A beam is supported at its ends by supports which are 12 m apart
Question: A beam is supported at its ends by supports which are 12 m apart. Since the load is concentrated at its center, there is a deflection of 3 cm at the center, and the deflected beam is in the shape of a parabola. How far from the center is the deflection 1 cm? Solution: Given: A beam is supported at its ends by supports which are 12 m apart. There is a deflection of 3 cm at the center, and the deflected beam is in the shape of a parabola. Need to find: How far from the center is the defl...
Read More →A rod of length 15 cm moves with its ends always touching the coordinate axes.
Question: A rod of length 15 cm moves with its ends always touching the coordinate axes. Find the equation of the locus of a point P on the rod, which is at a distance of 3 cm from the end in contact with the x-axis. Solution: Given: A rod of length 15 cm moves with its ends always touching the coordinate axes. A point P on the rod, which is at a distance of 3 cm from the end in contact with the x-axis Need to find: Find the equation of the locus of a point P Here AB is the rod making an angle w...
Read More →The point on the curve
Question: The point on the curve $y^{2}=4 x$ which is nearest to, the point $(2,1)$ is (a) $1,2 \sqrt{2}$ (b) $(1,2)$ (c) $(1,-2)$ (d) $(-2,1)$ Solution: $(b)(1,2)$ Let the required point be $(x, y)$. Then, $y^{2}=4 x$ $\Rightarrow x=\frac{y^{2}}{4}$ ......(1) Now, $d=\sqrt{(x-2)^{2}+(y-1)^{2}}$] Squaring both sides, we get $\Rightarrow d^{2}=(x-2)^{2}+(y-1)^{2}$ $\Rightarrow d^{2}=\left(\frac{y^{2}}{4}-2\right)^{2}+(y-1)^{2}$ $\Rightarrow d^{2}=\frac{y^{4}}{16}+4-y^{2}+y^{2}+1-2 y$ [From eq. (1...
Read More →The towers of bridge, hung in the form of a parabola, have their tops 30 m above the roadway, and are 200 m apart.
Question: The towers of bridge, hung in the form of a parabola, have their tops 30 m above the roadway, and are 200 m apart. If the cable is 5 m above the roadway at the center of the bridge, find the length of the vertical supporting cable, 30 m from the center Solution: Given: Top of the towers are 30 m above the roadway and are 200 m apart. Cable is 5 m above the roadway at center. Need to find: Length of the vertical supporting cable, 30 m from the center. $\mathrm{A}$ and $\mathrm{B}$ are t...
Read More →The maximum value
Question: The maximum value of $f(x)=\frac{x}{4-x+x^{2}}$ on $[-1,1]$ is (a) $-\frac{1}{4}$ (b) $-\frac{1}{3}$ (C) $\frac{1}{6}$ (d) $\frac{1}{5}$ Solution: Given : $f(x)=\frac{x}{4-x+x^{2}}$ $\Rightarrow f^{\prime}(x)=\frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}}=0$ $\Rightarrow 4-x+x^{2}-x(-1+2 x)=0$ $\Rightarrow 4-x+x^{2}+x-2 x^{2}=0$ $\Rightarrow ...
Read More →A parabolic reflector is 5 cm deep and its diameter is 20 cm.
Question: A parabolic reflector is 5 cm deep and its diameter is 20 cm. How far is its focus from the vertex? Solution: Given: Parabolic reflector is 5 cm deep its diameter is 20 cm Need to find: Distance of its focus from the vertex. Reflector is $5 \mathrm{~cm}$ deep, i.e., $O D=5 \mathrm{~cm}$ Diameter of the mirror is $20 \mathrm{com}$, i.e., $B C=20 \mathrm{~cm}$ Let, the equation of the parabola is $y^{2}=4 a x$, where a is the distance of the focus from the vertex. The $x$-coordinate of t...
