85427
The equation of normal of \(x^{2}+y^{2}-2 x+4 y-5\) \(=0\) at \((2,1)\) is
1 \(y=3 x-5\)
2 \(2 y=3 x-4\)
3 \(y=3 x+4\)
4 \(y=x+1\)
Explanation:
(A) : Given, \(x^{2}+y^{2}-2 x+4 y-5=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0 \Rightarrow(2 y+4) \frac{d y}{d x}=(2-2 x)\) \(\frac{d y}{d x}=\frac{(2-2 x)}{(2 y+4)} \Rightarrow \frac{d y}{d x}=\frac{2(1-x)}{2(y+2)}\) \(\frac{d y}{d x}=\frac{1-x}{y+2} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{(1-2)}{(1+2)}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,1)}=-\frac{1}{3}\) Thus, \(\mathrm{m}_{1}=-\frac{1}{3}\) We know that, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) On putting the value \(\mathrm{m}_{1}\) in above equation, \(\left(-\frac{1}{3}\right) \cdot \mathrm{m}_{2}=-1\) \(\mathrm{~m}_{2}=3\) The equation of the normal is, \(\left(\mathrm{y}-\mathrm{y}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \((\mathrm{y}-1)=3(\mathrm{x}-2)\) \(\mathrm{y}-1=3 \mathrm{x}-6\) \(\mathrm{y}=3 \mathrm{x}-6+1\) \(\mathrm{y}=3 \mathrm{x}-5\)
WB JEE-2010
Application of Derivatives
85428
The equation of one of the curves whose slope at any point is equal to \(y+2 x\) is
1 \(y=2\left(e^{x}+x-1\right)\)
2 \(y=2\left(e^{x}-x-1\right)\)
3 \(y=2\left(e^{x}-x+1\right)\)
4 \(y=2\left(e^{x}+x+1\right)\)
Explanation:
(B) : Given, \(y+2 x\) According to the question, \(\begin{aligned} & \frac{d y}{d x}=y+2 x \\ & \frac{d y}{d x}-y=2 x\end{aligned}\) It is a linear differential equation of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\) Where, \(\mathrm{P}=-1\) and \(\mathrm{Q}=2 \mathrm{x}\) \(\int \mathrm{pdx}=\int-1 \mathrm{dx}=-\mathrm{x}\) \(\therefore\) I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}\) \(=\mathrm{e}^{-\int \mathrm{Pdx}}=\mathrm{e}^{-\mathrm{x}}\) On multiplying both sides by \(\mathrm{e}^{-\mathrm{x}}\) of equation (i), \(\mathrm{e}^{-\mathrm{x}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) \(\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{e}^{-\mathrm{x}} \mathrm{y}=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) Integrating both sides with respect to \(x\), we get- \(\begin{aligned} & \mathrm{ye}^{-\mathrm{x}}=2 \int \mathrm{e}^{-\mathrm{x}} \mathrm{x} \cdot \mathrm{dx}+\mathrm{C} \\ & y e^{-x}=2 x \int e^{-x} d x-2 \int\left[\frac{d}{d x}(x) \int e^{-x} d x\right] d x+C \\ & \mathrm{ye}^{-\mathrm{x}}=-2 \mathrm{xe}^{-x}-2 \mathrm{e}^{-x}+\mathrm{C} \\ & \end{aligned}\) Since, the curve passes through origin, we have, \(0 \times \mathrm{e}^{\circ}=-2 \times 0 \times \mathrm{e}^{\circ}-2 \mathrm{e}^{\circ}+\mathrm{C}\) \(\mathrm{C}=2\) On putting the value of \(\mathrm{C}\) in equation (ii), we get \(y^{-x}=-2 x^{-x}-2 e^{-x}+2\) \(y=-2 x-2+2 e^{x}\) \(y=2\left(e^{x}-x-1\right)\)
WB JEE-2010
Application of Derivatives
85429
If the line \(a x+b y+c=0\) is a tangent to the curve \(x y=4\), then
1 a \(\lt 0, b>0\)
2 a \(\leq 0, b>0\)
3 \(\mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
4 a \(\leq 0\), b \(\lt 0\)
Explanation:
(C) : Given, \(x y=4\) \(y=\frac{4}{x}\) Therefore, \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4}{\mathrm{x}^{2}}\) Let, \(a x+b y+c=0\) \(y=\frac{-a}{b} x-\frac{c}{b}\) Slope of line \(=\frac{-a}{b}\) \(\text { i. e., }-\frac{4}{x^{2}}=-\frac{a}{b}\) \(x^{2}=\frac{4 b}{a}>0\) Which is true when, \(\mathrm{a}>0, \mathrm{~b}>0 \text { or } \mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
WB JEE-2010
Application of Derivatives
85430
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(f^{\prime}(3)\) i
1 1
2 -1
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given, \(y=f(x)\) \(\frac{d y}{d x}=f^{\prime}(x) \Rightarrow\left(\frac{d y}{d x}\right)=f^{\prime}(x)_{(3,4)}\) The slope of normal \(=-\frac{1}{f^{\prime}(x)_{(3,4)}}=-\frac{1}{f^{\prime}(3)}\) \(-\frac{1}{f^{\prime}(3)} =\tan \left(\frac{3 \pi}{4}\right)\) \(\frac{-1}{f^{\prime}(3)} =-1 \Rightarrow f^{\prime}(3)=1\)
85427
The equation of normal of \(x^{2}+y^{2}-2 x+4 y-5\) \(=0\) at \((2,1)\) is
1 \(y=3 x-5\)
2 \(2 y=3 x-4\)
3 \(y=3 x+4\)
4 \(y=x+1\)
Explanation:
(A) : Given, \(x^{2}+y^{2}-2 x+4 y-5=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0 \Rightarrow(2 y+4) \frac{d y}{d x}=(2-2 x)\) \(\frac{d y}{d x}=\frac{(2-2 x)}{(2 y+4)} \Rightarrow \frac{d y}{d x}=\frac{2(1-x)}{2(y+2)}\) \(\frac{d y}{d x}=\frac{1-x}{y+2} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{(1-2)}{(1+2)}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,1)}=-\frac{1}{3}\) Thus, \(\mathrm{m}_{1}=-\frac{1}{3}\) We know that, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) On putting the value \(\mathrm{m}_{1}\) in above equation, \(\left(-\frac{1}{3}\right) \cdot \mathrm{m}_{2}=-1\) \(\mathrm{~m}_{2}=3\) The equation of the normal is, \(\left(\mathrm{y}-\mathrm{y}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \((\mathrm{y}-1)=3(\mathrm{x}-2)\) \(\mathrm{y}-1=3 \mathrm{x}-6\) \(\mathrm{y}=3 \mathrm{x}-6+1\) \(\mathrm{y}=3 \mathrm{x}-5\)
WB JEE-2010
Application of Derivatives
85428
The equation of one of the curves whose slope at any point is equal to \(y+2 x\) is
1 \(y=2\left(e^{x}+x-1\right)\)
2 \(y=2\left(e^{x}-x-1\right)\)
3 \(y=2\left(e^{x}-x+1\right)\)
4 \(y=2\left(e^{x}+x+1\right)\)
Explanation:
(B) : Given, \(y+2 x\) According to the question, \(\begin{aligned} & \frac{d y}{d x}=y+2 x \\ & \frac{d y}{d x}-y=2 x\end{aligned}\) It is a linear differential equation of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\) Where, \(\mathrm{P}=-1\) and \(\mathrm{Q}=2 \mathrm{x}\) \(\int \mathrm{pdx}=\int-1 \mathrm{dx}=-\mathrm{x}\) \(\therefore\) I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}\) \(=\mathrm{e}^{-\int \mathrm{Pdx}}=\mathrm{e}^{-\mathrm{x}}\) On multiplying both sides by \(\mathrm{e}^{-\mathrm{x}}\) of equation (i), \(\mathrm{e}^{-\mathrm{x}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) \(\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{e}^{-\mathrm{x}} \mathrm{y}=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) Integrating both sides with respect to \(x\), we get- \(\begin{aligned} & \mathrm{ye}^{-\mathrm{x}}=2 \int \mathrm{e}^{-\mathrm{x}} \mathrm{x} \cdot \mathrm{dx}+\mathrm{C} \\ & y e^{-x}=2 x \int e^{-x} d x-2 \int\left[\frac{d}{d x}(x) \int e^{-x} d x\right] d x+C \\ & \mathrm{ye}^{-\mathrm{x}}=-2 \mathrm{xe}^{-x}-2 \mathrm{e}^{-x}+\mathrm{C} \\ & \end{aligned}\) Since, the curve passes through origin, we have, \(0 \times \mathrm{e}^{\circ}=-2 \times 0 \times \mathrm{e}^{\circ}-2 \mathrm{e}^{\circ}+\mathrm{C}\) \(\mathrm{C}=2\) On putting the value of \(\mathrm{C}\) in equation (ii), we get \(y^{-x}=-2 x^{-x}-2 e^{-x}+2\) \(y=-2 x-2+2 e^{x}\) \(y=2\left(e^{x}-x-1\right)\)
