NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85812
A hemispherical bowl of radius unity is filled up with water upto the depth \(\frac{1}{2}\). The volume of water in the bowl is
1 \(\frac{27 \pi}{24}\)
2 \(\frac{5 \pi}{24}\)
3 \(\frac{3 \pi}{4}\)
4 None of these
Explanation:
(B) : Hemispherical bowl's radius \(\mathrm{r}=1\) unit Hemispherical bowl's height up to where the water is filled, \(\mathrm{h}=\frac{1}{2}\) Now, in \(\mathrm{ABO}\) we have, \(\mathrm{AB}=\sqrt{\mathrm{OA}^{2}-\mathrm{OB}^{2}}\) \(\mathrm{AB}=\sqrt{1-\frac{1}{4}}\) \(\mathrm{AB}=\frac{\sqrt{3}}{2}\) so, the hemisphere's volume, \(\mathrm{V}=\frac{2 \pi}{3} \mathrm{r}^{3}\) Thus, the volume of water, \(\mathrm{V}=\frac{\pi \mathrm{h}}{6}\left(3 \mathrm{a}^{2}+\mathrm{h}^{2}\right)\) Now are have, \(\mathrm{a}=\mathrm{AB}=\frac{\sqrt{3}}{2}, \mathrm{~h}=\frac{1}{2}\) Put the values to calculate the volume of water as, \(\mathrm{V}=\frac{\pi}{2} \times \frac{1}{2}\left[3\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2}\left[3 \times \frac{3}{4}+\frac{1}{4}\right]=\frac{\pi}{6} \times \frac{1}{2}\left[\frac{10}{4}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2} \times \frac{5}{2}=\frac{5 \pi}{24}\) This is the required volume of water.
UPSEE-2011
Application of Derivatives
85813
The subtangent at any point of the curve \(\mathbf{x}^{m} \mathbf{y}^{\mathbf{n}}=\mathbf{a}^{\mathbf{m}+\mathbf{n}}\) varies as
1 \(\left(\right.\) abscissa) \({ }^{2}\)
2 (abscissa) \({ }^{3}\)
3 abscissa
4 ordinate
Explanation:
(C) : Given, \(x^{m} y^{n}=a^{m+n}\) \(m \log _{\mathrm{e}} \mathrm{x}+\mathrm{n} \log _{\mathrm{e}} \mathrm{y}=(\mathrm{m}+\mathrm{n}) \log _{\mathrm{e}} \mathrm{a}\) Differentiate both side with respect to ' \(x\) ' we get, \(\frac{\mathrm{m}}{\mathrm{x}}+\frac{\mathrm{n}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{my}}{\mathrm{nx}}\) Length the sub tangent, \(=\left|\frac{\frac{\mathrm{y}}{\mathrm{dy}}}{\frac{\mathrm{dx}}{}}\right|=\left|\frac{\frac{\mathrm{y}}{-\mathrm{my}}}{\mathrm{nx}}\right|=\frac{\mathrm{n}}{\mathrm{m}}|\mathrm{x}| \propto \mathrm{x}\)
[JCECE-2012]
Application of Derivatives
85814
If \(x=e^{t} \sin t, y=e^{t} \cos t, t\) is a parameter, then \(\frac{d^{2} y}{d x^{2}}\) at \((1,1)\) is equal \(t\)
85815
Let \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\). Then which of the following statements are true? \(P: x=0\) is a point of local minima of \(f\) \(Q: x=\sqrt{2}\) is point of inflection of \(f\) \(R: f^{\prime}\) is increasing for \(x>\sqrt{2}\)
1 Only P and Q
2 Only P and R
3 Only Q and R
4 All, P, Q and R
Explanation:
