CBSE 12th Maths Question Paper 2022 (Set 2)

Let $E$ be the event that the sum is even, and $F$ be the event that at least one die shows $3$. When the sum is even, both dice must show numbers of the same parity: either both odd or both even. Total outcomes with even sum = $3\times 3 + 3\times 3 = 18$. Now, for at least one die to show $3$ and the sum to be even, the other die must also be odd. The favorable outcomes are: $$(3,1),(3,3),(3,5),(1,3),(5,3)$$ So favorable outcomes = $5$. Therefore, $$P(F\mid E)=\frac{5}{18}.$$ Hence, the required probability is $\frac{5}{18}$.

Given $|2\vec a+3\vec b|=|3\vec a-2\vec b|$, where $\vec a$ and $\vec b$ are unit vectors. Squaring both sides: $$|2\vec a+3\vec b|^2=|3\vec a-2\vec b|^2$$ Now, $$|2\vec a+3\vec b|^2=4|\vec a|^2+9|\vec b|^2+12\vec a\cdot\vec b$$ $$|3\vec a-2\vec b|^2=9|\vec a|^2+4|\vec b|^2-12\vec a\cdot\vec b$$ Since $|\vec a|=|\vec b|=1$, we get $$4+9+12\vec a\cdot\vec b=9+4-12\vec a\cdot\vec b$$ $$13+12\vec a\cdot\vec b=13-12\vec a\cdot\vec b$$ $$24\vec a\cdot\vec b=0$$ $$\vec a\cdot\vec b=0$$ Thus, $$\cos\theta=0\implies \theta=90^\circ.$$ Hence, the angle between $\vec a$ and $\vec b$ is $90^\circ$.

We need to evaluate $$\int x\sin x\,dx.$$ Using integration by parts, let $$u=x,\quad dv=\sin x\,dx.$$ Then $$du=dx,\quad v=-\cos x.$$ So, $$\int x\sin x\,dx = uv-\int v\,du$$ $$=x(-\cos x)-\int(-\cos x)\,dx$$ $$=-x\cos x+\int \cos x\,dx$$ $$=-x\cos x+\sin x+C.$$ Hence, $$\boxed{\int x\sin x\,dx=-x\cos x+\sin x+C}.$$

Evaluate $$\int_0^1 e^{2x}\,e^x\,\log 2\,dx.$$ Since $e^{2x}e^x=e^{3x}$ and $\log 2$ is constant, $$\int_0^1 e^{2x}e^x\log 2\,dx = \log 2\int_0^1 e^{3x}\,dx.$$ Now, $$\int e^{3x}\,dx=\frac{1}{3}e^{3x}.$$ So, $$\log 2\int_0^1 e^{3x}\,dx = \log 2\left[\frac{1}{3}e^{3x}\right]_0^1$$ $$=\frac{\log 2}{3}(e^3-1).$$ Hence, the value of the integral is $$\boxed{\frac{(e^3-1)\log 2}{3}}.$$

The statement “probabilities of A and B solving a problem are $\frac{3}{2}$ and $\frac{5}{3}$” is not possible as probabilities must lie between $0$ and $1$. So the intended probabilities are most likely $\frac{2}{3}$ and $\frac{3}{5}$. Let $$P(A)=\frac{2}{3},\quad P(B)=\frac{3}{5}.$$ Then $$P(A')=1-\frac{2}{3}=\frac{1}{3},\quad P(B')=1-\frac{3}{5}=\frac{2}{5}.$$ Since they work independently, the probability that the problem is not solved by either is $$P(A'\cap B')=\frac{1}{3}\cdot\frac{2}{5}=\frac{2}{15}.$$ Therefore, the probability that the problem is solved is $$1-\frac{2}{15}=\frac{13}{15}.$$ Hence, the required probability is $\boxed{\frac{13}{15}}$.