triangles $AEF$ and $BFE$ share the same altitude on the line $AB.$ Hence, $\frac{u}{x}=\frac{AE}{EB}.$ Triangles $AEC$ and $BCE$ share the same altitude on the line $AB.$ Hence, $\frac{w+z}{x+y}=\frac{AE}{EB}.$ It follows that

$\frac{u}{x}=\frac{w+z}{x+y}.$

Similarly, considering triangles $AFD$ and $CFD,$ and $BAD$ and $BCD,$ we deduce that

$\frac{v}{z}=\frac{w+x}{z+y}.$

Hence,

$w=u+v=\frac{x(w+z)}{x+y}+\frac{z(w+x)}{z+y}.$

Further,

$w\left((x+y)(z+y)-x(z+y)-z(x+y)\right)=xz(x+z+2y),$

$w=\frac{xz(x+z+2y)}{y^2-xz}.$

Video Solution:

Problem 4.

Let $n_1,n_2,n_3,\cdots,n_{51}$ , be distinct natural numbers each of which has exactly 2023 positive integer factors. For instance, $2^{2022}$ has exactly 2023 positive integer factors $1,2,2^2, \cdots, 2^{2021}, 2^{2022}$ . Assume that no prime larger than 11 divides any of the $n_i$ ‘s. Show that there must be some perfect cube among the $n_i$ ‘s. You may use the fact that $2023=7 \times 17 \times 17$

Topic: Combinatorics
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 4 UGB Solution.

Solution:

Each number $n_i$ has at most five distinct prime divisors: $2,3,5,7,11.$ If $n_i=\prod^m_{j=1}p^{\alpha_j}_j,$ where $\{p_1,\ldots,p_m\}\subset \{2,3,5,7,11\},$ $\alpha_1,\ldots,\alpha_m\geq 1,$ then the number of divisors of $n_i$ equals $\prod^m_{j=1}(\alpha_j+1).$ On the other hand,

$\prod^m_{j=1}(\alpha_j+1)=2023=7\times 17\times 17.$

It follows that $m\leq 3.$ Assume that there are no perfect cubes among $n_1,\ldots,n_{51}.$ If $n_i=p^\alpha$ for some $i,$ then the number of divisors of $n_i$ equals

$\alpha+1=2023.$

Hence $\alpha=2022=3\times 674$ and $n_i$ is a perfect cube. If $n_i=p^{\alpha_1}_1p_2^{\alpha_2},$ where $p_1<p_2$ are prime numbers among $\{2,3,5,7,11\}$ and $\alpha_1,\alpha_2\geq 1,$ then

$(\alpha_1+1)(\alpha_2+1)=2023=7\times 17\times 17.$

If $\alpha_1+1=7,$ $\alpha_2+1=17\times 17,$ then $\alpha_1=6=3\times 2$ and $\alpha_2=288=3\times 96.$ Hence $n_i$ is a perfect cube. The same holds when $\alpha_1+1=17\times 17,$ $\alpha_2+1=7.$ It follows that either $\alpha_1+1=7\times 17,$ $\alpha_2+1=17,$ or $\alpha_1+1=17,$ $\alpha_2+1=7\times 17.$ Equivalently, either $\alpha_1=118,$ $\alpha_2=16,$ or $\alpha_1=16,$ $\alpha_2=118.$ The number of possible combinations for $n_i$ is ${5\choose 2}\times 2=20.$

If $n_i=p^{\alpha_1}_1p_2^{\alpha_2}p^{\alpha_3}_3,$ where $p_1<p_2<p_3$ are prime numbers among $\{2,3,5,7,11\}$ and $\alpha_1,\alpha_2,\alpha_3\geq 1,$ then

$(\alpha_1+1)(\alpha_2+1)(\alpha_3+1)=2023=7\times 17\times 17.$

Hence, values of $\alpha_1,\alpha_2,\alpha_3$ are $6,16,16.$ The number of possible combinations for $n_i$ is ${5\choose 3}\times 3=30.$

There are at most $50$ natural numbers that are not perfect cubes, that have exactly 2023 divisors, and whose prime divisors are $\leq 11.$ It follows that at least one of the numbers $n_1,\ldots,n_{51}$ is a perfect cube.

Video Solution:

Problem 5.

There is a rectangular plot of size $1 \times n$ . This has to be covered by three types of tiles – red, blue and black. The red tiles are of size $1 \times 1$ , the blue tiles are of size $1 \times 1$ and the black tiles are of size $1 \times 2$ . Let $t_n$ denote the number of ways this can be done. For example, clearly $t_1=2$ because we can have either a red or a blue tile. Also, $t_2=5$ since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.

(a) Prove that $t_{2 n+1}=t_n\left(t_{n-1}+t_{n+1}\right)$ for all $n>1$ .

