Hello, guys!

This contest is the first JEE contest on Brilliant. The objective of this game is to get well prepared for JEE-Advanced 2016 preparation having fun as well. There is so much to learn from each other, and the contest is an approach to accomplish the same. Each Brilliant user is invited to participate in the contest under the following rules:

- I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
- You may only post a solution of the problem
**below**the thread of problem and post your proposed problem in**a new thread**. Put them separately. - Please make a
**substantial comment**. - Make sure you
**know**how to solve your problem before posting it. In case there is no one can answer it within 24 hours, then you must post the solution, and you have a right to post another problem. - If the one who solves the last problem does not post his/her problem after solving it
**within a day**, then the one who has a right to post a problem is the last solver before him/her. - The scope of questions is only what is covered under JEE-Advanced syllabus. If you are unaware of the syllabus cover, You can check out the information brochure.
- You can use tricks/short methods/specific cases to post the solution of a problem only if they are logically valid.
- In the case of any argument or disagreement among the solvers, the decision/judgement is passed to me.
**DO NOT**ask the answer to the problem. Just post your**detailed solution**solution along with the answer.*It will be highly helpful to gain confidence in your problem solving and rock in JEE.*- Proof problems are not allowed.
You can post a problem only from Maths section.

**Please write the detailed solutions to the problems.**

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :[Post your solution here]

PROBLEM xxx (number of problem) :[Post your problem here]

Please try to post problems from all the spheres of JEE syllabus and **share** this note so that maximum users come to know about this contest and participate in it. (>‿◠)✌

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## Comments

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TopNewestHere is the first problem for the inauguration of the contest. It's an easy problem. All the best!

Problem 1:In a sequence of independent trials, the probability of success in one trial is \(\frac 14\). Find the probability that the second success takes place on or after the fourth trial.

Ritu Roy has solved this problem at the first place and provided the solution. Thanks!Log in to reply

Solution of problem 1

The required probability is

\(=_{ 1 }^{ 3 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 2 }.\frac { 1 }{ 4 } \quad +\quad _{ 1 }^{ 4 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 3 }.\frac { 1 }{ 4 } \quad +\quad _{ 1 }^{ 5 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 4 }.\frac { 1 }{ 4 } \quad +\quad ...\infty \\ =\frac { 9 }{ 256 } .24\\ =\boxed { \frac { 27 }{ 32 } } \)

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How did we get that 24? I used the reverse probability method 1 - 3C2.1/4.1/4.3/4 - 3C3.1/4.1/4.1/4 = 27/32

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Problem 10 :In \(\Delta \text{ABC}\), \(\text{AB} = \tan^{-1}\left(\sin \left(\sqrt{2} \right) \right)\) and \(\tan \left(\dfrac{\text{A}}{2} \right) = \ln(\pi) \tan \left(\dfrac{\text{B}}{2} \right) \)

Then the vertex \(\text{C}\) lies on

a) Ellipse

b) Parabola

c) Hyperbola

d) Straight Line

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@Aditya Kumar

I see that you've posted the next problem of contest i.e. Problem 11, but there is no solution of 10th problem by you. Can you clarify on this?

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I had posted the solution! Some careless mod has deleted it! I can't believe it.

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Can you please post that again?

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@Aditya Kumar ,your solution was incorrect.

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Problem 3:Let \(f\) be a twice differentiable function, such that \(f''(x)=-f(x)\) and \(f'(x)=g(x)\) , \(h(x)=[f(x)]^2+[g(x)]^2\). Given that \(h(5)=11\), evaluate \(h(10)\) .

Vighnesh Shenoy has solved this problem at the first place and provided the solution. Thanks!Log in to reply

Problem 51If \((a+\omega)^{-1}+(b+\omega)^{-1}+(c+\omega)^{-1}+(d+\omega)^{-1}=2\omega^{-1}\) and \((a+\omega')^{-1}+(b+\omega')^{-1}+(c+\omega')^{-1}+(d+\omega')^{-1}=2\omega'^{-1}\) ; where \(\omega\) and \(\omega'\) are the complex cube roots of unity, then what is the value of \((a+1)^{-1}+(b+1)^{-1}+(c+1)^{-1}+(d+1)^{-1}\) ?

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Problem 4:If \( n \in N \), evaluate the value of

\[ \dbinom{n}{0} + \dbinom{n}{4} + \dbinom{n}{8} + \ldots ,\]

where \( \dbinom{n}{r} =\dfrac{n!}{r!(n-r)!} \).

Akul Agrawal is the first person to solve this problem and provide the solution. Thanks!Log in to reply

\(\frac { { 2 }^{ \frac { n }{ 2 } +1 }\cos { \frac { n\pi }{ 4 } } +{ 2 }^{ n } }{ 4 } \)

put i, -i,1, -1 in expansion of (1+x)^n and add

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\( \displaystyle (1+x)^n = \sum_{r=0}^{n} \dbinom{n}{r} x^r\)

\( 1^{m} + (-1)^{m} + i^{m} + (-i)^{m} = 4 , m = 4k, \) else it is equal to \( 0 \)

Putting \( \displaystyle x=1, -1, i, -i\) and adding, we have,

\(\displaystyle \sum_{r=0}^{\infty} \dbinom{n}{4r} = \dfrac{(1+i)^n + (1-i)^n +2^n}{4}\)

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Problem 25:If \( p \) is an odd prime number, then \( \lfloor ( 2 + \sqrt{5})^{p} \rfloor - 2^{p+1} \) is always divisible by :

\( 1) 2p \)

\( 2) 3p \)

\( 3) p+1 \)

\( 4) 5p \)

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

\( (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \le (2+\sqrt{5})^{p} \le (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1 \)

Now,

\( (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \in I \)

\( (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1 \in I \)

\( \therefore \lfloor (2+\sqrt{5})^{p} \rfloor = (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \)

\( S(p) = \lfloor (2+\sqrt{5})^{p} \rfloor - 2^{p+1} = (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} - 2^{p+1} \)

\( \therefore S =2\displaystyle \sum_{r=0}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r} - 2^{p+1} \)

\( \therefore S = 2^{p+1} + \displaystyle 2\sum_{r=1}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r} - 2^{p+1} = \displaystyle 2\sum_{r=1}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r}\)

Clearly, \( S(p) \) is divisible by \( 2 \) and \( 5 \)

\( S(p) = 2\displaystyle \sum_{r=1}^{\frac{p-1}{2}}\frac{p}{2r}\dbinom{p-1}{2r-1}2^{p-2r}5^{r} \)

Now since p is a prime number, \( gcd(r,p) = 1 , r \in \left[1,p-1\right]\)

Therefore, \( S(p) \)is divisible by p.

Since \( S(p) \) is divisible by \( 2 , p ,5 \) it is divisible by \( 2p, 5p \)

\( S(5) = 1300 \)

\( S(5) \) is not divisible by \( 15 = 3\cdot5 \) and is also not divisible by \( 6 = 5 + 1 \)

Thus,

\( S(p) \) is not always divisible by \( p + 1 \) and \( 3p \)

The correct options are \( 1 \) and \( 4 \)

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Problem 36:Let \(a\) be a positive real number such that \(a^3 = 6(a + 1)\) then, find the nature of the roots of \(x^2 + ax + a^2 - 6 = 0\).

Both Mayank Chaturvedi and Vighnesh Shenoy have provided complete solution this problem. Thanks!Log in to reply

A graphical approach.

