IPM Indore 2024 Question Paper

Lets assume |X-5| = a & |Y-5| = b :
$$\left|a\right|\ +\ \left|b\right|\ \le\ 6$$
The graph with horizontal axis as "a" & vertical axis "b" will look like :

The points which satisfy this inequality will be (a,b) :

Case I : Points on the line/boundary : 
(0,$$\pm6$$) (1,$$\pm5$$) (2,$$\pm4$$) (3,$$\pm3$$) (4,$$\pm2$$) (5,$$\pm1$$) (6,0) 
(-1,$$\pm5$$) (-2,$$\pm4$$) (-3,$$\pm3$$) (-4,$$\pm2$$) (-5,$$\pm1$$) (-6,0)
Therefore , on the boundary  = 24

Case II : Points which lie inside the boundary :
(0,0)
($$\pm1$$,0) ($$\pm2$$,0) ($$\pm3$$,0), ($$\pm4$$,0) ,($$\pm5$$,0) 
(0,$$\pm1$$) (0,$$\pm2$$) (0,$$\pm3$$), (0,$$\pm4$$) ,(0,$$\pm5$$)
For X = $$\pm1$$ : 2x2x4 = 16
For X = $$\pm2$$ : 2x2x3 = 12
For X = $$\pm3$$ : 2x2x2 = 8
For X = $$\pm4$$ : 2x2x1 = 4
Therefore , inside no of points = 1+10+10+16+12+8+4 = 61

Hence total no of integral points which satisfy the inequality will be = 61+24 = 85.

Given total no of students 150. 
52 (liked tea) = only tea + (tea, coffee) + (tea, juice) + (tea, coffee, juice) .
48 (like juice) = only juice + (juice, tea) + (juice, coffee) + (coffee, juice, tea)
62 (like coffee) = only coffee + (coffee, juice) + (coffee, tea) + (coffee, juice, tea).

adding all three we get :
162 = Exactly 1 drink + 2(Exactly 2 drinks) + 3(Exactly all 3 drinks) . 
but given total 150 =  Exactly 1 drink + Exactly 2 drinks + Exactly all 3 drinks .
Substituting equation 2 in equation 1: 
162 = 150 + Exactly 2 drinks + 2(Exactly all 3 drinks) .

12 = {Exactly 2 drinks + Exactly 3 drinks} + {Exactly 3 drinks} .
12 = {More than 1 drink} + {Exactly 3 drinks}.

Therefore, the maximum of More than 1 drink occurs at exactly 3 drinks is 0. Hence, maximum of More than 1 drink is 12. 

Let the price of the chocolate be P.

It is given that the price of a chocolate is increased by x% and then reduced by x%..

The new price is 96.76% of the original price.

So, the equation will be

$$P\left(1+\dfrac{x}{100}\right)\left(1-\dfrac{x}{100}\right)=\dfrac{96.76}{100}P$$

=> $$1-\dfrac{x^2}{10000}=\dfrac{9676}{10000}$$

=> $$10000-x^2=9676$$

=> $$x^2=324$$

=> $$x\ =18$$

The value of x is equal to 18.

Given that f(a) + g(f(a)) = $$\dfrac{\ 13}{6}$$

f(a) + $$\dfrac{\ 1}{f\left(a\right)}$$ = $$\dfrac{\ 13}{6}$$ 

$$6(f(a))^2-13\left(f\left(a\right)\right)+6\ =\ 0$$

f(a) = $$\dfrac{13\pm\sqrt{13^2-24\left(6\right)\ }\ }{2\times6}$$ 

this implies , f(a) = $$\ \dfrac{\ 3}{2}\ (or)\ \dfrac{\ 2}{3}$$ 

Given that f(a) = |a+|a||
if $$a\le0\ $$, f(a) = |a-a| = 0
else if $$a\ge\ 0$$ , then f(a) = 2a.

