85401
If \(x+y=k\) is normal to \(y^{2}=12 x\), then \(k\) is equal to
1 3
2 9
3 -9
4 -3
Explanation:
(B) : Given, \(y^{2}=12 x \tag{i}\) On differentiating w.r.t \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{12}{2 y}\) \(\frac{d y}{d x}=\frac{6}{y}\) \(\therefore\) Slope of the normal \(=-\frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}=\frac{-\mathrm{y}}{6}\) Slope of the line \(\mathrm{x}+\mathrm{y}=\mathrm{k}\) is -1 \(\frac{-y}{6}=-1\) \(y=6\) Putting the value of \(y=6\) in equation (i), we get \(6^{2}=12 x\) \(36=12 x\) \(x=\frac{36}{12}\) \(x=3\) Putting the value of \(x\) and \(y\) in equation \(x+y=k\) \(x+y=k\) \(3+6=k\) \(k=9\)
MHT CET-2011
Application of Derivatives
85402
Length of the sub tangent at \((a, a)\) on the curve \(y^{2}=\frac{x^{2}}{2 a+x} \text { is equal to }\)
1 \(\frac{18}{5}\)
2 \(\frac{18 a}{5}\)
3 \(\frac{18 \mathrm{a}^{2}}{5}\)
4 \(-\frac{18 a^{2}}{5}\)
Explanation:
(C) : Given, On differentiating both sides, \(2 y \frac{d y}{d x}=\frac{(2 a+x)(2 x)-x^{2}(1)}{(2 a+x)^{2}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(2 a+a)(2 a)-a^{2}}{(2 a+a)^{2}}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(3 a)(2 a)-a^{2}}{2 a(3 a)^{2}}=\frac{6 a^{2}-a^{2}}{2 a \times 9 a^{2}}=\frac{5 a^{2}}{18 a^{3}}}\) Length of subtangent at \((a, a)\), \(=\frac{y}{\left(\frac{d y}{d x}\right)_{(a, a)}}=\frac{a}{\frac{5}{18 a}}=a \times \frac{18 a}{5}=\frac{18 a^{2}}{5}\)
COMEDK-2012
Application of Derivatives
85403
If the tangent to the curve \(2 y^{3}=a x^{2}+x^{3}\) at the point ( \(a\), a) cuts off intercepts \(\alpha\) and \(\beta\) on the coordinate axes where \(\alpha^{2}+\beta^{2}=61\), then the value of \(a\) is
1 25
2 36
3 \(\pm 30\)
4 \(\pm 40\)
Explanation:
(C) : Given, the curve - \(2 y^{3}=a x^{2}+x^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - \(6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2}\) \(\frac{d y}{d x}=\frac{2 a x+3 x^{2}}{6 y^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5 a^{2}}{6 a^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{6}\) The equation of the tangent at \((a, a)\) is, \(y-a=\frac{5}{6}(x-a)\) \(6 y-6 a=5 x-5 a\) \(5 x-6 y+a=0\) This intercepts lengths \(=\frac{-a}{5}\) and \(\frac{a}{6}\) with \(x\) and \(y\) axis respectively. \(\therefore \quad \alpha=\frac{-a}{5}, \beta=\frac{a}{6}\) Now, \(\alpha^{2}+\beta^{2}=61\) \(\left(\frac{-a}{5}\right)^{2}+\left(\frac{a}{6}\right)^{2}=61\) \(\frac{a^{2}}{25}+\frac{a^{2}}{36}=61\) \(\frac{36 a^{2}+25 a^{2}}{25 \times 36}=61\) \(61 a^{2}=61 \times 36 \times 25\) \(a^{2}=36 \times 25\) \(a=6 \times 5\) \(a= \pm 30\) The value of a is \(\pm 30\).
COMEDK-2012
Application of Derivatives
85404
If the radius of a sphere is measured as \(9 \mathrm{~cm}\) with an error of \(0.03 \mathrm{~cm}\), then find the approximating error in calculating its volume.