(b) Prove that $t_n=\sum_{d \geq 0}\left(\begin{array}{c}n-d \\ d\end{array}\right) 2^{n-2 d}$ for all $n>0$ .
Here,

$\left(\begin{array}{c} m \\ r \end{array}\right)= \begin{cases}\frac{m !}{r !(m-r) !}, & \text { if } 0 \leq r \leq m \\ 0, & \text { otherwise }\end{cases}$

for integers $m, r$ .

Topic: Combinatorics Recurrence
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 5 UGB Solution.

Solution:

(a) At first we note that $t_{n+1}=2t_n+t_{n-1}.$ Indeed, if the first square is covered either by a red tile or by a blue tile, then there remains $n$ squares more to cover. If the first two squares are covered by a black tile, then there remains $n-1$ squares to be covered.

Consider a plot of size $1\times (2n+1).$ If the middle square is covered either by a red tile or by a blue tile, then there remains $n$ squares on the left and $n$ squares on the right of it to be covered. This gives $2t^2_n$ possibilities. If the middle square is covered by a black tile, then either there remains $n-1$ squares on the left and $n$ squares on the right of this tile to be covered, or there remains $n$ squares on the left and $n-1$ squares on the right of this tile to be covered. This gives $2t_nt_{n-1}$ possibilities. Totally, we get

$t_{2n+1}=2t^2_n +2t_nt_{n-1}=t_n\left(t_{n-1}+2t_n+t_{n-1}\right)=t_n\left(t_{n-1}+t_{n+1}\right).$

(b) Consider covering with exactly $d$ black tiles (in particular, $2d\leq n$ ). Assume that these tiles occupy squares $n_1,n_1+1,n_2,n_2+1,\ldots,n_d,n_d+1.$ Remaining $n-2d$ squares are partitioned by black tiles into $d+1$ groups (possibly empty):

$\{1,\ldots,n_1-1\}, \{n_1+2,\ldots,n_2-1\}, \ldots, \{n_d+2,\ldots,n\}.$

Conversely, every such collection of $d+1$ groups define a configuration of black tiles. The number of such collections equals the number of non-negative integer solutions of the equation

$m_1+\ldots+m_{d+1}=n-2d,$

which is known to be equal to ${n-2d+d \choose d}={n-d \choose d}.$ When $d$ black tiles are placed, there remains $n-2d$ squares to be covered by red or blue tiles. There are $2^{n-2d}$ possibilities to do this. Totally, we get

$t_n=\sum_{d\geq 0}{n-d \choose d}2^{n-2d}.$

Video Solution:

Problem 6.

Let $\left\{u_n\right\}_{n \geq 1}$ be a sequence of real numbers defined as $u_1=1$ and

$u_{n+1}=u_n+\frac{1}{u_n} \text { for all } n \geq 1 \text {. }$

Prove that $u_n \leq \frac{3 \sqrt{n}}{2}$ for all $n$ .

Topic: Sequence, Real Analysis
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 6 UGB Solution.

Solution:

The function $f(x)=x+\frac{1}{x}$ is strictly increasing on $[1,\infty).$ Since $u_{n+1}>u_n$ for all $n\geq 1,$ and $u_1=1,$ it follows that $u_n\in [1,\infty)$ for all $n\geq 1.$ We will verify the inequality $u_n\leq \frac{3}{2}\sqrt{n}$ by induction on $n.$ If $n=1,$ the inequality is true. If $n=2$ the inequality is also true, as

$u_2=2<3\sqrt{2}=\frac{3}{2}\sqrt{2}.$

Assume the inequality was verified for $n\geq 2.$ Then

$u_{n+1}=f(u_{n})\leq f\left(\frac{3}{2}\sqrt{n}\right)=\frac{3}{2}\sqrt{n}+\frac{2}{3\sqrt{n}}=\frac{9n+4}{6\sqrt{n}}.$

It is enough to prove that $\frac{9n+4}{6\sqrt{n}}\leq \frac{3}{2}\sqrt{n+1}.$ Or, equivalently,

$9n+4\leq 9\sqrt{n(n+1)}.$

Taking squares of both sides we obtain, equivalently,

$81n^2+72n+16\leq 81n^2+81n$

or,

$9n\geq 16.$

The inequality is true, since $n\geq 2.$

Video Solution:

Problem 7.

(a) Let $n \geq 1$ be an integer. Prove that $X^n+Y^n+Z^n$ can be written as a polynomial with integer coefficients in the variables $\alpha=X+Y+Z$ , $\beta=X Y+Y Z+Z X$ and $\gamma=X Y Z$ .

(b) Let $G_n=x^n \sin (n A)+y^n \sin (n B)+z^n \sin (n C)$ . where $x, y, z, A, B, C$ are real numbers such that $A+B+C$ is an integral multiple of $\pi$ . Using (a) or otherwise, show that if $G_1=G_2=0$ , then $G_{\mathrm{n}}=0$ for all positive integers $n$ .