We have to see the nature of discriminant

\({ a }^{ 2 }-(4{ a }^{ 2 }-24)\\ or\quad 8-{ a }^{ 2 }\\ Now,\quad graph\quad of\quad { a }^{ 3 }-6a-6=f(a)\quad \\ takes\quad two\quad turns\quad at\quad a\quad =\sqrt { 2 } \quad and\quad -\sqrt { 2 } ,\quad \\ which\quad shows\quad a(root\quad of\quad function)\quad is\quad positive\quad and\quad at\quad least\quad \sqrt { 2 } \\remember\quad f(x)\quad is\quad continuous.\\ at\quad a=\sqrt { 8 } \quad the\quad f(a)\quad is\quad negative.\\ So\quad root\quad of\quad f(a)>\sqrt { 8 } .\quad \\ Hence\quad for\quad equation\quad { a }^{ 3 }-6a-6=0,\\ a>\sqrt { 8 } ;{ a }^{ 2 }>8\\ We\quad have\quad 8-{ a }^{ 2 }<0.\quad so\\the \quad roots\quad are\quad non\quad real \)

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\( \Delta = (a)^{2} -4(a^{2}-6) = 24-3a^{2} = 3(8-a^{2}) \)

Consider,

\( f(a) = a^{3} -6a - 6 \)

\( f'(a) = 3a^{2} - 6 = 3(a^{2}-2) \)

For \( 0 \le a \le \sqrt{2} \) , f(a) is decreasing, increasing for \( a \ge \sqrt{2} \)

\( f(0) = - 6 < 0 \)

\( f(\sqrt{2}) = 2\sqrt{2} -6\sqrt{2} - 6 < 0 \)

\( f(\sqrt{8}) = 8\sqrt{8}-6\sqrt{8} - 6 = 2\sqrt{8} - 6 < 0 \)

Thus, the positive root of \( f(a) \) is \( \ge \sqrt{8} \)

\( a \ge \sqrt{8} \)

\( \Delta < 0 \)

Thus, the roots are not real.

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Just completed solving the question and here you post the solution! But I think proof questions are not allowed, isn't it?

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@Vighnesh Shenoy. @Vighnesh Shenoy and @Mayank Chaturvedi,please make changes in your solutions accordingly.

I am changing the wording of the question ,as suggested byLog in to reply

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@Mayank Chaturvedi,please post the next question

OK.Log in to reply

A non-calculus approach.\({a}^{3}-6a-6=0\)

Let \( a=b+2/b \)

Therefore,\( { \left( b+2/b \right) }^{ 3 }-6(b+2/b)-6=0 \)

Simplifying we get that, \( {b}^{6}-6{b}^{3}+8=0 \)

Therefore, \( {b}^3=4\) or \(2\)

Substitute these values to get \(a\).

That time we see that only one real solution of \(a\) occurs which is, \(a={2}^{1/3}+{2}^{2/3} \)

We see that, \( {a}^{2}-6=6/a\)

Substituting this value in \({x}^{2}+ax+{a}^{2}-6=0 \) we get that,

\(a{x}^{2}+{a}^{2}x+6=0 \)

Assume that the roots of these quadratic equation are real, Then using formula for roots for quadratic equations,

\( x=\dfrac { -a\pm \sqrt { { a }^{ 2 }-24 } }{ 2 } \)

Then substituting the acquired value of \(a\) in this equation we get that \(x\) is a complex number. Hence, our assumption was wrong.

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Wondering, how did you thought of that a=b+(2/b).

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Problem 8:Find the locus of centers of the circles which touch the two circles\({x}^2+{y}^2={a}^2\) and \({x}^2+{y}^2=4ax\)

externally.

Akul Agrawal has solved this problem in the first place and posted the solution. Thanks!Log in to reply

Difference of distance of center of touching circle from the two given circles is constant. Hence equation will be hyperbola with (0,0) and (2a,0) as focii. Vertex is (a,0). difference between distance of center from focii is 2a-a=a.

Hence locus is

\[ \frac { { \left( x-a \right) }^{ 2 } }{ { \left( \frac { a }{ 2 } \right) }^{ 2 } } -\frac { { y }^{ 2 } }{ { \left( \sqrt { 3 } \frac { a }{ 2 } \right) }^{ 2 } } =1 \]

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Problem 41\(\displaystyle \int tanx. tan2x. tan3x\text{ }\,dx\)

Saarthak Marathe has posted the solution to the problem.Thanks!Log in to reply

\(\displaystyle tan3x=\frac{tan2x+tanx}{1-tanx.tan2x}\\ tanx.tan2x.tan3x=tan3x-tanx-tan2x\)

Now integrating we get,

\(\displaystyle \int tanx.tan2x.tan3x.dx=\frac{ln|sec3x|}{3}-\frac{ln|sec2x|}{2}-ln|secx|+constant \)

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Correct!Post the next question.

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Problem 2:Which real values of \(k\) exist such that the following system of equations has no solution? \[ \begin{cases} (k + 1)x + 8y = 4k \\ kx + (k + 3)y = 3k − 1 \end{cases}\]

Nihar Mahajan has solved this problem at the first place and has posted the solution. Thanks!Log in to reply

For the equations to have no solution, they should be parallel lines, hence their x-y coefficients are in proportion but not in proportion when ratio of the constant terms is included.

\[\dfrac{k+1}{k} = \dfrac{8}{k+3}\neq \dfrac{4k}{3k-1}\]

First solving the quadratic from \(\dfrac{k+1}{k} = \dfrac{8}{k+3}\) we have \(k=1,3\). However when \(k=3\) are substituted in \(\dfrac{4k}{3k-1}\) the ratio does not remain consistent. Hence, \(k=3\) is the solution and number of solutions is \(1\).

(Feel free to correct me)

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Problem 5:\[\displaystyle \sum _{ r=1 }^{ 7 }{ { \tan }^{ 2 }\left( \frac { r\pi }{ 16 } \right) } =?\]

Samuel Jones is the first person to solve this problem and provide the solution. Thanks!Log in to reply

The sum can be written as \(\displaystyle -8 +\sum_{r=1}^{7} \sec^2 \left(\dfrac{r \pi}{16}\right)\)

\(\displaystyle =-8 + \sum_{r=1}^{7} \dfrac{1}{ \cos^2 \left(\dfrac{r \pi}{16}\right)}\)

Consider the \( \displaystyle 15^{\text{th}}\) Chebyshev Polynomials of the Second kind

\(\displaystyle U_{15} (x) = 2^{15}\prod_{k=1}^{15} \left(x-\cos \left(\dfrac{r \pi}{n} \right)\right)\)

Taking logarithm and differentiating two times at \(x=0\), we have

\(\displaystyle S = -8 + 43 =\boxed{35} \)

This way, we can generalize to \(n\) terms also.

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Problem 6:Evaluate the sum of the non - real roots of the equation

\[x^4+x^3-5x^2-12x-6=0.\]

Aditya Kumar solved the problem at the first place and provided the solution. Thanks!Log in to reply

The equation can be written as \[\left( { x }^{ 2 }-2x-2 \right) \left( { x }^{ 2 }+3x+3 \right) =0\]

The non- real solutions are: \(x=\frac { -1 }{ 2 } \left( \sqrt { 3 } i+3 \right) \) and \(x=\frac { 1 }{ 2 } \left( \sqrt { 3 } i-3 \right) \quad \)

Hence the sum is: \(-3\)

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Just for enlightening, I'm posting my solution (again).

Let \(\displaystyle x = t-1\) the equation converts to

\(\displaystyle t^4-3t^3-2t^2-3t+1 = 0\)

\(\displaystyle \implies t^2 - 3t -2 -\dfrac{3}{t} + \dfrac{1}{t^2} = 0 \)

Now, let \(\displaystyle y = t + \dfrac{1}{t}\), we have,

\(\displaystyle y^2 -3y - 4 = 0\)

\(\displaystyle \implies (y-4)(y+1) = 0 \)

\(\displaystyle \implies (t^2-4t+1)(t^2-t+1) = 0 \quad \left( \because y = t + \dfrac{1}{t} \right)\)

\( \displaystyle \implies (x^2 - 2x -2)( x^2 + 3x + 3) = 0 \quad \left( \because t = x+1 \right) \)

Only the second factor gives non - real roots since it's discriminant is negative, therefore sum is \(\boxed{-3}\)

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The equation factors out as,

\( ({x}^{2}-2x-2)({x}^{2}+3x+3)=0\)

Now solve \({x}^{2}-2x-2=0\) and \({x}^{2}+3x+3=0 \).

