Put z=x+iy
x+iy+α∣x+iy−1∣+2i=0
⇒x+α(x−1)2+y2+i(y+2)=0+0i
⇒y+2=0 and x+α(x−1)2+y2=0
⇒y=−2 and α2=x2−2x+5x2
Now x2−2x+5x2∈[0,45]
∴α2∈[0,45]⇒α∈[−25,25]
∴p=−25;q=25
⇒4(p2+q2)=4(45+45)=10
JEE Main 2021 — Mathematics Algebra
If the least and the largest real values of α, for which the equation z+α∣z−1∣+2i=0 (z∈Candi=−1) has a solution, are p and q respectively; then 4(p2+q2) is equal to_______.
Held on 24 Feb 2021 · Verified 6 Jul 2026.
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.