NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87307
The solution of the differential equation \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) is
1 \(y \sin 2 x=e^{x}+C\)
2 \(y \cos 2 x=e^{x}+C\)
3 \(y=e^{x} \cos 2 x+C\)
4 \(y \cos 2 x+e^{x}=C\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) Which is a linear differential equation, \(P=-2 \tan 2 x \text { and } Q=e^{x} \sec 2 x\) I.F, \(\quad=\mathrm{e}^{\int p \mathrm{pdx}}=\mathrm{e}^{-\int 2 \tan 2 x d x}=\mathrm{e}^{-\left[\frac{\log \sec 2 \mathrm{x}}{2}\right]}\) \(=(\sec 2 x)^{-1}=\cos 2 x\) Now for general solution \(y \cdot \cos 2 x=\int \cos 2 x \cdot e^{x} \sec 2 x d x\) \(y \cos 2 x=\int e^{x} d x\) \(y \cos 2 x=e^{x}+C\)
AP EAMCET-2013
Differential Equation
87308
If \(X\) is a poisson variation with \(P(X=0)=0.8\), then the variance of \(X\) is
1 \(\log _{\mathrm{e}} 20\)
2 \(\log _{10} 20\)
3 \(\log _{\mathrm{e}} 5 / 4\)
4 0
Explanation:
(C) : We have- \(\mathrm{P}(\mathrm{X}=0)=0.8\) We have that, \(P(X=0)=\frac{e^{-\lambda} \lambda^{x}}{x !} \Rightarrow x=0\) Then, \(\frac{\mathrm{e}^{-\lambda}(\lambda)^{0}}{0 !}=0.8 \Rightarrow \mathrm{e}^{-\lambda}=0.8\) \(\mathrm{e}^{-\lambda}=4 / 5\) \(\mathrm{e}^{-\lambda}=5 / 4\) \(\lambda=\log _{\mathrm{e}} 5 / 4\)
AP EAMCET-2004
Differential Equation
87309
The differential equation of the family of parabolas with vertex at \((0,-1)\) and having axis along the \(\mathbf{Y}\)-axis is
1 \(y y^{\prime}+2 x y+1=0\)
2 \(x y^{\prime}+y+1=0\)
3 \(x y^{\prime}-2 y-2=0\)
4 \(x y^{\prime}-y-1=0\)
Explanation:
(C) : We have the family of parabola having axis along the y axis \((\mathrm{x}-\mathrm{h})^{2}=4 \mathrm{a}(\mathrm{y}-\mathrm{k})\) Where \((h, k)\) be the vertex of parabola, Given vertex \((\mathrm{h}, \mathrm{k}) \rightarrow(0,-1)\) The equation become \((x-0)^{2}=4 a(y+1)\) \(x^{2}=4 a(y+1)\) differentiating w.r.t \(x\) we get- \(2 x=4 a \frac{d y}{d x} \Rightarrow a=\frac{x}{2} \frac{d x}{d y}\) Substituting he value of 'a' in equation we get- \(x^{2}=4 \cdot \frac{x}{2} \frac{d x}{d y}(y+1) \Rightarrow x \frac{d y}{d x}=2(y+1)\) \(x \frac{d y}{d x}-2 y-2=0\) \(x y^{\prime}-2 y-2=0\)
AP EAMCET-2014
Differential Equation
87310
The solution of \(\frac{d y}{d x}+\frac{1}{x}=\frac{e^{y}}{x^{2}}\) is
87307
The solution of the differential equation \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) is
1 \(y \sin 2 x=e^{x}+C\)
2 \(y \cos 2 x=e^{x}+C\)
3 \(y=e^{x} \cos 2 x+C\)
4 \(y \cos 2 x+e^{x}=C\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) Which is a linear differential equation, \(P=-2 \tan 2 x \text { and } Q=e^{x} \sec 2 x\) I.F, \(\quad=\mathrm{e}^{\int p \mathrm{pdx}}=\mathrm{e}^{-\int 2 \tan 2 x d x}=\mathrm{e}^{-\left[\frac{\log \sec 2 \mathrm{x}}{2}\right]}\) \(=(\sec 2 x)^{-1}=\cos 2 x\) Now for general solution \(y \cdot \cos 2 x=\int \cos 2 x \cdot e^{x} \sec 2 x d x\) \(y \cos 2 x=\int e^{x} d x\) \(y \cos 2 x=e^{x}+C\)
AP EAMCET-2013
Differential Equation
87308
