∣11112313λ∣=λ−5
Δ1=∣12313λ5μ1∣=10λ+3μ−λμ−44
Δ2=∣11113λ5μ1∣=5λ+μ−λμ−13
Δ3=∣1111235μ1∣=−2μ+6
For unique solution
λ=5,μ∈1,2,3,4,5,6
⇒p=65×1=65
For No Solution
λ=5,μ=3
⇒q=61×65=365
JEE Main 2021 — Mathematics Algebra
Two fair dice are thrown. The numbers on them are taken as λ and μ, and a system of linear equations
x+y+z=5
x+2y+3z=μ
x+3y+λz=1
is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then:
Held on 26 Aug 2021 · Verified 6 Jul 2026.
p=61 and q=365
p=65 and q=361
p=61 and q=361
p=65 and q=365
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:
If $\alpha=1$ and $\beta=1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation $x^3+ax^2+bx+c=0$, $a,b,c \in \mathbb{R}$, then $\int_{-1}^{1}(x^3+ax^2+bx+c)dx$ is equal to:
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
$\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^{2}}{3^{23}}+\ldots+\frac{10 \cdot 2^{24}}{3}$ is equal to :
Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If $P_{n}=C_{0}-C_{1}+\frac{2^{2}}{3} C_{2}-\frac{2^{3}}{4} C_{3}+\ldots. .+\frac{(-2)^{n}}{n+1} C_{n}$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.
Work through every JEE Main Algebra PYQ, year by year.