CBSE 10th Maths Question Paper 2022 (Set 2)

Given A.P. is $12x, 10x, 8x, ..., 2x$. Here, the first term $a = 12x$ and the common difference $d = 10x - 12x = -2x$. The last term $a_n = 2x$. Using the formula $a_n = a + (n-1)d$, we have $2x = 12x + (n-1)(-2x)$. Subtracting $12x$ from both sides, $-10x = (n-1)(-2x)$. Dividing by $-2x$, we get $5 = n-1$, which implies $n = 6$. Thus, there are 6 terms.

The integers between 2 and 100 divisible by 3 are $3, 6, 9, ..., 99$. This forms an A.P. where $a = 3$, $d = 3$, and $a_n = 99$. Using $a_n = a + (n-1)d$, we get $99 = 3 + (n-1)3$. Dividing by 3, we get $33 = 1 + (n-1)$, so $33 = n$. Thus, there are 33 such integers.

For the quadratic equation $kx^2 - 5x + k = 0$ to have real and equal roots, the discriminant $D$ must be 0. $D = b^2 - 4ac = (-5)^2 - 4(k)(k) = 25 - 4k^2$. Setting $D = 0$, we have $4k^2 = 25$, so $k^2 = \frac{25}{4}$. Therefore, $k = \pm \frac{5}{2}$.

Since PQ and PR are tangents from point P to the circle with center O, the radius is perpendicular to the tangent at the point of contact. Therefore, $\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$. In quadrilateral PQOR, the sum of opposite angles $\angle OQP + \angle ORP = 90^\circ + 90^\circ = 180^\circ$. Since the sum of a pair of opposite angles is $180^\circ$, the quadrilateral PQOR is cyclic.

The equation is $y^2 + \frac{2\sqrt{3}}{5}y - 5 = 0$. Multiplying by 5 to simplify: $5y^2 + 2\sqrt{3}y - 25 = 0$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $a=5, b=2\sqrt{3}, c=-25$. $D = (2\sqrt{3})^2 - 4(5)(-25) = 12 + 500 = 512$. Thus, $y = \frac{-2\sqrt{3} \pm \sqrt{512}}{10} = \frac{-2\sqrt{3} \pm 16\sqrt{2}}{10} = \frac{-\sqrt{3} \pm 8\sqrt{2}}{5}$.