1 \(2.46 \pi \mathrm{cm}^{3}\)
2 \(8.62 \pi \mathrm{cm}^{3}\)
3 \(9.72 \pi \mathrm{cm}^{3}\)
4 \(7.46 \pi \mathrm{cm}^{3}\)
Explanation:
(C) : We have, Radius of sphere \((\mathrm{r})=9 \mathrm{~cm}\) and Error \(\Delta \mathrm{r}=0.03 \mathrm{~cm}\) Volume of sphere, \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) \(\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^{2}\) \(\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{(\mathrm{r}=9)}=4 \pi \times 9^{2}=324 \pi\) Let, \(\Delta \mathrm{V}\) be the error in \(\mathrm{V}\) due to error \(\Delta \mathrm{r}\) in \(\mathrm{r}\). Then, \(\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta \mathrm{r}\) \(\Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}\) Thus, the approximate error in calculating the volume is \(9.72 \pi \mathrm{cm}^{3}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85401
If \(x+y=k\) is normal to \(y^{2}=12 x\), then \(k\) is equal to
1 3
2 9
3 -9
4 -3
Explanation:
(B) : Given, \(y^{2}=12 x \tag{i}\) On differentiating w.r.t \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{12}{2 y}\) \(\frac{d y}{d x}=\frac{6}{y}\) \(\therefore\) Slope of the normal \(=-\frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}=\frac{-\mathrm{y}}{6}\) Slope of the line \(\mathrm{x}+\mathrm{y}=\mathrm{k}\) is -1 \(\frac{-y}{6}=-1\) \(y=6\) Putting the value of \(y=6\) in equation (i), we get \(6^{2}=12 x\) \(36=12 x\) \(x=\frac{36}{12}\) \(x=3\) Putting the value of \(x\) and \(y\) in equation \(x+y=k\) \(x+y=k\) \(3+6=k\) \(k=9\)
MHT CET-2011
Application of Derivatives
85402
Length of the sub tangent at \((a, a)\) on the curve \(y^{2}=\frac{x^{2}}{2 a+x} \text { is equal to }\)
1 \(\frac{18}{5}\)
2 \(\frac{18 a}{5}\)
3 \(\frac{18 \mathrm{a}^{2}}{5}\)
4 \(-\frac{18 a^{2}}{5}\)
Explanation:
(C) : Given, On differentiating both sides, \(2 y \frac{d y}{d x}=\frac{(2 a+x)(2 x)-x^{2}(1)}{(2 a+x)^{2}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(2 a+a)(2 a)-a^{2}}{(2 a+a)^{2}}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(3 a)(2 a)-a^{2}}{2 a(3 a)^{2}}=\frac{6 a^{2}-a^{2}}{2 a \times 9 a^{2}}=\frac{5 a^{2}}{18 a^{3}}}\) Length of subtangent at \((a, a)\), \(=\frac{y}{\left(\frac{d y}{d x}\right)_{(a, a)}}=\frac{a}{\frac{5}{18 a}}=a \times \frac{18 a}{5}=\frac{18 a^{2}}{5}\)
COMEDK-2012
Application of Derivatives
85403
If the tangent to the curve \(2 y^{3}=a x^{2}+x^{3}\) at the point ( \(a\), a) cuts off intercepts \(\alpha\) and \(\beta\) on the coordinate axes where \(\alpha^{2}+\beta^{2}=61\), then the value of \(a\) is
1 25
2 36
3 \(\pm 30\)
4 \(\pm 40\)
Explanation:
(C) : Given, the curve - \(2 y^{3}=a x^{2}+x^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - \(6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2}\) \(\frac{d y}{d x}=\frac{2 a x+3 x^{2}}{6 y^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5 a^{2}}{6 a^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{6}\) The equation of the tangent at \((a, a)\) is, \(y-a=\frac{5}{6}(x-a)\) \(6 y-6 a=5 x-5 a\) \(5 x-6 y+a=0\) This intercepts lengths \(=\frac{-a}{5}\) and \(\frac{a}{6}\) with \(x\) and \(y\) axis respectively. \(\therefore \quad \alpha=\frac{-a}{5}, \beta=\frac{a}{6}\) Now, \(\alpha^{2}+\beta^{2}=61\) \(\left(\frac{-a}{5}\right)^{2}+\left(\frac{a}{6}\right)^{2}=61\) \(\frac{a^{2}}{25}+\frac{a^{2}}{36}=61\) \(\frac{36 a^{2}+25 a^{2}}{25 \times 36}=61\) \(61 a^{2}=61 \times 36 \times 25\) \(a^{2}=36 \times 25\) \(a=6 \times 5\) \(a= \pm 30\) The value of a is \(\pm 30\).
