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?

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

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

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

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

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

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

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

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

A graphical approach.

We have to see the nature of discriminant

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

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

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.

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

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

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

Now integrating we get,

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

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)

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.

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: $-3$

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 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

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

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

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

$\displaystyle = Ax +B$ (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 $0$.

@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 .

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

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

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

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

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

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

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.

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

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

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.

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.

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

Hence the answer is $\boxed 4$

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

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