Find the equation of the hyperbola whose vertices are (0, ±3) and the
Question: Find the equation of the hyperbola whose vertices are (0, 3) and the eccentricity is $\frac{4}{3}$. Also, find the coordinates of its foci. Solution: Given: Vertices are $(0, \pm 3)$ and the eccentricity is $\frac{4}{3}$ Need to find: The equation of the hyperbola and coordinates of foci. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ Vertices are $(\pm 3,0)$, that means, $a=3$ And also given, the eccentricity, $e=\frac{4}{3}$ We know that, $e=\sqrt{...
Read More →The maximum value
Question: The maximum value of $x^{1 / x}, x0$ is (a) $e^{1 / e}$ (b) $\left(\frac{1}{e}\right)^{e}$ (c) 1 (d) none of these Solution: (a) $e^{\frac{1}{e}}$ Given : $f(x)=x^{\frac{1}{x}}$ Taking log on both sides, we get $\log f(x)=\frac{1}{x} \log x$ Differentiating w.r.t. $x$, we get $\frac{1}{f(x)} f^{\prime}(x)=\frac{-1}{x^{2}} \log x+\frac{1}{x^{2}}$ $\Rightarrow f^{\prime}(x)=f(x) \frac{1}{x^{2}}(1-\log x)$ $\Rightarrow f^{\prime}(x)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \lo...
Read More →Find the equation of the hyperbola with eccentricity
Question: Find the equation of the hyperbola with eccentricity $\sqrt{2}$ and the distance between whose foci is 16. Solution: Given: Eccentricity is $\sqrt{2}$, and the distance between foci is 16 Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ Distance between the foci is 16, i.e., $2 \mathrm{ae}=16$ And also given, the eccentricity, e = $\sqrt{2}$ Therefore, $2 \mathrm{a} \sqrt{2}=16$ $a=\frac{16}{2 \sqrt{2}}=\fra...
Read More →Write the maximum value
Question: Write the maximum value of $\mathrm{f}(\mathrm{x})=\frac{\log x}{x}$, if it exists. Solution: Given : $f(x)=\frac{\log x}{x}$ $\Rightarrow f^{\prime}(x)=\frac{1-\log x}{x^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{1-\log x}{x^{2}}=0$ $\Rightarrow 1-\log x=0$ $\Rightarrow \log x=1$ $\Rightarrow \log x=\log e$ $\Rightarrow x=e$ Now, $f^{\prime \prime}(x)=\frac{-x-2 x(1-\log x)}{x^{4}}=\frac{-3 x-2 x \log x}{x^{4}}$ At $x=e:$ $f^{\prime \...
Read More →Write the maximum value
Question: Write the maximum value of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{1 / \mathrm{x}}$. Solution: Given: $f(x)=x^{\frac{1}{x}}$ Taking $\log$ on both sides, we get $\log f(x)=\frac{1}{x} \log x$ Differentiating w.r.t. $x$, we get $\frac{1}{f(x)} f^{\prime}(x)=\frac{-1}{x^{2}} \log x+\frac{1}{x^{2}}$ $\Rightarrow f^{\prime}(x)=f(x) \frac{1}{x^{2}}(1-\log x)$ $\Rightarrow f^{\prime}(x)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)$ ...........(1) $\Rightarrow f^{\prime}(x)=...
Read More →Find the equation of the hyperbola, the length of whose latus rectum is 4
Question: Find the equation of the hyperbola, the length of whose latus rectum is 4 and the eccentricity is 3. Solution: Given: The length of latus rectum is 4, and the eccentricity is 3 Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ The length of the latus rectum is 4 units. Therefore, $\frac{2 b^{2}}{a}=4 \Rightarrow b^{2}=2 a^{-\cdots}$ (1) And also given, the eccentricity, $e=3$ We know that, $e=\sqrt{1+\frac{b^...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are $(\pm \sqrt{5} 0)$ and the eccentricity is $\sqrt{\frac{5}{3}}$. Solution: Given: Foci are $(\pm \sqrt{5}, 0)$, and the eccentricity is $\sqrt{\frac{5}{3}}$ Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ The eccentricity, $\mathrm{e}=\sqrt{\frac{5}{3}}$ And also given, foci are $(\pm \sqrt{5}, 0)$ That means, ae $=\sqrt{5}$ $\Rightarrow a=\frac{\sqrt{5}}{e}...
Read More →Write the minimum value
Question: Write the minimum value of $f(x)=x^{x}$. Solution: Given: $f(x)=x^{x}$ Taking log on both sides, we get $\log f(x)=x \log x$ Differentiating w.r.t. $x$, we get $\frac{1}{f(x)} f^{\prime}(x)=\log x+1$ $\Rightarrow f^{\prime}(x)=f(x)(\log x+1)$ $\Rightarrow f^{\prime}(x)=x^{x}(\log x+1)$ .......(1) For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow x^{x}(\log x+1)=0$ $\Rightarrow \log x=-1$ $\Rightarrow x=\frac{1}{e}$ Now, $f^{\prime \prime}(x)=x^{x}(\log x...
