The 22nd week of the Kankinara Faculty and Training Programme, held on March 8th, 2026, focused on establishing a strong foundation in core academic competencies. The week’s sessions were designed to help faculty assess and reinforce fundamental skills through structured testing and consistent verbal practice, ensuring that the learning journey remains both measurable and progressive.

All participants engaged in a daily English introductory session to begin the day. This recurring practice is designed to build verbal fluency and confidence through repetition, creating a lively and interactive environment. By normalizing English conversation as a daily habit, the programme helps students and teachers alike overcome communication barriers, keeping them actively involved and motivated to express their ideas.

A key highlight of the week was the comprehensive Literacy and Numeracy Assessment. Both students and teachers from Kankinara participated in this evaluation, which utilized a formally structured question paper to gauge proficiency levels. This session was crucial for reinforcing the importance of foundational knowledge, demonstrating how complex progress can be tracked through contextual understanding and data-driven evaluation. It provided faculty with a clear roadmap for future teaching strategies while encouraging students to take pride in their academic growth.

Following the academic evaluation, the participants returned to their basic embroidery training. Continuing the established tradition of craft-based learning, this session allowed students to further refine their needlework skills. The repetition of these stitching techniques encourages patience, creativity, and fine motor precision. By integrating this manual craft alongside rigorous testing, the programme shows how needlework can be used as a calming and productive balance to academic pressure.

Overall, the 22nd week of the Kankinara Faculty Training Programme was a period of focused evaluation and steady refinement. The participants felt a sense of purpose through the literacy and numeracy challenges, and the atmosphere remained energetic and positive throughout the day. The week successfully supported faculty development by proving that consistent testing, combined with creative practice, makes for a well-rounded and effective classroom experience.

The 18th week of the Kankinara Faculty and Training Programme focused on collaborative classroom engagement and creative learning through activity-based sessions. The week’s sessions were designed to help faculty experience how to conduct integrated classes where language learning and hands-on activities go together, keeping students actively involved and motivated.

Two faculty members from Kankinara attended the session and brought two of their students as well. All participants attended the class together, creating a lively and interactive learning environment. During the session, they selected a short English story, prepared a set of question–answer items from it, and conducted a light grammar-focused discussion. This helped reinforce how reading comprehension can be effectively linked with structured Q&A practice and basic grammatical understanding.

A key highlight of the week was the special session conducted by Ms. Ujjoyini, who introduced participants to simple handcraft activities. Students were guided to create jewellery using ghoongroos and other decorative materials, along with colourful paper-based craft work. The session encouraged creativity, patience, and fine motor skills, while also showing how craft-based learning can be incorporated into classroom engagement.

Overall, the 18th week of the Kankinara Faculty Training Programme brought a refreshing blend of academic and creative learning. The participants, especially the students, felt joyful and happy during the sessions, and the atmosphere remained energetic and positive throughout. The week supported faculty development by demonstrating how innovative activities and interactive teaching can make classroom learning more enjoyable and effective.

The 17th week of the Kankinara Faculty and Training Programme focused on strengthening digital documentation skills through practical training on Google Docs. The sessions were designed to help participants create structured academic materials for classroom use in an organised and professional manner.

Participants were guided on how to use Google Docs to prepare question–answer papers for their students. The training covered document formatting, typing and editing text, inserting headings, and organising content clearly. Special emphasis was placed on creating well-structured Q&A papers that are easy for students to understand and follow.

The sessions also included guidance on setting appropriate question paper patterns. Participants learned how to design balanced question papers by including different types of questions, such as short answers, descriptive questions, and grammar-based exercises. This helped them understand how to align question patterns with learning objectives and student levels.

Overall, the 17th week of the Kankinara Faculty Training Programme contributed to participants’ ability to design effective academic resources. By learning to use Google Docs for question paper creation and understanding question-setting patterns, the sessions supported their professional growth and classroom readiness.

The 16th week of the Kankinara Faculty and Training Programme focused on developing social awareness, values, and language skills through lessons on basic etiquette and moral learning. The sessions were designed to help participants understand appropriate behaviour in everyday and professional settings while strengthening their reading and comprehension abilities.

Participants were introduced to basic etiquette, including respectful communication, polite language use, and appropriate conduct in academic and social environments. These discussions encouraged reflection on how manners and behaviour influence personal growth, relationships, and workplace interactions.

The programme also included the study of moral stories from the prescribed textbook. Through guided reading and discussion, participants explored the central messages of the stories and reflected on values such as honesty, responsibility, and empathy. This activity supported the development of reading comprehension while also promoting ethical thinking and self-awareness.

In addition, trainers conducted a review of students’ notebooks to assess regular practice, understanding, and handwriting. This copy-checking exercise helped identify learning gaps, reinforce discipline, and encourage consistent effort among participants.

