IPM Indore 2022 Question Paper

MANAGEMENT has 2 M's, 2 A's, 2 N's, 1 G, 2 E's, 1 T

Vowels are A and E.

No two vowels are together. So, between every vowel there has to be atleast a consonant.

So, if we write the 6 consonants like:

$$\times\ M\times\ M\times\ N\times\ N\times\ G\times\ T\times\ $$

There are 7 gaps, out of which we need to place 4 vowels.

No. of ways of selecting 4 places out of 7 and then placing the 4 vowels are $$^7C_4\times\ \dfrac{4!}{2!\cdot2!}$$

Number of ways in which the 6 consonants can be arranged =$$\dfrac{6!}{2!\cdot2!}$$

So, total number of permutations =$$^7C_4\times\ \dfrac{4!}{2!\cdot2!}\times\ \dfrac{6!}{2!\cdot2!}$$

=$$\dfrac{7!}{4!\cdot3!}\times\ \dfrac{4!}{2!\cdot2!}\times\ \dfrac{6!}{2!\cdot2!}=\dfrac{7!}{3!}\times\ \dfrac{6!}{\left(2!\right)^4}=\dfrac{5040}{6}\times\ \dfrac{720}{16}=315\times\ 120=37800$$

Given,

$$81^x+81^{f\left(x\right)}=3$$

Now, $$81^{f\left(x\right)}=3-81^x$$

or, $$81^{f\left(x\right)}=3-3^{4x}$$

Now, $$81^{f\left(x\right)}$$ is an exponential term which is always greater than zero

So, $$3-3^{4x}>0$$

or, $$3>3^{4x}$$

or, $$4x<1$$

or, $$x<0.25$$

So, x will lie in the interval $$(-\infty, 0.25)$$

The lines are concurrent, so area of triangle formed by the lines will be zero.

So, $$\begin{vmatrix}1&-1 & -1 \\ 2 &3&-12\\ 2 & -3 & k \end{vmatrix}$$=0

Applying transformation, C2-->C1+C2 and C3-->C3+C1,

$$\begin{vmatrix}1&0 & 0 \\ 2 &5&-10\\ 2 & -1 & (k+2) \end{vmatrix}$$=0

Expanding along R1,

$$1\cdot$$ $$\begin{vmatrix}5 & -10 \\ -1&(k+2)\end{vmatrix}$$=0

So, $$5\left(k+2\right)-\left(-1\right)\cdot\left(-10\right)=0$$

or, $$5k+10-10=0$$

or, $$5k=0$$

or, $$k=0$$

The standard ellipse is in the form of $$\ \dfrac{\ X^2}{\left(A\right)^2}+\dfrac{\ Y^2}{\left(B\right)^2}=1$$, where the major axis is known by comparing the values of A and B. For instance, If A>B , then major axis is X-axis and focus lie on the X axis, else if B>A, then major axis is Y axis and focus lie on Y axis.

Now, first lets see if a and b , both positive or not :

a = $$Sin\sqrt{\ 2}-Sin\sqrt{\ 3}$$, b =$$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$

$$\sqrt{\ 2}\ =\ 1.414\ \ ,\ \sqrt{\ 3}\ =1.732,\ \ \ \dfrac{\ \pi\ }{2}\ =\ 1.5714\ \ $$

Now, Lets use Sin A - Sin B = 2$$Cos\dfrac{\ A+B}{2}Sin\ \dfrac{\ A-B}{2}$$

$$Sin\sqrt{\ 2}-Sin\sqrt{\ 3}\ =\ 2Cos\dfrac{\ \sqrt{\ 2}+\sqrt{\ 3}}{2}Sin\ \dfrac{\ \sqrt{\ 2}\ -\sqrt{\ 3}}{2}$$

                                         = $$2Cos\left(1.573\right)Sin\left(-0.159\right)$$

Now, as 1.573 > $$\ \frac{\ \pi\ }{2}$$ , this implies Cos(1.573) is negative value & Sin(-0.159) is also negative .. Therefore (-Ve)x(-Ve)  is positive . Hence we can conclude 'a' is positive .. Therefore it can be written in the form of $$A^2$$ ..

Similarly, we can check for b :

b =$$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$

Now, lets use$$CosA-CosB=\ -2\ Sin\ \dfrac{\ A+B}{2}Sin\ \dfrac{\ A-B}{2}$$    

$$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$ =$$-2Sin\left(1.573\right)Sin\left(-0.159\right)$$ 

As both the values 1.573 and 0.159 are less than $$\pi\ $$ , both sin values are positive , but there are two negative signs , hence again positive number. Therefore b > 0.. So this can also be written in the format of $$B^2$$.

Therefore, we can conclude that its a ellipse. Now lets find the major axis by comparing a & b.

Now lets check for a > b :

  $$Sin\sqrt{\ 2}-Sin\sqrt{\ 3}$$  >  $$Cos\sqrt{\ 2}-Cos\sqrt{\ 3}$$,

 $$Sin\sqrt{\ 2}-Cos\sqrt{\ 2}$$  >  $$Sin\sqrt{\ 3}-Cos\sqrt{\ 3}$$,

now, squaring on both sides : { $$Sin^2\left(\theta\ \right)+Cos^2\left(\theta\ \right)\ =\ 1$$,    $$\left(Sin(2\theta\right)\ =\ 2Sin\left(\theta\ \right)Cos\left(\theta\ \right)$$ }

1 - 2($$Sin\sqrt{\ 2}Cos\sqrt{\ 2}$$)  >  1 - 2($$Sin\sqrt{\ 3}Cos\sqrt{\ 3}$$)   

-$$\left(Sin(2\sqrt{\ 2}\right)$$  >  -$$\left(Cos(2\sqrt{\ 3}\right)$$  

This implies ,$$Sin2\sqrt{\ 2}\ \ <\ \ Sin2\sqrt{\ 3}$$

$$2\sqrt{\ 2}\ =\ 2.828\ ,\ which\ is\ less\ than\ \pi\ $$

$$2\sqrt{\ 3}\ =\ 3.464 \ ,\ which\ is\ greater\ than\ \pi\ $$

Hence, we can say that the $$Sin2\sqrt{\ 2}\ >\ 0,\ Sin2\sqrt{\ 3}\ <\ 0$$

Therefore , we can say our assumption of a > b is wrong, and the b > a is true relation.