(D) : \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\) \(P: x=0\) is a point of local minima of \(f\). \(\mathrm{Q}: \mathrm{x}=\sqrt{2}\) is point of inflection of \(\mathrm{f}\). \(\mathrm{R}: \mathrm{f}^{\prime}\) is increasing for \(\mathrm{x}>\sqrt{2}\) Then, \(f(x)=3^{4} \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(f(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(\therefore \quad f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \frac{d}{d x}\left(x^{2}-2\right)^{3}\) \(f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right) \times 2 x\) \(f^{\prime}(x)=243 \times 2 \times x\left(x^{2}-2\right)^{2} \log 3\) \(f^{\prime}(x)=(81 \times 6) \times x\left(x^{2}-2\right)^{2} \log 3\) Then, \(\mathrm{x}=0\) is point of local minimum. \(f^{\prime}(x)=\frac{(486 \cdot \log 3)}{k} \cdot \frac{3^{\left(x^{2}-2\right)^{3}} x\left(x^{2}-2\right)^{2}}{g(x)}\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)^{2}+x \cdot 3^{\left(x^{2}-2\right)^{3}} \cdot 4 x\left(x^{2}-2\right)\) \(+x\left(x^{2}-2\right)^{2} \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right)^{2} \cdot 2 x\) \(\mathrm{g}^{\prime}(\mathrm{x})=3^{\left(\mathrm{x}^{2}-2\right)^{3}}\left(\mathrm{x}^{2}-2\right)\left[\mathrm{x}^{2}-2+4 \mathrm{x}^{2}+6 \mathrm{x}^{2} \log \left(\mathrm{x}^{2}-2\right)^{3}\right]\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)\left[5 x^{2}-2+6 x^{2} \log 3\left(x^{2}-2\right)^{3}\right]\) \(f^{\prime}(x)=\mathrm{kg}^{\prime}(\mathrm{x})\) \(f^{\prime \prime}(x)=f^{\prime \prime}(\sqrt{2})=0, f^{\prime \prime}\left(\sqrt{2}^{+}\right)>0\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{2}^{-}\right)\lt 0\) \(x=\sqrt{2}\) is point of inflection. \(\mathrm{f}^{\prime \prime}(\mathrm{x})>0\) for \(\mathrm{x}>\sqrt{2}\), so \(\mathrm{f}^{\prime}(\mathrm{x})\) is increa sing So, all P, Q and R statements are true.
85812
A hemispherical bowl of radius unity is filled up with water upto the depth \(\frac{1}{2}\). The volume of water in the bowl is
1 \(\frac{27 \pi}{24}\)
2 \(\frac{5 \pi}{24}\)
3 \(\frac{3 \pi}{4}\)
4 None of these
Explanation:
(B) : Hemispherical bowl's radius \(\mathrm{r}=1\) unit Hemispherical bowl's height up to where the water is filled, \(\mathrm{h}=\frac{1}{2}\) Now, in \(\mathrm{ABO}\) we have, \(\mathrm{AB}=\sqrt{\mathrm{OA}^{2}-\mathrm{OB}^{2}}\) \(\mathrm{AB}=\sqrt{1-\frac{1}{4}}\) \(\mathrm{AB}=\frac{\sqrt{3}}{2}\) so, the hemisphere's volume, \(\mathrm{V}=\frac{2 \pi}{3} \mathrm{r}^{3}\) Thus, the volume of water, \(\mathrm{V}=\frac{\pi \mathrm{h}}{6}\left(3 \mathrm{a}^{2}+\mathrm{h}^{2}\right)\) Now are have, \(\mathrm{a}=\mathrm{AB}=\frac{\sqrt{3}}{2}, \mathrm{~h}=\frac{1}{2}\) Put the values to calculate the volume of water as, \(\mathrm{V}=\frac{\pi}{2} \times \frac{1}{2}\left[3\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2}\left[3 \times \frac{3}{4}+\frac{1}{4}\right]=\frac{\pi}{6} \times \frac{1}{2}\left[\frac{10}{4}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2} \times \frac{5}{2}=\frac{5 \pi}{24}\) This is the required volume of water.
