CAT 2023 Slot 3 Question Paper

Given that AB = AC => Angle C = Angle B (1)

AD and BE are altitudes => they make 90 degrees with the sides

Angle AOB = 105 => Angle EOD = 105 (Vertically Opposite Angles)

In quadrilateral DOEC

Angle C = 360 - 105 - 90 - 90 = 75 => Angle B = 75 (from 1)

We know that from the area of the triangle AD * BC = BE * AC

=> $$\dfrac{AD}{BE}=\dfrac{AC}{BC}=\dfrac{2R\sin B}{2R\sin A}=\dfrac{\sin\left(75\right)}{\sin\left(30\right)}=2\sin\left(75\right)$$ = 2cos(15)

[sin(x) = cos(90-x)]

Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> $$\left(\frac{l}{2}\right)^2+b^2=2^2$$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of $$\left(\frac{l}{2}\right)^2$$ and $$b^2$$.

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> $$\left(\frac{l}{2}\right)^2=b^2$$ => l = 2b.
=> l/b = 2/1.

The sum of the interior angles of a polygon of 'n' sides is given by $$\left(2n-4\right)\times\ 90$$, and the sum of the exterior angles of a polygon is 360 degrees.

So, the difference between them will be 120 * n

=> $$\left(2n-4\right)90-360=120n$$

=> 60n = 720 => n = 12.

We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

The given sequence can be written as:

$$1\left(1+\ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)+\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+...\right)+\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+...\right)+..$$

We know that the sum of an infinite G.P. is $$\dfrac{a}{1-r}$$, where a is the first term and r is the common ratio.

=> The first term = $$\frac{1}{1-\frac{1}{4}}=\dfrac{4}{3}$$

=> The second term = $$\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{9}$$

=> The third term = $$\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{108}$$

Observing these three terms, we see that they are in G.P. with a common ratio of $$\dfrac{1}{12}$$

=> Sum of this infinite G.P. = $$\dfrac{\left(\dfrac{4}{3}\right)}{1-\left(\dfrac{1}{12}\right)}=\dfrac{16}{11}$$

The first series goes as follows:

46, 54, 62, 70, 78, 86, 94, 102,....

The second series goes as follows:

98, 102, 106, 110,...

The first common term is 102 (first term of the common terms) and the common difference between them will be hcf(4,8) = 8

=> The required sequence is 102, 110, 118,..... (last term should be less than 468 (100th term of second series))

=> 102 + (n-1)(8) $$\le$$ 498

=> n is less than or equal to 50.5  =>  n = 50

Using the summation of A.P. formula:

Required sum = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)=\dfrac{50}{2}\left(2\times\ 102+49\times\ 8\right)=14900$$

Given that f(3x + 2y, 2x - 5y) = 19x.

Let us assume the function f(a,b) is a linear combination of a and b.

=> f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x

=> 3m + 2n = 19 and 2m - 5n = 0

Solving we get m = 5 and n = 2

=> f(a,b) = 5a+2b

=> f(x,2x) = 5x + 2(2x) = 9x = 27 => x = 3.