Read More →Find the equation of the hyperbola whose vertices are
Question: Find the equation of the hyperbola whose vertices are (2, 0) and the eccentricity is 2. Solution: Given: Vertices are (2, 0) and the eccentricity is 2 Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ Vertices are $(\pm 2,0)$, that means, $a=2$ And also given, the eccentricity, $e=2$ We know that, $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$ Therefore, $\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}=2$ $\Rightarrow 1+\frac{\ma...
Read More →Find the least value
Question: Find the least value of $\mathrm{f}(\mathrm{x})=a x+\frac{b}{x}$, where $\mathrm{a}0, \mathrm{~b}0$ and $\mathrm{x}0$. Solution: We have, $f(x)=a x+\frac{b}{x}$ $\Rightarrow f^{\prime}(x)=a-\frac{b}{x^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow a-\frac{b}{x^{2}}=0$ $\Rightarrow x^{2}=\frac{b}{a}$ $\Rightarrow x=\sqrt{\frac{b}{a}},-\sqrt{\frac{b}{a}}$ But, $x0$ $\Rightarrow x=\sqrt{\frac{b}{a}}$ Now, $f^{\prime \prime}(x)=\frac{2 b}{x^{3}}$ At...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are $(\pm 3 \sqrt{5}, 0)$ and the length of the latus rectum is 8 units. Solution: Given: Foci are $(\pm 3 \sqrt{5}, 0)$ the length of the latus rectum is 8 units Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ The length of the latus rectum is 8 units. Therefore, $\frac{2 b^{2}}{a}=8 \Rightarrow b^{2}=4 a \cdots$ (1) The foci are given at $(\pm 3 \sqrt{5}, 0)$ ...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are $(\pm 3 \sqrt{5}, 0)$ and the length of the latus rectum is 8 units. Solution: Given: Foci are $(\pm 3 \sqrt{5}, 0)$ the length of the latus rectum is 8 units Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ The length of the latus rectum is 8 units. Therefore, $\frac{2 b^{2}}{a}=8 \Rightarrow b^{2}=4 a \cdots$ (1) The foci are given at $(\pm 3 \sqrt{5}, 0)$ ...
Read More →Prove the following
Question: $\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}$ Solution: Given $\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}$ Multiply and divide both numerator and denominator by $4 x^{2} / 16 x^{2}$ then we get $\Rightarrow \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}=\lim _{x \rightarrow 0} \frac{\left(\sin ^{2} 2 x\right) / 4 x^{2}}{\left(\sin ^{2} 4 x\right) / 16 x^{2}} \times \frac{4 x^{2}}{16 x^{2}}$ On simplifying $\Rightarrow \lim _{x \rightarrow 0...
Read More →Write the point where
Question: Write the point where $f(x)=x \log , x$ attains minimum value. Solution: Given : $f(x)=x \log _{e} x$ $\Rightarrow f^{\prime}(x)=\log _{e} x+1$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \log _{e} x+1=0$ $\Rightarrow \log _{e} x=-1$ $\Rightarrow x=\frac{1}{e}$ $\Rightarrow f\left(\frac{1}{e}\right)=\frac{1}{e} \log _{e}\left(\frac{1}{e}\right)=-\frac{1}{e}$ Now, $f^{\prime \prime}(x)=\frac{1}{x}$ $f^{\prime \prime}\left(\frac{1}{e}\right)=\frac{1}...
Read More →Prove the following
Question: Find ' $n$ ', if $\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80, n \in \mathbf{N}$ Solution: Given $\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80$ We know that $\lim _{x \rightarrow a} \frac{x^{n^{n}}-a^{n}}{x-a}=n a^{n-1}$ By using this formula we get $\Rightarrow \mathrm{n} 2^{\mathrm{n}-1}=80=5 \times 2^{4}=5 \times 2^{5-1}$ $\Rightarrow \mathrm{n}=5$...
Read More →Find the equation of the hyperbola whose foci are (±5, 0) and the conjugate
Question: Find the equation of the hyperbola whose foci are (5, 0) and the conjugate axis is of the length 8. Also, find its eccentricity. Solution: Given: Foci are (5, 0), the conjugate axis is of the length 8 Need to find: The equation of the hyperbola and eccentricity. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ The conjugate axis is of the length 8, i.e., $2 b=8$ Therefore, $b=4$ The foci are given at $(\pm 5,0)$ That means, $\mathrm{ae}=5$, where $\mat...