We see that the discriminant of \({x}^{2}+3x+3=0 \) is negative.

Therefore the sum of the non-real roots are

-3Log in to reply

Problem 7:If \({ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+...+{ C }_{ n }{ x }^{ n }\).

Then find a closed of: \[{ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }+...+\left( n+1 \right) { C }_{ n }^{ 2 }\]

Rohit Ner has solved this problem at the first place and posted the solution. Thanks!Log in to reply

\(x{(1+x)}^{n}=C_0 x+C_1 {x}^2 +C_2 {x}^3+\cdots +C_n {x}^{n+1} \)

Differentiating both sides, \({(1+x)}^{n-1}(nx+1+x)=C_0 +2C_1 x +3C_2 {x}^2 +\cdots + (n+1)C_n {x}^{n}\)

Also \({\left(1+\dfrac{1}{x}\right)}^{n}=C_0 + \dfrac{C_1}{x} + \dfrac{C_2}{{x}^2} +\cdots +\dfrac{C_n}{{x}^{n}}\)

The given series is sum of all constant terms obtained in the expansion of

\({(1+x)}^{n-1}((n+1)x+1).{\left(1+\dfrac{1}{x}\right)}^{n}\)

which is same as coefficient of \({x}^{n}\) in \({(1+x)}^{2n-1}\) i.e. \(\binom{2n-1}{n}\)

plus coefficient of \({x}^{n-1}\) in \((n+1){(1+x)}^{2n-1}\) i.e. \((n+1)\binom{2n-1}{n-1}\)

So the closed form is \(\binom{2n-1}{n}+(n+1)\binom{2n-1}{n-1}\)

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Problem 9:(On behalf of Akul Agarwal)If \(\displaystyle f(x)=x+\int _{ 0 }^{ 1 }{ t(x+t)f(t).dt } \), then find the value of the definite integral,

\[\int _{ 0 }^{ 1 }{ f(x)dx } \]

Ishan Singh has provided the complete solution to the problem. Thanks!Log in to reply

\( \displaystyle f(x) = x \left(1+\int_{0}^{1} t f(t) \ \mathrm{d}t \right) +\int_{0}^{1} t^2 f(t) \ \mathrm{d}t \)

\(\displaystyle = Ax +B\) (say)

\(\displaystyle \implies f(x) = x\left(1+\int_{0}^{1} t (At+B) \ \mathrm{d}t \right) + \int_{0}^{1} t^2 (At+B) \ \mathrm{d}t\)

\(\displaystyle = x \left(1+\frac{A}{3}+\frac{B}{2} \right) + \frac{A}{4} + \frac{B}{3}\)

\(\displaystyle \implies Ax+B = x \left(1+\frac{A}{3}+\frac{B}{2} \right) + \frac{A}{4} + \frac{B}{3} \)

Comparing coefficients, we have,

\(\displaystyle A = \frac{65}{23} \)

\(\displaystyle \implies \int_{0}^{1} f(t) dt = A - 1 = \boxed{\dfrac{42}{23}} \)

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Problem 11:If the medians of a \(\Delta ABC\) make angles \(\alpha ,\beta ,\gamma \) with each other, then find the value of:\[\cot { \left( \alpha \right) } +\cot { \left( \beta \right) } +\cot { \left( \gamma \right) } +\cot { \left( A \right) } +\cot { \left( B \right) } +\cot { \left( C \right) } \]

Consider \(\alpha ,\beta ,\gamma \) to be acute.

Saarthak Marathe is the first person to solve this problem and provide the solution. Thanks!Log in to reply

The answer is \(0\).

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Problem 12:Evaluate:

\[\large \displaystyle \lim_{ x\rightarrow 0 }{ \left\{ \frac { sinx }{ x } \right\} }^{ \frac { 1 }{ \left\{ \frac { tanx }{ x } \right\} } } \]

where,\( \{ . \} \) is fractional part function.

Rishabh Cool has provided a complete solution to this problem. Thanks!Log in to reply

Problem 13:\[ f^{ 2 }\left( n+1 \right) f\left( n \right) +2f\left( n+1 \right) =f\left( n \right) \]

Given \[f\left( 0 \right) =1\]

Find \[\lim_{n\to\infty} { 2 }^{ n }f\left( n \right) \]

Ashu Dablo has solved this problem at the first place and has posted the solution. Thanks!Log in to reply

@Akul Agrawal Simplify the given equation to the form:

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

Now we say that for \(f\left(n \right) = \tan \theta \), \(f\left(n+1 \right)=\tan \frac{\theta} {2}\)

As f(1) =1, from above relation, we can say that \(f\left(n \right)=\tan \frac{\pi}{2^{(n+1)}}\) which for n \(-> \infty , -> 0\)

multiply divide by \(tan \frac{\pi}{2^{(n+1)}}\)

now for \(x->0\) \(\frac {\tan x}{x} =1\)

So answer is \( \frac{\pi}{4} \)

Sorry for my latex!

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The equation becomes inconsistent for n=0 . Please check .

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Problem 14 :If the sum of the first n terms of an AP is \( cn^2\), then find the sum of the cubes of these n terms.

Saakshi Singh has posted the solution first. Thanks!Log in to reply

Using sum formula of AP, we get \[2cn= 2a + (n-1)d\]

\[ \Rightarrow 2cn - a = a + (n-1)d \\ \Rightarrow (2cn - a)^3 = ( a + (n-1)d)^3 .\]

We have to find \(\displaystyle \sum_{n=1}^{ n} \left( a + (n-1)d \right)^3 =\displaystyle \sum_{n=1}^{ n}(2cn-a)^3 \). Nos putting \(n=1\), we get \(a=c\).

So we have sum of the cubes of \(n\) terms

\[= c^3 \cdot \displaystyle \sum_{ n=1}^{ n} (2n-1)^3 \]

\[ = c^3 \left( n^2(2n^2 - 1) \right) .\]

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I've cleaned up your solution for now. But it's very easy to type an equation using latex which you can learn from this note. @Saakshi Singh

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This one is a fairly easy one:

Problem 16:The maximum value of the function \(f(x) = 2x^{3} - 15 x^{2} + 36x -48\)

for {\( x | x^{2} + 20 \leq 9x\)}

Vignesh Shenoy has posted the correct solution. Thanks!Log in to reply

\( x^{2} - 9x + 20 \le 0 \)

\( (x-4)(x-5) \le 0 \rightarrow x \in \text{[}4,5\text{]} \)

\( f(x) = 2x^{3} - 15x^{2} + 36x - 48 \)

Differentiating,

\( f(x) = 6x^{2} - 30x + 36 \)

\( f'(x) = 6(x^{2}-5x+6) \)

\( f'(x) = 6(x-2)(x-3) \)

For \( x \in \text{[}4,5\text{]} \) , \( f'(x) > 0 \rightarrow f(x) \) is increasing.

\( f(x)_{max} = f(5) = 7 \)

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Problem 18:Consider the ellipse :

\( \dfrac{x^{2}}{25} + \dfrac{y^{2}}{16} = 1 \)

Let L be the length of perpendicular drawn from the origin to any normal of the ellipse. Find the maximum value of L.

Bonus : Solve it without calculus and generalize for any ellipse

Aditya Kumar has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Solution to Problem 18:Consider the point \((5\cos(\theta),4\sin(\theta))\).

The equation of normal would be: \(5x\sec(\theta)-4y\csc(\theta)\)

Hence, distance \(d=\left| \frac { 9 }{ \sqrt { 25\sec ^{ 2 }{ \left( \theta \right) } +16\csc ^{ 2 }{ \left( \theta \right) } } } \right| \)

To maximise \(d\), we have to minimise the denominator. By minimising the denominator(did that by calculus), the denominator comes out to be 9.

Hence \(\boxed{d_{max}=1}\)

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Absolutely Correct.

