NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85821
If a curve passes through the origin and the slope of the tangent to it at any point \((x, y)\) is \(\frac{x^{2}-4 x+y+8}{x-2}\), then this curve also passes through the point
1 \((5,4)\)
2 \((4,5)\)
3 \((4,4)\)
4 \((5,5)\)
Explanation:
(D) : Given, Slope \(=\frac{x^{2}-4 x+y+8}{x-2}\) \(\frac{d y}{d x}=\frac{x^{2}-4 x+y+8}{x-2} \Rightarrow \frac{d y}{d x}=\frac{(x-2)^{2}+(x+4)}{(x-2)}\) \(\frac{d y}{d x}=(x-2)+\frac{y+4}{x-2}\) Let, \(\quad(\mathrm{x}-2)=\mathrm{t}\) \(\mathrm{dx}=\mathrm{dt}\) and \(\quad(x+4)=u\) \(\mathrm{dy}=\mathrm{du} \Rightarrow \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dt}}\) Now, \(\frac{d y}{d x}=(x-2)+\frac{(y+4)}{(x-2)}\) \(\frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{t}+\frac{\mathrm{u}}{\mathrm{t}}, \quad \frac{\mathrm{du}}{\mathrm{dt}}-\frac{\mathrm{u}}{\mathrm{t}}=\mathrm{t}\) Here, integrating factor \((\mathrm{IF})=\frac{1}{\mathrm{t}}\) \(u \cdot\left(\frac{1}{t}\right)=\int t\left(\frac{1}{t}\right) d t \Rightarrow \frac{u}{t}=t+C\) \(\frac{(y+4)}{(x-2)}=(x-2)+C\) \(\because\) It passes through origin \((0,0)\). \(\therefore \quad \frac{4}{-2}=-2+\mathrm{C}\) \(-2+2=\mathrm{C}\) \(\mathrm{C}=0\) \(\text { Hence, } \frac{(\mathrm{y}+4)}{(\mathrm{x}-2)}=(\mathrm{x}-2)+0 \quad[\because \mathrm{C}=0]\) \(\mathrm{x}+4=(\mathrm{x}-2)^{2}\) Clearly, this curve passes through \((5,5)\) as it satisfies the equation.
JEE Main-2021-25.02.2021
Application of Derivatives
85822
If \(\frac{d y}{d x}=4\) and \(\frac{d^{2} y}{d x^{2}}=-3\) at a point \(P\) on the curve \(y=f(x)\), then \(\left(\frac{d^{2} x}{d y^{2}}\right)_{P}=\)
85811
The part of circle \(x^{2}+y^{2}=9\) in between \(y=0\) and \(\mathbf{y}=\mathbf{2}\) is revolved about \(\mathbf{y}\)-axis. The volume of generating solid will be
1 \(\frac{46}{3} \pi\)
2 \(12 \pi\)
3 \(16 \pi\)
4 \(28 \pi\)
Explanation:
(A) : Given, \(\mathrm{x}^{2}+\mathrm{y}^{2}=9, \mathrm{y}=0\) and \(\mathrm{y}=2\) The volume of this frustum \(=\pi \int_{a}^{b} x^{2} d y\) \(=\pi \int_{0}^{2}\left(9-y^{2}\right) d y\) \(=\pi\left[9 \mathrm{y}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{2}=\pi\left[18-\frac{8}{3}-0\right]=\pi \cdot \frac{46}{3}=\frac{46}{3} \pi\)
85821
If a curve passes through the origin and the slope of the tangent to it at any point \((x, y)\) is \(\frac{x^{2}-4 x+y+8}{x-2}\), then this curve also passes through the point
1 \((5,4)\)
2 \((4,5)\)
3 \((4,4)\)
4 \((5,5)\)
Explanation:
(D) : Given, Slope \(=\frac{x^{2}-4 x+y+8}{x-2}\) \(\frac{d y}{d x}=\frac{x^{2}-4 x+y+8}{x-2} \Rightarrow \frac{d y}{d x}=\frac{(x-2)^{2}+(x+4)}{(x-2)}\) \(\frac{d y}{d x}=(x-2)+\frac{y+4}{x-2}\) Let, \(\quad(\mathrm{x}-2)=\mathrm{t}\) \(\mathrm{dx}=\mathrm{dt}\) and \(\quad(x+4)=u\) \(\mathrm{dy}=\mathrm{du} \Rightarrow \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dt}}\) Now, \(\frac{d y}{d x}=(x-2)+\frac{(y+4)}{(x-2)}\) \(\frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{t}+\frac{\mathrm{u}}{\mathrm{t}}, \quad \frac{\mathrm{du}}{\mathrm{dt}}-\frac{\mathrm{u}}{\mathrm{t}}=\mathrm{t}\) Here, integrating factor \((\mathrm{IF})=\frac{1}{\mathrm{t}}\) \(u \cdot\left(\frac{1}{t}\right)=\int t\left(\frac{1}{t}\right) d t \Rightarrow \frac{u}{t}=t+C\) \(\frac{(y+4)}{(x-2)}=(x-2)+C\) \(\because\) It passes through origin \((0,0)\). \(\therefore \quad \frac{4}{-2}=-2+\mathrm{C}\) \(-2+2=\mathrm{C}\) \(\mathrm{C}=0\) \(\text { Hence, } \frac{(\mathrm{y}+4)}{(\mathrm{x}-2)}=(\mathrm{x}-2)+0 \quad[\because \mathrm{C}=0]\) \(\mathrm{x}+4=(\mathrm{x}-2)^{2}\) Clearly, this curve passes through \((5,5)\) as it satisfies the equation.