Overall, the 16th week of the Kankinara Faculty Training Programme was centred on character development, language practice, and learning discipline. By combining etiquette lessons, moral storytelling, and academic review, the sessions contributed to the participants’ holistic growth and continued progress.

UGA

1. (A)	2. (C)	3. (B)	4. (B)	5. (A)	6. (D)
7. (D)	8. (D)	9. (B)	10. (B)	11. (C)	12. (D)
13. (C)	14. (D)	15. (B)	16. (D)	17. (D)	18. (C)
19. (C)	20. (A)	21. (A)	22. (C)	23. (B)	24. (A)
25. (C)	26. (B)	27. (A)	28. (B)	29. (D)	30. (B)

Problem 1

In the \(x y\)-plane, the curve \(3 x^3 y+6 x y+2 x y^3=0\) represents

(A) a pair of straight lines
(B) an ellipse
(C) a pair of straight lines and an ellipse
(D) a hyperbola

Solution

\(3 x^3 y+6 x y+2 x y^3=0\)

\(x y\left(3 x^2+2 y^2+6\right)=0\)

\(\Rightarrow x y=0\)

OR

\(3 x^2+2 y^2+6=0\)

no solution as \(3 x^2+2 y^2 \geqslant 0\)

\(\therefore \quad x y=0\)

\(\Rightarrow x=0\) or \(y=0\)

Answer: Pair of Straight Lines.

Problem 2

Let \(I=\int_3^5 \frac{1}{1+x^3} d x\). Then

(A) \(I<\frac{1}{64}\)

(B) \(I>\frac{1}{13}\)
(C) \(\frac{1}{63}<I<\frac{1}{14}\)

(D) \(I>\frac{1}{2}\left(\frac{1}{14}+\frac{1}{63}\right)\)

Solution

We know that

\(m(a-b) \leqslant \int_b^a p(x) d x \leqslant M(a-b)\)

Where \(m=\min . f(x)\) and

\[
B \leqslant x \leqslant a
\]

\(M=m o x \quad f(x)\)

\(b \leqslant x<a\)

So, \(2 x \frac{1}{126}<\int_3^5 f(x) d x \quad \leqslant 2 x \frac{1}{28}\)

\(\Rightarrow \frac{1}{ 63}<\int_3^5 f(x) d x<\frac{1}{14}\)

Problem 3

The coefficient of \(x^8\) in \((1-3 x)^6\left(1+9 x^2\right)^6(1+3 x)^6\) is

(A) \(-3^9 \times 5\)
(B) \(3^9 \times 5\)
(C) \(-3^8 \times 5\)
(D) \(3^8 \times 5\)

Solution:

\((1-3 x)^6\left(1+9 x^2\right)^6(1+3 x)^6\)

=\(\left(1-9 x^2\right)^6\left(1+9 x^2\right)^6\)

=\(\left(1-81 x^4\right)^6\)

\(\therefore\binom{6}{0} 1^0 \cdot\left(-81 x^4\right)^6+\cdots+\binom{6}{4} 1^4\left(-81 x^4\right)^2+\cdots\)

=\(5 \times 3^9 \times x^8\)

Problem 4

Consider two events \(A\) and \(B\) with probabilities \(P(A)\) and \(P(B)\) respectively such that \(0<P(A), P(B)<1\). Define

\[
P(A \mid B)=\frac{P(A \cap B)}{P(B)}
\]

Consider the following statements.

(I) \(P\left(A \mid B^c\right)+P(A \mid B)=1\).
(II) \(P\left(A^c \mid B\right)+P(A \mid B)=1\).

Then, in general,

(A) (I) is true and (II) is false
(B) (I) is false and (II) is true
(C) both (I) and (II) are true
(D) both (I) and (II) are false

Solution

Problem 5

Let \(f(x)=7 x^{11}+4 x^3-3\). Then \(f\) has

(A) exactly 1 real root
(B) exactly 3 real roots
(C) exactly 5 real roots
(D) 11 real roots

Solution

\(f(x)=7 x^{11}+4 x^5-3\)

\(f^{\prime}(x)=77 x^{10}+20 x^4\)

\(\therefore f^{\prime}(x)>0 \quad) if (\quad x \neq 0\)

\[
\lim_{x \to \infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to -\infty} f(x) = -\infty
\]

By intermediate value theorem there is at least 1 root

\(\therefore f\) is (almost) monotone hence there is exactly 1 root.