Generalisation for any ellipse x^2 / a^2 + y^2/b^2 = 1

Maximum distance of normal from origin = Modulus (b-a)

Also you could have avoided calculus by using AM Gm in last step by writing sec and cosec in terms of tan and cot

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Problem 26:Let \( ^{n}a = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_{n \text{times}} \)

\( f(x) = \displaystyle \sum_{r=1}^{x} (^{r}r) \)

Find the last digit of \( f(15) \)

Mayank Chaturvedi has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

The last digit is 8. Short methods are always welcome. I went to find out last digit of individual terms. Yes!. This looks lengthy but when we note down last terms, last digits of 1,2,4,5,6,10,11,14,15 are easy to find. For rest terms, dividing exponents by 4, using modular arithmetic would do.

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Correct.

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Problem 39:Solve\(\displaystyle {x}^{3}\frac{dy}{dx}={y}^{3}+{y}^{2}\sqrt{{y}^{2}-{x}^{2}} \)

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The answer is \(xy=c(y-\sqrt{y^{2}-x^{2}})\) The above equation is a homogeneous equation and can be written of the form \[\frac{dy}{dx}={\frac{y}{x}}^{3}+{\frac{y}{x}}^{2}*\sqrt{y^{2}-x^{2}}\] Making the substitution \(y=vx\) and we see that \(\dfrac{dy}{dx}=(v+\dfrac{dv}{dx}*x)\). On simplification we get \[ \frac{dv}{v(v^{2}-1+v\sqrt{v^{2}-1})}=\frac{dx}{x}\]. Taking \(\sqrt{v^{2}-1}\) and then Multiplying and dividing by \(v-\sqrt{v^{2}-1}\) we have \[\frac{dv}{\sqrt{v^{2}-1}} -\frac{dv}{v}=\frac{dx}{x}\]. On integrating both sides we have \[\ln{\frac{v-\sqrt{v^{2}-1}} {v}}=\ln{x}+c\]. This on substituting back \(v\) gives us the above answer.

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Problem 43:Find the remainder when \( 32^{32^{32}} \) is divided by 7.

Mayank Chaturvedi has solved this problem with the correct solution.

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\(32\equiv -3 (mod 7)\\ {3}^{3}\equiv -1 (mod 7)\\ {32}^{3*10+2}\equiv {(-3)}^{3*10+2} \equiv (1)*9 \equiv 9 (mod 7)\\ {32}^{32^{32}}\equiv {(9)}^{3*10+2} \equiv (1)*81 \equiv 4 (mod 7)\\ \)

Hence the answer is \(\boxed 4\)

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The answer is right, however you used

\( 32^{32^{32}} \equiv (32^{32})^{32} \equiv 9^{32} \equiv 4 \pmod{7} \)

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agreewith you. Have a check @Saarthak Marathe the answer is a coincidence this time.Let me try once.

Reference-euler's theorem

\(\varphi (7)=6\quad and\quad gcd(32,7)=1\\ { 32 }^{ 6 }\equiv 1mod(7)............me-1\\ Now\quad we\quad find\quad { 32 }^{ 32 }mod\quad 6\\ { 32 }^{ 32 }\equiv { 2 }^{ 32 }\equiv 4^{ 4 }\equiv 4mod(6)......me-2\\ Using\quad me-1\quad and\quad me-2\\ { 32 }^{ { 32 }^{ 32 } }\equiv { 32 }^{ 6x+4 }\equiv { 32 }^{ 4 }mod(7)\\ { 32 }^{ 4 }\equiv 4^{ 4 }\equiv 256=4mod(7)\\ So\quad 4\quad is\quad the\quad remainder\quad :)\)

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Problem 44:Find the locus of point of intersection of tangents to an ellipse \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \) at two points,whose eccentric angles differ by a constant angle \(\alpha\) .

Kunal Verma has provided a complete solution to this problem.Thanks!Log in to reply

Point of intersection of tangents at points who's eccentric angles are \( i \) and \( j \) :-

x= \( a \times \frac{\cos \frac{i+j}{2}}{\cos \frac{i-j}{2}} \) and y= \( b \times \frac{\sin \frac{i+j}{2}}{\cos \frac{i-j}{2}} \)

Thus \(\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{i-j}{2} \)

\(\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{\alpha }{2} \)

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Correct! But prove your 1st statement in your solution and post the next question

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Problem 45The coefficient of \(x^{n-6} \) in the expansion:

\( n! \times \ [ \ x \ - \frac{\binom{n}{0} \ + \binom{n}{1}}{\binom{n}{0}} \ ] \ [ \frac{x}{2} \ - \frac{\binom{n}{1} \ + \binom{n}{2}}{\binom{n}{1}} \ ] \ .... \ [ \frac{x}{n} \ - \frac{\binom{n}{n-1} \ + \binom{n}{n}}{\binom{n}{n-1}} \ ] \)

is equal to \( \binom{x}{y} \times \ z^y \)

Find x,y,z if all are integers( x, y and z can be in terms of \( n \) )

Sarvesh Nalawade has provided a complete solution to the problem.Log in to reply

Let S = \( n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{{n \choose i-1} + {n \choose i}}{{n \choose i-1}}) \)

\( Therefore, S= n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{{n+1 \choose i}}{{n \choose i-1}} ) \)

\( S = n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{n+1}{i} ) \)

\( Therefore, S = (x-n-1)^{n} \) .

Co-efficient of \(x^{n-6} = {n \choose 6} (n+1)^{6} \)

Therefore x=n , y=6, z= n+1

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You are required to find the individual values of \(x \) , \( y \) and \( z \), not the sum ,although it is correct. You may post the next problem after posting the solution.

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Problem 46If \( M = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \) is an orthogonal matrix with real entries , what is the minimum value of abc ?

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For an orthogonal matrix M,

\( MM^{T} = I \) where I is the identity matrix.

Multiplying the matrices on LHS , and comparing with the identity matrix we get,

\( a^{2} + b^{2} + c^{2} = 1 \)

\( ab + bc + ca = 0 \)

From these two equations we get,

\( a + b + c = \pm 1 \)

\( f(x) = x^{3} - (a+b+c)x^{2} + (ab+bc+ca)x - d \) where d = abc )

Thus, a,b,c are roots of \( f(x) = 0 \)

\( f(x) = x^{3} - (a+b+c)x^{2} - d \)

Let \( a + b + c = p \)

\( f(x) =x^{3} - px^{2} - d \)

Differentiating,

\( f'(x) = 3x^{2} - 2px \)

The roots of \( f'(x) = 0 \) are,

\( x = 0, \dfrac{2p}{3} \)

For the equation to have 3 real roots, the first turn of the graph should be above x-axis, and second must be below x-axis.

Consider case of \( p =1 \)

The roots are \( x = 0 , \dfrac{2}{3} \)

\( \therefore f(0)f\left(\dfrac{2}{3}\right) \le 0 \)

\( \left(-d\right)\left(\dfrac{-4}{27}-d\right) \le 0 \)

\( \left(d\right)\left(d+\dfrac{4}{27}\right) \le 0 \)

\( d \in \left[\dfrac{-4}{27},0 \right] \)

When \( p = -1 \)

The roots of \( f'(x) = 0 \) are \( x = 0 , \dfrac{-2}{3} \)

\( f(0)f\left(\dfrac{-2}{3}\right) \le 0 \)

\( \left(-d\right)\left(\dfrac{4}{27}-d\right)\le 0 \)

\( d \in \left[0,\dfrac{4}{27}\right] \)

Range of d \( \left[\dfrac{-4}{27}, \dfrac{4}{27} \right] \)

The mininum value occurs when \( a,b,c \) are a permutation of \( \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{-1}{3} \).

