We are given a real-valued function f(x) of real numbers that satisfies f(x + xy) = f(x) + f(xy) for all real x, y.
f(2020) = 1. 
We need to find f(2021)
f(x + xy) = f(x) + f(xy)
We know that when f(x+y) = f(x) + f(y) 
Then, f(x) = kx
So, here, assuming xy = w
f(x+w) = f(x) + f(w) = kx
We know that f(2020) = k*2000 = 1
k = $\dfrac{1}{2000}$
So, f(2021) = k*2021 = $\dfrac{2021}{2020}$

We know that$\log_{2} [\log_{3}(\log_{4}a)] = \log_{3} [\log_{4} (\log_{2}b)] = \log_{4} [\log_{2} (\log_{3}c)] = 0$Now Equating,$\log_2[\log_3(\log_4a)]=0$If we take 2 on RHS, it will be$[\log_3(\log_4a)]=2^0$$\log_3(\log_4a)=1$Now, again taking 3 on the RHS,$(\log_4a)=3^1$Now taking 4 on RHS$a=4^3=64$Similarly, equating$\log_3[\log_4(\log_2b)]=0$If we take 3 on RHS, it will be $\log_4(\log_2b)=3^0$
$\log_4(\log_2b)=1$Now taking 4 on RHS$\log_2b=4^1$Now taking 2 on the RHS$b=2^4=16$Similarly, equating$\log_4[\log_2(\log_3c)]=0$Now taking 4 on RHS$\log_2(\log_3c)=1$Now, taking 2 on RHS$\log_3c=2$Thus,$c=9$Therefore, a+b+c = 64+16+9 = 89

We are given that$S_{5} = S_{9}$We know the sum of n terms of a given AP is$S_n=\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$Where a is the first term, d is the common difference, and n is the number of terms.Equating$S_{5} = S_{9}$$\dfrac{5}{2}\left[2a+4d\right]=\dfrac{9}{2}\left[2a+8d\right]$$10a+20d=18a+72d$$-8a=52d$$a=-\dfrac{52d}{8}=-\dfrac{13d}{2}$Now we need to find the ratio of $a_3:a_5$$\dfrac{a_3}{a_5}=\dfrac{a+2d}{a+4d}=\dfrac{-\dfrac{13d}{2}+2d}{-\dfrac{13d}{2}+4d}$$\dfrac{a_3}{a_5}=\dfrac{-\dfrac{9d}{2}}{-\dfrac{5d}{2}}=\dfrac{9}{5}$9:5

We are give $BA=AB$.We will post-multiply this by $A^{-1}$=$BAA^{-1}=ABA^{-1}$= $B=ABA^{-1}$ [$AA^{-1}=I$]Now, we will pre-mutliply the equation with $A^{-1}$= $A^{-1}B=A^{-1}ABA^{-1}$=$A^{-1}B=BA^{-1}$Adding $A^{-1}A$ both sides -=$A^{-1}B+A^{-1}A=BA^{-1}+AA^{-1}$=$A^{-1}\left(B+A\right)=\left(B+A\right)A^{-1}$Taking the inverse on both sides [$(AB)^{-1}=B^{-1}A^{-1}$]=$\left(A+B\right)^{-1}A=A\left(A+B\right)^{-1}\rightarrow1$Similarly, we will get =$\left(A+B\right)^{-1}B=B\left(A+B\right)^{-1}\rightarrow2$Now, we need to find the value of$2A — B — A(A + B)^{−1}A + B(A + B)^{−1}B$Substituting the value of$\left(A+B\right)^{-1}A=A\left(A+B\right)^{-1}$ from eq. 1, and the value of$\left(A+B\right)^{-1}B=B\left(A+B\right)^{-1}$ from eq. 2 -=$2A—B—AA(A+B)^{−1}+BB(A+B)^{−1}$=$2A—B—A^2(A+B)^{−1}+B^2(A+B)^{−1}$=$2A—B—(A+B)^{−1}\left(A^2-B^2\right)$Now, $(A+B)(A-B)=A^2+AB-BA-B^2$, and since $AB=BA$, thus$(A+B)(A-B)=A^2-B^2$=$2A—B—(A+B)^{−1}\left(A+B\right)\left(A-B\right)$=$2A—B—\left(A-B\right)$ [$(A+B)(A+B)^{-1}=I$]=$2A—B—A+B$=$A$

If the angles A, B,C of a triangle are in arithmetic progression such 
that $\sin(2A + B) = 1/2$ then $\sin(B + 2C)$ is equal to
The sum of all angles in a triangle = 180$^{\circ\ }$
Angles A, B and C are in AP.
Thus, 2B = A + C
B + 2B = 180
B = 60
Also we are given that$\sin(2A + B) = 1/2$ 
This is only possible if 2A+B = 30 degree or 150 degree
2A+B cannot be equal to 30 as B = 60 and A cannot be negative.
Therefore, 2A+B = 150
Since, B = 60
A = 45 and C = 75
Therefore,$\sin(B + 2C)$ = 
B + 2C = 60+150 = 210 
$\sin(210^{\circ\ })$ = -1/2

The unit digit in $(743)^{85}$ = 3 (since the cyclicity for powers of numbers with unit digit3 is 4)The unit digit in $(525)^{37}$ = 5The unit digit in $(987)^{96}$ = 1 (since the cyclicity for powers of numbers with unit digit 7 is 4)The unit digit of$(743)^{85} −(525)^{37} + (987)^{96}$ will be = (.......3) - (.......5) + (.......1)Now, (3-5+1) will give us a negative number, thus we will take a carry from the second last digit of the first number.Therefore, the unit digit will be = 13 - 5 + 1 = 9

