For finding sum of coefficients, wehave to put the value of x as 1.So, sum of coefficients =$(5-9)^4=\left(-4\right)^4=\left(4\right)^4=256$

Given, $A=\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ0ㅤ1 \\ 0ㅤ1ㅤ0 \end{bmatrix}$ Now, $A^2=A\cdot A=\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ0ㅤ1 \\ 0ㅤ1ㅤ0 \end{bmatrix}\times\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ0ㅤ1 \\ 0ㅤ1ㅤ0 \end{bmatrix}$ Upon performing matrix multiplication, $A^2=\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ1ㅤ0 \\ 0ㅤ0ㅤ1 \end{bmatrix}$ So, $A^2$ is identity matrix $I$ So, $A^3=A^2\cdot A=I\cdot A=A$ $A^6=(A^2)^3=I^3=I$ $A^9=(A^2)^4\cdot A=(I)^4\cdot A=I\cdot A=A$ So, $A^9+A^6+A^3+A=A+I+A+A=3A+I$ Now, $3A+I=3\cdot\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ0ㅤ1 \\ 0ㅤ1ㅤ0 \end{bmatrix}+\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ1ㅤ0 \\ 0ㅤ0ㅤ1 \end{bmatrix}$ $=\begin{bmatrix}3ㅤ0ㅤ0 \\ 0ㅤ0ㅤ3 \\ 0ㅤ3ㅤ0 \end{bmatrix}+\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ1ㅤ0 \\ 0ㅤ0ㅤ1 \end{bmatrix}$ $=\begin{bmatrix}4ㅤ0ㅤ0 \\ 0ㅤ1ㅤ3 \\ 0ㅤ3ㅤ1 \end{bmatrix}$ Now, determinant of $(A^9+A^6+A^3+A)=\begin{vmatrix}4ㅤ0ㅤ0 \\ 0ㅤ1ㅤ3 \\ 0ㅤ3ㅤ1 \end{vmatrix}$ Breaking along $R_1$, $=4\cdot\begin{vmatrix}1ㅤ3 \\ 3ㅤ1 \end{vmatrix}$ $=4(1\times 1-3\times 3)$ $=4(1-9)$ $=-32$ So, absolute value of determinant is $32$.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/Screenshot-2025-06-17-095759.png">The shaded region above represents the area enclosed by$2|x| + 3|y| \leq 6$Now area of$\triangle\ OAD\ =\ \dfrac{1}{2}\cdot3\cdot2=3$ sq. unitLike$\triangle\ OAD$, we are also having$\triangle\ OAB$,$\triangle\ OBC$,$\triangle\ OCD$So, total area $=4\times\ 3=12$ sq. unit

$\sin \alpha + \sin \beta = \dfrac{\sqrt{2}}{\sqrt{3}}$
Squaring this equation we get,$\sin^2\ \alpha\ +\sin^2\beta\ +2\sin\ \alpha\ \sin\ \beta\ =\dfrac{2}{3}$ ------(1)
$\cos \alpha + \cos \beta = \dfrac{1}{\sqrt{3}}$
Squaring this equation we get,$\cos^2\alpha\ +\cos^2\beta\ +2\cos\ \alpha\ \cos\ \beta\ =\dfrac{1}{3}$ --------(2)
Adding equation (1) and (2),
$\left(\sin^2\alpha\ +\cos^2\alpha\right)\ +\left(\sin^2\beta\ +\cos^2\beta\right)\ +2\left(\sin\alpha\ \sin\beta\ +\cos\alpha\ \cos\beta\ \right)=\dfrac{2}{3}+\dfrac{1}{3}=1$ 
or,$1+1+2\cos\left(\alpha\ -\beta\ \right)=1$
or,$2\cos\left(\alpha\ -\beta\ \right)=-1$
or,$\cos\left(\alpha\ -\beta\ \right)=-\dfrac{1}{2}$
Now,$\cos\left(\alpha\ -\beta\ \right)=2\cos^2\left(\dfrac{\alpha\ -\beta\ }{2}\right)-1$
or,$-\dfrac{1}{2}=2\cos^2\left(\dfrac{\alpha\ -\beta\ }{2}\right)-1$
or,$2\cos^2\dfrac{\alpha\ -\beta\ }{2}=1-\dfrac{1}{2}=\dfrac{1}{2}$
or,$\cos^2\dfrac{\alpha\ -\beta\ }{2}=\dfrac{1}{4}$
So,$\left(20\cos\dfrac{\alpha\ -\beta\ }{2}\right)^2=400\cdot\cos^2\left(\dfrac{\alpha\ -\beta\ }{2}\right)=400\cdot\dfrac{1}{4}=100$