UPSEE-2011
Application of Derivatives
85813
The subtangent at any point of the curve \(\mathbf{x}^{m} \mathbf{y}^{\mathbf{n}}=\mathbf{a}^{\mathbf{m}+\mathbf{n}}\) varies as
1 \(\left(\right.\) abscissa) \({ }^{2}\)
2 (abscissa) \({ }^{3}\)
3 abscissa
4 ordinate
Explanation:
(C) : Given, \(x^{m} y^{n}=a^{m+n}\) \(m \log _{\mathrm{e}} \mathrm{x}+\mathrm{n} \log _{\mathrm{e}} \mathrm{y}=(\mathrm{m}+\mathrm{n}) \log _{\mathrm{e}} \mathrm{a}\) Differentiate both side with respect to ' \(x\) ' we get, \(\frac{\mathrm{m}}{\mathrm{x}}+\frac{\mathrm{n}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{my}}{\mathrm{nx}}\) Length the sub tangent, \(=\left|\frac{\frac{\mathrm{y}}{\mathrm{dy}}}{\frac{\mathrm{dx}}{}}\right|=\left|\frac{\frac{\mathrm{y}}{-\mathrm{my}}}{\mathrm{nx}}\right|=\frac{\mathrm{n}}{\mathrm{m}}|\mathrm{x}| \propto \mathrm{x}\)
[JCECE-2012]
Application of Derivatives
85814
If \(x=e^{t} \sin t, y=e^{t} \cos t, t\) is a parameter, then \(\frac{d^{2} y}{d x^{2}}\) at \((1,1)\) is equal \(t\)
85815
Let \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\). Then which of the following statements are true? \(P: x=0\) is a point of local minima of \(f\) \(Q: x=\sqrt{2}\) is point of inflection of \(f\) \(R: f^{\prime}\) is increasing for \(x>\sqrt{2}\)
1 Only P and Q
2 Only P and R
3 Only Q and R
4 All, P, Q and R
Explanation:
(D) : \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\) \(P: x=0\) is a point of local minima of \(f\). \(\mathrm{Q}: \mathrm{x}=\sqrt{2}\) is point of inflection of \(\mathrm{f}\). \(\mathrm{R}: \mathrm{f}^{\prime}\) is increasing for \(\mathrm{x}>\sqrt{2}\) Then, \(f(x)=3^{4} \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(f(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(\therefore \quad f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \frac{d}{d x}\left(x^{2}-2\right)^{3}\) \(f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right) \times 2 x\) \(f^{\prime}(x)=243 \times 2 \times x\left(x^{2}-2\right)^{2} \log 3\) \(f^{\prime}(x)=(81 \times 6) \times x\left(x^{2}-2\right)^{2} \log 3\) Then, \(\mathrm{x}=0\) is point of local minimum. \(f^{\prime}(x)=\frac{(486 \cdot \log 3)}{k} \cdot \frac{3^{\left(x^{2}-2\right)^{3}} x\left(x^{2}-2\right)^{2}}{g(x)}\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)^{2}+x \cdot 3^{\left(x^{2}-2\right)^{3}} \cdot 4 x\left(x^{2}-2\right)\) \(+x\left(x^{2}-2\right)^{2} \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right)^{2} \cdot 2 x\) \(\mathrm{g}^{\prime}(\mathrm{x})=3^{\left(\mathrm{x}^{2}-2\right)^{3}}\left(\mathrm{x}^{2}-2\right)\left[\mathrm{x}^{2}-2+4 \mathrm{x}^{2}+6 \mathrm{x}^{2} \log \left(\mathrm{x}^{2}-2\right)^{3}\right]\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)\left[5 x^{2}-2+6 x^{2} \log 3\left(x^{2}-2\right)^{3}\right]\) \(f^{\prime}(x)=\mathrm{kg}^{\prime}(\mathrm{x})\) \(f^{\prime \prime}(x)=f^{\prime \prime}(\sqrt{2})=0, f^{\prime \prime}\left(\sqrt{2}^{+}\right)>0\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{2}^{-}\right)\lt 0\) \(x=\sqrt{2}\) is point of inflection. \(\mathrm{f}^{\prime \prime}(\mathrm{x})>0\) for \(\mathrm{x}>\sqrt{2}\), so \(\mathrm{f}^{\prime}(\mathrm{x})\) is increa sing So, all P, Q and R statements are true.