Problem 6

Let \(A\) be an \(m \times n\) matrix with the \((i, j)\) th entry given by the real number \(a_{i j}, 1 \leq i \leq m, 1 \leq j \leq n\). Let

\[
a = \max_{1 \leq j \leq n} \left( \min_{1 \leq i \leq m} a_{ij} \right)
\quad \text{and} \quad
\beta = \min_{1 \leq j \leq n} \left( \max_{1 \leq i \leq m} a_{ij} \right).
\]

Then

(A) \(\alpha \leq \beta\) but not necessarily \(\alpha=\beta\)
(B) \(\beta \leq \alpha\) but not necessarily \(\alpha=\beta\)
(C) \(\alpha=\beta\)
(D) nothing can be said in general

Solution

Problem 7

Consider the cyclic quadrilateral \(A B C D\) given below.

Assume that \(A B=B C, A D=C D\), and \(\frac{A B}{A D}=\frac{1}{3}\). Let \(\theta=\angle A D C\). Then \(\cos \theta\) is equal to

(A) \(\frac{1}{5}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{3}{5}\)
(D) \(\frac{4}{5}\)

Solution

\(x^2+x^2-2 x \cdot x \cdot \cos (180-\theta)\)

=\((3 x)^2+(3 x)^2-2 \cdot 3 x \cdot 3 x \cos \theta\)

\(\Rightarrow 2 x^2+2 x^2 \cos \theta=18 x^2-18 x^2 \cos \theta\)

\(\Rightarrow 20 x^2 \cos \theta=16 x^2\)

\(\cos \theta=\frac{16 x^2}{20 x^2}\)

=\(\frac{4}{5}\)

Problem 8

Let \(A={(x, y): x, y \in[0,1]}\) and \(B={(x, y): x, y \in[0,2]}\). Define \(f: A \rightarrow B\) by \(f(x, y)=\left(x^2+y, x+y^2\right)\). Then \(f\) is

(A) one-to-one but not onto
(B) onto but not one-to-one
(C) both one-to-one and onto
(D) neither one-to-one nor onto

Solution

\(A={(x, y): x, y \in[0,1]}\)

\(B={(x, y): x, y \in[0,2]}\)

\(f: A \rightarrow B\)

\(f(x, y)=\left(x^2+y ; x+y^2\right)\)

Not one -on - one

as \(\begin{array}{r}(1,0) \ (0,1)\end{array}\) --> both map to \((1,1)\)

Onto, Examina the presimage of \([0,2]\)

\(x^2+y=0\)

\(x+y^2=2\).

\(\because \quad 0 \leqslant x, y \leqslant 1\)

\(\therefore x^2+y=0 \Rightarrow \quad x=0 \quad y=0\)

But then \(x+y^2=2\) is not true hence no solution.

Problem 9

The number of ordered pair \((a, b)\) of positive integers with \(a<b\) satisfying \(a^2+b^2=2025\) is

(A) 0
(B) 1
(C) 2
(D) 6

Solution

\(a^2+b^2=2.25\)

\(a<b\)

\(\therefore \quad 2025=5^2 \cdot 3^4=45^2\)

\(0,1,4,7 \quad \bmod 9\)

None of these work except \(0+0\)

\(\therefore \quad a^2 \equiv 0 \mathrm{mod} 9\)

\(b^2 \equiv 0\) mod 9

\(\therefore a \leq 0 \bmod a\)

or \(a \equiv 6 \bmod 9\).

\(\therefore\) Both a and 6 are divisible by 3

\(\left(3 a_1\right)^2+\left(3 b_1\right)^2=25 \times 81\)

\(\Rightarrow a_1^2+b_1^2=9 \times 25\)

\(\Rightarrow \quad a_2{ }^2+b_2{ }^2=25\)

\(\therefore a_2=3, b_2=9\)

\(\therefore a=27 \quad b=36\)

Problem 10

Twelve boxes are placed along a circle. In each box, \(1,2,3\) or 4 balls are put such that the total number of balls in any 4 consecutive boxes is same. The number of ways this can be done is

(A) 4 !
(B) \(4^4\)
(C) \((4!)^3\)
(D) \((4!)^4\)

Solution

once we choose number of balls in first 4 boxes, the remaining choices become fixed.

For each of the four boxes, we have four choices,

\(\therefore 4^4\) cases in total

Problem 11

Let \(a_0=0, a_1=1\) and \(a_n=5 a_{n-1}+a_{n-2}\) for \(n \geq 2\). Then the value of the determinant

is

(A) -1
(B) \(-5^{101}\)
(C) 1
(D) \(5^{101}\)

Solution

Problem 12

The lengths of the three sides of a right angled triangle are geometric progression. The smallest angle of the triangle is

(A) \(\tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\)
(B) \(\cos ^{-1}\left(\frac{2}{\sqrt{5}-1}\right)\)
(C) \(\sin ^{-1}\left(\frac{2}{\sqrt{5}-1}\right)\)
(D) \(\sin ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\)

Solution

Problem 13

Consider the following statements about two similar triangles \(\Delta_1), and (\Delta_2\).