Therefore , 2a = $$\ \dfrac{\ 3}{2}\ (or)\ \dfrac{\ 2}{3}$$ 
that is, a = $$\ \dfrac{\ 3}{4}\ (or)\ \dfrac{\ 1}{3}$$

g($$\dfrac{\ 3}{4}$$) = $$\dfrac{\ 4}{3}$$

g($$\dfrac{\ 1}{3}$$) = $$\dfrac{\ 3}{1}$$

Therefore f(g(a)) = $$\dfrac{\ 8}{3}$$ or 6.

Maximum value is 6.

Given a series in GP which are positive real numbers ..Let the first term be "a" and common ratio be "r".
$$p^{th}$$ term =$$a\times r^{p-1}$$ = q 
$$q^{th}$$ term =$$a\times r^{q-1}$$ = p
By dividing both, we get $$r=\ \left(\dfrac{\ q}{p}\right)^{\ \dfrac{\ 1}{p-q}}$$
By substituting this in 2nd equation , we get : $$a\left(\ \dfrac{\ q}{p}\right)^{\ \dfrac{q-1\ }{p-q}}=p$$
Now applying log on both sides :
Log a + $$\dfrac{\ q-1}{p-1}\left(\log\dfrac{\ q}{p}\right)$$ = log p
Therefore , log a = log p $$-$$ $$\ \dfrac{\ q-1}{p-q}\left(\log\ q-\log\ p\right)$$

Hence, log a = $$\ \dfrac{\ \left(p-1\right)\log\ p\ +\ \left(q-1\right)\log\ q}{p-q}$$

Lets assume the distance b/w point and centre of circle as "d" . let radius of circle be R .

There are two cases , one case is in which point is inside the circle and the other one is point lies totally outside circle :

Case I : Point lying outside the circle 
  

The shortest distance is PB = PO - BO = d - r = 4
The longest distance is PA = PO + OA = d + r = 9
Solving this we get r = 2.5 .

Case II : Point lying inside the circle 

The shortest distance is PB = OB - OP = r - d = 4
The longest distance is PA = PO + OA = r + d  = 9
Solving this we get r = 6.5

Therefore the radius of circle can be either 2.5 cm or 6.5 cm

If $$\mid x + 1 \mid + (y + 2)^2 = 0$$

That means x = -1 and y = -2

Then, only LHS can be equal to zero.

And, $$ax - 3ay = 1$$

$$-a +6a = 1$$

$$5a=1$$

Thus, $$a=\dfrac{1}{5}$$

The number of real solutions of the equation $$x^2 - 10 \mid x \mid - 56 = 0$$ is

We need to find the solutions to the equation - $$x^2 - 10 \mid x \mid - 56 = 0$$

Taking two cases

Case 1: x > 0

$$x^2-10x-56=0$$

$$x^2-14x+4x-56=0$$

$$x\left(x-14\right)+4\left(x-14\right)=0$$

$$\left(x+4\right)\left(x-14\right)=0$$

x = -4 and x = 14

Since, x > 0

x = 14 is the only possible value.

Case 2: x < 0

$$x^2+10x-56=0$$

$$x^2+14x-4x-56=0$$

=> $$x\left(x+14\right)-4\left(x+14\right)=0$$

=> $$\left(x-4\right)\left(x+14\right)=0$$

x = 4 and x = -14

Since, x < 0

x = -14 is the only possible value.

Therefore, there are 2 possible real solutions.

$$4^{100}$$ can be written as $$2^{200}$$ which is less than $$2^{300}$$ .

$$2^{100}+3^{100}$$ < $$2(3^{100})$$ < $$3^{200}$$

Now we have to see which is greatest among $$2^{300}$$ & $$3^{200}$$ .
$$2^{300}$$ = $$2^{3^{100}}$$ = $$8^{100}$$
$$3^{200}$$ = $$3^{2^{100}}$$ = $$9^{100}$$

Therefore, $$3^{200}$$ is greatest among all.