The maximum value occurs when \( a,b,c \) are a permutation of \( \dfrac{-2}{3} , \dfrac{-2}{3} , \dfrac{1}{3} \)

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Great! Please post the next problem. @Vighnesh Shenoy

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An alternate solution to@Vighnesh Shenoy 'salready well written and beautiful solution:\(a^2+b^2+c^2=1...(1) \\ab+bc+ca=0...(2)\)

Let

\(a=cos\alpha\\b=cos\beta\\c=cos\gamma\)

From \((1)\) we can write \(cos^2\alpha+cos^2\beta=sin^2\gamma...(3)\)

Now, using \((1)\) and \((2)\) we can write

\(cos\alpha+cos\beta+cos\gamma=\pm 1\\ cos\alpha+cos\beta=\pm 1-cos\gamma\)

Squaring both sides we get\(cos^2\alpha+cos^2\beta+2cos\alpha .cos\beta=1+cos^2\gamma \mp 2cos\gamma\)

From \((3)\) we can write \(cos\alpha .cos\beta=cos^2\gamma \mp cos\gamma\)

Now multiplying \(cos\gamma\) on both sides we get:

\(cos\alpha .cos\beta .cos\gamma=cos^3\gamma\mp cos^2\gamma\)

Critical pointsof the expression on the right-hand side of the above equation(1.)\(\{0,\frac{2}{3}\}\) when \(cos\alpha .cos\beta .cos\gamma=cos^3\gamma- cos^2\gamma\)(2.)\(\{0,\frac{-2}{3}\}\) when \(cos\alpha .cos\beta .cos\gamma=cos^3\gamma+cos^2\gamma\)On checking these values, minima occurs at \(cos\gamma = \frac{2}{3}\) and the minimum value is \(\frac{-4}{27}\), and maximum occur maxima occurs at \(cos\gamma = \frac{-2}{3}\) and the maximum value is \(\frac{4}{27}\)

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Problem 47: In a \( \Delta ABC \) the ratio of side \(BC \) and \( AC \) to the circumradius is \( 2 \) and \( \dfrac{3}{2} \) respectively. If the ratio of length of angle bisectors of angle \( B \) to length of angle bisector of angle \( C \) is given by \( \dfrac{\alpha(\sqrt{\alpha}-1)}{\beta\sqrt{\gamma}} \) find\( \dfrac{\alpha + \beta + \gamma}{3} \)

Note :

\( \alpha, \beta, \gamma \) are positive integers with \( gcd(\alpha, \beta ) = gcd(\alpha,\gamma) = gcd( \beta, \gamma = 1 \)

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Firstly Using Sine Rule we get,

\( \frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =2R \)

Hence, we get \( A = \frac{\pi}{2} and \sin{B} = \frac{3}{4} \)

Let \( BC = 4k , AC = 3k , AB = \sqrt{7}k \)

Let BD and CE be the respective angle bisectors.

\( \frac { AD }{ CD } =\frac { BA }{ BC } \quad and\quad \frac { AE }{ EB } =\frac { AC }{ BC } \)

\( AD=\frac { 3\sqrt { 7 } }{ 4+\sqrt { 7 } }k \quad and\quad AE=\frac { 3 }{ \sqrt { 7 } }k \)

Then using Pythagoras Theorem , we get values of BD and CE as :

\(BD=\frac { 4\sqrt { 7 } }{ 1+\sqrt { 7 } } k\quad and\quad CE=\frac { 6\sqrt { 2 } }{\sqrt{ 7 }} k\quad \)

Their ratio will be : \( \dfrac{7(\sqrt{7} - 1 )}{9\sqrt2} \)

Hence, \( \alpha =7,\quad \beta =9,\quad \gamma =2 \)

Therefore, final answer is 6

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Getting the same thing @Sarvesh Nalawade. I think you should go ahead and post the solution

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Correct.

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Problem 15:If N is the number of ways in which 3 distinct numbers can be selected from the set {3, 3^2 ,3^3 ,...... ,3^10 } , so that they form a GP, then find the value of N/5.

Ashu Dablo has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

@Saakshi Singh

The problem is similar to having to get number of possible Ap's from the set of {1,2,3..10}

So number of possible Ap's is (8+6+4+2)=20

8 for common dif =1 (1,2,3) , (2,3,4) ........(8,9,10)

6 for common dif =2 (1,3,5) , (2,4,6) ........(6,8,10)

4 for common dif =3 (1,4,7) , (2,5,8) ........(4,7,10)

2 for common dif = 4 (1,5,9) and (2,6,10)

So the answer to your question is

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the number of possibilities in that order is 8+6+4+2 but the the set containing these in reverse order can also be in gp ie(4,7,10) &(10,7,4) similar is the case with all other elements so the So number of possible GP's is 2

(8+6+4+2)=220=40 so N/5 =8 this was also my solutionLog in to reply

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My answer is 8 (NoteThe following is a table) Difference in power. No terms. Set(of powers) 1. 8. (123,234,...89 10) 2. 6. (135,246....68 10) 3. 4. (147....47 10) 4. 2. (159,2710) Total = 20 But r can be 3^{1,2,3} or 3^{-1,-2,-3} so N = 2x20 = 40 So N/5 =8

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Problem 17:\( z \) is a complex number in the complex plane such that \(\Im(z) \ne 0 \).

If \( \dfrac{z^{2} + z + 1}{z^{2}-z+1} \in \Re \) find the value of \( 10|z| \)

Details:\( \Im(z) \) denotes imaginary part of \( z \)

\( \Re \) denotes real numbers.

Prakhar Bindal has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Problem 19:If \[{ \left( 1-{ x }^{ 3 } \right) }^{ n }=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r }{ \left( 1-x \right) }^{ 3n-2r } } \] then find \(a_r\).

Saarthak Marathe has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Write \( {(1-{x}^{3})}^{n} = {(3x(1-x)+{(1-x)}^{3})}^{n}=\sum _{ r=0 }^{ n }{ { 3 }^{ r } } \binom{ n }{ r }{ (x(1-x)) }^{ r }{ (1-x) }^{ 3n-3r } \)

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Answer is \({3}^{r}\cdot \dbinom{n}{r} \)

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Problem 20:A series of chords of a parabola \({y}^{2}=4ax\) are drawn so that their projections on the straight line,which is inclined at an angle \(\alpha\) to the axis,are of constant length \(c\). Find the locus of the midpoints of these chords.

Prakhar Bindal has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Assume end point of chord in parametric form (at^2 , 2at) and (as^2,2as)

Let The equation of line be y = xtan(alpha)+b

Now find foot of perpendicular from end points onto the line. Using distance formula between them and equate it to c .

Let midpoint be (h,k)

2h = a(t^2 + s^2)

2k = 2a(t+s)

From above two relation find value of t-s and t+s as they will come in expression of distance formula.

Substitute these into that expression and on simplifying further and replacing h and k by x and y.

We obtain locus as

p = alpha (inclination of line)

(y^2 - 4ax) (ycosp+2asinp)^2 + (ac)^2 = 0

Alternate : Find out Length of chord using distance of formula. Then find angle between chord and line by using standard results. And finally take component of length of chord along line and equate to c . That would reduce some of the calculation work .

Credits @Samarth Agarwal for alternate way

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Problem 21

Find the equation of the director circle to the circle circumscribing the quadrilateral formed by the lines in order 2x+3y-2 = 0 , x+y =0 , 2x+5y-1 = 0 , 5x+27y-1 = 0

Samarth Agarwal has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Circle curcumscribing quad is l1

l2+(lambda)l3l4=0 lambda can be found by cofficient of xy=0Here lambda came out to be - 0.5

So circle is (x-11/6)^2+(y-25/6)^2=698/36 and directer circle is (x-11/6)^2+(y-25/6)^2=698/18

Is this correct.?

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The equation is \( {(x-11/6)}^{2}+{(y-25/6)}^{2}=349/9 \). The steps are same as that of Samarth's solution

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PROBLEM 22Let \(a(n)=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1} {2^n-1} \).. Then:

A: \(a(100) \leq 100 \)B: \(a(100)>100 \)C: \(a(200)\leq 100 \)D: \(a(200)> 100 \)For sake of clarification \(\displaystyle a\left( n \right) =\sum _{ r=1 }^{ { 2 }^{ n }-1 }{ \frac { 1 }{ r } } \)

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Solution to problem 22:Clearly \(a(n)\leq n\). Hence, \(a(100)\leq100\)

Now, for \(a(200)\), we need to do the following steps: \[a(n)>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{2^n}+...+\frac{1}{2^n}\right)\]

This is an AGP.