\(S_1\): Lengths of the sides of \(\Delta_1\) are in arithmetic progression
\(S_2\). Lengths of the sides of \(\Delta_1\) are in geometric progression.

\(S_3\). Lengths of the sides of \(\Delta_2\) are in arithmetic progression.

\(S_4\). Lengths of the sides of \(\Delta_2\) are in geometric progression.

Then

(A) \(S_1\) implies \(S_3\), but \(S_2\) does not imply \(S_4\)
(B) \(S_1\) does not imply \(S_3\), but \(S_2\) implies \(S_4\)
(C) \(S_1\) implies \(S_3\), and \(S_2\) implies \(S_4\)
(D) \(S_1\) does not imply \(S_3\), and \(S_2\) does not imply \(S_4\)

Solution

Let \((\sigma, b, c)\) and \(\left(a^{\prime}, b^{\prime}, c^{\prime}\right)\) are

Sides of two similar triangles sit.

\(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}\)

and \(b=a r\) and \(a-a x^2\)

Ther, \(\frac{b^{\prime}}{a^{\prime}}=\frac{c\prime}{b\prime}=r\) also

So, \(S_2\) implips \(S_4\)

Now, let

\(b=a+d\) \(c=a+d\)

so, \(a^{\prime}: a k, b^{\prime}:(a+d) k\).

So, \(a^{\prime}, b^{\prime}, c^{\prime}\), are also in AP

so, \(S_2\) implies \(S_4\)

Problem 14

For each \(n \geq 1\), let \(a_n\) and \(b_n\) be real numbers such that \(a_n \neq 0\) \(\frac{a n d}{c} b_n \neq 0\). Let

\(\left(a_n+i b_n\right)^n=n\left(a_n+i b_n\right) \quad\) for all \(n \geq 6\).

Then

(A) no such (a_n, b_n) exist
(B) \(x_n=n^{\frac{1}{n+1}} \cos \frac{2 \pi}{n+1}, b_n=n^{\frac{1}{n+1}} \sin \frac{2 \pi}{n+1}\)
(C) \(a_n=n^{\frac{1}{n}} \cos \frac{2 \pi}{n}, b_n=n^{\frac{1}{n}} \sin \frac{2 \pi}{n}\)
(D) \(a_n=n^{\frac{1}{n-1}} \cos \frac{2 \pi}{n-1}, b_n=n^{\frac{1}{n-1}} \sin \frac{2 \pi}{n-1}\)

Solution

\(\left(a_n+i \cdot b_n\right)^n=n\left(a_n+i \cdot b_n\right)\).

\(\Rightarrow\left(a_n+i \cdot b_n^m\right)^{n-1}=n\).

As, \(\quad a_n+i b_n \neq 0\)

So, \(a_n+i b_n=\)

\(\frac{1}{n-1}\left[\cos \left(\frac{2 \pi k}{n-1}\right)\right.\) \(fi (\sin \left(\frac{2 \pi k}{n-1}]\right)\)

where \(0 \leqslant K<(n-1)\)

So, option (D)

Problem 15

The lengths of the two adjacent sides of a parallelogram are 2 cm and 3 cm . The length of one diagonal is \(\sqrt{19} \mathrm{~cm}\). Then the length of the other diagonal is

(A) \(\sqrt{5} \mathrm{~cm}\)
(B) \(\sqrt{7} \mathrm{~cm}\)
(C) \(\sqrt{15} \mathrm{~cm}\)
(D) \(\sqrt{21} \mathrm{~cm}\)

Solution

\(d_1^2=a_b^2+b_b^2-2 a b \cos \theta\)

\(d_2^2=a^2+b^2-2 a b \cos \left(180^{\circ}-\theta\right)\)

\(\therefore d_1^2+d_2^2=2\left(a^2+b^2\right)\)

\(\Rightarrow 19+d_2^2=2(4+9)\)

\(\Rightarrow \quad d_2^2=26-19 \Rightarrow d_2^2=7\)

\(=d_2=\sqrt{7}\)

Problem 16

Let \(f\) \((x) = \frac{1}{1+|x-1|}+\frac{1}{1+|x+1|}\). Then the function \(f\) has
(A) neither a local maximum nor a local minimum
(B) a local minimum at \(x=0\), but no local maximum
(C) local maxima at \(x= \pm 1\), but no local minimum
(D) a local minimum at \(x=0\) and local maxima at \(x= \pm 1\)

Solution

\(f(x)=\frac{1}{1+|x-1|}+\frac{1}{1+|x+1|}\)

when, \(x<-1\),

\(f(x)=\frac{1}{2-x}-\frac{1}{x} \Rightarrow f^{\prime}(x)=\frac{1}{(2-x)^2}+\frac{1}{x^2}>0\)

Hence, (f(x)) is increasing

\(f(x)=\frac{1}{x+2}+\frac{1}{2-x} \Rightarrow f^{\prime}(w)=\frac{1}{(2-w)^2}-\frac{1}{(x+2)^2}\)

Critical point, at \(x=0\)

\(f^{\prime}(n)>0) for (0 \leqslant n<1\).