JEE Main-2021-25.02.2021
Application of Derivatives
85822
If \(\frac{d y}{d x}=4\) and \(\frac{d^{2} y}{d x^{2}}=-3\) at a point \(P\) on the curve \(y=f(x)\), then \(\left(\frac{d^{2} x}{d y^{2}}\right)_{P}=\)
85811
The part of circle \(x^{2}+y^{2}=9\) in between \(y=0\) and \(\mathbf{y}=\mathbf{2}\) is revolved about \(\mathbf{y}\)-axis. The volume of generating solid will be
1 \(\frac{46}{3} \pi\)
2 \(12 \pi\)
3 \(16 \pi\)
4 \(28 \pi\)
Explanation:
(A) : Given, \(\mathrm{x}^{2}+\mathrm{y}^{2}=9, \mathrm{y}=0\) and \(\mathrm{y}=2\) The volume of this frustum \(=\pi \int_{a}^{b} x^{2} d y\) \(=\pi \int_{0}^{2}\left(9-y^{2}\right) d y\) \(=\pi\left[9 \mathrm{y}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{2}=\pi\left[18-\frac{8}{3}-0\right]=\pi \cdot \frac{46}{3}=\frac{46}{3} \pi\)
85821
If a curve passes through the origin and the slope of the tangent to it at any point \((x, y)\) is \(\frac{x^{2}-4 x+y+8}{x-2}\), then this curve also passes through the point
1 \((5,4)\)
2 \((4,5)\)
3 \((4,4)\)
4 \((5,5)\)
Explanation:
(D) : Given, Slope \(=\frac{x^{2}-4 x+y+8}{x-2}\) \(\frac{d y}{d x}=\frac{x^{2}-4 x+y+8}{x-2} \Rightarrow \frac{d y}{d x}=\frac{(x-2)^{2}+(x+4)}{(x-2)}\) \(\frac{d y}{d x}=(x-2)+\frac{y+4}{x-2}\) Let, \(\quad(\mathrm{x}-2)=\mathrm{t}\) \(\mathrm{dx}=\mathrm{dt}\) and \(\quad(x+4)=u\) \(\mathrm{dy}=\mathrm{du} \Rightarrow \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dt}}\) Now, \(\frac{d y}{d x}=(x-2)+\frac{(y+4)}{(x-2)}\) \(\frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{t}+\frac{\mathrm{u}}{\mathrm{t}}, \quad \frac{\mathrm{du}}{\mathrm{dt}}-\frac{\mathrm{u}}{\mathrm{t}}=\mathrm{t}\) Here, integrating factor \((\mathrm{IF})=\frac{1}{\mathrm{t}}\) \(u \cdot\left(\frac{1}{t}\right)=\int t\left(\frac{1}{t}\right) d t \Rightarrow \frac{u}{t}=t+C\) \(\frac{(y+4)}{(x-2)}=(x-2)+C\) \(\because\) It passes through origin \((0,0)\). \(\therefore \quad \frac{4}{-2}=-2+\mathrm{C}\) \(-2+2=\mathrm{C}\) \(\mathrm{C}=0\) \(\text { Hence, } \frac{(\mathrm{y}+4)}{(\mathrm{x}-2)}=(\mathrm{x}-2)+0 \quad[\because \mathrm{C}=0]\) \(\mathrm{x}+4=(\mathrm{x}-2)^{2}\) Clearly, this curve passes through \((5,5)\) as it satisfies the equation.
JEE Main-2021-25.02.2021
Application of Derivatives
85822
If \(\frac{d y}{d x}=4\) and \(\frac{d^{2} y}{d x^{2}}=-3\) at a point \(P\) on the curve \(y=f(x)\), then \(\left(\frac{d^{2} x}{d y^{2}}\right)_{P}=\)
85811
The part of circle \(x^{2}+y^{2}=9\) in between \(y=0\) and \(\mathbf{y}=\mathbf{2}\) is revolved about \(\mathbf{y}\)-axis. The volume of generating solid will be
1 \(\frac{46}{3} \pi\)
2 \(12 \pi\)
3 \(16 \pi\)
4 \(28 \pi\)
Explanation:
(A) : Given, \(\mathrm{x}^{2}+\mathrm{y}^{2}=9, \mathrm{y}=0\) and \(\mathrm{y}=2\) The volume of this frustum \(=\pi \int_{a}^{b} x^{2} d y\) \(=\pi \int_{0}^{2}\left(9-y^{2}\right) d y\) \(=\pi\left[9 \mathrm{y}-\frac{\mathrm{y}^{3}}{3}\right]_{0}^{2}=\pi\left[18-\frac{8}{3}-0\right]=\pi \cdot \frac{46}{3}=\frac{46}{3} \pi\)
