$\log_ab=\frac{1}{\log_ba}$$\log_{\cos x}(\sin x)=\frac{1}{\log_{\sin x}(\cos x)}$$\log_{\cos x}(\sin x)=t$$t+\frac{1}{t}=2$The minimum value of$t+\frac{1}{t}$ is 2. And this is when t = 1.So,$\log_{\cos x}(\sin x) = 1$cos x = sin xx = $n\pi+\frac{\pi}{4}$Because cos and sin are in the base of log, they can't be negative. They will takea positive value only when x =$2n\pi+\frac{\pi}{4}$

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/image_QELjNOy.png">Let's draw the point on the coordinate axis. For the first point, we would go +32 on the x-axis.For the second point, we will go -16 on y asix.For the third point, we go -8 on the x-axis. And for the next point, we go +4 on the y-axis. For the next one, we go +2 on the x-axis, and for the next one, we go -1 on the y-axis. For the next point, we go -1/2 on the x-axis, and for the next one, we go +1/4 on the y-axis.Let's observe the moment wrt the x-axis.+32, -8, +2, -1/2 ...This is a G.P series with -1/4 as the standard ratio.The final point's x coordinate will be the sum of all the moments with respect to the x-axis.Sum to infinite terms of GP =$\frac{a}{1-r}=\frac{32}{1-\left(-\frac{1}{4}\right)}=\frac{32}{\frac{5}{4}}=\frac{128}{5}$Similarly, let's see the moment wrt the y-axis.-16, +4, -1, +1/4This is a G.P series with -1/4 as the standard ratio.The final point's y coordinate will be the sum of all the moments with respect to the y-axis.Sum to infinite terms of GP = $\frac{a}{1-r}=\frac{-16}{1-\left(-\frac{1}{4}\right)}=\frac{-16}{\frac{5}{4}}=\frac{-64}{5}$The distance from the origin to this point =$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$ =$\sqrt{\left(\frac{128}{5}-0\right)^2+\left(-\frac{64}{5}-0\right)^2}=\sqrt{\left(\frac{128}{5}\right)^2+\left(\frac{64}{5}\right)^2}$$\sqrt{\left(\frac{64}{5}\right)^2\left(2^2+1^2\right)}=\sqrt{\left(\frac{64}{5}\right)^2\left(5\right)}$=$\frac{64}{5}\sqrt{5}$

$x^{2} + y^{2} − 2x − 4y + 5 = 0$This can be rewritten as $x^2+y^2−2x+1−4y+4=0$$\left(x-1\right)^2+\left(y−2\right)^2=0$This is a circle with radius 0. So, this is a point.

We are told that the radius of the small bead is 1/5th of the actual ball.Let us assume that the radius of the initial sphere is r, the radius of the small bead will be r/5.Number of small beads =$\frac{\left(\frac{4}{3}\pi r^3\right)}{\frac{4}{3}\pi\frac{\ r^3}{125}}=125$So, the cost price of each bead =$\frac{1000000}{125}=8000$We are told that he got a profit of 20%. So, the selling price should be 1.2*8000 = 9600.And the selling price is after giving 205 discount. So, the marked price 0.8 MP=9600MP = 12000

The time taken to travel each side:1st side:$T\ =\ \frac{100}{100}=1hr$2nd side:$T\ =\ \frac{100}{200}=0.5hr$3rd side:$T\ =\ \frac{100}{300}=0.33hr$4th side:$T\ =\ \frac{100}{400}=0.25hr$Average speed =$\frac{\text{total distance}}{\text{total time}}$ =$\frac{400}{1+0.5+0.33+0.25}=192.3$