\(f^{\prime}(x)<0) for (-1<x<0\)

when, \(n>1\).

\(f(n)=\frac{1}{n}+\frac{1}{n+2}\)

\(=-\frac{1}{n^2}-\frac{1}{(n+2)^2}<0\)

\(f(x)\) is decreasing.

at \(x= \pm 1 \quad f(x)\) is changing it's sign from positive to

& at \(x=0\) it is changing from negative to positive.

Hence, option (D)

Problem 17

\(f(x) =
\begin{cases}
-1, & \text{if } x < 0,
\ 0, & \text{if } x = 0,
\ 1, & \text{if } x > 0.
\end{cases}\)

Then the function \(F\) defined by \(F(x)=\int_{-5}^x f(t) d t\) is

(A) not continuous
(B) continuous, but nowhere differentiable
(C) differentiable everywhere
(D) differentiable everywhere except at 0

Solution

Problem 18

Let
\(L=\lim _{n \rightarrow \infty}(n+100)^{\frac{5}{5 g_e(n-50)}}\).
Then
(A) \(2 \leq L \leq 16\)
(B) \(16 \leq L \leq 32\)
(C) \(32 \leq L<243\)
(D) \(L>243\)

Solution

\(L=\lim _{n \rightarrow \infty}(n+100) \frac{5}{\ln (n-\infty)}\)

\(\Rightarrow \ln L=\frac{5 \ln (n+100)}{\left.\lim _{n \rightarrow \infty} \frac{\ln (n-50)}{\ln (n-5}\right)}=5\)

\(L=e^5\)

\(2<e<3\)

\(32<L<243\)

Problem 19

Let \(a, b, c, d\) be positive integers such that the product abcd (=999). Then the number of different ordered 4 -tuples \((a, b, c, d)\) is

(A) 20
(B) 48
(C) 80
(D) 84

Solution

\(999=3^3 \times 27\)

Hence the number of non-negetive integer solution

\(\binom{3+4-1}{4-1}\binom{1+4-1}{4-1}\)

\(=20 \times 4\)

\(=80\)

Problem 20

Let \(|x|\) denote the greatest integer less than or equal to \(x \in \mathbb{R}\) and \(|x|\) has its usual meaning, that is, \(|x|=x\) if \(x \geq 0\), and \(|x|=-x\), if \(x<0\). Then the value of the integral

\(\int_{-2}^1([x]+2)^{|x|} d x\)

is
(A) \(1+\frac{1}{\log _e 2}\)
(B) \(1+\log _e \frac{1}{2}\)
(C) \(2-\log _e 2\)
(D) none of the above

Solution

\(\int_{-2}^1([x]+2)^{|x|} d x\)

\(=\int_0^1 2^x d x+\int_{-1}^0 1^{-x} d x\)

\(f \int_{-2}^{-1} 0 \cdot d x\)

\(=\left[\frac{2 x}{\ln 2}\right]_0^1+1.1\)

\(\left(\frac{1}{\ln^2}+1\right)\)

Problem 21

For a real number \(x\), let \(f(x)=\int_{-20}^{20} g(t) g(x-t) d t\), where
\(g(x)= \begin{cases}1, & \text { if } x \in[0,1] \ 0, & \text { otherwise }\end{cases}\)

Then \(f(x)\) is equal to
(A) \(\begin{cases}x, & \text { if } x \in[0,1], \ 2-x, & \text { if } x \in[1,2], \ 0, & \text { otherwise }\end{cases}\)
(B) \(\begin{cases}1+x, & \text { if } x \in[0,1), \ 1-x, & \text { if } x \in[1,2), \ 0, & \text { otherwise }\end{cases}\)
(C) \(\begin{cases}1, & \text { if } x \in(-20,20), \ 0, & \text { otherwise }\end{cases}\)
(D) none of the above

Solution

Now, if \(x<0\) or \(x>2\)then the integration becomes 0 .

if \(\quad 0 \leq x \leq 1\)

Then \(f(x)\)

\(=\int_{x-1}^x g(t) d t\)

\(=\int_0^x d t=x\)

\(=\int_{-20}^{20} \theta(t) g(x-t) d t\)

\(=\int_0^1 g(x-t) d t\)

\(=\int_0^1 g(t+x-1) d t\)

\(=\int_{x-1}^x g(t) d t\)

it \(10 \leqslant x \leqslant 2\)

\(f(x): \int_{x-1}^x g(t) d t\)