The number of ways to arrange 'n' distinct entities in a circular table = (n-1)!
Now, the number of ways in which Mr and Mrs Sharma (1 entity) and 4 persons along with Mr and Mrs Ahuja (4+2=6 entities) can sit together in the round table is = $6!\cdot2!$
(Since there are 1+6=7 entities, no. of ways of arranging them are (7-1)!=6! ways
Also, Mr and Mrs Sharma can be arranged in 2! ways)
But in these arrangements, there are cases in which Mr and Mrs Ahuja are also together. We need to eliminate such cases.
Now, the number of ways in which Mr and Mrs Sharma (1 entity), Mr and Mrs Ahuja (1 entity) and 4 persons (4 entities) can sit together in the round table are = $5!\cdot2!\cdot2!$
(Since there are 6 entities, no. of ways of arranging them is 5! ways
Also, Mr and Mrs Sharma can be arranged in 2! ways, and Mr and Mrs Ahuja can be arranged in 2! ways)
So, total number of seating arrangements = $6!\cdot2!-5!\cdot2!\cdot2!=5!(6\cdot2-2\cdot2)=5!(12-4)=120\times\ 8=960$ ways

By definition, if set X has n elements, P(X) has $2^n$ elementsNow, A is the null set. So, number of elements in set A is zero.So, P(A) will have$2^0=1$ elementor, P(P(A)) will have$2^1=2$ elementsor, P(P(P(A))) will have$2^2=4$ elementsor,P(P(P(P(A)))) will have$2^4=16$ elements

For the maximum possible sum of all the 50 integers, we should maximise the largest 25 integers first.
The largest integer givenis 100
So, the maximum possible value of thelargest 25 integers can be 100,99,98,.......,76
Average of these 25 integers =$\dfrac{100+76}{2}=88$
Now, average of these 25 integers exceeds average of other 25 integers by 50
So, average of the other 25 integers = 88-50=38
So, maximum possible value of sum = $25\ \times\ 88+25\times\ 38=25\times\ 126=3150$

Let, Aruna buys 'a' number of apples and 'm' number of mangoes
So, total C.P. =$20a+25m$
In the first case, whenshe sells all the apples at 10% profit and all the mangoes at 20% loss, total S.P.= $20a\times\ 1.1+25m\times\ 0.8$
In this case, she is neither making profit nor loss.
So, total C.P. = total S.P.
or, $20a+25m=20a\times\ 1.1+25m\times\ 0.8$
or,$20a+25m=22a+20m$
so,$2a=5m$ ----(1)
In the second case,whenshe sells all the apples at 20% loss and all the mangoes at 10% profit, total S.P. =$20a\times\ 0.8+25m\times\ 1.1$
or, Total S.P. =$16a+27.5m$
In this case, she made a loss of INR 150.
So, Total S.P.+150 = Total C.P.
or,$16a+27.5m+150=20a+25m$
or,$2.5m+150=4a$ -----(2)
Now, from equation (1),
$2a=5m$or,$a=2.5m$
Putting this in equation (2),$a+150=4a$
or,$3a=150$
or,$a=\dfrac{150}{3}=50$So, number of apples purchased by Aruna is 50