$10^{2023}=2^{2023}\cdot5^{2023}$$10^{2001}=2^{2001}\cdot5^{2001}$$10^{2023}$ can be written as$10^{2001}*2^{22}\cdot5^{22}$The number of factors of $10^{2023}$ that are multiple of$10^{2001}$ will be the number of factors of$2^{22}\cdot5^{22}$Number of factors of $2^{a}\cdot5^{b} = (a+1)(b+1)$ 
Number of factors of $2^{22}\cdot5^{22}$ = 23*23 = 529And the total number of factors for$10^{2023}=2^{2023}\cdot5^{2023}$ is (2023+1)*(2023+1) = $2024^2$Probability =$\frac{529}{2024^2}$

We can draw a figure like this:<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/Screenshot-2025-06-20-141542.png">Let us consider the lengths of AB,BC and CA be 2 units, 4 units and 3 units respectively.As D is the mid point of BC, AD is the median. Also, BD=CD= 2 units.Using Apollonius theorem,$AB^2+AC^2=2\left(AD^2+BD^2\right)$or,$2^2+3^2=2\left(AD^2+2^2\right)$or,$AD^2=\dfrac{5}{2}$or,$AD=\dfrac{\sqrt{\ 10}}{2}$ ------(1)Now,$AM$ is perpendicular to $BC$.So,$\triangle\ AMB$ and$\triangle\ AMC$ both are right angled triangles.Using Pythagoras theorem in$\triangle\ AMB$, we can write,$AM^2=AB^2-BM^2$or,$AM^2=2^2-BM^2$ -----(2)Similarly,using Pythagoras theorem in $\triangle\ AMC$, we can write,$AM^2=AC^2-CM^2$or,$AM^2=3^2-\left(4-BM\right)^2$ -------(3)From equation (2) and (3),$3^2-\left(4-BM\right)^2=2^2-BM^2$or,$3^2-2^2=\left(4-BM\right)^2-BM^2$or,$5=\left(4-2BM\right)\cdot4$or,$BM=\dfrac{11}{8}$ -----(4)From equation (1) and (4),$\dfrac{BM}{AD}=\dfrac{\dfrac{11}{8}}{\dfrac{\sqrt{\ 10}}{2}}=\dfrac{11}{4\sqrt{\ 10}}$So,$BM:AD=11:4\sqrt{\ 2}$

Given, det ($A^{3}$ − $3A^{2}$ − 5A) = 0So, det ($A^3$- $3A^2$ - $5A$) = det ((A)($A^2$-$3A$-$5I$)) = det $A\times det $($A^2$-$3A$-$5I$)=0So, det $A$=0or, $\begin{vmatrix}1ㅤ2 \\ 3ㅤa\end{vmatrix}$=0or, $1\times\ a-2\times\ 3=0$or, $a=6$

The equation of circle is $x^{2} + y^{2} - 6x + 4y - 12 = 0$We can compare this with standard equation of circle, $x^2+y^2+2gx+2fy+c=0$so, $g=-3, f=2, c=-12$Now, centre of circle is $\left(-g,-f\right)$. So, centre will be $\left(3,-2\right)$Radius of circle will be r = $\sqrt{g^2+f^2-c}=\sqrt{3^2+\left(-2\right)^2-\left(-12\right)}=\sqrt{25}=5$ unitsNow for a line to be tangent to a circle, perpendicular distance from the centre of the circle to the line, should be equal to its radius.For the line $4x+3y+19=0$, perpendicular distance from the centre =$\dfrac{\left|4\left(3\right)+3\left(-2\right)+19\right|}{\sqrt{\left(4^2+3^2\right)}}=\dfrac{25}{5}=5$ units = radiusSo,$4x+3y+19=0$ is a tangent to the circle.For the line$4x+3y-31=0$,perpendicular distance from the centre = $\dfrac{\left|4\left(3\right)+3\left(-2\right)-31\right|}{\sqrt{\left(4^2+3^2\right)}}=\dfrac{25}{5}=5$ units = radiusSo, $4x+3y-31=0$ is also a tangent to the circle.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/Screenshot-2025-06-19-173723.png">Centre of circle is$\left(3,-2\right)$Purple colour line is $4x+3y-31=0$Green colour line is$4x+3y+19=0$OA and OB are radii perpendicular to the respective tangents.