85812
A hemispherical bowl of radius unity is filled up with water upto the depth \(\frac{1}{2}\). The volume of water in the bowl is
1 \(\frac{27 \pi}{24}\)
2 \(\frac{5 \pi}{24}\)
3 \(\frac{3 \pi}{4}\)
4 None of these
Explanation:
(B) : Hemispherical bowl's radius \(\mathrm{r}=1\) unit Hemispherical bowl's height up to where the water is filled, \(\mathrm{h}=\frac{1}{2}\) Now, in \(\mathrm{ABO}\) we have, \(\mathrm{AB}=\sqrt{\mathrm{OA}^{2}-\mathrm{OB}^{2}}\) \(\mathrm{AB}=\sqrt{1-\frac{1}{4}}\) \(\mathrm{AB}=\frac{\sqrt{3}}{2}\) so, the hemisphere's volume, \(\mathrm{V}=\frac{2 \pi}{3} \mathrm{r}^{3}\) Thus, the volume of water, \(\mathrm{V}=\frac{\pi \mathrm{h}}{6}\left(3 \mathrm{a}^{2}+\mathrm{h}^{2}\right)\) Now are have, \(\mathrm{a}=\mathrm{AB}=\frac{\sqrt{3}}{2}, \mathrm{~h}=\frac{1}{2}\) Put the values to calculate the volume of water as, \(\mathrm{V}=\frac{\pi}{2} \times \frac{1}{2}\left[3\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2}\left[3 \times \frac{3}{4}+\frac{1}{4}\right]=\frac{\pi}{6} \times \frac{1}{2}\left[\frac{10}{4}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2} \times \frac{5}{2}=\frac{5 \pi}{24}\) This is the required volume of water.
UPSEE-2011
Application of Derivatives
85813
The subtangent at any point of the curve \(\mathbf{x}^{m} \mathbf{y}^{\mathbf{n}}=\mathbf{a}^{\mathbf{m}+\mathbf{n}}\) varies as
1 \(\left(\right.\) abscissa) \({ }^{2}\)
2 (abscissa) \({ }^{3}\)
3 abscissa
4 ordinate
Explanation:
(C) : Given, \(x^{m} y^{n}=a^{m+n}\) \(m \log _{\mathrm{e}} \mathrm{x}+\mathrm{n} \log _{\mathrm{e}} \mathrm{y}=(\mathrm{m}+\mathrm{n}) \log _{\mathrm{e}} \mathrm{a}\) Differentiate both side with respect to ' \(x\) ' we get, \(\frac{\mathrm{m}}{\mathrm{x}}+\frac{\mathrm{n}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{my}}{\mathrm{nx}}\) Length the sub tangent, \(=\left|\frac{\frac{\mathrm{y}}{\mathrm{dy}}}{\frac{\mathrm{dx}}{}}\right|=\left|\frac{\frac{\mathrm{y}}{-\mathrm{my}}}{\mathrm{nx}}\right|=\frac{\mathrm{n}}{\mathrm{m}}|\mathrm{x}| \propto \mathrm{x}\)
[JCECE-2012]
Application of Derivatives
85814
If \(x=e^{t} \sin t, y=e^{t} \cos t, t\) is a parameter, then \(\frac{d^{2} y}{d x^{2}}\) at \((1,1)\) is equal \(t\)
85815
Let \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\). Then which of the following statements are true? \(P: x=0\) is a point of local minima of \(f\) \(Q: x=\sqrt{2}\) is point of inflection of \(f\) \(R: f^{\prime}\) is increasing for \(x>\sqrt{2}\)
1 Only P and Q
2 Only P and R
3 Only Q and R
4 All, P, Q and R
Explanation:
(D) : \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\) \(P: x=0\) is a point of local minima of \(f\). \(\mathrm{Q}: \mathrm{x}=\sqrt{2}\) is point of inflection of \(\mathrm{f}\). \(\mathrm{R}: \mathrm{f}^{\prime}\) is increasing for \(\mathrm{x}>\sqrt{2}\) Then, \(f(x)=3^{4} \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(f(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(\therefore \quad f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \frac{d}{d x}\left(x^{2}-2\right)^{3}\) \(f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right) \times 2 x\) \(f^{\prime}(x)=243 \times 2 \times x\left(x^{2}-2\right)^{2} \log 3\) \(f^{\prime}(x)=(81 \times 6) \times x\left(x^{2}-2\right)^{2} \log 3\) Then, \(\mathrm{x}=0\) is point of local minimum. \(f^{\prime}(x)=\frac{(486 \cdot \log 3)}{k} \cdot \frac{3^{\left(x^{2}-2\right)^{3}} x\left(x^{2}-2\right)^{2}}{g(x)}\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)^{2}+x \cdot 3^{\left(x^{2}-2\right)^{3}} \cdot 4 x\left(x^{2}-2\right)\) \(+x\left(x^{2}-2\right)^{2} \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right)^{2} \cdot 2 x\) \(\mathrm{g}^{\prime}(\mathrm{x})=3^{\left(\mathrm{x}^{2}-2\right)^{3}}\left(\mathrm{x}^{2}-2\right)\left[\mathrm{x}^{2}-2+4 \mathrm{x}^{2}+6 \mathrm{x}^{2} \log \left(\mathrm{x}^{2}-2\right)^{3}\right]\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)\left[5 x^{2}-2+6 x^{2} \log 3\left(x^{2}-2\right)^{3}\right]\) \(f^{\prime}(x)=\mathrm{kg}^{\prime}(\mathrm{x})\) \(f^{\prime \prime}(x)=f^{\prime \prime}(\sqrt{2})=0, f^{\prime \prime}\left(\sqrt{2}^{+}\right)>0\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{2}^{-}\right)\lt 0\) \(x=\sqrt{2}\) is point of inflection. \(\mathrm{f}^{\prime \prime}(\mathrm{x})>0\) for \(\mathrm{x}>\sqrt{2}\), so \(\mathrm{f}^{\prime}(\mathrm{x})\) is increa sing So, all P, Q and R statements are true.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85812
A hemispherical bowl of radius unity is filled up with water upto the depth \(\frac{1}{2}\). The volume of water in the bowl is
1 \(\frac{27 \pi}{24}\)
2 \(\frac{5 \pi}{24}\)
3 \(\frac{3 \pi}{4}\)
4 None of these
Explanation:
(B) : Hemispherical bowl's radius \(\mathrm{r}=1\) unit Hemispherical bowl's height up to where the water is filled, \(\mathrm{h}=\frac{1}{2}\) Now, in \(\mathrm{ABO}\) we have, \(\mathrm{AB}=\sqrt{\mathrm{OA}^{2}-\mathrm{OB}^{2}}\) \(\mathrm{AB}=\sqrt{1-\frac{1}{4}}\) \(\mathrm{AB}=\frac{\sqrt{3}}{2}\) so, the hemisphere's volume, \(\mathrm{V}=\frac{2 \pi}{3} \mathrm{r}^{3}\) Thus, the volume of water, \(\mathrm{V}=\frac{\pi \mathrm{h}}{6}\left(3 \mathrm{a}^{2}+\mathrm{h}^{2}\right)\) Now are have, \(\mathrm{a}=\mathrm{AB}=\frac{\sqrt{3}}{2}, \mathrm{~h}=\frac{1}{2}\) Put the values to calculate the volume of water as, \(\mathrm{V}=\frac{\pi}{2} \times \frac{1}{2}\left[3\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2}\left[3 \times \frac{3}{4}+\frac{1}{4}\right]=\frac{\pi}{6} \times \frac{1}{2}\left[\frac{10}{4}\right]\) \(\mathrm{V}=\frac{\pi}{6} \times \frac{1}{2} \times \frac{5}{2}=\frac{5 \pi}{24}\) This is the required volume of water.