If \(X\) is a poisson variation with \(P(X=0)=0.8\), then the variance of \(X\) is
1 \(\log _{\mathrm{e}} 20\)
2 \(\log _{10} 20\)
3 \(\log _{\mathrm{e}} 5 / 4\)
4 0
Explanation:
(C) : We have- \(\mathrm{P}(\mathrm{X}=0)=0.8\) We have that, \(P(X=0)=\frac{e^{-\lambda} \lambda^{x}}{x !} \Rightarrow x=0\) Then, \(\frac{\mathrm{e}^{-\lambda}(\lambda)^{0}}{0 !}=0.8 \Rightarrow \mathrm{e}^{-\lambda}=0.8\) \(\mathrm{e}^{-\lambda}=4 / 5\) \(\mathrm{e}^{-\lambda}=5 / 4\) \(\lambda=\log _{\mathrm{e}} 5 / 4\)
AP EAMCET-2004
Differential Equation
87309
The differential equation of the family of parabolas with vertex at \((0,-1)\) and having axis along the \(\mathbf{Y}\)-axis is
1 \(y y^{\prime}+2 x y+1=0\)
2 \(x y^{\prime}+y+1=0\)
3 \(x y^{\prime}-2 y-2=0\)
4 \(x y^{\prime}-y-1=0\)
Explanation:
(C) : We have the family of parabola having axis along the y axis \((\mathrm{x}-\mathrm{h})^{2}=4 \mathrm{a}(\mathrm{y}-\mathrm{k})\) Where \((h, k)\) be the vertex of parabola, Given vertex \((\mathrm{h}, \mathrm{k}) \rightarrow(0,-1)\) The equation become \((x-0)^{2}=4 a(y+1)\) \(x^{2}=4 a(y+1)\) differentiating w.r.t \(x\) we get- \(2 x=4 a \frac{d y}{d x} \Rightarrow a=\frac{x}{2} \frac{d x}{d y}\) Substituting he value of 'a' in equation we get- \(x^{2}=4 \cdot \frac{x}{2} \frac{d x}{d y}(y+1) \Rightarrow x \frac{d y}{d x}=2(y+1)\) \(x \frac{d y}{d x}-2 y-2=0\) \(x y^{\prime}-2 y-2=0\)
AP EAMCET-2014
Differential Equation
87310
The solution of \(\frac{d y}{d x}+\frac{1}{x}=\frac{e^{y}}{x^{2}}\) is
87307
The solution of the differential equation \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) is
1 \(y \sin 2 x=e^{x}+C\)
2 \(y \cos 2 x=e^{x}+C\)
3 \(y=e^{x} \cos 2 x+C\)
4 \(y \cos 2 x+e^{x}=C\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) Which is a linear differential equation, \(P=-2 \tan 2 x \text { and } Q=e^{x} \sec 2 x\) I.F, \(\quad=\mathrm{e}^{\int p \mathrm{pdx}}=\mathrm{e}^{-\int 2 \tan 2 x d x}=\mathrm{e}^{-\left[\frac{\log \sec 2 \mathrm{x}}{2}\right]}\) \(=(\sec 2 x)^{-1}=\cos 2 x\) Now for general solution \(y \cdot \cos 2 x=\int \cos 2 x \cdot e^{x} \sec 2 x d x\) \(y \cos 2 x=\int e^{x} d x\) \(y \cos 2 x=e^{x}+C\)
AP EAMCET-2013
Differential Equation
87308
If \(X\) is a poisson variation with \(P(X=0)=0.8\), then the variance of \(X\) is
1 \(\log _{\mathrm{e}} 20\)
2 \(\log _{10} 20\)
3 \(\log _{\mathrm{e}} 5 / 4\)
4 0
Explanation:
(C) : We have- \(\mathrm{P}(\mathrm{X}=0)=0.8\) We have that, \(P(X=0)=\frac{e^{-\lambda} \lambda^{x}}{x !} \Rightarrow x=0\) Then, \(\frac{\mathrm{e}^{-\lambda}(\lambda)^{0}}{0 !}=0.8 \Rightarrow \mathrm{e}^{-\lambda}=0.8\) \(\mathrm{e}^{-\lambda}=4 / 5\) \(\mathrm{e}^{-\lambda}=5 / 4\) \(\lambda=\log _{\mathrm{e}} 5 / 4\)
AP EAMCET-2004
Differential Equation
87309
The differential equation of the family of parabolas with vertex at \((0,-1)\) and having axis along the \(\mathbf{Y}\)-axis is
1 \(y y^{\prime}+2 x y+1=0\)
2 \(x y^{\prime}+y+1=0\)
3 \(x y^{\prime}-2 y-2=0\)
4 \(x y^{\prime}-y-1=0\)