Given,$a_{1} a_{2}, a_{3}$ are in G.P.
Let the common ratio of the G.P. be$r$
So,$a_2=a_1r,\ a_3=a_1r^2$
Putting the values in equation,$a_{1}x^{2} + 2a_{2} x + a_{3} = 0$
So,$a_1x^2+2a_1xr+a_1r^2=0$
or,$a_1\left(x^2+2xr+r^2\right)=0$
or,$a_1\left(x+r\right)^2=0$
Now,$a_1\ne\ 0$
So,$\left(x+r\right)^2=0$
or,$x=-r$ is the only root
So the common root is also$x=-r$
Putting this in the second equation,$b_{1}x^{2} + 2b_{2}x + b_{3} = 0$
or,$b_{1}r^{2} - 2b_{2}r + b_{3} = 0$
Now we can replace$r^2=\dfrac{a_3}{a_1},\ r=\dfrac{a_2}{a_1}$
So,$b_1\cdot\dfrac{a_3}{a_1}+b_3=2b_2\cdot\dfrac{a_2}{a_1}$
Dividing all terms by$a_3$,
$\dfrac{b_1}{a_1}+\dfrac{b_3}{a_3}=2\cdot\dfrac{b_2a_2}{a_1a_3}=\dfrac{2b_2a_2}{a_2^2}=\dfrac{2b_2}{a_2}$
(Since,$a_1,a_2,a_3$ are in G.P.,$a_1\cdot a_3=a_2^2$)
so,$\dfrac{b_1}{a_1}+\dfrac{b_3}{a_3}=\dfrac{2b_2}{a_2}$
or,$\dfrac{b_1}{a_1},\dfrac{b_2}{a_2},\dfrac{b_3}{a_3}$ are in A.P.

ipmat-indore 2023 Complete Paper Solution | MCQ

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Question 5.

Question 6.

Question 7.

Question 8.

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Question 10.

ipmat-indore 2023 Complete Paper Solution | MCQ

Question 1.

<div><p>If <span>$\log_{cos x} (\sin x) + \log_{\sin x}(cos x) = 2$</span>, then the value of x is</p></div>

Question 2.

Question 3.

<div><p>The equation $x^{2}$ + $y^{2}$ &minus; $2x$ &minus; $4y$ + $5$ = $0$ represents</p></div>

Question 4.

Question 5.

<div><p>A helicopter flies along the sides of a square field of side length 100 kms. The first side is covered at a speed of 100 kmph, and for each subsequent side the speed is increased by 100 kmph till it covers all the sides. The average speed of the helicopter is</p></div>

Question 6.

<div><p>The probability that a randomly chosen positive divisor of <span>$10^{2023}$</span> is an integer multiple of <span>$10^{2001}$</span> is</p></div>

Question 7.

<div><p>In a triangle ABC, let D be the mid-point of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be</p></div>

Question 8.

<div><p>If $A =\begin{bmatrix}1ㅤ2 \\3ㅤa \end{bmatrix}$ where as is a real number and det ($A^{3}$ &minus; $3A^{2}$ &minus; 5A) = 0 then one of the value of a can be</p></div>

Question 9.

<div><p>Which of the following straight lines are both tangent to the circle $x^{2}$ + $y^{2}$ &minus; $6x$ + $4y$ &minus; $12$ = $0$</p></div>

Question 10.

<div><p>Let <span>$a_{1} a_{2}, a_{3}$</span> be three distinct real numbers in geometric progression. If the equations <span>$a_{1}x^{2} + 2a_{2} x + a_{3} = 0$</span> and <span>$b_{1}x^{2} + 2b_{2}x + b_{3} = 0$</span> has a common root,then which of the following is necessarily true?</p></div>