Hence, \(a(200)>\left(1-\frac{1}{2^200}\right)+100>100\).

Therefore options

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Problem 23:Find the closed form of \[\int _{ 0 }^{ \infty }{ \left\lfloor \frac { n }{ { e }^{ x } } \right\rfloor dx } \]

Statutory warning: Do not relate this to Gamma Function.

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

\( I = \displaystyle \int_{0}^{\infty} \left \lfloor \dfrac{n}{e^{x}} \right \rfloor dx \)

\( e^{-x} = t \rightarrow dx = \dfrac{-dt}{t} \)

\( I = \displaystyle \int_{0}^{1} \left \lfloor nt \right \rfloor \dfrac{dt}{t} \)

\( I = \displaystyle \sum_{k=1}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} \dfrac{k}{t} dt \)

\( I = \displaystyle \sum_{k=1}^{n-1} k\log\left(\dfrac{k+1}{k}\right) \)

\( S = \displaystyle \sum_{k=1}^{n-1} \left[(k+1)\log(k+1)-k\log(k) - \log(k+1)\right]\)

\( S = n\log(n) - log(n!)\)

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Time for an easy question.

Problem. 24 :Let \( A_{1}, A_{2}, A_{3}, A_{4} \) be the areas of four faces of a tetrahedron, and \( h_{1}, h_{2}, h_{3}, h_{4} \) be the corresponding altitudes. Give that the volume of tetrahedron is 5 cubic units . Find the minimum value of the expression

\( \dfrac{(A_{1} + A_{2} + A_{3} + A_{4})(h_{1}+h_{2}+h_{3}+h_{4})}{5!} \)

Sarvesh Nalawade has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Volume of tetrahedron = \( \dfrac{A \cdot h}{3} \)

\( \therefore 15 = A_{1} \cdot h_{1} = A_{2} \cdot h_{2} = A_{3} \cdot h_{3} = A_{4} \cdot h_{4} \)

We have to minimize :

\( S = \dfrac{15(A_{1}+A_{2}+A_{3}+A_{4})\left(\dfrac{1}{A_{1}} + \dfrac{1}{A_{2}} + \dfrac{1}{A_{3}} + \dfrac{1}{A_{4}}\right)}{5!} \)

Using AM GM ineqaulity :

\( \displaystyle \sum_{i=1}^{4}A_{i} \geq 4\left({A_{1} \cdot A_{2} \cdot A_{3} \cdot A_{4}}\right)^{\frac{1}{4}} \)

\( \displaystyle \sum_{i=1}^{4} \dfrac{1}{A_{i}} \geq 4\left(\dfrac{1}{A_{1} \cdot A_{2}\cdot A_{3}\cdot A_{4}}\right)^{\frac{1}{4}} \)

Multiplying these two we get :

\( \displaystyle \sum_{i=1}^{4}A_{i} \times \displaystyle \sum_{i=1}^{4} \dfrac{1}{A_{i}} \geq 16 \)

Hence,

\( S_{min} = \dfrac{15 \times 16}{5!} = 2 \)

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Correct!

I used the Cauchy-Schwartz inequality instead of the AM-GM.

\( \left(\sqrt{A_{1}}^{2}+\sqrt{A_{2}}^{2}+\sqrt{A_{3}}^{2}+\sqrt{A_{4}}^{2}\right)\left(\dfrac{1}{\sqrt{A_{1}}^{2}} + \dfrac{1}{\sqrt{A_{2}}^{2}} + \dfrac{1}{\sqrt{A_{3}}^{2}} + \dfrac{1}{\sqrt{A_{4}}^{2}}\right) \geq \left(\sqrt{A_{1}}\dfrac{1}{\sqrt{A_{1}}} + \sqrt{A_{2}}\dfrac{1}{\sqrt{A_{2}}} + \sqrt{A_{3}}\dfrac{1}{\sqrt{A_{3}}} + \sqrt{A_{4}}\dfrac{1}{\sqrt{A_{4}}}\right)^{2} \)

Equality holds when :

\( A_{1} = A_{2} =A_{3} = A_{4} \)

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PROBLEM 27:\(\csc ^{ -1 }{ \sqrt { 5 } } +\csc ^{ -1 }{ \sqrt { 65 } } +\csc ^{ -1 }{ \sqrt { 325 } } +\ldots =\frac { n\pi }{ 32 } \)

If you were given a chance to visit n (its value to be found from above expression) towns, in any way you want irrespective of order. I tell you that you can do so in \({ A }^{ B }\) ways.A & B are naturals.Find minimum value of A+B.

\( A\quad 10\\ B\quad 8\\ C\quad 6\\ D\quad 4\).

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

\( S = \csc^{-1}(\sqrt{5}) + \csc^{-1}(\sqrt{65}) + \csc^{-1}(\sqrt{325}) \ldots \)

\( S = \sin^{-1}\left(\dfrac{1}{\sqrt{5}}\right) + \sin^{-1}\left(\dfrac{1}{\sqrt{65}}\right) + \sin^{-1}\left(\dfrac{1}{\sqrt{325}}\right) + \ldots \)

\(S = \tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{8} + \tan^{-1}\dfrac{1}{18} + \ldots \)

\( S = \displaystyle \sum_{k=1}^{\infty} \tan^{-1}\dfrac{1}{2k^{2}} = \sum_{k=1}^{\infty} \tan^{-1}(2k+1) - \tan^{-1}(2k-1) = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4} \)

\( \therefore n = 8 \)

I can visit 0 towns, 1 town, 2 towns, and so forth till 8 towns irrespective of order.

Number of ways I can do this = \( \displaystyle \sum_{r=0}^{8}\dbinom{8}{r} = 2^{8} \)

\( 2^{8} = 4^{4} = 16^{2} \)

Minimum value of A + B = 8.

Next time you give a summation, please give more terms. I am still not sure that the series I used is correct.

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I think you mean \(\tan ^{ -1 }{ \frac { 1 }{ 2 } } \) in the third line, and not \(\tan ^{ -1 }{ \frac { 1 }{ 4 } } \)

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Correct . And yes i will take care of more terms in the series next time.

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Problem 28:(On behalf of Vighnesh Shenoy)If \(\vec a\) and \(\vec b\) are two vectors such that \(|\vec{a}|=1, |\vec b|=4\) and \( \vec a \cdot \vec b =2\), then find the angle between \(\vec b\) and \(\vec c\) given that \(\vec c=\left( 2 \vec a \times \vec b \right) - 3 \vec b\).

Mayank Chaturvedi has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!Log in to reply

Sir please extend this page by making a new note as it is becoming slower to load the contents and rendering the latex.

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Yes sir @Sandeep Bhardwaj. The note is lagging too much and taking too much time to load.

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\(PROBLEM \quad 29\):

Time for an easy question.

Find the range of \(\beta\) such that \( (0,2\beta-1) \) lies on or inside the triangles formed by the lines. \( y +3x +2 = 0 \)

\( 3y- 2x- 5=0 \)

\( 4y+x-14=0 \)

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Simply by drawing the graph I got \( \frac{4}{3} \leq \beta \leq \frac{9}{4} \)

. The triangle intersects Y-axis at points \( \left(0,\frac{5}{3}\right) \) and \( \left(0,\frac{7}{2}\right) \)

\( \therefore \frac{5}{3} \leq 2\beta-1 \leq \frac{7}{2} \)

\( \therefore \frac{4}{3} \leq \beta \leq \frac{9}{4} \)

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CorrectLog in to reply

Great! Can you please add the graph for the sake of solution? Thanks!

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Problem 30:Let P(x) be a polynomial of degree 11 such that : \( P(x) = \dfrac{1}{x+1} , 0 \le x \le 11 \)

Find the value of \( P(12) \) .

