MCQ
$y = c{e^{{{\sin }^{ - 1}}x}}$ નું વિકલ સમીકરણ મેળવો.
- ✓$\frac{{dy}}{{dx}} = \frac{y}{{\sqrt {1 - {x^2}} }}$
- B$\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}$
- C$\frac{{dy}}{{dx}} = \frac{x}{{\sqrt {1 - {x^2}} }}$
- Dએકપણ નહી.
$\frac{{dy}}{{dx}} = c{e^{{{\sin }^{ - 1}}x}}.\frac{1}{{\sqrt {1 - {x^2}} }} = \frac{y}{{\sqrt {1 - {x^2}} }}$ or $\frac{{dy}}{{dx}} = \frac{y}{{\sqrt {1 - {x^2}} }}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$
માં સૌથી મહતમ હોય તો . . .