COMEDK-2012
Application of Derivatives
85404
If the radius of a sphere is measured as \(9 \mathrm{~cm}\) with an error of \(0.03 \mathrm{~cm}\), then find the approximating error in calculating its volume.
1 \(2.46 \pi \mathrm{cm}^{3}\)
2 \(8.62 \pi \mathrm{cm}^{3}\)
3 \(9.72 \pi \mathrm{cm}^{3}\)
4 \(7.46 \pi \mathrm{cm}^{3}\)
Explanation:
(C) : We have, Radius of sphere \((\mathrm{r})=9 \mathrm{~cm}\) and Error \(\Delta \mathrm{r}=0.03 \mathrm{~cm}\) Volume of sphere, \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) \(\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^{2}\) \(\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{(\mathrm{r}=9)}=4 \pi \times 9^{2}=324 \pi\) Let, \(\Delta \mathrm{V}\) be the error in \(\mathrm{V}\) due to error \(\Delta \mathrm{r}\) in \(\mathrm{r}\). Then, \(\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta \mathrm{r}\) \(\Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}\) Thus, the approximate error in calculating the volume is \(9.72 \pi \mathrm{cm}^{3}\).
85401
If \(x+y=k\) is normal to \(y^{2}=12 x\), then \(k\) is equal to
1 3
2 9
3 -9
4 -3
Explanation:
(B) : Given, \(y^{2}=12 x \tag{i}\) On differentiating w.r.t \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{12}{2 y}\) \(\frac{d y}{d x}=\frac{6}{y}\) \(\therefore\) Slope of the normal \(=-\frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}=\frac{-\mathrm{y}}{6}\) Slope of the line \(\mathrm{x}+\mathrm{y}=\mathrm{k}\) is -1 \(\frac{-y}{6}=-1\) \(y=6\) Putting the value of \(y=6\) in equation (i), we get \(6^{2}=12 x\) \(36=12 x\) \(x=\frac{36}{12}\) \(x=3\) Putting the value of \(x\) and \(y\) in equation \(x+y=k\) \(x+y=k\) \(3+6=k\) \(k=9\)
MHT CET-2011
Application of Derivatives
85402
Length of the sub tangent at \((a, a)\) on the curve \(y^{2}=\frac{x^{2}}{2 a+x} \text { is equal to }\)
1 \(\frac{18}{5}\)
2 \(\frac{18 a}{5}\)
3 \(\frac{18 \mathrm{a}^{2}}{5}\)
4 \(-\frac{18 a^{2}}{5}\)
Explanation:
(C) : Given, On differentiating both sides, \(2 y \frac{d y}{d x}=\frac{(2 a+x)(2 x)-x^{2}(1)}{(2 a+x)^{2}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(2 a+a)(2 a)-a^{2}}{(2 a+a)^{2}}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(3 a)(2 a)-a^{2}}{2 a(3 a)^{2}}=\frac{6 a^{2}-a^{2}}{2 a \times 9 a^{2}}=\frac{5 a^{2}}{18 a^{3}}}\) Length of subtangent at \((a, a)\), \(=\frac{y}{\left(\frac{d y}{d x}\right)_{(a, a)}}=\frac{a}{\frac{5}{18 a}}=a \times \frac{18 a}{5}=\frac{18 a^{2}}{5}\)