$3 \sin^{2}x + 12 \cos x - 3 = p$$3\left(1-\cos^2x\right)+12\cos x-3=p$$3-3\cos^2x+12\cos x-3=p$$-3\cos^2x+12\cos x=p$Adding and subtracting 12 in LHS$-3\cos^2x+12\cos x-12+12=p$$-3\left(\cos^2x-4\cos x+4\right)+12=p$$-3\left(\cos x-2\right)^2+12=p\rightarrow1$Now, we know that -$-1\le\cos x\le1$$-3\le\cos x-2\le-1$$1\le\left(\cos x-2\right)^2\le9$$-27\le-3\left(\cos x-2\right)^2\le-3$$-15\le-3\left(\cos x-2\right)^2+12\le9$Substituting it as p, from eq. 1-$-15\le p\le9$

ABCD is a quadrilateral whose diagonals AC and BD intersect at O. Iftriangles AOB and COD have areas 4 and 9 respectively, then the minimumarea that ABCD can have isIt is given thatABCD is a quadrilateral whose diagonals AC and BD intersect at O.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/jhdjdom-2025-06-19-11-37-34_onJNS09.png">Now, Triangles AOB and COD have areas 4 and 9, respectively.We need to find outthe minimum area that ABCD can have isWe know ar ABCD = ar AOB + ar COD + ar BOC + ar AODIn any quadrilateral, when the diagonals are drawn, they divide the quadrilateral into four triangles. The product of the areas of two opposite triangles is equal to the product of the areas of the other two opposite trianglesThus,(ar AOB)*(ar COD) =(ar BOC)*(ar AOD)(ar BOC)*(ar AOD) = 4*9 = 36The sum has to be minimum, and the product of ar BOC and ar AOD is 36.This will happen when both areas are equalThus, ar BOC = 6 = ar AODTherefore, the minimum area that ABCD can have is = 4+9+6+6 = 25

We need to find the highest possible value of the ratio of a four digit number and the sum of its four digits is:Let us see the minimum four-digit number1000 The ratio will be$\frac{1000}{1+0+0+0}=1000$Similarly, for the maximum four-digit number9999The ratio will be$\frac{9999}{9+9+9+9}=277.75$The ratio will be maximum when the numerator is maximum and the denominator is minimum.But as we increase the numerator, the denominator is also increasing.This results in a decrease in the overall ratio.Also, from the given options, we see that the maximum ratio can be 1000.Thus, it is the correcr answer.

$g(x)=-2x$, and it passes through (a,b) = $b=-2a\rightarrow1$$f(x) = ax^{2} + bx + c$, and it cuts the x-axis at (-2,0) = $0=a(4)+b(-2)+c$. We will substitute the value of $b=-2a$ from eq. 1 = $0=4a+4a+c$ = $c=-8a$=$f(x) = ax^{2} + (-2a)x + (-8a)$=$f(x) = ax^{2} -2ax -8a$minimum value of f(x) + 9a + 1 will be -=$f(x)+9a+1 = ax^{2} -2ax -8a+9a+1$=$f(x)+9a+1 = ax^{2} -2ax +a+1$= $f(x)+9a+1 = a(x^{2} -2x + 1) + 1$=$f(x)+9a+1=a\left(x-1\right)^2+1$The minimum value of $a(x-1)^2$ is 0; thus, the minimum value of f(x) + 9a + 1 will be 1.

ipmat-indore 2021 Complete Paper Solution | MCQ

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

ipmat-indore 2021 Complete Paper Solution | MCQ

Question 1.

<div><p>Suppose that a real-valued function f(x) of real numbers satisfies f(x + xy) = f(x) + f(xy) for all real x, y, and that f(2020) = 1. Compute f(2021)</p></div>

Question 2.

<div><p>Suppose that <span>$\log_{2} [\log_{3}(\log_{4}a)] = \log_{3} [\log_{4} (\log_{2}b)] = \log_{4} [\log_{2} (\log_{3}c)] = 0$</span>. Then the value of a + b + c is</p></div>

Question 3.

<div><p>Let <span>$S_{n}$</span> be sum of the first n terms of an A.P. If <span>$S_{5} = S_{9}$</span>, what is the ratio of <span>$a_3:a_5$</span></p></div>

Question 4.

<div><p>If A, B and A + B are non-singular matrices and AB = BA, then <span>$2A — B — A(A + B)^{−1}A + B(A + B)^{−1}B$</span> equals</p></div>

Question 5.

<div><p>If the angles A, B,C of a triangle are in arithmetic progression such that <span>$\sin(2A + B) = 1/2$</span> then <span>$\sin(B + 2C)$</span> is equal to</p></div>

Question 6.

<div><p>The unit digit in <span>$(743)^{85} −(525)^{37} + (987)^{96}$</span> IS ________</p></div>

Question 7.

<div><p>The set of all real values of p for which the equation <span>$3 \sin^{2}x + 12 \cos x - 3 = p$</span> has at least one solution is</p></div>

Question 8.

<div><p>ABCD is a quadrilateral whose diagonals AC and BD intersect at O. If triangles AOB and COD have areas 4 and 9 respectively, then the minimum area that ABCD can have is</p></div>

Question 9.

<div><p>The highest possible value of the ratio of a four digit number and the sum of its four digits is:</p></div>

Question 10.

<div><p>Consider the polynomials <span>$f(x) = ax^{2} + bx + c$</span>, where a 0, b, c are real, and g(x) = -2x. If f(x) cuts the x-axis at (-2,0) and g(x) passes through (a, b), then the minimum value of f(x) + 9a + 1 is</p></div>