Topic: Polynomials, Mathematical Induction, Complex Numbers
Difficulty Level: Hard

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 7 UGB Solution.

Solution:

(a) We prove the result by induction on $n.$ If $n=1,$ then $X+Y+Z=\alpha.$ If $n=2,$ then

$X^2+Y^2+Z^2=\left(X+Y+Z\right)^2-2XY-2YZ-2XZ=\alpha^2-2\beta.$

If $n=3,$ then

$X^3+Y^3+Z^3=\alpha^3-3\alpha\beta+3\gamma.$

Assume that for every $k\in \{1,\ldots,n\},$ $n\geq 3,$ there exists a polynomial with integer coefficients $P_k$ in three variables, such that

$X^k+Y^k+Z^k=\alpha^k+P_k(\alpha,\beta,\gamma).$

Then

$\begin{aligned} \alpha^{n+1}&=\alpha^n\times \alpha=\left(X^n+Y^n+Z^n-P_n(\alpha,\beta,\gamma)\right)\left(X+Y+Z\right)\\ &=X^{n+1}+Y^{n+1}+Z^{n+1}+X^n Y+X^nZ+XY^n+Y^nZ+XZ^n+YZ^n\\ &-\alpha P_n(\alpha,\beta,\gamma) \\ &=X^{n+1}+Y^{n+1}+Z^{n+1}+XY(X^{n-1}+Y^{n-1})+XZ(X^{n-1}+Z^{n-1})+YZ(Y^{n-1}+Z^{n-1})\\ &-\alpha P_n(\alpha,\beta,\gamma)\\ &=X^{n+1}+Y^{n+1}+Z^{n+1}+\left(XY+XZ+YZ\right) \left(X^{n-1}+Y^{n-1}+Z^{n-1}\right)\\ &-XYZ\left(X^{n-2}+Y^{n-2}+Z^{n-2}\right)-\alpha P_n(\alpha,\beta,\gamma)\\ &=X^{n+1}+Y^{n+1}+Z^{n+1}+\beta \left(\alpha^{n-1}+P_{n-1}(\alpha,\beta,\gamma)\right)-\gamma \left(\alpha^{n-2}+P_{n-2}(\alpha,\beta,\gamma)\right)-\alpha P_n(\alpha,\beta,\gamma) \end{aligned}$

Hence,

$X^{n+1}+Y^{n+1}+Z^{n+1}=\alpha^{n+1}-\beta \alpha^{n-1}-\beta P_{n-1}(\alpha,\beta,\gamma)+\gamma \alpha^{n-2}+\gamma P_{n-2}(\alpha,\beta,\gamma)+\alpha P_{n}(\alpha,\beta,\gamma).$

Hence, for every $n\geq 1,$ $X^n+Y^n+Z^n$ can be expressed as a polynomial in $\alpha,\beta,\gamma$ with integer coefficients.

(b) According to part (a), for every $n\geq1$ there exists a polynomial $Q_n$ in three variables with integer coefficients, such that

$X^n+Y^n+Z^n=Q_n\left(X+Y+Z,XY+YZ+XZ,XYZ\right).$

Moreover, $Q_1(\alpha,\beta,\gamma)=\alpha,$ $Q_2(\alpha,\beta,\gamma)=\alpha^2-2\beta.$

We note that

$G_n=\frac{x^n\left(e^{inA}-e^{-inA}\right)+y^n\left(e^{inB}-e^{-inB}\right)+z^n\left(e^{inC}-e^{-inC}\right)}{2i}=$

$=\frac{\left((xe^{iA})^n+(ye^{iB})^n+(ze^{iC})^n\right)-\left((xe^{-iA})^n+(ye^{-iB})^n+(ze^{-iC})^n\right)}{2i}.$

The first summand in the numerator is expressed as a polynomial $Q_n$ applied to quantities

$xe^{iA}+ye^{iB}+ze^{iC}=:\alpha, xye^{i(A+B)}+xze^{i(A+C)}+yze^{i(B+C)}=:\beta,$

$xyze^{i(A+B+C)}=:\gamma.$

The second summand in the numerator is expressed as a polynomial $Q_n$ applied to quantities

$xe^{-iA}+ye^{-iB}+ze^{-iC}=\overline{\alpha}, xye^{-i(A+B)}+xze^{-i(A+C)}+yze^{-i(B+C)}=\overline{\beta},$

$xyze^{-i(A+B+C)}=\overline{\gamma},$

where $\overline{z}$ is a conjugate of a complex number $z.$

Hence, we get a representation

$G_n=\frac{Q_n(\alpha,\beta,\gamma)-Q_n(\overline{\alpha},\overline{\beta},\overline{\gamma})}{2i}.$

