JEE-Advanced Maths Contest '16

Solution of problem 1

The required probability is

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

@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?

A graphical approach.

We have to see the nature of discriminant

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

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.

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

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

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

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

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

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

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

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 \((a+1)^{-1}+(b+1)^{-1}+(c+1)^{-1}+(d+1)^{-1}\) ?

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

Now integrating we get,

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

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

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

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

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

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 \( \dfrac{-2}{3} , \dfrac{-2}{3} , \dfrac{1}{3} \)

An alternate solution to @Vighnesh Shenoy 's already 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 points of 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 \(\frac{4}{27}\)

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

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

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

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.

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.

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

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

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

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.

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

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

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

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

@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!

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

The answer is \(0\).

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

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

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 \(\boxed{-3}\)

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: \(-3\)

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

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.

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)

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!

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.

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

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

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

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

\( p = 4 , q = \dfrac{3}{2} \)
\( pq = 6 \)

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

Hence the answer is \(\boxed 4\)

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

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

The above question can be written in the following manner:

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

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 \(\boxed {108545}\)

@Saarthak Marathe

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

Thanks!

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

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

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

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

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

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 event to the area of sample space(\(=3600\)) gives the answer \(\frac{5}{9}\)

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

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 \(\boxed{2}\)

Is this the answer?

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

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

I got the answer 5/2. Applying cauchy-schwarz inequality

\(\sqrt { ({ a }^{ 2 }+b^{ 2 })(x^{ 2 }{ y }^{ 2 }+y^{ 2 }z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(x^{ 2 }+z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(R^{ 2 }-y^{ 2 }) } \ge (axy+byz)\\ \frac { { R }^{ 2 } }{ 2 } \sqrt { ({ a }^{ 2 }+b^{ 2 }) } \ge (axy+byz)\)

\( P(x)(x+1) = 1 \)
\( P(x)(x+1) - 1 = g(x) \)
\( g(x) \) 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 \( x = - 1 \),
\(-1 = a(-1)(-2)\ldots(-12) \)
\( -1 = 12!a \)
\( a = -\dfrac{1}{12!} \)
\( P(12)(13) - 1 = -\dfrac{1}{12!}\times 12! \)
\( P(12)(13) - 1 = -1 \)
\( P(12) = 0 \)

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 \( \left(0,\frac{7}{2}\right) \)
\( \therefore \frac{5}{3} \leq 2\beta-1 \leq \frac{7}{2} \)

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

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

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

\(S = \tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{8} + \tan^{-1}\dfrac{1}{18} + \ldots \)
\( S = \displaystyle \sum_{k=1}^{\infty} \tan^{-1}\dfrac{1}{2k^{2}} = \sum_{k=1}^{\infty} \tan^{-1}(2k+1) - \tan^{-1}(2k-1) = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4} \)
\( \therefore n = 8 \)

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.

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

\( I = \displaystyle \int_{0}^{\infty} \left \lfloor \dfrac{n}{e^{x}} \right \rfloor dx \)
\( e^{-x} = t \rightarrow dx = \dfrac{-dt}{t} \)
\( I = \displaystyle \int_{0}^{1} \left \lfloor nt \right \rfloor \dfrac{dt}{t} \)
\( I = \displaystyle \sum_{k=1}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} \dfrac{k}{t} dt \)
\( I = \displaystyle \sum_{k=1}^{n-1} k\log\left(\dfrac{k+1}{k}\right) \)
\( S = \displaystyle \sum_{k=1}^{n-1} \left[(k+1)\log(k+1)-k\log(k) - \log(k+1)\right]\)
\( S = n\log(n) - log(n!)\)

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 A and D are correct.

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

Circle curcumscribing quad is l1l2+(lambda)l3l4=0 lambda can be found by cofficient of xy=0

Here 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.?

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

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

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

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

@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 4

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

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

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

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

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

The answer is 270 Let \[f(x)=9^{a}+2\cdot 9^{b}+3\cdot 9^{c}+4\cdot 9^{d}\] 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 \[\inf{f(x)}=27×(1+2+3+4)=270\]