CUET PG 2023 Question Paper COQP12 Shift-2

In triangles, the length of the sides is proportional to the sine of the angle opposite to the. 

The angle opposite AB is 55 degrees while the angle opposite BC will be 45 degrees. 

Since sin 55 > sin 45, AB>Bc

Therefore, Option A is the correct answer. 

The volume of a cone can be written as $$\pi r^2h$$. 
Where r is the radius of the base of the cone and h is the height of the cone.

The heights of two right circular cones are in the ratio 1 : 2, 
Let us say, H1 is 1X and H2 is 2X

Perimeters of their bases are in the ratio 3 : 4, 
Let us say, $$\2pi r_1$$ is 3Y and $$\2pi r_1$$ is 4Y. 

$$r1=\frac{3Y}{2\pi\ }$$ and $$r2=\frac{4Y}{2\pi\ }$$

Volume of cone 1 will be, $$\pi\left(\frac{3Y}{2\pi\ }\right)^2\left(X\right)$$

Volume of cone 2 will be, $$\pi\left(\frac{4Y}{2\pi\ }\right)^2\left(2X\right)$$

The ratio will hence be 9:32. 

The surface area of a sphere is, A = $$4\pi r^2$$

Let the radii of the two spheres be a and b. Given that their surface areas are in the ratio:

A1:A2= $$\dfrac{4}{25}$$

Substituting the formula for surface area:

$$4\pi a^2$$:$$4\pi b^2$$ = 4:25

Canceling 4π from both sides:

$$a^2:b^2=4:25$$

Taking the square root on both sides:

a:b = 2:5

Now, the volume of a sphere is V = $$\dfrac{4}{3}\pi r^3$$

The ratio of their volumes is:

V1:V2= $$a^3:b^3$$ = 8:125

Let's rewrite the equation:

$$a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}=4$$

We know that $$x^2+\dfrac{1}{x^2}\ge\ 2$$

If we compare, we get,

$$a^2+\dfrac{1}{a^2}=\ 2$$ and $$b^2+\dfrac{1}{b^2}=\ 2$$

This is possible when $$a^2=1\ \&\ b^2=1$$

Hence, $$a^2+b^2\ =\ 1\ +\ 1\ =\ 2$$

$$2x^{2} + 3x - 35 = 0$$

This can be written as

$$2x^2+10x-7x-35=0$$

$$2x\left(x+5\right)-7\left(x+5\right)=0$$

$$\left(2x-7\right)\left(x+5\right)=0$$

Thus, x = 7/2 or -5

Similarly, for $$4y^{2} + 10y - 104 = 0$$

This can be written as

$$4y^2-16y+26y-104=0$$

$$4y\left(y-4\right)+26\left(y-4\right)=0$$

y = 4 or -26/4

Therefore, x and y are unequal, and no relation can be established between them.

The given information can be shown in an image as follows,

We can see that for the outer square, the length of the side is equal to the diameter of the circle. So, the side of the outer square is 2a, and the area of the outer square is $$2a\ \times\ 2a\ =\ 4a^2$$

We can see that for the inner square, the length of the diagonal is equal to the diameter of the circle. So, the area of the inner square is $$\dfrac{1}{2}\ \times\ 2a\ \times\ 2a\ =\ 2a^2$$.

So, the difference between the areas of the outer and the inner squares  = $$4a^2\ -\ 2a^2\ =\ 2a^2$$

Hence, the correct answer is option B.

If PQ and RS are two diameters of a circle of radius r, then they are mutually perpendicular.

RS = 2r (Where r is the radius of the circle)

RP = RS = $$r\sqrt{2}$$

Area of circle = $$\pi r^2$$

Triangle SPR is a right-angled triangle, right-angled at P.  (Angle RPS is 90 degrees as RS is the diameter)

Area of $$\triangle PRS$$ = $$\dfrac{1}{2}\times PR\times\ PS$$ = $$\dfrac{1}{2}\times r\sqrt{2}\times\ r\sqrt{2}$$ = $$r^2$$

Thus, the ratio is $$\pi r^2$$:$$r^2$$ = $$\pi:1$$

The value of $$\dfrac{1}{4 \times 5} + \dfrac{1}{5 \times 6} + \dfrac{1}{6 \times 7} + ....... + \dfrac{1}{10 \times 11}$$:

The terms can be written as

$$\Sigma\ \dfrac{1}{\left(n+3\right)\left(n+4\right)}$$

=> Where sigma means the summation of all terms where n = 1 to n = 7

$$\Sigma\ \dfrac{1}{\left(n+3\right)\left(n+4\right)}$$

=> $$\Sigma\ \dfrac{\left(n+4\right)-\left(n+3\right)}{\left(n+3\right)\left(n+4\right)}$$

=> $$\Sigma\ \left[\dfrac{1}{\left(n+3\right)}-\dfrac{1}{\left(n+4\right)}\right]$$

=> $$\dfrac{1}{4}-\dfrac{1}{5}$$ + $$\dfrac{1}{5}-\dfrac{1}{6}$$ + .....$$\dfrac{1}{10}-\dfrac{1}{11}$$

=> All other terms will get cancelled, and only 1/4 and -1/11 will left.

Thus, $$\dfrac{1}{4}-\dfrac{1}{11}=\dfrac{7}{44}$$

We know that $$\left(a\ +\ b\right)^2\ -\ \left(a\ -\ b\right)^2\ =\ 4ab$$

So,

$$\left(x\ +\ \frac{1}{x}\right)^2\ -\ \left(x\ -\ \frac{1}{x}\right)^2\ =\ 4\times\ x\ \times\ \frac{1}{x}\ =\ 4$$

$$\left(x\ -\ \frac{1}{x}\right)^2\ =\ \left(\frac{17}{4}\right)^2\ -\ 4$$

$$\ \left(x\ -\ \frac{1}{x}\right)^2\ =\ \frac{\left(289\ -\ 64\right)}{16}\ =\ \frac{225}{16}$$

$$\ x\ -\ \frac{1}{x}\ =\ \frac{15}{4}$$

Hence, the correct answer is option B.