\(=\int_{x-1}^1 d t=(2-x)\)

Problem 22

Let \(n \geq 3\). There are \(n\) straight lines in a plane, no two of which are parallel and no three pass through a common point. Their points of intersection are joined. Then the number of fresh line segments thus created is
(A) \(\frac{n(n-1)(n-2)}{8}\)
(B) \(\frac{n(n-1)(n-2)(n-3)}{6}\)
(C) \(\frac{n(n-1)(n-2)(n-3)}{8}\)
(D) none of the above

Solution

The lines intersect at \(\binom{n}{2}\) different points But there are also \((n-1)\) points in part of line

So, Total fresh line

\(=\binom{\binom{n}{2}}{2}-n\binom{n-1}{2}\)

\(=\left(\frac{n(n-1)}{2}\right)-n \cdot \frac{(n-1)(n-2)}{2}\)

\(=\frac{n(n-1)(n-2)(n+1)}{8}-\frac{n(n-1)(n-2)}{2}\)

\(=\frac{n(n-1)}{8} \cdot\left(n^2-n-2-4 n+\theta\right)\)

\(=\frac{n(n-1)(n-2)(n-3)}{8}\)

Problem 23

In a certain test there are \(n\) questions. At least \(i\) questions were wrongly answered by \(2^{n-i}\) students, where \(i=1,2, \ldots, n\). If the total number of wrong answers given by all students is 2047 , then (n) is equal to
(A) 10
(B) 11
(C) 12
(D) 13

Solution

At least \(i\) questions were wrongly answered by \(2^{n-i}\) students.

\(\therefore\) At least wrong answers

\(=\) Exactly \(n-3\) wrong answers +

Exactly \(n-2\) wrong answers+

Exactly \(n-1\) wrong answers+

Exactly \(n\) wrong answers

\(\therefore\) Exactly: questions wrong \(=2^{n-1}-2^{n-2}\) Exactly 2 questions wrong \(=2^{n-2}-2^{n-3}\)

\(\therefore\) Total number of wrongs

\[
\begin{aligned}
& 1\left(2^{n-1}-2^{n-2}\right)+2\left(2^{n-2}-2^{n-3}\right)+\cdots+n\left(2^1-2^0\right) \
& =2^{n-1}+2^{n-2}+2^{n-3}+\cdots+2^1+2^0
\end{aligned}
\]

\(\therefore 2^0+2^1+2^2+\cdots+2^{n-3}+2^{n-2}+2^{n-1}=2047\)

\(\Rightarrow \frac{2^0\left(2^n-1\right)}{2-1}=2047\)

\(\Rightarrow \quad 2^n-1=2047 \Rightarrow n=11\)

Problem 24

Let \(n\) be a positive integer. The value of \(\sum_{k=0}^n \tan ^{-1} \frac{1}{k^2+k+1}\) is
(A) \(\tan ^{-1}(n+1)\)
(B) \(\tan ^{-1}\left(\frac{1}{n+1}\right)\)
(C) \(\tan ^{-1} n\)
(D) \(\tan ^{-1}\left(\frac{1}{n}\right)\)

Solution

\(=\sum_{k=0}^n \tan ^{-1} \frac{1}{k^2+k+1}\)

\(=\sum_{k=0}^n \tan ^{-1} \frac{1}{k(k+1)+1}\)

\(=\sum_{k=0}^n \tan ^{-1} \frac{(k+1)-k}{k(k+1)+1}\)

\(=\sum_{k=0}^n\left[\tan ^{-1}(k+1)-\tan ^{-1}(k)\right]\)

\(=\tan ^{-1}(n+1)\)

Problem 25

Let \(d\) be the side length of the largest possible equilateral triangle that can be put inside a square of side length 1 . Then
(A) \(d<1\)
(B) \(d=1\)
(C) \(1<d<\frac{2}{3^{1 / 4}}\)
(D) \(d \geq \frac{2}{3^{1 / 4}}\)

Problem 26

Let \(f(x)=\left(x^2+18\right)(x-4) x(x+4)-2\). Then
(A) \(f\) has exactly one real root
(B) \(\int\) has exactly 3 distinct real roots
(C) \(f\) has 5 distinct real roots
(D) \(f\) has a repeated root

Solution

\(f(x)=\left(x^2+18\right) x(x-4)(x+4)-2\)

\(f^{\prime}(x)=5 x^4+6 x^2-288\)

\(f^{\prime}\) has two real roots

But \(f\) can have at most three real roots. But \(f(-4)<0 \quad f(-3)>0 \quad f(0)<0 \quad f(4)<0 \quad f(5)>0\)

\( \therefore \) it has 3 real roots (B).