Hence, we can say Its an ellipse with major axis as Y - axis , therefore the focus lies on Y axis .

Set A has 3 elements and set B has 2 elements.

Number of relations from set A to B = $$2^{\left(3\times\ 2\right)}=2^6$$

(Number of relations from a set having $$m$$ elements to a set having $$n$$ elements is given by the formula $$2^{mn}$$)

Number of functions from set A to set B =$$2^3$$

(Number of functions from a set having $$m$$ elements to a set having $$n$$ elements is given by the formula $$n^m$$)

So, probability that a relation is also a function = $$\dfrac{2^3}{2^6}=\dfrac{1}{2^3}=\dfrac{1}{8}$$

Let speed of Ashok be denoted as $$S_A$$, speed of Bipin be denoted by $$S_B$$, speed of Chandan be denoted by $$S_C$$.

Also let Ashok takes $$t$$ seconds to complete the race.

So, $$\dfrac{S_A}{S_B}=\dfrac{t+15}{t}$$ ------>(1)

Also, $$\dfrac{S_A}{S_C}=\dfrac{t+25}{t}$$ ------>(2)

Also, it is given, Ashok beats Bipin by 150 metres.

So, $$\dfrac{S_A}{S_B}=\dfrac{400}{400-150}=\dfrac{400}{250}=\dfrac{8}{5}$$ ------>(3)

From equation (1) and (3),

$$\dfrac{t+15}{t}=\dfrac{8}{5}$$

or, $$5t+75=8t$$

or, $$3t=75$$

or, $$t=25$$

Dividing equation (2) by equation (1),

$$\dfrac{S_B}{S_C}=\dfrac{t+25}{t+15}=\dfrac{25+25}{25+15}=\dfrac{50}{40}=\dfrac{5}{4}=\dfrac{400}{320}$$

So, by the time Bipin has covered 400 metres (completed the race), C covers 320 metres.

So, Bipin beats Chandan by (400-320) metres = 80 metres

Given equation is,

$$x^{2} + |x + 4| + |x − 4| − 35 = 0$$

i.) When $$x>4$$,

$$x^2+x+4+x-4-35=0$$

or, $$x^2+2x-35=0$$

or, $$x^2+7x-5x-35=0$$

or, $$\left(x+7\right)\left(x-5\right)=0$$

so, $$x=5,-7$$

But $$x=-7$$ is not acceptable as $$x>4$$

So, only acceptable value is $$x=5$$

ii.) When $$-4\le x\le4$$

$$x^2+x+4-x+4-35=0$$

or, $$x^2-27=0$$

or, $$x=\pm\ \sqrt{\ 27}$$

But $$\sqrt{\ 27}=5.196$$

So, the obtained values of $$x$$ does not lie in this range. Hence, the case is rejected.

iii.) When $$x<-4$$,

$$x^2-x-4+4-x-35=0$$

or, $$x^2-2x-35=0$$

or, $$x^2-7x+5x-35=0$$

or, $$\left(x-7\right)\left(x+5\right)=0$$

or, $$x=7,-5$$

But $$x=7$$ is not acceptable as $$x<-4$$

So, only acceptable value is $$x=-5$$

So, the roots of the equation are $$x=5,-5$$

Sum of square of roots =$$\left(5\right)^2+\left(-5\right)^2=25+25=50$$

Putting $$x=0$$ in the given equation, we get the following relation,

$$f(f(y)) = y$$

Thus, the function produces the same output when it is iterated twice, once on $$y$$ and then on $$f(y)$$. Since $$f(x)=x$$ is a function that satisfies this condition, we can assume $$f(x)$$ to be equal to $$x$$ for every real number $$x$$. 

Therefore $$f(0)=0$$, and putting this value in the final equation, we get $$[f(0)]^2 + f(0) = 0+0= 0$$

Lets use the properties of logarithms :

$$\log_ba\ =\dfrac{\ 1}{\log_ab}\ ,\ \ n\log_ba\ =\log_ba^n\ \&\ \ \log_{b^n}a\ =\frac{\ 1}{n}\log_ba,\ \ \ b^{\log_ba}=a$$

Now, the inequality can be written as :

$$\log_{27}8\le\log_3x<9^{\ \frac{\ 1}{\log_23}}$$

$$\log_{3^3}2^3\ \le\ \log_3X\ \ <\ 9^{\log_32}$$

this implies : $$\log_32\ \le\ \log_3X\ \ <\ 3^{2\left(\log_32\right)}$$

$$\log_32\ \le\ \log_3X\ \ <\ 3^{\log_34}$$

$$\log_32\ \le\ \log_3X\ \ <\ 4$$

This implies that, $$X\ \ge\ 2\ \&\ X\ <\ 3^4$$

Therefore, X lies in [2,81) .

As already given $$X\ne0\ ,\ we\ can\ say\ either\ X\ can\ be\ positive\ or\ negative\ ,\ that\ is\ 2\ cases\ :$$

If x>0 :

|x| = x , Hence f(x) = 1 for x > 0.

Else if, x < 0:

|x| = -x , Hence f(x) = -1 for x < 0.

Therefore, {1,-1} are possible values of f(x).