It is enough to show that $\alpha=\overline{\alpha},$ $\beta=\overline{\beta},$ $\gamma=\overline{\gamma}$ in order to deduce that all $G_n=0.$

Since $G_1=0,$ and $Q_1=\alpha,$ it follows that $\alpha=\overline{\alpha}.$ Since $G_2=0,$ it follows that

$Q_2(\alpha,\beta,\gamma)=Q_2(\overline{\alpha},\overline{\beta},\overline{\gamma}),$

$\alpha^2-2\beta=\overline{\alpha}^2-2\overline{\beta}, \ \beta=\overline{\beta}.$

Finally, since $A+B+C$ is an integral multiple of $\pi,$ $A+B+C=m\pi,$ then

$\gamma=xyz e^{im\pi}=xyz e^{im\pi-2im\pi}=xyz e^{-im\pi}=\overline{\gamma}.$

From relations $\alpha=\overline{\alpha},$ $\beta=\overline{\beta},$ $\gamma=\overline{\gamma}$ it follows that $G_n=0$ for all $n\geq 1.$

Video Solution:

Problem 8.

Let $f:[0,1] \rightarrow R$ be a continuous function which is differentiable on $(0,1)$ . Prove that either $f$ is a linear function $f(x)=a x+b$ or there exists $t \in(0,1)$ such that $|f(1)-f(0)|<\left|f^{\prime}(t)\right|$ .

Topic: Real Analysis, Differentiation
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 8 UGB Solution.

Solution :

Assume that $|f'(t)|\leq |f(1)-f(0)|$ for all $t\in (0,1).$ If $f(1)=f(0),$ then $f$ is constant. Assume that $f(1)>f(0).$ By the mean value theorem,

$|f(t)-f(0)|\leq (f(1)-f(0))t, \ t\in [0,1].$

Assume that $f(t)-f(0)<(f(1)-f(0))t$ for some $t\in (0,1).$ Consider the function $g(s)=f(s)-f(0),$ $s\in [t,1].$ Since $|g'(s)|\leq f(1)-f(0),$ $s\in [t,1),$ then by the mean value theorem

$|g(1)-g(t)|\leq (f(1)-f(0))(1-t).$

Hence,

$|f(1)-f(t)|\leq (f(1)-f(0))(1-t)$

and

$f(1)-f(0)=f(1)-f(t)+f(t)-f(0)<(f(1)-f(0))(1-t)+(f(1)-f(0))t=f(1)-f(0).$

This is a contradiction. Hence, $f(t)-f(0)\geq (f(1)-f(0))t$ for all $t\in (0,1),$ and

$(f(1)-f(0))t\leq f(t)-f(0)\leq (f(1)-f(0))t, \ t\in [0,1].$

It follows that $f(t)=f(0)+(f(1)-f(0))t.$

If $f(1)<f(0),$ it is enough to apply the proved result to the fucntion $-f(t).$

Indian Statistical Institute, ISI BStat & BMath 2023 UGB Solutions & Discussions

ISI BMath & BSTAT 2023 Subjective Questions UGB: solutions and discussions

Past Year’s BMATH/BSTAT Papers & Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2023 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Answer Keys: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Question Paper: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGB Question Paper: Click Here

Problem 1.

Topic: Number Theory Difficulty level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 1 UGB Solution

Solution:

Video Solution:

Problem 2.

Topic: Trigonometric Sequences Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 2 UGB Solution.

Solution:

Video Solution:

Problem 3.

Topic: Geometry Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 3 UGB Solution.

Solution:

Video Solution:

Problem 4.

Topic: Combinatorics Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 4 UGB Solution.

Solution:

Video Solution:

Problem 5.

Topic: Combinatorics Recurrence Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 5 UGB Solution.

Solution:

Video Solution:

Problem 6.

Topic: Sequence, Real Analysis Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 6 UGB Solution.

Solution:

Video Solution:

Problem 7.

Topic: Polynomials, Mathematical Induction, Complex Numbers Difficulty Level: Hard

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 7 UGB Solution.

Solution:

Video Solution:

Problem 8.

Topic: Real Analysis, Differentiation Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2023 Problem 8 UGB Solution.

Solution :

Video Solution:

Past Year’s BMATH/BSTAT Papers & Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2023 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Answer Keys: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Question Paper: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGB Question Paper: Click Here

Topic: Number Theory
Difficulty level: Medium

Topic: Trigonometric Sequences
Difficulty Level: Medium

Topic: Geometry
Difficulty Level: Medium

Topic: Combinatorics
Difficulty Level: Easy

Topic: Combinatorics Recurrence
Difficulty Level: Medium

Topic: Sequence, Real Analysis
Difficulty Level: Medium

Topic: Polynomials, Mathematical Induction, Complex Numbers
Difficulty Level: Hard

Topic: Real Analysis, Differentiation
Difficulty Level: Medium