Question
The area of circle $x^2+y^2=4$ is

Answer

self

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Choose the correct answer from the given four options.Let $f : R \rightarrow R $ be defined by $f(x) = 3x^2 – 5$ and $g : R \rightarrow R$ by $\text{g}(\text{x})=\frac{\text{x}}{\text{x}^2+1}.$ Then gof is:
Find the principal values of: $\cot ^{-1}(1)$
The direction ratios of the line x - y + z - 5 = 0 = x - 3y - 6 are proportional to:
  1. $3,1,-2$
  2. $2,-4,1$
  3. $\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}$
  4. $\frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}}$
If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
  1. f(a) = 0
  2. f(b) = 0
  3. f(0) = 0
  4. f(1) = 0
If $\begin{bmatrix}\text{r}+4&\text{amp; 6}\\3&\text{amp; 3}\end{bmatrix}=\begin{bmatrix}{5}&\text{amp;}\text{ r}+5\\\text{r+2}&\text{amp; 4}\end{bmatrix}$ then $\text{r}=$
  1. 1
  2. 2
  3. 3
  4. -1
If $ \displaystyle \begin{vmatrix}\text{a} &\text{amp; }\text{b} &\text{amp; 0}\\ 0 &\text{amp; a} &\text{amp; b}\\\text{b}&\text{amp; a}&\text{amp; 0}\end{vmatrix}=0,$ then the order is:
  1. 3 × 3
  2. 2 × 3
  3. 2 × 2
  4. None of these
Let $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}.$ Then,
The point which does not lie in the half - plane 2x + 3y -12 < 0 is:
  1. (2, 1)
  2. (1, 2)
  3. (-2, 3)
  4. (2, 3)
Choose the correct answer: Area lying between the curves $y^2 = 4x$ and $y = 2x$ is:
If $x=t^2+1, y=2 a t$, then $\frac{d^2 y}{d x^2}$ at $t=a$ is