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:

1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
2. 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.
3. Please make a substantial comment.
4. 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.
5. 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.
6. 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.
7. You can use tricks/short methods/specific cases to post the solution of a problem only if they are logically valid.
8. In the case of any argument or disagreement among the solvers, the decision/judgement is passed to me.
9. 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.
10. Proof problems are not allowed.
11. 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) :

PROBLEM xxx (number of problem) :

Note by Sandeep Bhardwaj
3 years, 6 months ago

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Here 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!

- 3 years, 6 months ago

Solution of problem 1

The required probability is

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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?

- 3 years, 6 months ago

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

- 3 years, 6 months ago

Ooops, sorry for that! I don't know who did that.

Can you please post that again?

- 3 years, 6 months ago

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

Problem 51

If $(a+\omega)^{-1}+(b+\omega)^{-1}+(c+\omega)^{-1}+(d+\omega)^{-1}=2\omega^{-1}$ 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 $(a+1)^{-1}+(b+1)^{-1}+(c+1)^{-1}+(d+1)^{-1}$ ?

- 1 year, 11 months ago

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!

- 3 years, 6 months ago

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

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

- 3 years, 6 months ago

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

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

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

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

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

$(2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \le (2+\sqrt{5})^{p} \le (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1$
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 $4$

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

A graphical approach.

We have to see the nature of discriminant

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

Oh yes.

- 3 years, 6 months ago

I am changing the wording of the question ,as suggested by @Vighnesh Shenoy. @Vighnesh Shenoy and @Mayank Chaturvedi,please make changes in your solutions accordingly.

- 3 years, 6 months ago

I guess he should have asked if the roots were real or complex.

- 3 years, 6 months ago

Mayank should post the new question.

- 3 years, 6 months ago

OK. @Mayank Chaturvedi,please post the next question

- 3 years, 6 months ago

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.

- 3 years, 6 months ago

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

- 3 years, 6 months ago

One way of solving cubic equation of the type $ax^3+bx+c=0$ is to take $x=d+y/d$ and manipulate the value of $y$ to get a solvable $6th$ degree equation in $d$

- 3 years, 6 months ago

By solvable, do you mean quadratic type equations with higher degrees?

- 3 years, 6 months ago

yes

- 3 years, 6 months ago

Great!!!

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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

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

- 3 years, 6 months ago

Problem 41

$\displaystyle \int tanx. tan2x. tan3x\text{ }\,dx$

Saarthak Marathe has posted the solution to the problem.Thanks!

- 3 years, 5 months ago

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

Now integrating we get,

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

- 3 years, 5 months ago

Correct!Post the next question.

- 3 years, 5 months ago

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!

- 3 years, 6 months ago

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)

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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.

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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: $-3$

- 3 years, 6 months ago

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 $\boxed{-3}$

- 3 years, 6 months ago

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 -3

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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

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 $\binom{2n-1}{n}+(n+1)\binom{2n-1}{n-1}$

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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

$\displaystyle = Ax +B$ (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,

$\displaystyle A = \frac{65}{23}$

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

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

The answer is $0$.

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

@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!

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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

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

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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

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

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

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.

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

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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 $\boxed{d_{max}=1}$

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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!

- 3 years, 6 months ago

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.

- 3 years, 6 months ago

Correct.

- 3 years, 6 months ago

Problem 39:

Solve

$\displaystyle {x}^{3}\frac{dy}{dx}={y}^{3}+{y}^{2}\sqrt{{y}^{2}-{x}^{2}}$

- 3 years, 6 months ago

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.

- 3 years, 6 months ago

Problem 43 :

Find the remainder when $32^{32^{32}}$ is divided by 7.
Mayank Chaturvedi has solved this problem with the correct solution.

- 3 years, 5 months ago

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

Hence the answer is $\boxed 4$

- 3 years, 5 months ago

The answer is right, however you used
$32^{32^{32}} \equiv (32^{32})^{32} \equiv 9^{32} \equiv 4 \pmod{7}$

- 3 years, 5 months ago

Yes i agree with you. Have a check @Saarthak Marathe the answer is a coincidence this time.

Let me try once.

Reference-euler's theorem

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

- 3 years, 5 months ago

Perfect.

- 3 years, 5 months ago

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!

- 3 years, 5 months ago

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

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

The coefficient of $x^{n-6}$ in the expansion:
$n! \times \ [ \ x \ - \frac{\binom{n}{0} \ + \binom{n}{1}}{\binom{n}{0}} \ ] \ [ \frac{x}{2} \ - \frac{\binom{n}{1} \ + \binom{n}{2}}{\binom{n}{1}} \ ] \ .... \ [ \frac{x}{n} \ - \frac{\binom{n}{n-1} \ + \binom{n}{n}}{\binom{n}{n-1}} \ ]$