WB JEE-2010
Application of Derivatives
85429
If the line \(a x+b y+c=0\) is a tangent to the curve \(x y=4\), then
1 a \(\lt 0, b>0\)
2 a \(\leq 0, b>0\)
3 \(\mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
4 a \(\leq 0\), b \(\lt 0\)
Explanation:
(C) : Given, \(x y=4\) \(y=\frac{4}{x}\) Therefore, \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4}{\mathrm{x}^{2}}\) Let, \(a x+b y+c=0\) \(y=\frac{-a}{b} x-\frac{c}{b}\) Slope of line \(=\frac{-a}{b}\) \(\text { i. e., }-\frac{4}{x^{2}}=-\frac{a}{b}\) \(x^{2}=\frac{4 b}{a}>0\) Which is true when, \(\mathrm{a}>0, \mathrm{~b}>0 \text { or } \mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
WB JEE-2010
Application of Derivatives
85430
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(f^{\prime}(3)\) i
1 1
2 -1
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given, \(y=f(x)\) \(\frac{d y}{d x}=f^{\prime}(x) \Rightarrow\left(\frac{d y}{d x}\right)=f^{\prime}(x)_{(3,4)}\) The slope of normal \(=-\frac{1}{f^{\prime}(x)_{(3,4)}}=-\frac{1}{f^{\prime}(3)}\) \(-\frac{1}{f^{\prime}(3)} =\tan \left(\frac{3 \pi}{4}\right)\) \(\frac{-1}{f^{\prime}(3)} =-1 \Rightarrow f^{\prime}(3)=1\)
85427
The equation of normal of \(x^{2}+y^{2}-2 x+4 y-5\) \(=0\) at \((2,1)\) is
1 \(y=3 x-5\)
2 \(2 y=3 x-4\)
3 \(y=3 x+4\)
4 \(y=x+1\)
Explanation:
(A) : Given, \(x^{2}+y^{2}-2 x+4 y-5=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0 \Rightarrow(2 y+4) \frac{d y}{d x}=(2-2 x)\) \(\frac{d y}{d x}=\frac{(2-2 x)}{(2 y+4)} \Rightarrow \frac{d y}{d x}=\frac{2(1-x)}{2(y+2)}\) \(\frac{d y}{d x}=\frac{1-x}{y+2} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{(1-2)}{(1+2)}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,1)}=-\frac{1}{3}\) Thus, \(\mathrm{m}_{1}=-\frac{1}{3}\) We know that, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) On putting the value \(\mathrm{m}_{1}\) in above equation, \(\left(-\frac{1}{3}\right) \cdot \mathrm{m}_{2}=-1\) \(\mathrm{~m}_{2}=3\) The equation of the normal is, \(\left(\mathrm{y}-\mathrm{y}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \((\mathrm{y}-1)=3(\mathrm{x}-2)\) \(\mathrm{y}-1=3 \mathrm{x}-6\) \(\mathrm{y}=3 \mathrm{x}-6+1\) \(\mathrm{y}=3 \mathrm{x}-5\)
WB JEE-2010
Application of Derivatives
85428
The equation of one of the curves whose slope at any point is equal to \(y+2 x\) is
1 \(y=2\left(e^{x}+x-1\right)\)
2 \(y=2\left(e^{x}-x-1\right)\)
3 \(y=2\left(e^{x}-x+1\right)\)
4 \(y=2\left(e^{x}+x+1\right)\)
Explanation:
(B) : Given, \(y+2 x\) According to the question, \(\begin{aligned} & \frac{d y}{d x}=y+2 x \\ & \frac{d y}{d x}-y=2 x\end{aligned}\) It is a linear differential equation of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\) Where, \(\mathrm{P}=-1\) and \(\mathrm{Q}=2 \mathrm{x}\) \(\int \mathrm{pdx}=\int-1 \mathrm{dx}=-\mathrm{x}\) \(\therefore\) I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}\) \(=\mathrm{e}^{-\int \mathrm{Pdx}}=\mathrm{e}^{-\mathrm{x}}\) On multiplying both sides by \(\mathrm{e}^{-\mathrm{x}}\) of equation (i), \(\mathrm{e}^{-\mathrm{x}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) \(\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{e}^{-\mathrm{x}} \mathrm{y}=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) Integrating both sides with respect to \(x\), we get- \(\begin{aligned} & \mathrm{ye}^{-\mathrm{x}}=2 \int \mathrm{e}^{-\mathrm{x}} \mathrm{x} \cdot \mathrm{dx}+\mathrm{C} \\ & y e^{-x}=2 x \int e^{-x} d x-2 \int\left[\frac{d}{d x}(x) \int e^{-x} d x\right] d x+C \\ & \mathrm{ye}^{-\mathrm{x}}=-2 \mathrm{xe}^{-x}-2 \mathrm{e}^{-x}+\mathrm{C} \\ & \end{aligned}\) Since, the curve passes through origin, we have, \(0 \times \mathrm{e}^{\circ}=-2 \times 0 \times \mathrm{e}^{\circ}-2 \mathrm{e}^{\circ}+\mathrm{C}\) \(\mathrm{C}=2\) On putting the value of \(\mathrm{C}\) in equation (ii), we get \(y^{-x}=-2 x^{-x}-2 e^{-x}+2\) \(y=-2 x-2+2 e^{x}\) \(y=2\left(e^{x}-x-1\right)\)
WB JEE-2010
Application of Derivatives
85429
If the line \(a x+b y+c=0\) is a tangent to the curve \(x y=4\), then
1 a \(\lt 0, b>0\)
2 a \(\leq 0, b>0\)
3 \(\mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
4 a \(\leq 0\), b \(\lt 0\)
Explanation:
(C) : Given, \(x y=4\) \(y=\frac{4}{x}\) Therefore, \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4}{\mathrm{x}^{2}}\) Let, \(a x+b y+c=0\) \(y=\frac{-a}{b} x-\frac{c}{b}\) Slope of line \(=\frac{-a}{b}\) \(\text { i. e., }-\frac{4}{x^{2}}=-\frac{a}{b}\) \(x^{2}=\frac{4 b}{a}>0\) Which is true when, \(\mathrm{a}>0, \mathrm{~b}>0 \text { or } \mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
WB JEE-2010
Application of Derivatives
85430
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(f^{\prime}(3)\) i
1 1
2 -1
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given, \(y=f(x)\) \(\frac{d y}{d x}=f^{\prime}(x) \Rightarrow\left(\frac{d y}{d x}\right)=f^{\prime}(x)_{(3,4)}\) The slope of normal \(=-\frac{1}{f^{\prime}(x)_{(3,4)}}=-\frac{1}{f^{\prime}(3)}\) \(-\frac{1}{f^{\prime}(3)} =\tan \left(\frac{3 \pi}{4}\right)\) \(\frac{-1}{f^{\prime}(3)} =-1 \Rightarrow f^{\prime}(3)=1\)
85427
The equation of normal of \(x^{2}+y^{2}-2 x+4 y-5\) \(=0\) at \((2,1)\) is
1 \(y=3 x-5\)
2 \(2 y=3 x-4\)
3 \(y=3 x+4\)
4 \(y=x+1\)
Explanation:
(A) : Given, \(x^{2}+y^{2}-2 x+4 y-5=0\) On differentiating both sides w.r.t.x, we get- \(2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0 \Rightarrow(2 y+4) \frac{d y}{d x}=(2-2 x)\) \(\frac{d y}{d x}=\frac{(2-2 x)}{(2 y+4)} \Rightarrow \frac{d y}{d x}=\frac{2(1-x)}{2(y+2)}\) \(\frac{d y}{d x}=\frac{1-x}{y+2} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{(1-2)}{(1+2)}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,1)}=-\frac{1}{3}\) Thus, \(\mathrm{m}_{1}=-\frac{1}{3}\) We know that, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) On putting the value \(\mathrm{m}_{1}\) in above equation, \(\left(-\frac{1}{3}\right) \cdot \mathrm{m}_{2}=-1\) \(\mathrm{~m}_{2}=3\) The equation of the normal is, \(\left(\mathrm{y}-\mathrm{y}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \((\mathrm{y}-1)=3(\mathrm{x}-2)\) \(\mathrm{y}-1=3 \mathrm{x}-6\) \(\mathrm{y}=3 \mathrm{x}-6+1\) \(\mathrm{y}=3 \mathrm{x}-5\)
