$$ \text { The particular solution of the differential equation }(x-y)(d x+d y)=(d x-d y) \text { when } y=-1 \text { and } x=0 \text { is } $$
$$ \log |x+y|=x-y+1 $$
$$ \log \left|\frac{x-y}{x+y}\right|=1 $$
$$ \log |x-y|=x+y+1 $$
$$ \log |x-y|=x-y+1 $$
$$ \int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x= $$
$$ \frac{\pi}{4} x-\frac{x^2}{4}+C $$
$$ \frac{\pi}{4} x-\frac{x^2}{2}+C $$
$$ \frac{\pi}{2} x-\frac{x^2}{4}+C $$
$$ \frac{\pi}{4}-\frac{x}{4}+C $$
The area enclosed by the curve $y=-x^2$ and the line $x+y+2=0$ is
$4.5 $ sq units
$3 .5$ sq units
$4 $ sq units
$5 .5$ sq units
Let $A=\left[a_{i j}\right]$ be a square matrix of order $3 \times 3$, where the elements are defined as $a_{i j}=\left\{\begin{array}{ll}i-2 j & \text { if } i=j \\ 0 & \text { if } i> j \\ 1 & \text { if } i < j\end{array} \quad\right.$ then the value of $\left|A^t\right|$ is
-6
1
-5
-11
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