IPM Indore 2019 Question Paper

Since the other two circles drawn on the chord are touching the midpoint of the chord, the centre of all three circles lies on the same line. 

Let O be the centre of the main circle (Green Coloured), and the radius of that be 2R. Thus, AO = 2R.

Now, AB is the chord such that the length of the chord is equal to the radius of the circle. Thus, AB = 2R => AC = R

Now, $$OC^2+AC^2=OA^2$$ => $$OC^2+R^2=4R^2$$ => $$OC=\sqrt{3}R$$

Let the radius of the smaller circle below the chord be 'a', and the radius of the bigger circle above the chord be 'A'. 

Then, CD = 2a and OD = 2R

=> $$OC+CD=OD$$

=> $$\sqrt{3}R+2a=2R$$

=> $$a=\dfrac{\left(2-\sqrt{3}\right)R}{2}$$

Also, $$OC+OE=2A$$, and OE = 2R

=> $$\sqrt{3}R+2R=2A$$

=> $$A=\dfrac{\left(\sqrt{3}+2\right)R}{2}$$

=> Ratio of area of larger circle to area of smaller circle = $$\dfrac{\pi A^2}{\pi a^2}$$

=> Ratio of area of larger circle to area of smaller circle = $$\dfrac{\left[\frac{R\left(2+\sqrt{3}\right)}{2}\right]^{^2}}{\left[\frac{R\left(2-\sqrt{3}\right)}{2}\right]^{^2}}$$

=> Ratio of area of larger circle to area of smaller circle = $$\dfrac{7+4\sqrt{3}}{7-4\sqrt{3}}$$

Rationalising the numerator and the denominator - 

=> Ratio of area of larger circle to area of smaller circle = $$\dfrac{7+4\sqrt{3}}{7-4\sqrt{3}}\times\dfrac{7+4\sqrt{3}}{7+4\sqrt{3}}$$

=> Ratio of area of larger circle to area of smaller circle = $$\dfrac{49+48+56\sqrt{3}}{49-48}$$

=> Ratio of area of larger circle to area of smaller circle = $$97+56\sqrt{3}:1$$

Let the side of square be $$(n+1)$$ units.

So we can make a diagram like:

Now $$\triangle\ QRC$$ is a right angled triangle.

So, $$RQ=\sqrt{\ QC^2+RC^2}=\sqrt{\ n^2+1}$$

So, area of square $$PQRS$$=$$\left(\sqrt{\ n^2+1}\right)^2$$ = $$n^2+1$$

Area of ABCD = $$\left(n+1\right)^2$$

So, required ratio = $$\dfrac{n^2+1}{\left(n+1\right)^2}$$

Let the distance AC be $$x$$ metres

Also, let the chord AC subtend an angle $$\theta\ $$ at the centre of the circle.

So, angle subtended by the chord AC on minor arc i.e. at point B will be $$180-\dfrac{\theta}{2}\ $$

Now, applying cosine rule in $$\triangle\ OAC$$,

$$\cos\theta\ =\dfrac{6^2+6^2-x^2}{2\times\ 6\times\ 6}=\dfrac{72-x^2}{72}$$ ------>(1)

Also, applying cosine rule in $$\triangle\ BAC$$,

$$\cos\left(180-\dfrac{\theta}{2}\right)\ =\dfrac{2^2+3^2-x^2}{2\times\ 2\times\ 3}$$

or, $$-\cos\dfrac{\theta}{2}\ \ =\dfrac{2^2+3^2-x^2}{2\times\ 2\times\ 3}=\dfrac{13-x^2}{12}$$ -------->(2)