Problem 27

Yet \(k\) be a positive integer and \(f(x)=e^x-1\). Then

\[\lim _{x \rightarrow 0} \frac{f(x)+f\left(\frac{x}{2}\right)+f\left(\frac{x}{2^2}\right)+\cdots+f\left(\frac{x}{2^k}\right)}{x}\]

(x) \(2-\frac{1}{2^k}\)
(B) \(2-\frac{1}{2^{k+1}}\)
(C) \(k\)
(D) \(2^{k+1}-1\)

Solution

\(f(x)=e^x-1\)

\(\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\frac{e^{x / 2}-1}{x}+\frac{e^{x / 2^2}-1}{x}+\cdots+\frac{e^{x / 2^x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\frac{e^{x / 2}-1}{x / 2} \times \frac{1}{2}+\frac{e^{x / 2^2}-1}{x / 2^2} \times \frac{1}{2^2}+\cdots+\frac{e^{x / 2^x}-1}{x / 2^x} \times \frac{1}{2^x}\)

=\(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^x}=-\frac{1\left(\frac{1}{2^{x+1}}-1\right)}{1 / 2}\)

\(=-2\left(\frac{1}{2^{x+1}}-1\right)\)

\(=-\left(\frac{1}{2^k}-2\right)=2-\frac{1}{2^k}\)

Problem 28

Let

\[a_n=\frac{n^2}{\sqrt{n^6+1}}+\frac{n^2}{\sqrt{n^6+2}}+\cdots+\frac{n^2}{\sqrt{n^6+n}}, \quad n \geq 1\]

Then \(\lim _{n \rightarrow \infty} a_n\)
(A) does not exist
(B) is equal to 1
(C) is equal to \(e\)
(D) is equal to \(\frac{1}{e}\)

Solution

\(a_n=\frac{n^2}{\sqrt{n^6+1}}+\frac{n^2}{\sqrt{n^6+2}}+\cdots+\frac{n^2}{\sqrt{n^6+n}}\)

\(\sqrt{n^6+0} \leq \sqrt{n^6+r} \leq \sqrt{n^6+n}\)\

\(\frac{1}{\sqrt{n^6}} \leqslant \frac{1}{\sqrt{n^6+r}} \leqslant \frac{1}{\sqrt{n^6+n}}\)

\(\frac{n^2}{n^3}-\leqslant \frac{n^2}{\sqrt{n^6+r}} \leqslant \frac{n^2}{\sqrt{n^6+n}}\)

\(\sum_{r=1}^n \frac{n^2}{n^3} \leqslant \sum_{r=1}^n \frac{n^2}{\sqrt{n^6+r}} \leqslant \sum_{n=1}^n \frac{n^2}{\sqrt{n^6+n}}\)

\(\frac{n^3}{n^3} \leqslant \sum_{r=1}^n \frac{n^2}{\sqrt{n^6+r}} \leqslant \frac{n^3}{\sqrt{n^6+n}}\)

\(1 \leqslant \sum_{r=1}^n \frac{n^2}{\sqrt{n^6+r}} \leqslant \frac{n^{B^3}}{\sqrt{n^6+n}}\)

\[
1 \leq \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n^2}{\sqrt{n^6 + r}} \leq 1 = \lim_{n \to \infty} \frac{n^3}{\sqrt{n^6 + n}}
\]

So, \(\lim _{n \rightarrow \infty} a_n=1\)

Problem 29

A subset ${u_1, u_2, u_3, u_4, u_5}$ of the first 90 positive integers can be selected in $\binom{90}{5}$ ways. Let $u_{\text{max}} = \max{u_1, u_2, u_3, u_4, u_5}$ and $u_{\text{min}} = \min{u_1, u_2, u_3, u_4, u_5}$. Then the arithmetic mean of $u_{\text{max}} + u_{\text{min}}$ over all such subsets is

(A) 45
(B) 46
(C) 89
(D) 91

Solution

No. of subsets whose max element 90 is \(\quad 90\binom{89}{4}\)

When \(u_{\text {max }}=89 \rightarrow 89\binom{88}{4}\)

When \(u_{\text {max }}^{\text {max }}=88 \rightarrow 88\binom{87}{4}\)

" " "

" " "

" " "

when \(u_{\max }=5 \rightarrow 5\binom{4}{4}\)

Now, when \(u_{\min }=1 \rightarrow 1\binom{89}{4}\).

when \(u_{\min }^{\min }=2 \rightarrow 2\binom{88}{4}\)

when \(u_{\text {min }}=3 \rightarrow 3\binom{87}{4}\)

" " "

" " "

when \(u_{\min }=86 \rightarrow 86\binom{4}{4}\)

\(\therefore\) Sum of all possible values

\(90\binom{89}{4}+89\binom{88}{4}+\left(88\binom{87}{4}+\cdots+5\binom{4}{4}\right.\)