Explanation:
(C) : We have the family of parabola having axis along the y axis \((\mathrm{x}-\mathrm{h})^{2}=4 \mathrm{a}(\mathrm{y}-\mathrm{k})\) Where \((h, k)\) be the vertex of parabola, Given vertex \((\mathrm{h}, \mathrm{k}) \rightarrow(0,-1)\) The equation become \((x-0)^{2}=4 a(y+1)\) \(x^{2}=4 a(y+1)\) differentiating w.r.t \(x\) we get- \(2 x=4 a \frac{d y}{d x} \Rightarrow a=\frac{x}{2} \frac{d x}{d y}\) Substituting he value of 'a' in equation we get- \(x^{2}=4 \cdot \frac{x}{2} \frac{d x}{d y}(y+1) \Rightarrow x \frac{d y}{d x}=2(y+1)\) \(x \frac{d y}{d x}-2 y-2=0\) \(x y^{\prime}-2 y-2=0\)
AP EAMCET-2014
Differential Equation
87310
The solution of \(\frac{d y}{d x}+\frac{1}{x}=\frac{e^{y}}{x^{2}}\) is
87307
The solution of the differential equation \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) is
1 \(y \sin 2 x=e^{x}+C\)
2 \(y \cos 2 x=e^{x}+C\)
3 \(y=e^{x} \cos 2 x+C\)
4 \(y \cos 2 x+e^{x}=C\)
Explanation:
(B) : We have differential equation- \(\frac{d y}{d x}-2 y \tan 2 x=e^{x} \sec 2 x\) Which is a linear differential equation, \(P=-2 \tan 2 x \text { and } Q=e^{x} \sec 2 x\) I.F, \(\quad=\mathrm{e}^{\int p \mathrm{pdx}}=\mathrm{e}^{-\int 2 \tan 2 x d x}=\mathrm{e}^{-\left[\frac{\log \sec 2 \mathrm{x}}{2}\right]}\) \(=(\sec 2 x)^{-1}=\cos 2 x\) Now for general solution \(y \cdot \cos 2 x=\int \cos 2 x \cdot e^{x} \sec 2 x d x\) \(y \cos 2 x=\int e^{x} d x\) \(y \cos 2 x=e^{x}+C\)
AP EAMCET-2013
Differential Equation
87308
If \(X\) is a poisson variation with \(P(X=0)=0.8\), then the variance of \(X\) is
1 \(\log _{\mathrm{e}} 20\)
2 \(\log _{10} 20\)
3 \(\log _{\mathrm{e}} 5 / 4\)
4 0
Explanation:
(C) : We have- \(\mathrm{P}(\mathrm{X}=0)=0.8\) We have that, \(P(X=0)=\frac{e^{-\lambda} \lambda^{x}}{x !} \Rightarrow x=0\) Then, \(\frac{\mathrm{e}^{-\lambda}(\lambda)^{0}}{0 !}=0.8 \Rightarrow \mathrm{e}^{-\lambda}=0.8\) \(\mathrm{e}^{-\lambda}=4 / 5\) \(\mathrm{e}^{-\lambda}=5 / 4\) \(\lambda=\log _{\mathrm{e}} 5 / 4\)
AP EAMCET-2004
Differential Equation
87309
The differential equation of the family of parabolas with vertex at \((0,-1)\) and having axis along the \(\mathbf{Y}\)-axis is
1 \(y y^{\prime}+2 x y+1=0\)
2 \(x y^{\prime}+y+1=0\)
3 \(x y^{\prime}-2 y-2=0\)
4 \(x y^{\prime}-y-1=0\)
Explanation:
(C) : We have the family of parabola having axis along the y axis \((\mathrm{x}-\mathrm{h})^{2}=4 \mathrm{a}(\mathrm{y}-\mathrm{k})\) Where \((h, k)\) be the vertex of parabola, Given vertex \((\mathrm{h}, \mathrm{k}) \rightarrow(0,-1)\) The equation become \((x-0)^{2}=4 a(y+1)\) \(x^{2}=4 a(y+1)\) differentiating w.r.t \(x\) we get- \(2 x=4 a \frac{d y}{d x} \Rightarrow a=\frac{x}{2} \frac{d x}{d y}\) Substituting he value of 'a' in equation we get- \(x^{2}=4 \cdot \frac{x}{2} \frac{d x}{d y}(y+1) \Rightarrow x \frac{d y}{d x}=2(y+1)\) \(x \frac{d y}{d x}-2 y-2=0\) \(x y^{\prime}-2 y-2=0\)
AP EAMCET-2014
Differential Equation
87310
The solution of \(\frac{d y}{d x}+\frac{1}{x}=\frac{e^{y}}{x^{2}}\) is