Now we know,

$$\cos\theta\ =2\cos^2\dfrac{\theta}{2}-1$$

So from (1) and (2) we can say,

$$\dfrac{\left(72-x^2\right)}{72}=2\left(\dfrac{13-x^2}{12}\right)^2-1$$

or, $$\dfrac{\left(72-x^2\right)}{72}=\dfrac{\left(13-x^2\right)^2}{72}-1$$

or, $$72-x^2=\left(13-x^2\right)^2-72$$

or, $$72-x^2=169-26x^2+x^4-72$$

or, $$x^4-25x^2+25=0$$

or, $$x^2=\dfrac{25+\sqrt{\ \left(25^2-4\times\ 25\times\ 1\right)}}{2}=\dfrac{25+\sqrt{\ 525}}{2}$$ (we can't take negative root as $$x^2$$ is always positive)

or,$$x^2=\dfrac{50+2\sqrt{\ 525}}{4}=\dfrac{\left(\sqrt{\ 15}\right)^2+\left(\sqrt{\ 35}\right)^2+2\sqrt{\ 15}\times\ \sqrt{\ 35}}{4}=\left(\dfrac{\sqrt{\ 15}+\sqrt{\ 35}}{2}\right)^2$$

or, $$x=\dfrac{\sqrt{\ 15}\ +\ \sqrt{\ 35}}{2}$$ m

We can draw the graph like this:

We have to basically find the area of ABCD

$$\triangle\ OAB$$ is a right angled triangle

Area of $$\triangle\ OAB=\dfrac{1}{2}\times\ OA\times\ OB=\dfrac{1}{2}\times\ 2\times\ 3=3$$ square units

Now we can see, ABCD is made by four such triangles $$\triangle\ OAB,\triangle\ OBC,\triangle\ OAD$$ and $$\triangle\ ODC$$

So, area of square ABCD = $$4\times\ 3$$ = $$12$$ square units.

Let us consider the two points which are 1 m apart to be$$A(0,0)$$ and $$B(1,0)$$

Let at any instant the position of cow be denoted by coordinates $$C(x,y)$$

So, AC=$$\sqrt{\ x^2+y^2}$$

BC=$$\sqrt{\left(\ x-1\right)^2+y^2}$$

Given,$$\dfrac{AC}{BC}=\dfrac{3}{2}$$

or, $$\dfrac{AC^2}{BC^2}=\dfrac{9}{4}$$

or, $$\dfrac{x^2+y^2}{\left(x-1\right)^2+y^2}=\dfrac{9}{4}$$

or, $$4\left(x^2+y^2\right)=9\left(\left(x-1\right)^2+y^2\right)$$

or, $$4x^2+4y^2=9x^2-18x+9+9y^2$$

or, $$5x^2+5y^2-18x+9=0$$

This a second degree equation in $$x$$ and $$y$$ with coefficients of $$x$$ and $$y$$ being equal.

Therefore, this equation represents a circle.

Hence, we can conclude the cow moves in a circle.

Given,

$$\cos x + \cos y = 1$$

Squaring both sides, $$\cos^2x+\cos^2y+2\cos x\cos y=1$$

or, $$1-\sin^2x+1-\sin^2y+2\cos x\cos y=1$$

or, $$1+2\cos x\cos y=\sin^2x+\sin^2y$$

Adding $$-2\sin x\sin y$$ on both sides,

$$1+2\cos x\cos y-2\sin x\sin y=\sin^2x+\sin^2y-2\sin x\sin y$$

or, $$1+2\cos\left(x+y\right)=\left(\sin x-\sin y\right)^2$$

Now, the maximum value of $$\cos\left(x+y\right)$$ can be 1

So, the maximum value of $$\left(\sin x-\sin y\right)^2$$ will be $$1+2\times\ 1=3$$

So, the maximum and minimum possible values of  $$\sin x-\sin y$$ will be $$\sqrt{\ 3}$$ and $$-\sqrt{\ 3}$$ respectively

So, the range of $$\sin x-\sin y$$ is $$[-\surd{3}, \surd{3}]$$

$$\sin \theta + \cos \theta = m$$

Squaring both sides, we get

$$\sin^2\theta\ +\cos^2\theta\ +2\sin\theta\ \cos\theta\ \ =\ m^2$$

We know that $$\sin^2\theta\ +\cos^2\theta\ =1$$

$$1+2\sin\theta\ \cos\theta\ \ =\ m^2$$

$$\sin\theta\ \cos\theta\ \ =\ \dfrac{m^2-1}{2}$$  -----(1)

Now, using the algebra property $$a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$$