\(+1\binom{89}{4}+2\binom{88}{4}+\cdots+86\binom{4}{4}\)

\(=91\left[\binom{89}{4}+\binom{88}{4}+\binom{87}{4}+\cdots+\binom{4}{4}\right]\)

\(=91\binom{90}{5} \quad\) [Hockey-Stick Identity]

\(\therefore A M\) of all \(u_{\text {max }}+u_{\text {min }}=\frac{91\binom{90}{5}}{\binom{90}{5}}\)

\(=91\)

Problem 30

Let

\[a_n=\frac{2^3-1}{2^3+1} \times \frac{3^3-1}{3^3+1} \times \cdots \times \frac{n^3-1}{n^3+1}, \quad n \geq 2\]

Then \(\lim _{n \rightarrow \infty} a_n\)
(A) does not exist
-(B) is equal to \(\frac{2}{3}\)
(C) is equal to 1
(D) is equal to \(\frac{1}{2}\)

Solution

\(\frac{r^3-1}{r^3+1}=\frac{r-1}{r+1} \times \frac{r^2+r+1}{r^2-r+1}\)

\(\left(\frac{r-1}{r+1}\right) \times \frac{(r+1)^2-(r+1)+1}{\left(r^2-r+1\right)}\)

\(\prod_{r=2}^n\left(\frac{r-1}{r+1}\right) \prod_{r=2}^n \frac{(r+1)^2-(r+1)+1}{r^2-r+1}\)

\(\prod_{r=2}^n\left(\frac{r-1}{r+1}\right)=\frac{1}{3} \times \frac{2}{4} \times \frac{8}{5} \times \frac{4}{6} \times \frac{5}{7} \times \frac{8}{8} \cdots=i\)

\(\prod_{r=2}^n \frac{(r+1)^2-(r+1)+1}{r^2-r+1}=\frac{3^2-3+1}{2^2-3+1} \times \frac{4^2-4+1}{3^2-3+1} \times \frac{5^2-5+1}{4^2-4+1} \times \cdots\)

= \(\frac{1}{4-2+1}=\frac{1}{3}\) (ii)

So from (i) and (ii) we get \(\frac{2}{3}\)

UGB

Problem 1

Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) is differentiable and \(\left|f^{\prime}(x)\right|<\frac{1}{2}\) for all \(x \in \mathbb{R}\). Show that for some \(x_0 \in \mathbb{R}, f\left(x_0\right)=x_0\).

Solution

Problem 2

If the interior angles of a triangle (A B C) satisfy the equality,

\[
\sin ^2 A+\sin ^2 B+\sin ^2 C=2\left(\cos ^2 A+\cos ^2 B+\cos ^2 C\right)
\]

prove that the triangle must have a right angle.

Problem 3

Suppose \(f:[0,1] \rightarrow \mathbb{R}\) is differentiable with \(f(0)=0\). If \(\left|f^{\prime}(x)\right| \leq f(x)\) for all \(x \in[0,1]\), then show that \(f(x)=0\) for all \(x\).

Problem 4

Let \(S^1={\{z \in \mathbb{C}| | z \mid=1}\}\) be the unit circle in the complex plane. Let Let \(f: S^1 \rightarrow S^1\) be the map given by \( f(z)=z^2 \). We define \(f^{(1)}:=f\) and \(f^{(k+1)}:=f \circ f^{(k)}\) for \(k \geq 1\). The smallest positive integer \(n\) such that \(f^{(n)}(z)=z\) is called the period of \(z\). Determine the total number of points in \(S^1\) of period 2025.
(Hint: \(2025=3^4 \times 5^2) \)

Problem 5

Let \(a, b, c\) be nonzero real numbers such that \(a+b+c \neq 0\). Assume that

\[
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}
\]

Show that for any odd integer \(k\),

\[
\frac{1}{a^k}+\frac{1}{b^k}+\frac{1}{c^k}=\frac{1}{a^k+b^k+c^k}
\]

Problem 6

Let \(\mathbb{N}\) denote the set of natural numbers, and let \(\left(a_i, b_i\right)\), \(1 \leq i \leq 9\), be nine distinct tuples in \(\mathbb{N} \times \mathbb{N}\). Show that there are three distinct elements in the set \({2^{a_i} 3^{b_i}: 1 \leq i \leq 9}\) whose product is a perfect cube.

Problem 7

Consider a ball that moves inside an acute-angled triangle along a straight line, until it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence (=) angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

Problem 8

Let \(n \geq 2\) and let \(a_1 \leq a_2 \leq \cdots \leq a_n\) be positive integers such that \(\sum_{i=1}^n a_i=\Pi_{i=1}^n a_i\). Prove that \(\sum_{i=1}^n a_i \leq 2 n\) and determine when equality holds.