WB JEE-2010
Application of Derivatives
85428
The equation of one of the curves whose slope at any point is equal to \(y+2 x\) is
1 \(y=2\left(e^{x}+x-1\right)\)
2 \(y=2\left(e^{x}-x-1\right)\)
3 \(y=2\left(e^{x}-x+1\right)\)
4 \(y=2\left(e^{x}+x+1\right)\)
Explanation:
(B) : Given, \(y+2 x\) According to the question, \(\begin{aligned} & \frac{d y}{d x}=y+2 x \\ & \frac{d y}{d x}-y=2 x\end{aligned}\) It is a linear differential equation of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\) Where, \(\mathrm{P}=-1\) and \(\mathrm{Q}=2 \mathrm{x}\) \(\int \mathrm{pdx}=\int-1 \mathrm{dx}=-\mathrm{x}\) \(\therefore\) I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}\) \(=\mathrm{e}^{-\int \mathrm{Pdx}}=\mathrm{e}^{-\mathrm{x}}\) On multiplying both sides by \(\mathrm{e}^{-\mathrm{x}}\) of equation (i), \(\mathrm{e}^{-\mathrm{x}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) \(\mathrm{e}^{-\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{e}^{-\mathrm{x}} \mathrm{y}=\mathrm{e}^{-\mathrm{x}} \cdot 2 \mathrm{x}\) Integrating both sides with respect to \(x\), we get- \(\begin{aligned} & \mathrm{ye}^{-\mathrm{x}}=2 \int \mathrm{e}^{-\mathrm{x}} \mathrm{x} \cdot \mathrm{dx}+\mathrm{C} \\ & y e^{-x}=2 x \int e^{-x} d x-2 \int\left[\frac{d}{d x}(x) \int e^{-x} d x\right] d x+C \\ & \mathrm{ye}^{-\mathrm{x}}=-2 \mathrm{xe}^{-x}-2 \mathrm{e}^{-x}+\mathrm{C} \\ & \end{aligned}\) Since, the curve passes through origin, we have, \(0 \times \mathrm{e}^{\circ}=-2 \times 0 \times \mathrm{e}^{\circ}-2 \mathrm{e}^{\circ}+\mathrm{C}\) \(\mathrm{C}=2\) On putting the value of \(\mathrm{C}\) in equation (ii), we get \(y^{-x}=-2 x^{-x}-2 e^{-x}+2\) \(y=-2 x-2+2 e^{x}\) \(y=2\left(e^{x}-x-1\right)\)
WB JEE-2010
Application of Derivatives
85429
If the line \(a x+b y+c=0\) is a tangent to the curve \(x y=4\), then
1 a \(\lt 0, b>0\)
2 a \(\leq 0, b>0\)
3 \(\mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
4 a \(\leq 0\), b \(\lt 0\)
Explanation:
(C) : Given, \(x y=4\) \(y=\frac{4}{x}\) Therefore, \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4}{\mathrm{x}^{2}}\) Let, \(a x+b y+c=0\) \(y=\frac{-a}{b} x-\frac{c}{b}\) Slope of line \(=\frac{-a}{b}\) \(\text { i. e., }-\frac{4}{x^{2}}=-\frac{a}{b}\) \(x^{2}=\frac{4 b}{a}>0\) Which is true when, \(\mathrm{a}>0, \mathrm{~b}>0 \text { or } \mathrm{a}\lt 0, \mathrm{~b}\lt 0\)
WB JEE-2010
Application of Derivatives
85430
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(f^{\prime}(3)\) i
1 1
2 -1
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given, \(y=f(x)\) \(\frac{d y}{d x}=f^{\prime}(x) \Rightarrow\left(\frac{d y}{d x}\right)=f^{\prime}(x)_{(3,4)}\) The slope of normal \(=-\frac{1}{f^{\prime}(x)_{(3,4)}}=-\frac{1}{f^{\prime}(3)}\) \(-\frac{1}{f^{\prime}(3)} =\tan \left(\frac{3 \pi}{4}\right)\) \(\frac{-1}{f^{\prime}(3)} =-1 \Rightarrow f^{\prime}(3)=1\)