Assuming $$a=\sin^2\theta\ $$ and $$b=\cos^2\theta\ $$

$$\sin^6\theta\ +\cos^6\theta\ =\left(\sin^2\theta\ +\cos^2\theta\ \right)^3-3\sin^2\theta\ \cos^2\theta\ \left(\sin^2\theta\ +\cos^2\theta\ \right)$$

$$\sin^6\theta\ +\cos^6\theta\ =\ 1-3\sin^2\theta\ \cos^2\theta\ \left(1\ \right)$$

Substituting the value from equation (1)

$$\sin^6\theta\ +\cos^6\theta\ =\ 1-3\left(\dfrac{m^2-1}{2}\right)^2$$

Thus, option D is the correct answer.

We know that $$AA^{-1}=I$$

=> $$\begin{bmatrix}2 & -0.5 \\-1 & x \end{bmatrix}\begin{bmatrix}1 & 1 \\2 & 4 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$$

=> $$\begin{bmatrix}1 & 0 \\ -1+2x & -1+4x \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$$

=> $$-1+4x=1$$ => $$x=0.5$$

Given, $$f(x) = \dfrac{x^{3} - 5x^{2} - 8x}{3}$$

so, $$f(x)=\dfrac{x\left(x^2-5x-8\right)}{3}$$

Let's find the roots of the quadratic $$x^2-5x-8=0$$

Roots of this equation are $$\dfrac{5+\sqrt{\ 5^2-4\times\ \left(-8\right)}}{2}$$ and $$\dfrac{5-\sqrt{\ 5^2-4\times\ \left(-8\right)}}{2}$$

or, roots of this equation are $$\dfrac{5+\sqrt{57\ }}{2}$$ and $$\dfrac{5-\sqrt{57\ }}{2}$$

So, $$f\left(x\right)=x\left(x-\dfrac{5-\sqrt{\ 57}}{2}\right)\left(x-\dfrac{5+\sqrt{\ 57}}{2}\right)$$

So, roots of the equation $$f(x)=0$$ are 0,$$\dfrac{5+\sqrt{57\ }}{2}$$ and $$\dfrac{5-\sqrt{57\ }}{2}$$

Approximately, $$\dfrac{5+\sqrt{57\ }}{2}=6.27$$ and $$\dfrac{5-\sqrt{57\ }}{2}=-1.27$$

Taking a value greater than $$\dfrac{5+\sqrt{57\ }}{2}$$, we can see $$f(x)>0$$

Taking a value in between 0 and $$\dfrac{5+\sqrt{57\ }}{2}$$, we can see $$f(x)<0$$

Similarly, taking a value between $$\dfrac{5-\sqrt{57\ }}{2}$$ and 0, we can see $$f(x)>0$$

Taking a value lesser than $$\dfrac{5-\sqrt{57\ }}{2}$$. we can see $$f(x)<0$$

So, we can draw the graph of $$f(x)$$ like:

We can see for $$x$$ lesser than $$\dfrac{5-\sqrt{\ 57}}{2}$$ the function is monotonically decreasing and negative.

For $$x$$ greater than $$\dfrac{5+\sqrt{\ 57}}{2}$$ the function is monotonically increasing and positive.

So, option C is the correct answer.

For a > b > c > 0, the minimum value of the function f(x) = |x - a| + |x - b| + |x - c| is

We know a > b > c > 0

Let us put x = a in the f(x)

f(x) = |x - a| + |x - b| + |x - c|

f(a) = 0 + a - b + a - c = (a-b)+(a-c)  [ Since, a > b > c > 0]

f(a) = (a-b)+(a-c)  --------(1)

Similarly, put x = b in the f(x)

f(b) = a - b + 0 + b - c  [ Since, a > b so |b - a | = a-b   ]

f(b) = (a - b) + (b - c) = (a - c)   ----------(2)

Similarly, put x = c in the f(x)

f(c) = a - c + b - c + 0 

f(c) = (a-c) + (b-c)    ---------(3)

Comparing (1), (2) and (3)

We can say that (2) is the minimum value.

Thus, the minimum value of the function f(x) = |x - a| + |x - b| + |x - c| is a-c