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\( P(x)(x+1) = 1 \)

\( P(x)(x+1) - 1 = g(x) \)

\( g(x) \) has roots \( 0,1,2,3,4\ldots 11 \)

\( \therefore g(x) = ax(x-1)(x-2)\ldots(x-11) \)

\( P(x)(x+1)-1 = ax(x-1)(x-2)\ldots(x-11) \)

Substituting \( x = - 1 \),

\(-1 = a(-1)(-2)\ldots(-12) \)

\( -1 = 12!a \)

\( a = -\dfrac{1}{12!} \)

\( P(12)(13) - 1 = -\dfrac{1}{12!}\times 12! \)

\( P(12)(13) - 1 = -1 \)

\( P(12) = 0 \)

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That's correct !!

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Problem 31:Given that,

\( x^{2} + y^{2} + z^{2} = R^{2} \)

Let,

\( P = axy + byz \)

If \( P_{max} = R^{2}f(a,b) \)

Find \( f(3,4) \)

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Let \(x=R\sin\alpha\sin\beta,y=R\cos\beta,z=R\cos\alpha\sin\beta\)

\(\begin{align} P&=y\left(ax+bz\right)\\&=\dfrac{{R}^2}{2}\sin 2\beta \left(a\sin\alpha+b\cos\alpha\right)\\{P}_{max}&=\dfrac{{R}^2}{2}\sqrt{{a}^2+{b}^2}\\\Rightarrow f(a,b)&=\dfrac{\sqrt{{a}^2+{b}^2}}{2}\\f(3,4)&=\frac{5}{2}\end{align}\)

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Correct.

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I got the answer

5/2. Applying cauchy-schwarz inequality\(\sqrt { ({ a }^{ 2 }+b^{ 2 })(x^{ 2 }{ y }^{ 2 }+y^{ 2 }z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(x^{ 2 }+z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(R^{ 2 }-y^{ 2 }) } \ge (axy+byz)\\ \frac { { R }^{ 2 } }{ 2 } \sqrt { ({ a }^{ 2 }+b^{ 2 }) } \ge (axy+byz)\)

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You guys decide who is going to post next.

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Problem 32:The area bounded by the curve \(y=x-{x}^2\) and the line \(y=mx\) equals \(\large\frac{9}{2}\).

Find the sum of all possible values of \(m\).

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Let the points of intersection between \(y=x-x^2\) and \(y'=mx\) be \(x_1\) and \(x_2\). Therefore \(x_1, x_2\) are the roots of the equation \(x^2 +(m-1)x=0\)...\((1)\)

Given:

\(\displaystyle \int_{x_1}^{x_2} y-y'\, dx =\frac{9}{2}\)

\(\displaystyle \int_{x_1}^{x_2} x(1-m)-x^3\, dx=\frac{9}{2}\)

\(\large \frac{(1-m)(x_2-x_1)(x_2+x_1)}{2}-\frac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{3}=\frac{9}{2}\)

Now using equation \((1)\) we can write \(x_2+x_1=1-m\); \(x_1x_2=0\) and \(x_2-x_1=|1-m|\)

Therefore finally,

\(m=-2\) or \(m=4\)

Therefore, the required answer is \(\boxed{2}\)

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Correct. :). Waiting for the next problem.

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Problem 33:Two persons \(A\) and \(B\) agree to meet at a place between \(11\) to \(12\) noon. The first one to arrive waits for \(20\) minutes and then leaves. If the time of their arrival be independent and random, what is the probability that \(A\) and \(B\) shall meet?

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Just an alternate solution:

Let the time of arrival of \(A\) be denoted along the \(x-axis\) and that of \(B\) along the \(y-axis\).

Therefore, the required event is denoted graphically be the area enclosed in \(|x-y|\le 20\). Now the ratio of the

area of the eventto thearea of sample space(\(=3600\))gives the answer \(\frac{5}{9}\)Log in to reply

Probability of meeting If the first person comes between 0-40 minutes=\(2/3*1/3\)=\(2/9\)

Probability of meeting if first person comes between 40-60 minutes=\(1/3\)

Therefore,total probability=\(5/9\)

Is this correct? @Miraj Shah

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Ya! The answer is correct.

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Problem 34:Solve:\(\displaystyle ({x}^{2}+y)\frac{dy}{dx}=6x \)

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We can write the question as follows:

\((x^2+y)dy=3\times 2xdx\)

Now let \(x^2=\lambda\)

\((\lambda+y)=3\large \frac{d\lambda}{dy}\)

We'll do another substitution here, let \(\lambda+y=t\)

Therefore,

\(t+3=3\large \frac{dt}{dy}\)

Now after rearranging the terms and integrating we get:

\(3ln|x^2+y+3| = y+c\)

Is this correct? @Saarthak Marathe

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It is incorrect. You missed a number in your final answer

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Problem 35:It can be proved that the areas \(S_0,S_1,S_2,S_3,...\) bounded by the \(x-axis\) and the half waves of the curve \(y=e^{-\alpha x}sin\beta x\), \(x\ge 0\) form a geometric progression. Find the common ratio of this geometric progression.

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Answer:\({e}^{-\frac{\alpha\pi}{\beta}}\)First I wrote the general integral for \({S}_{j}\) and then divided by \({S}_{j-1}\)

Is it correct? @Miraj Shah

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The answer is correct! Can you post a clearer picture of your working if possible?

Thanks!

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PROBLEM 37In how many ways 12 different books can be distributed among 5 children so that 2 get three books each, and 3 get two books each.

Saarthak Marathe and Miraj Shah had solved this questionLog in to reply

\(\dbinom{12}{6}\dbinom{6}{3}\dbinom{6}{2}\dbinom{4}{2}\dbinom{5}{3}=16632000 \)

First choose 2 people of 5 which get 3 books each. Select 6 books(3*2=6) out of 12 books for these two 2 people. Then distribution of these 6 books by letting the first person choose 3 books out of 6 and the rest goes to 2nd person. Now for the remaining 3 people,there are 6 books. 1st person selects 2 books out of 6 books and 2nd person chooses 2 out of 4 books and the remaining goes to the 3rd person. I hope it is clear now. @Mayank Chaturvedi

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Can you please tell what is the numerical figure you are getting after solving your expression?

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As you asked, your solution is

Crystal clearBut it would be interesting if we can find out an explanation for solution by @Miraj Shah.Which I feel is as good approach as yours.Log in to reply

The number of ways distributing \(n\) distinct things into \(r\) groups and arranging them among \(r\) people such that \(s_1, s_2,s_3,...s_r\) denotes the number of things in the respective groups is :

\(\large \frac{(n!)( r!)}{(s_1!)( s_2!) ...(s_r!)}\)

Note in the above formula \(s_1\neq s_2\neq s_3\neq ... s_r\)

For the given question \(n=12\); \(r=5\); \(s_1=s_2=3\) and \(s_3=s_4=s_5=2\).

Now, the above formula can be used but, with a slight modification. Since groups are only identified by the number of objects it has, therefore there is no difference between \(s_1\) and \(s_2\) and there is no difference between \(s_3, s_4\) and \(s_5\). Hence the number of cases will reduce by a factor of \(2!\) with respect to \(s_1\) and \(s_2\) and by \(3!\) with respect to \(s_3, s_4\) and \(s_5\). Hence, in totality the required answer is:

\(\large \frac{12!\times 5!}{(3!)^2(2!)^3(2!)(3!)} = 16632000\)

Is the above answer correct? @Mayank Chaturvedi

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Problem 38:Find the number of ways to go from \( (0,0) \) to \( (8,7) \) in a rectangle formed by vertices \( (0,0) , (8,0) , (0,7), (8,7) \). The person can only move from \( (i,j) \) to \( (i+1,j) \)

OR\((i,j+1)\)OR\((i+1,j+1)\) in one step.Log in to reply

@Saarthak Marathe

No one has solved the problem in the time limit. Can you please post the solution and the next problem?