Given, $\log_{(x^{2})}y + \log_{(y^{2})} x= 1$
or,$\dfrac{\log\ y}{\log\ x^2}+\dfrac{\log\ x}{\log\ y^2}=1$ or,$\dfrac{\log\ y}{2\log\ x}+\dfrac{\log\ x}{2\log\ y}=1$
or,$\dfrac{\log\ y}{\log\ x}+\dfrac{\log\ x}{\log\ y}=2$
or,$\left(\log\ y\right)^2+\left(\log\ x\right)^2=2\left(\log y\right)\left(\log x\right)$
or,$\left(\log\ y\right)^2+\left(\log\ x\right)^2-2\left(\log y\right)\left(\log x\right)=0$
or,$\left(\log\ y-\ \log\ x\right)^2=0$
or,$\log\ y=\log\ x$
or,$x=y$
Now,$y=x^2-30$
or,$x=x^2-30$
or,$x^2-x-30=0$
or,$x^2-6x+5x-30=0$
or,$\left(x-6\right)\left(x+5\right)=0$
or,$x=6$ or$x=-5$
But for$\log x$ to be defined,$x$ has to be positive
So,$x=6$
So,$x^2+y^2=6^2+6^2=36+36=72$

We know, if a,b,c are in A.P. then 2b=a+c
So,$2\cdot(2^{x+3}-2^{2x-1}-16)=2^{2x-1}+16+(-16)$
or,$2\cdot(2^{x+3}-2^{2x-1}-16)=2^{2x-1}$
Multiplying both sides with 2,
$2^2\cdot(2^{x+3}-2^{2x-1}-16)=2^{2x}$
or,$2^{x+5}-2^{2x+1}-64=2^{2x}$
or,$2^{x+5}-2^{2x}\cdot2-64=2^{2x}$
or,$2^{x+5}-64=2^{2x}\cdot3$
or,$2^{2x}\cdot3-2^{x+5}+64=0$
or,$\left(2^{2x}\right)\cdot3-\left(2^x\right)\cdot32+64=0$
let, $2^x=z$
So,$3z^2-32z+64=0$
or,$3z^2-24z-8z+64=0$
or,$3z\left(z-8\right)-8\left(z-8\right)=0$
or,$\left(z-8\right)\left(3z-8\right)=0$
or,$z=8$ or$z=\dfrac{8}{3}$Considering only integral values of $z$, $z=8$So, $2^x=8$or, $x=3$.

ipmat-indore 2022 Complete Paper Solution | SA

Instructions

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

ipmat-indore 2022 Complete Paper Solution | SA

Instructions

<div><p>For the following questions answer them individually</p></div>

Question 1.

<div><p>The sum of the coefficients of all the terms in the expansion of $(5x - 9)^{4}$ is __________.</p></div>

Question 2.

<div><p>$\begin{bmatrix}1ㅤ0ㅤ0 \\ 0ㅤ0ㅤ1\\ 0ㅤ1ㅤ0 \end{bmatrix}$, then the absolute value of the determinant of $(A^9+A^6+A^3+A)$ is ________.</p></div>

Question 3.

<div><p>The area enclosed by $2|x| + 3|y| \leq 6$ is____________ sq. units</p></div>

Question 4.

<div><p>If <span>$\sin \alpha + \sin \beta = \frac{\sqrt{2}}{\sqrt{3}}$</span> and <span>$\cos \alpha + \cos \beta = \frac{1}{\sqrt{3}}$</span>, then the value of <span>$(20 \cos (\frac{\alpha - \beta}{2}))^{2}$</span> is_____</p></div>

Question 5.

Question 6.

<div><p>Let P(X) denote power set of a set X. If A is the null set, then the number of elements in P(P(P(P(A)))) is</p></div>

Question 7.

<div><p>Let 50 distinct positive integers be chosen such that the highest among them is 100 , and the average of the largest 25 integers among them exceeds the average of the remaining integers by 50 . Then the maximum possible value of the sum of all the 50 integers is _________.</p></div>

Question 8.

Question 9.

<div><p>If <span>$\log_{(x^{2})}y + \log_{(y^{2})} x= 1$</span> and <span>$y = x^{2} - 30$</span>, then the value of <span>$x^{2} + y^{2}$</span> is _________</p></div>

Question 10.

<div><p>The numbers <span>$-16, 2^{x+3} -2^{2x -1}-16, 2^{2x-1} + 16$</span> are in an arithmetic progression. Then x equals _________.</p></div>