CBSE 12th Maths Question Paper 2022 (Set 1)

When two dice are thrown, total outcomes $=36$. 1) Sum is even: This happens when both numbers are even or both are odd. - Even-even outcomes: $3\times 3=9$ - Odd-odd outcomes: $3\times 3=9$ So favourable outcomes $=18$. Hence, $$P(\text{sum even})=\frac{18}{36}=\boxed{\frac12}.$$ 2) At least one die shows 3: Use the complement. Outcomes with no 3 on either die: $5\times 5=25$. So favourable outcomes $=36-25=11$. Hence, $$P(\text{at least one 3})=\boxed{\frac{11}{36}}.$$

The given probabilities are written as $\frac{3}{2}$ and $\frac{5}{3}$, but probabilities cannot exceed 1. So these are clearly intended to be $\frac{3}{5}$ and $\frac{2}{3}$ (the standard CBSE-type values). Let $$P(A)=\frac35, \qquad P(B)=\frac23.$$ Since A and B work independently, the probability that the problem is not solved by either of them is $$\left(1-\frac35\right)\left(1-\frac23\right)=\frac25\cdot\frac13=\frac{2}{15}.$$ Therefore, the probability that the problem is solved is $$1-\frac{2}{15}=\boxed{\frac{13}{15}}.$$

The line passing through $P(2,2,1)$ and $Q(5,1,2)$ has direction vector $$\overrightarrow{PQ}=(5-2,1-2,2-1)=(3,-1,1).$$ Hence its cartesian equation is $$\frac{x-2}{3}=\frac{y-2}{-1}=\frac{z-1}{1}.$$ Now, for the point on the line whose $z$-coordinate is 2, let the parameter be $t$: $$x=2+3t,\quad y=2-t,\quad z=1+t.$$ Given $z=2$, $$1+t=2 \Rightarrow t=1.$$ Therefore, $$y=2-1=1.$$ So the required $y$-coordinate is $$\boxed{1}.$$

Let $\overrightarrow{AB}=\vec b$ and $\overrightarrow{AD}=\vec d$. Then, in a parallelogram, $$\overrightarrow{AC}=\vec b+\vec d,$$ and $$\overrightarrow{BD}=\vec d-\vec b.$$ Given $$\vec b+\vec d=\hat i+\hat j \quad ...(1)$$ $$\vec d-\vec b=2\hat i+\hat j+\hat k \quad ...(2)$$ Adding (1) and (2), $$2\vec d=3\hat i+2\hat j+\hat k$$ so $$\overrightarrow{AD}=\vec d=\frac{3}{2}\hat i+\hat j+\frac{1}{2}\hat k.$$ Subtracting (2) from (1), $$2\vec b=(\hat i+\hat j)-(2\hat i+\hat j+\hat k)=-\hat i-\hat k,$$ so $$\overrightarrow{AB}=\vec b=-\frac{1}{2}\hat i-\frac{1}{2}\hat k.$$ Now the area of the parallelogram is $$|\overrightarrow{AB}\times\overrightarrow{AD}|.$$ Here, $$\overrightarrow{AB}=\left(-\frac12,0,-\frac12\right),\quad \overrightarrow{AD}=\left(\frac32,1,\frac12\right).$$ So $$\overrightarrow{AB}\times\overrightarrow{AD}=\left(\frac12,-\frac12,-\frac12\right),$$ and hence $$\text{Area}=\sqrt{\left(\frac12\right)^2+\left(-\frac12\right)^2+\left(-\frac12\right)^2} =\sqrt{\frac34}=\frac{\sqrt3}{2}.$$ Therefore, $$\boxed{\overrightarrow{AB}=-\frac12\hat i-\frac12\hat k,\quad \overrightarrow{AD}=\frac32\hat i+\hat j+\frac12\hat k,\quad \text{Area}=\frac{\sqrt3}{2}.}$$

Using integration by parts, let $$u=\tan^{-1}x,\quad dv=x\,dx.$$ Then $$du=\frac{1}{1+x^2}\,dx,\quad v=\frac{x^2}{2}.$$ So $$\int_0^1 x\tan^{-1}x\,dx =\left[\frac{x^2}{2}\tan^{-1}x\right]_0^1-\frac12\int_0^1 \frac{x^2}{1+x^2}\,dx.$$ Now $$\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2},$$ so $$\int_0^1 x\tan^{-1}x\,dx =\frac12\cdot\frac{\pi}{4}-\frac12\int_0^1 \left(1-\frac{1}{1+x^2}\right)dx.$$ That is, $$=\frac{\pi}{8}-\frac12\left[ x-\tan^{-1}x \right]_0^1 =\frac{\pi}{8}-\frac12\left(1-\frac{\pi}{4}\right).$$ Hence $$\int_0^1 x\tan^{-1}x\,dx=\boxed{\frac{\pi}{4}-\frac12}.$$