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Chapter 5: Problem 27
\(y=2 x^{2}-x+1\) \(\frac{d y}{d x}=4 x_{1}-1\) \(4 x_{1}-1=3\) \(x_{1}=1\) \(y_{1}=2\)
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\(\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2\) $\frac{n}{a}\left(\frac{x}{a}\right)^{n-1}+\frac{n}{b}\left(\frac{y}{b}\right)^{n-t} \frac{d y}{d x}=0$ \(\frac{d y}{d x}=-\frac{b^{n} x^{n-1}}{a^{n} y^{n-1}}=-\frac{b}{a}\)
\(f_{1}^{\prime}(x)=2 x-1 \quad \& \quad f_{2}^{\prime}(x)=3 x^{2}-2 x-2\) \(\Rightarrow 2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2\) \(\Rightarrow 3 x_{2}^{2}-2 x_{2}-2 x_{1}-1=0\) For \(\mathrm{x}_{2}\) be real, \(\mathrm{D} \geq 0\) \(4-4\left(2 x_{1}+1\right)(3)\) \(4-24 x_{1}-12\) \(\Rightarrow-\left(24 x_{1}+8\right)\) There can be infinite such values of \(\mathrm{x}_{\mathrm{l}}\)
\(y=\frac{a x}{1+x}=a-\frac{a}{1+x}\) \(\frac{d y}{d x}=\frac{a}{(1+x)^{2}}=-1\) \(a=-\left(1+x_{1}\right)^{2}\) eqn of tangent \(y-\frac{a x_{1}}{1+x_{1}}=-1\left(x-x_{1}\right)\) \(\mathrm{y}+\mathrm{x}=\mathrm{x}_{1}+\frac{\mathrm{ax}_{1}}{1+\mathrm{x}_{1}}\) \& \(y-3=-x \Rightarrow y+x=3\) \(\Rightarrow \mathrm{x}_{1}+\frac{\mathrm{ax}_{1}}{1+\mathrm{x}_{1}}=3\) \(\Rightarrow \mathrm{x}_{1}-\mathrm{x}_{1}\left(1+\mathrm{x}_{1}\right)=3\)
\(x+y=x^{y}\) cuts \(x\)-axis at \((1,0)\) Now, \(1+y^{\prime}=x^{y}\left(\frac{y}{x}+y^{\prime} \ln x\right)\) \(y^{\prime}=-1\) Slope of normal \(=1\) eqn of normal \(y=x-1\)
Ellipse \(\rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) \(\Rightarrow \frac{x}{8} \frac{d x}{d t}+\frac{2 y}{9} \frac{d y}{d t}=1\) $\Rightarrow \frac{4 \sqrt{1-y^{2} / 9}}{8} \frac{d x}{d t}+\frac{2 y}{9} \times \frac{d y}{d x} \times \frac{d x}{d t}=1$ Now put \(\frac{d y}{d x}=1 \quad \& \quad y=1\) $\Rightarrow \frac{4}{6 \sqrt{2}} \frac{d x}{d t}+\frac{2}{9} \frac{d x}{d t}=1$ \(\Rightarrow \frac{d x}{d t}\left(\frac{\sqrt{2}}{3}+\frac{2}{9}\right)=1\) \(\Rightarrow \frac{d x}{d t}=\frac{9}{3 \sqrt{2}+2}\)
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