Thanks!

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Let the person travel \(d\) diagonals in one trip. So the remaining \(15-2d\) sides,he travels up or right, which can be done in \(\dbinom{15-2d}{8-d}\) ways. The \(d\) diagonals can be placed in \(\dbinom{15-d}{d}\) ways.

So the total number of ways=\(\displaystyle \sum_{d=0}^{7} \dbinom{15-2d}{8-d}\dbinom{15-d}{d} \).

This summation turns out as \(\boxed {108545}\)

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Problem no. 40.

An easy question:Find the following \[\dbinom{n}{0}+3\dbinom{n}{1}+5\dbinom{n}{2}+........+(2n+1)\dbinom{n}{n}\] Miraj Shah has posted the solution to the problem. Thanks!

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The above question can be written in the following manner:

\(\displaystyle \sum_{r=0}^{n} (2r+1)\dbinom{n}{r}\) \( =2\displaystyle\sum_{r=0}^{n}r\dbinom{n}{r} +\displaystyle\sum_{r=0}^{n}\dbinom{n}{r} = n2^n +2^n= \boxed{(n+1)2^n}\)

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Please post the next question

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Problem 42:Evaluate\(\displaystyle \lim_{n \rightarrow \infty} \left[\frac{{n}^{1/2}}{{n}^{3/2}}+\frac{{n}^{1/2}}{{(n+3)}^{3/2}}+\frac{{n}^{1/2}}{{(n+6)}^{3/2}}+........+\frac{{n}^{1/2}}{{(n+3(n-1))}^{3/2}}\right] \)

Vighnesh Shenoy has provided solution to this question. Thanks!Log in to reply

This problem can be easily solved by using the concept of definite integral as the sum of limits, which is applied as

\( \displaystyle \lim_{n \to \infty} \sum_{k=0}^{n-1} \dfrac{n^{\frac{1}{2}}}{(n+3k)^{\frac{3}{2}}} = \displaystyle \int_{0}^{1} \dfrac{1}{(1+3x)^{\frac{3}{2}}}dx = \left[ \dfrac{2}{3(\sqrt{1+3x})}\right]_{1}^{0} = \dfrac{2}{3} - \dfrac{2}{6} = \dfrac{2}{6} = \dfrac{1}{3} \)

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Great! It would be awesome if you can make your solution more detailed. You can post the next problem. Thanks!

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Problem 48: If \( \displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ \left( a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } \right) ^{ 2 } } } =\quad \frac { \pi \left( a+b \right) }{ p{ \left( ab \right) }^{ q } } \) ;where p and q are positive rational numbers and ab>0 , then find the value of pq

Vighnesh Shenoy has provided a complete solution to this problem.Log in to reply

\( I = \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} \)

Put, \( \tan(x) = t \)

\(I = \displaystyle \int_{0}^{\infty} \dfrac{dt}{a+bt^{2}} = \dfrac{\pi}{2\sqrt{ab}} \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} =\dfrac{\pi}{2}(ab)^{\frac{-1}{2}} \)

For a continuous and differentiable function,

\(\displaystyle \dfrac{d}{dy} \int_{a}^{b} f(x,y)dx = \int_{a}^{b} \dfrac{\partial}{\partial y} f(x,y) dx \)

Differentiating with respect to a,

\( \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{\partial}{\partial a} \dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} = \dfrac{\partial}{\partial a} \left( \dfrac{\pi}{2}(ab)^{\frac{-1}{2}}\right) \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{-\cos^{2}dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{-\pi}{4}b^{\frac{-1}{2}}a^{\frac{-3}{2}} \)

Differentiate with respect to b,

\( \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{\partial}{\partial b} \dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} = \dfrac{\partial}{\partial b} \left( \dfrac{\pi}{2}(ab)^{\frac{-1}{2}}\right) \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{-\sin^{2}dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{-\pi}{4}b^{\frac{-3}{2}}a^{\frac{-1}{2}} \)

Adding both the integrals,

\( \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{\pi}{4\sqrt{ab}}\left(\dfrac{1}{a} + \dfrac{1}{b} \right) \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{\pi(a+b)}{4(ab)^{\frac{3}{2}}} \)

\( p = 4 , q = \dfrac{3}{2} \)

\( pq = 6 \)

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Last problem for this page:

Problem No. '50'A cevian \(AQ\) of a equilateral \(\Delta\) \(ABC\) is extended to meet circumcircle at \(P\). If \(PB=50\) and \(PC=45\), find \(PQ\) upto 3 decimal places.

Deeparaj Bhat has provided the answerLog in to reply

Great! This is the last problem of this contest on this note. The solution poster will post the next problem on the continued part of the contest: JEE-Advanced Maths Contest (Continued).

Thanks!

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23.684. I'll post the soln in some time...

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Waiting for the solution.

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Solution by Deeparaj Bhat:Extend \(CP\) to \(D\) such that \(\Delta BDP\) is equilateral. Then, \[ \begin{align*}\angle BCD&=\angle QCP \\\angle QPC&=\angle ABC\quad \left(\because \text{ angles in the same segment are equal }\right)\\&=60^{\circ}\\&=\angle BDP\\\implies \Delta DCB &\sim \Delta PCQ\\\implies \frac{ DB }{PQ}&=\frac{ DC}{PC }\\&=1+\frac{PB}{PC}\quad \left( \because DC=DP+PC=BP+PC \right)\\\implies \frac{1}{PC}+\frac{1}{PB}&=\frac{1}{PQ}\quad \left( \because BP=DB \right)\end{align*} \]

Substituting the given values the answer comes out to be

23.684.Log in to reply

If the equation x^4-Px^2+9=0 has 4 real roots then P lies in the interval:- A} (0,infinity) B} (6,infinity) C} (-infinty,-6)

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If a^b^b^a=a√2, find a^2/b

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Problem 49:\( a,b,c,d \) are real numbers such that,

\( a + 2b + 3c + 4d = 15 \)

Find the minimum value of

\( 9^{a} + 2\cdot9^{b} + 3\cdot9^{c} + 4\cdot 9^{d} \)

Vignesh S has posted the solution to this problem. Feel free to add new approaches. Thanks!Log in to reply

The answer is

270Let \[f(x)=9^{a}+2\cdot 9^{b}+3\cdot 9^{c}+4\cdot 9^{d}\] and \[g(x)=a+2b+3c+4d\]. Using Lagrange Multipliers we have \[\dfrac{\partial{f(a,b,c,d)}}{\partial x}=k \times \dfrac{\partial{g(a,b,c,d)}}{\partial x}\] where \(x=a,b,c,d\) \[\dfrac{\partial{(9^{a}+2\cdot 9^{b}+3 \cdot 9^{c}+4 \cdot 9^{d})}}{\partial x}=k \dfrac{\partial (a+2b+3c+4d)}{\partial x}\] \[\ln(9) \cdot 9^{a}=k \times1\] Similarly doing for b,c,d we have \[2\ln(9)\cdot 9^{b}=k \times 2\] \[3\ln(9)\cdot9^{c}=k \times 3\] \[4\ln(9) \cdot 9^{d}=k \times4\] \[\implies a=b=c=d\]. So substituting in \(g\) the above condition we get \[10a=15\implies a=1.5\] Therefore \[\inf{f(x)}=27×(1+2+3+4)=270\]Log in to reply

the given expression (S) whose minimum value is to be calculated can be written as

S = 9^a + 9^b + 9^b + 9^c + 9^c + 9^c + 9^d + 9^d + 9^d + 9^d

Now applying AM - GM inequality

S/10 >= ( 9 ^(a+2b+3c+4d) )^1/10

S/10 >= 9^(15/10)

S/10 >= 9^3/2 = 27

Therefore S >= 270

and hence minimum value of S is 270

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The answer is correct.

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@Vignesh S

Please post the next problem(Problem 50) of the contest. And after this problem on this note, the contest will be continued on JEE-Advanced Maths Contest(Continued) note.

Thanks!

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