IPM Indore 2023 Question Paper

Given quadratic equation is $$(5 + \sqrt{2})x^{2} − bx + 8 + 2\sqrt{5} = 0$$

Let the roots of the equation be $$\alpha\ $$ and $$\beta\ $$

So, harmonic mean of roots = $$\dfrac{2\alpha\ \beta\ }{\alpha\ +\beta\ }$$

Now, $$\alpha\cdot\beta\ $$ = Product of roots = $$\dfrac{\left(8+2\sqrt{\ 5}\right)}{\sqrt{\ 2}+5}$$

$$\alpha\ +\beta\ $$ = Sum of roots = $$-\dfrac{\left(-b\right)}{\sqrt{\ 2}+5}=\dfrac{b}{\sqrt{\ 2}+5}$$

So, Harmonic mean = $$2\cdot\dfrac{\dfrac{\left(8+2\sqrt{\ 5}\right)}{\sqrt{\ 2}+5}}{\dfrac{b}{\sqrt{\ 2}+5}}=4$$

or, $$2\cdot\dfrac{8+2\sqrt{\ 5}}{b}=4$$

or, $$2\cdot\dfrac{2\left(4+\sqrt{\ 5}\right)}{b}=4$$

or, $$b=4+\sqrt{\ 5}$$ 

Given,

$$2[x]=x+nx$$

or, $$2[x]=x+n(x-[x])$$

or, $$2[x]=x+nx-n[x]$$

or, $$[x](2+n)=x(n+1)$$

or, $$x=\dfrac{\left[x\right]\cdot\left(n+2\right)}{\left(n+1\right)}$$

Now, $$\left[x\right]$$ can take integral values from 1 to n.

So sum of all values of $$x$$ = $$\dfrac{n+2}{n+1}$$ $$\cdot$$ (Sum of all possible values of $$\left[x\right]$$)

=$$\dfrac{n+2}{n+1}\left(1+2+3+4+....+n\right)$$

=$$\dfrac{n+2}{n+1}\times\ \dfrac{n\left(n+1\right)}{2}$$

=$$\dfrac{n\left(n+2\right)}{2}$$

Given, $$\dfrac{a+b}{b+c} = \dfrac{c+d}{d+a}$$

From, the properties of ratio, $$\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}=\dfrac{a+b+c+d}{a+b+c+d}$$

Now there are two possibilities,

i.) $$a+b+c+d\ne\ 0$$

When, $$a+b+c+d\ne\ 0$$, $$\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}=\dfrac{a+b+c+d}{a+b+c+d}=1$$

So, $$\dfrac{a+b}{b+c}=1$$

or, $$a+b=b+c$$

or, $$a=c$$

ii.) $$a+b+c+d=\ 0$$

In this case $$\dfrac{a+b+c+d}{a+b+c+d}$$ won't give a finite value.

So, either $$a=c$$ or $$a+b+c+d=0$$ is always true.

Given inequality is,

$$\dfrac{x^{2}(x + 1)}{(x - 1)(2x + 1)^{3}}> 0$$ is

or, $$x^2\left(x+1\right)\left(x-1\right)\left(2x+1\right)^3>0$$

Now the terms in square are always positive. So we can eliminate them.

So, the inequality will be, 

$$\left(x+1\right)\left(x-1\right)\left(2x-1\right)>0$$

Plotting the critical points $$x=-1,1,\frac{1}{2}$$ in number line and applying wavy-curve method we get,

The black shaded region is the required domain of $$x$$

So, option A is the correct answer

It is given, 5 contestants are participating in the tournament and each player plays against all the other players exactly once.

So, we can say each player plays 4 matches.

Suppose a player wins all the 4 matches, he will get 4*1=4 points.

In option (c), 2 players are getting 4 points.

This case is not possible because if one player is getting 4 points, he/she has to win all 4 matches. Now when he/she wins all 4 matches, the other players will definitely lose one match for sure. So, other player(s) can't have 4 points.

So, two players getting 4 points is not possible.