COMEDK-2012
Application of Derivatives
85403
If the tangent to the curve \(2 y^{3}=a x^{2}+x^{3}\) at the point ( \(a\), a) cuts off intercepts \(\alpha\) and \(\beta\) on the coordinate axes where \(\alpha^{2}+\beta^{2}=61\), then the value of \(a\) is
1 25
2 36
3 \(\pm 30\)
4 \(\pm 40\)
Explanation:
(C) : Given, the curve - \(2 y^{3}=a x^{2}+x^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - \(6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2}\) \(\frac{d y}{d x}=\frac{2 a x+3 x^{2}}{6 y^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5 a^{2}}{6 a^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{6}\) The equation of the tangent at \((a, a)\) is, \(y-a=\frac{5}{6}(x-a)\) \(6 y-6 a=5 x-5 a\) \(5 x-6 y+a=0\) This intercepts lengths \(=\frac{-a}{5}\) and \(\frac{a}{6}\) with \(x\) and \(y\) axis respectively. \(\therefore \quad \alpha=\frac{-a}{5}, \beta=\frac{a}{6}\) Now, \(\alpha^{2}+\beta^{2}=61\) \(\left(\frac{-a}{5}\right)^{2}+\left(\frac{a}{6}\right)^{2}=61\) \(\frac{a^{2}}{25}+\frac{a^{2}}{36}=61\) \(\frac{36 a^{2}+25 a^{2}}{25 \times 36}=61\) \(61 a^{2}=61 \times 36 \times 25\) \(a^{2}=36 \times 25\) \(a=6 \times 5\) \(a= \pm 30\) The value of a is \(\pm 30\).
COMEDK-2012
Application of Derivatives
85404
If the radius of a sphere is measured as \(9 \mathrm{~cm}\) with an error of \(0.03 \mathrm{~cm}\), then find the approximating error in calculating its volume.
1 \(2.46 \pi \mathrm{cm}^{3}\)
2 \(8.62 \pi \mathrm{cm}^{3}\)
3 \(9.72 \pi \mathrm{cm}^{3}\)
4 \(7.46 \pi \mathrm{cm}^{3}\)
Explanation:
(C) : We have, Radius of sphere \((\mathrm{r})=9 \mathrm{~cm}\) and Error \(\Delta \mathrm{r}=0.03 \mathrm{~cm}\) Volume of sphere, \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) \(\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^{2}\) \(\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{(\mathrm{r}=9)}=4 \pi \times 9^{2}=324 \pi\) Let, \(\Delta \mathrm{V}\) be the error in \(\mathrm{V}\) due to error \(\Delta \mathrm{r}\) in \(\mathrm{r}\). Then, \(\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta \mathrm{r}\) \(\Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}\) Thus, the approximate error in calculating the volume is \(9.72 \pi \mathrm{cm}^{3}\).