Read More →The least value of the function
Question: The least value of the function $\mathrm{f}(\mathrm{x})=x 3-18 x 2+96 x$ in the interval $[0,9]$ is (a) 126 (b) 135 (c) 160 (d) 0 Solution: (d) 0 Given: $f(x)=x^{3}-18 x^{2}+96 x$ $\Rightarrow f^{\prime}(x)=3 x^{2}-36 x+96$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}-36 x+96=0$ $\Rightarrow x^{2}-12 x+32=0$ $\Rightarrow(x-4)(x-8)=0$ $\Rightarrow x=4,8$ So, $f(8)=(8)^{3}-18(8)^{2}+96(8)=512-1152+768=128$ $f(4)=(4)^{3}-18(4)^{2}+96(4)=64-288+...
Read More →The least value of the function
Question: The least value of the function $\mathrm{f}(\mathrm{x})=x 3-18 x 2+96 x$ in the interval $[0,9]$ is (a) 126 (b) 135 (c) 160 (d) 0 Solution: (d) 0 Given: $f(x)=x^{3}-18 x^{2}+96 x$ $\Rightarrow f^{\prime}(x)=3 x^{2}-36 x+96$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}-36 x+96=0$ $\Rightarrow x^{2}-12 x+32=0$ $\Rightarrow(x-4)(x-8)=0$ $\Rightarrow x=4,8$ So, $f(8)=(8)^{3}-18(8)^{2}+96(8)=512-1152+768=128$ $f(4)=(4)^{3}-18(4)^{2}+96(4)=64-288+...
Read More →The focus of a parabolic mirror is at a distance of 6 cm from its vertex
Question: The focus of a parabolic mirror is at a distance of 6 cm from its vertex. If the mirror is 20 cm deep, find its diameter. Solution: Given: The focus of a parabolic mirror is at a distance of 6 cm from its vertex. And the mirror is 20 cm deep. Need to find: Diameter of the mirror Here $O$ is the vertex and $A$ is the Focus. So, $O A=a=6 \mathrm{~cm}$. OD is the deep of the mirror $=20 \mathrm{~cm}$ $B C$ is the diameter of the mirror. Equation of the parabola is, $\mathrm{y}^{2}=4 \math...
Read More →If the solve the problem
Question: If $x$ lies in the interval $[0,1]$, then the least value of $x 2+x+1$ is (a) 3 (b) $\frac{3}{4}$ (C) 1 (d) none of these Solution: (c) 1 Given : $f(x)=x^{2}+x+1$ $\Rightarrow f^{\prime}(x)=2 x+1$ For a local maxima or a local minima, we must have' $f^{\prime}(x)=0$ $\Rightarrow 2 x+1=0$ $\Rightarrow 2 x=-1$ $\Rightarrow x=\frac{-1}{2} \notin[0,1]$ At extreme points : $f(0)=0$ $f(1)=1+1+1=30$ So, $x=1$ is a local minima....
Read More →If the solve the problem
Question: At $x=\frac{5 \pi}{6}, f(x)=2 \sin 3 x+3 \cos 3 x$ is (a) 0 (b) maximum (c) minimum (d) none of these Solution: (d) none of these Given: $f(x)=2 \sin 3 x+3 \cos 3 x$ $\Rightarrow f^{\prime}(x)=6 \cos 3 x-9 \sin 3 x$ For a local minima or a local maxima, we must have $f^{\prime}(x)=0$ $\Rightarrow 6 \cos 3 x-9 \sin 3 x=0$ $\Rightarrow 6 \cos 3 x=9 \sin 3 x$' $\Rightarrow \frac{\sin 3 x}{\cos 3 x}=\frac{2}{3}$ $\Rightarrow \tan 3 x=\frac{2}{3}$ ......(1) At $x=\frac{5 \pi}{6}:$ $\tan 3 x...