UPSEE-2011
Application of Derivatives
85813
The subtangent at any point of the curve \(\mathbf{x}^{m} \mathbf{y}^{\mathbf{n}}=\mathbf{a}^{\mathbf{m}+\mathbf{n}}\) varies as
1 \(\left(\right.\) abscissa) \({ }^{2}\)
2 (abscissa) \({ }^{3}\)
3 abscissa
4 ordinate
Explanation:
(C) : Given, \(x^{m} y^{n}=a^{m+n}\) \(m \log _{\mathrm{e}} \mathrm{x}+\mathrm{n} \log _{\mathrm{e}} \mathrm{y}=(\mathrm{m}+\mathrm{n}) \log _{\mathrm{e}} \mathrm{a}\) Differentiate both side with respect to ' \(x\) ' we get, \(\frac{\mathrm{m}}{\mathrm{x}}+\frac{\mathrm{n}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{my}}{\mathrm{nx}}\) Length the sub tangent, \(=\left|\frac{\frac{\mathrm{y}}{\mathrm{dy}}}{\frac{\mathrm{dx}}{}}\right|=\left|\frac{\frac{\mathrm{y}}{-\mathrm{my}}}{\mathrm{nx}}\right|=\frac{\mathrm{n}}{\mathrm{m}}|\mathrm{x}| \propto \mathrm{x}\)
[JCECE-2012]
Application of Derivatives
85814
If \(x=e^{t} \sin t, y=e^{t} \cos t, t\) is a parameter, then \(\frac{d^{2} y}{d x^{2}}\) at \((1,1)\) is equal \(t\)
85815
Let \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\). Then which of the following statements are true? \(P: x=0\) is a point of local minima of \(f\) \(Q: x=\sqrt{2}\) is point of inflection of \(f\) \(R: f^{\prime}\) is increasing for \(x>\sqrt{2}\)
1 Only P and Q
2 Only P and R
3 Only Q and R
4 All, P, Q and R
Explanation:
(D) : \(f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in R\) \(P: x=0\) is a point of local minima of \(f\). \(\mathrm{Q}: \mathrm{x}=\sqrt{2}\) is point of inflection of \(\mathrm{f}\). \(\mathrm{R}: \mathrm{f}^{\prime}\) is increasing for \(\mathrm{x}>\sqrt{2}\) Then, \(f(x)=3^{4} \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(f(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}}\) \(\therefore \quad f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \frac{d}{d x}\left(x^{2}-2\right)^{3}\) \(f^{\prime}(x)=81 \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right) \times 2 x\) \(f^{\prime}(x)=243 \times 2 \times x\left(x^{2}-2\right)^{2} \log 3\) \(f^{\prime}(x)=(81 \times 6) \times x\left(x^{2}-2\right)^{2} \log 3\) Then, \(\mathrm{x}=0\) is point of local minimum. \(f^{\prime}(x)=\frac{(486 \cdot \log 3)}{k} \cdot \frac{3^{\left(x^{2}-2\right)^{3}} x\left(x^{2}-2\right)^{2}}{g(x)}\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)^{2}+x \cdot 3^{\left(x^{2}-2\right)^{3}} \cdot 4 x\left(x^{2}-2\right)\) \(+x\left(x^{2}-2\right)^{2} \cdot 3^{\left(x^{2}-2\right)^{3}} \log 3 \cdot 3\left(x^{2}-2\right)^{2} \cdot 2 x\) \(\mathrm{g}^{\prime}(\mathrm{x})=3^{\left(\mathrm{x}^{2}-2\right)^{3}}\left(\mathrm{x}^{2}-2\right)\left[\mathrm{x}^{2}-2+4 \mathrm{x}^{2}+6 \mathrm{x}^{2} \log \left(\mathrm{x}^{2}-2\right)^{3}\right]\) \(g^{\prime}(x)=3^{\left(x^{2}-2\right)^{3}}\left(x^{2}-2\right)\left[5 x^{2}-2+6 x^{2} \log 3\left(x^{2}-2\right)^{3}\right]\) \(f^{\prime}(x)=\mathrm{kg}^{\prime}(\mathrm{x})\) \(f^{\prime \prime}(x)=f^{\prime \prime}(\sqrt{2})=0, f^{\prime \prime}\left(\sqrt{2}^{+}\right)>0\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{2}^{-}\right)\lt 0\) \(x=\sqrt{2}\) is point of inflection. \(\mathrm{f}^{\prime \prime}(\mathrm{x})>0\) for \(\mathrm{x}>\sqrt{2}\), so \(\mathrm{f}^{\prime}(\mathrm{x})\) is increa sing So, all P, Q and R statements are true.