85401
If \(x+y=k\) is normal to \(y^{2}=12 x\), then \(k\) is equal to
1 3
2 9
3 -9
4 -3
Explanation:
(B) : Given, \(y^{2}=12 x \tag{i}\) On differentiating w.r.t \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{12}{2 y}\) \(\frac{d y}{d x}=\frac{6}{y}\) \(\therefore\) Slope of the normal \(=-\frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}=\frac{-\mathrm{y}}{6}\) Slope of the line \(\mathrm{x}+\mathrm{y}=\mathrm{k}\) is -1 \(\frac{-y}{6}=-1\) \(y=6\) Putting the value of \(y=6\) in equation (i), we get \(6^{2}=12 x\) \(36=12 x\) \(x=\frac{36}{12}\) \(x=3\) Putting the value of \(x\) and \(y\) in equation \(x+y=k\) \(x+y=k\) \(3+6=k\) \(k=9\)
MHT CET-2011
Application of Derivatives
85402
Length of the sub tangent at \((a, a)\) on the curve \(y^{2}=\frac{x^{2}}{2 a+x} \text { is equal to }\)
1 \(\frac{18}{5}\)
2 \(\frac{18 a}{5}\)
3 \(\frac{18 \mathrm{a}^{2}}{5}\)
4 \(-\frac{18 a^{2}}{5}\)
Explanation:
(C) : Given, On differentiating both sides, \(2 y \frac{d y}{d x}=\frac{(2 a+x)(2 x)-x^{2}(1)}{(2 a+x)^{2}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(2 a+a)(2 a)-a^{2}}{(2 a+a)^{2}}}\) \({\left[\frac{d y}{d x}\right]_{(a, a)}=\frac{(3 a)(2 a)-a^{2}}{2 a(3 a)^{2}}=\frac{6 a^{2}-a^{2}}{2 a \times 9 a^{2}}=\frac{5 a^{2}}{18 a^{3}}}\) Length of subtangent at \((a, a)\), \(=\frac{y}{\left(\frac{d y}{d x}\right)_{(a, a)}}=\frac{a}{\frac{5}{18 a}}=a \times \frac{18 a}{5}=\frac{18 a^{2}}{5}\)
COMEDK-2012
Application of Derivatives
85403
If the tangent to the curve \(2 y^{3}=a x^{2}+x^{3}\) at the point ( \(a\), a) cuts off intercepts \(\alpha\) and \(\beta\) on the coordinate axes where \(\alpha^{2}+\beta^{2}=61\), then the value of \(a\) is
1 25
2 36
3 \(\pm 30\)
4 \(\pm 40\)
Explanation:
(C) : Given, the curve - \(2 y^{3}=a x^{2}+x^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - \(6 y^{2} \frac{d y}{d x}=2 a x+3 x^{2}\) \(\frac{d y}{d x}=\frac{2 a x+3 x^{2}}{6 y^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5 a^{2}}{6 a^{2}}\) \(\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{6}\) The equation of the tangent at \((a, a)\) is, \(y-a=\frac{5}{6}(x-a)\) \(6 y-6 a=5 x-5 a\) \(5 x-6 y+a=0\) This intercepts lengths \(=\frac{-a}{5}\) and \(\frac{a}{6}\) with \(x\) and \(y\) axis respectively. \(\therefore \quad \alpha=\frac{-a}{5}, \beta=\frac{a}{6}\) Now, \(\alpha^{2}+\beta^{2}=61\) \(\left(\frac{-a}{5}\right)^{2}+\left(\frac{a}{6}\right)^{2}=61\) \(\frac{a^{2}}{25}+\frac{a^{2}}{36}=61\) \(\frac{36 a^{2}+25 a^{2}}{25 \times 36}=61\) \(61 a^{2}=61 \times 36 \times 25\) \(a^{2}=36 \times 25\) \(a=6 \times 5\) \(a= \pm 30\) The value of a is \(\pm 30\).
COMEDK-2012
Application of Derivatives
85404
If the radius of a sphere is measured as \(9 \mathrm{~cm}\) with an error of \(0.03 \mathrm{~cm}\), then find the approximating error in calculating its volume.
1 \(2.46 \pi \mathrm{cm}^{3}\)
2 \(8.62 \pi \mathrm{cm}^{3}\)
3 \(9.72 \pi \mathrm{cm}^{3}\)
4 \(7.46 \pi \mathrm{cm}^{3}\)
Explanation:
(C) : We have, Radius of sphere \((\mathrm{r})=9 \mathrm{~cm}\) and Error \(\Delta \mathrm{r}=0.03 \mathrm{~cm}\) Volume of sphere, \(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}\) \(\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^{2}\) \(\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{(\mathrm{r}=9)}=4 \pi \times 9^{2}=324 \pi\) Let, \(\Delta \mathrm{V}\) be the error in \(\mathrm{V}\) due to error \(\Delta \mathrm{r}\) in \(\mathrm{r}\). Then, \(\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta \mathrm{r}\) \(\Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}\) Thus, the approximate error in calculating the volume is \(9.72 \pi \mathrm{cm}^{3}\).