So, option (c) is the correct choice

Given,

$$\cos \alpha + \cos \beta = 1$$

Squaring both sides, $$\left(\cos\ \alpha\ +\cos\ \beta\ \right)^2=1$$

or, $$\cos\ ^2\alpha\ +\cos\ ^2\beta\ +2\cos\ \alpha\ \cdot\cos\ \beta\ =1$$

or, $$1-\sin\ ^2\alpha\ +1-\sin\ ^2\beta\ +2\cos\ \alpha\ \cdot\cos\ \beta\ =1$$

or, $$1+2\cos\ \alpha\ \cdot\cos\ \beta\ =\sin^2\alpha\ +\sin^2\beta\ $$

Adding, $$-2\sin\alpha\ \sin\beta\ $$ on both sides,

$$1+2\cos\ \alpha\ \cdot\cos\ \beta-2\sin\alpha\ \sin\beta\ \ =\sin^2\alpha\ +\sin^2\beta\ -2\sin\ \alpha\ \sin\ \beta\ $$

 or, $$1+2\cos\ \left(\alpha+\beta\ \right)\ \ \ =\left(\sin\alpha\ -\sin\beta\ \right)^2$$

Now, for maximum value of $$\left(\sin\alpha−\sin\beta\right)$$, $$\cos\left(\alpha\ +\beta\ \right)$$ should be 1 (its maximum value)

So,$$\left(\sin\alpha\ -\sin\beta\right)\ ^2=1+2\times\ 1=3$$

or, $$\left(\sin\alpha\ -\sin\beta\right)=\sqrt{\ 3}$$

So, maximum value of $$\sin \alpha − \sin \beta$$ is $$\sqrt{\ 3}$$

Given, polynomial $$P(x)$$ leaves a remainder 2 when divided by $$(x-1)$$

or, $$P(1)=2$$

similarly, $$P(x)$$ leaves a remainder 1 when divided by $$(x-2)$$

or, $$P(2)=1$$

Now, when divided by $$(x-1)(x-2)$$, which is a second degree polynomial, remainder must be linear (of the form $$Ax+B$$)

(As remainders always are a degree less than the divisor)

So, let the remainder be $$Ax+B$$, when P(x) is divided by (x-1)(x-2),

Now, $$P(1)=2$$

or, $$A+B=2$$ ----->(i)

also,$$P(2)=1$$

or, $$2A+B=1$$ ------>(ii)

Upon solving equations (i) and (ii), we get, $$A=-1,B=3$$

So the remainder is $$(-1)\cdot x+3=3-x$$

Total students are 120, out of which 80 students are from the Science stream. So, 120-80=40 students must be from commerce stream.

Let us tabulate the data:

$$x$$ denotes number of students from science stream who are MI supporters.

$$80-x$$ denotes number of students from science stream who are CSK supporters.

$$x-30$$ denotes number of students from commerce stream who are CSK supporters.

$$70-x$$ denotes number of students from commerce stream who are MI supporters.

We are asked to find the value of x.

The only constraints on x are:

i.) $$x-30\ge\ 0$$, or, $$x\ge\ 30$$

ii.) $$70-x\ge\ 0$$, or, $$x\le70$$

So, overall, $$30\le\ x\le70$$

Option A (30 or more) satisfies this condition.

So, it is the correct choice.

$$p^3$$ is having unit digit of 4

So, $$p$$ must also be having unit digit of 4

(Since, $$4^3=64$$ is the only case among first 10 natural numbers where unit digit of a perfect cube is 4)

So, $$\left(p+3\right)$$ must have unit digit $$4+3=7$$

Then, unit digit of $$\left(p+3\right)^3$$ will be same as unit digit in case of $$7^3$$ which is 3.       ($$7^3=343$$)

So, unit digit of $$\left(p+3\right)^3$$ is 3.

There are total 900, 3-digit numbers.

So, number of 3-digit numbers, divisible by 3 =$$\left[\dfrac{900}{3}\right]=300$$

Similarly, number of 3-digit numbers, divisible by 4 =$$\left[\dfrac{900}{4}\right]=225$$

Now, number of 3-digit numbers, divisible by both 3 and 4 (means divisible by 12)=$$\left[\dfrac{900}{12}\right]=75$$

Number of 3-digit numbers divisible by either 3 or 4 = $$300+225-75=450$$

So, number of 3-digit numbers divisible by neither 3 nor 4 = $$900-450=450$$

So, probability that the randomly selected number is neither divisible by 3 nor by 4 = $$\dfrac{450}{900}=\dfrac{1}{2}$$