- A$(2y - 1){{dy} \over {dx}} - \sin x = 0$
- B$(2y - 1)\cos x + {{dy} \over {dx}} = 0$
- C$(2y - 1)\cos x - {{dy} \over {dx}} = 0$
- ✓$(2y - 1)\frac{{dy}}{{dx}} - \cos x = 0$
$ \Rightarrow y = \sqrt {\sin x + y} \Rightarrow {y^2} = \sin x + y$
On differentiating both sides, we get
$2y\frac{{dy}}{{dx}} = \cos x + \frac{{dy}}{{dx}} $
$\Rightarrow \frac{{dy}}{{dx}}(2y - 1) = \cos x$.
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where $[x]$ denotes step up function then at $x = 2$ function
$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$
$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$
$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$
$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$
$2 x+y-z=3$
$x-y-z=\alpha$
$3 x+3 y+\beta z=3$
has infinitely many solution, then $\alpha+\beta-\alpha \beta$ is equal to .... .
$\lim _{n \rightarrow 0^{+}} \int_n^{1-n} t^{-3}(1-t)^{a-1} d t$
exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0,1)$.
$1.$ The value of $g\left(\frac{1}{2}\right)$ is
$(A)$ $\pi$ $(B)$ $2 \pi$ $(C)$ $\frac{\pi}{2}$ $(D)$ $\frac{\pi}{4}$
$2.$ The value of $g ^{\prime}\left(\frac{1}{2}\right)$ is
$(A)$ $\frac{\pi}{2}$ $(B)$ $\pi$ $(C)$ $-\frac{\pi}{2}$ $(D)$ $0$
Give the answer question $1$ and $2.$