Read More →Write the maximum value
Question: Write the maximum value of $\mathrm{f}(\mathrm{x})=x+\frac{1}{x}, x0$. Solution: Given: $f(x)=x+\frac{1}{x}$ $\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 1-\frac{1}{x^{2}}=0$ $\Rightarrow x^{2}=1$ $\Rightarrow x=1,-1$ But $x0$ $\Rightarrow x=-1$ Now, $f^{\prime \prime}(x)=\frac{1}{x^{3}}$ At $x=-1:$ $f^{\prime \prime}(-1)=\frac{2}{(-1)^{3}}=-20$ So, $x=-1$ is a point of local maximum. Thus, the local max...
Read More →Prove the following
Question: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{3}}-\sqrt{1-x^{3}}}{x^{2}}$ Solution: Given $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{x^{2}}$ Now to rationalize the denominator by multiplying the given equation by its rationalizing factor we get $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{3}}-\sqrt{1-x^{3}}}{x^{2}}=\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{3}}-\sqrt{1-x^{3}}}{x^{2}} \times\left(\frac{\sqrt{1+x^{3}}+\sqrt{1-x^{3}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)$ On si...
Read More →Prove the following
Question: $\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2}$ Solution: This question can be easily solved using LH rule that is L. Hospital's rule which is given below $\Rightarrow$ if $\lim _{x \rightarrow a g(x)} \frac{f(x)}{g}=\frac{0}{0}$ then $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f r(x)}{g t(x)}$ Given $\lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2}$ If we apply the limit we will get in determinant form $\lim _{x \rightarrow...
Read More →Write the minimum value
Question: Write the minimum value of $\mathrm{f}(\mathrm{x})=x+\frac{1}{x}, x0$ Solution: Given : $f(x)=x+\frac{1}{x}$ $\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 1-\frac{1}{x^{2}}=0$ $\Rightarrow x^{2}=1$ $\Rightarrow x=1,-1$ But $x0$ $\Rightarrow x=1$ Now, $f^{\prime \prime}(x)=\frac{1}{x^{3}}$ At $x=1:$ $f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=20$ So, $x=1$ is a point of local minimum. Thus, the local minimum v...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are $(\pm \sqrt{29}, 0)$ and the transverse axis is of the length 10 . Solution: Given: Foci are $(\pm \sqrt{29}, 0)$, the transverse axis is of the length 10 Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ The transverse axis is of the length 10, i.e., $2 \mathrm{a}=10$ Therefore, $a=5$ The foci are given at $(\pm \sqrt{29}, 0)$ That means, ae $=\pm \sqrt{29}$,...
Read More →Prove the following
Question: $\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{3 x-2}-\sqrt{x+2}}$ Solution: $\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{3 x-2}-\sqrt{x+2}}$ Now to rationalize the denominator by multiplying the given equation by its rationalizing factor we get $\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{3 x-2}-\sqrt{x+2}}=\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{3 x-2}-\sqrt{x+2}} \times\left(\frac{\sqrt{3 x-2}+\sqrt{x+2}}{\sqrt{3 x-2}+\sqrt{x+2}}\right)$ Taking $(x-2)(x+2)$ as common we get $\...
Read More →A particle is moving in a straight line such that its distance at any time t is given by
Question: A particle is moving in a straight line such that its distance at any time $t$ is given by $S=\frac{t^{4}}{4}-2 t^{3}+4 t^{2}-7$. Find when its velocity is maximum and acceleration minimum. Solution: Given : $s=\frac{t^{4}}{4}-2 t^{3}+4 t^{2}-7$ $\Rightarrow v=\frac{d s}{d t}=t^{3}-6 t^{2}+8 t$ $\Rightarrow a=\frac{d v}{d t}=3 t^{2}-12 t+8$ For maximum or minimum values of $v$, we must have $\frac{d v}{d t}=0$ $\Rightarrow 3 t^{2}-12 t+8=0$ On solving the equation, we get $t=2 \pm \fra...
Read More →Prove the following
Question: $\lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}$ Solution: Given $\lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}$ Now to rationalize the denominator by multiplying the given equation by its rationalizing factor we get $\Rightarrow \lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}=\lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1} \times\left(\frac{\sqrt{x}+1}{\sqrt{x}+1}\right)$ On simplifying the above equation we get $\Rightarrow \lim _{x \rightar...
Read More →Find the equation of the hyperbola with vertices at
Question: Find the equation of the hyperbola with vertices at $(0, \pm 5)$ and foci at $(0, \pm 8)$. Solution: Given: Vertices at $(0, \pm 5)$ and foci at $(0, \pm 8)$ Need to find: The equation of the hyperbola. Let, the equation of the parabola be: $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ Vertices of the parabola are at $(0, \pm 5)$ That means $\mathrm{a}=5$ The foci are given at $(0, \pm 8)$ $A$ and $B$ are the vertices. $C$ and $D$ are the foci. That means, ae $=8$, where $e$ is the eccen...
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