Read More →The function
Question: The function $f(x)=\sum_{r=1}^{5}(x-r)^{2}$ assumes minimum value at $x=$ (a) 5 (b) $\frac{5}{2}$ (c) 3 (d) 2 Solution: (c) 3 Given : $f(x)=\sum_{r=1}^{5}(x-r)^{2}$ $\Rightarrow f(x)=(x-1)^{2}+(x-2)^{2}+(x-3)^{2}+(x-4)^{2}+(x-5)^{2}$ $\Rightarrow f^{\prime}(x)=2(x-1+x-2+x-3+x-4+x-5)$ $\Rightarrow f^{\prime}(x)=2(5 x-15)$ For a local maxima and a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 2(5 x-15)=0$ $\Rightarrow 5 x-15=0$ $\Rightarrow 5 x=15$ $\Rightarrow x=3$ Now, $f^{...
Read More →The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is
Question: The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{1}{8}$ (d) none of these Solution: $(b) \frac{1}{2}$ Let the two non-zero numbers be $\mathrm{x}$ and $\mathrm{y}$. Then, $x+y=8$ $\Rightarrow y=8-x$ .......(1) Now, $f(x)=\frac{1}{x}+\frac{1}{y}$ $\Rightarrow f(x)=\frac{1}{x}+\frac{1}{8-x}$ [From eq. (1)] $\Rightarrow f^{\prime}(x)=\frac{-1}{x^{2}}+\frac{1}{(8-x)^{2}}$ For a local minima or a local ma...
Read More →The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is
Question: The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{1}{8}$ (d) none of these Solution: $(b) \frac{1}{2}$ Let the two non-zero numbers be $\mathrm{x}$ and $\mathrm{y}$. Then, $x+y=8$ $\Rightarrow y=8-x$ .......(1) Now, $f(x)=\frac{1}{x}+\frac{1}{y}$ $\Rightarrow f(x)=\frac{1}{x}+\frac{1}{8-x}$ [From eq. (1)] $\Rightarrow f^{\prime}(x)=\frac{-1}{x^{2}}+\frac{1}{(8-x)^{2}}$ For a local minima or a local ma...
Read More →If the solve the problem
Question: Let $f(x)=(x-a)^{2}+(x-b)^{2}+(x-c)^{2}$. Then, $f(x)$ has a minimum at $x=$ (a) $\frac{a+b+c}{3}$ (b) $\sqrt[3]{a b c}$ (c) $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ (d) none of these Solution: (a) $\frac{a+b+c}{3}$ Given : $f(x)=(x-a)^{2}+(x-b)^{2}+(x-c)^{2}$ $\Rightarrow f^{\prime}(x)=2(x-a)+2(x-b)+2(x-c)$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 2(x-a)+2(x-b)+2(x-c)=0$ $\Rightarrow 2 x-2 a+2 x-2 b+2 x-2 c=0$ $\Rightarrow 6 x=2(a+b+c)$ ...
Read More →Find the equation of the hyperbola having its foci at
Question: Find the equation of the hyperbola having its foci at $(0, \pm \sqrt{14})$ and passing through the point P(3, 4). Solution: Given: Foci at $(0, \pm \sqrt{14})$ and passing through the point $\mathrm{P}(3,4)$ Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ It passes through the point $P(3,4)$ So putting the values of $(x, y)$ we get, $\frac{3^{2}}{a^{2}}-\frac{4^{2}}{b^{2}}=1 \Rightarrow \frac{9}{a^{2}}-\fra...
Read More →The number which exceeds its square by
Question: The number which exceeds its square by the greatest possible quantity is (a) $\frac{1}{2}$ (b) $\frac{1}{4}$ (C) $\frac{3}{4}$ (d) none of these Solution: (a) $\frac{1}{2}$ Let the required number be $x$. Then, $f(x)=x-x^{2}$ $\Rightarrow f^{\prime}(x)=1-2 x$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 1-2 x=0$ $\Rightarrow 2 x=1$ $\Rightarrow x=\frac{1}{2}$ Now, $f^{\prime \prime}(x)=-20$ So, $x=\frac{1}{2}$ is a local maxima. Hence, the required ...
Read More →The minimum value
Question: The minimum value of $\mathrm{f}(\mathrm{x})=x 4-x 2-2 x+6$ is (a) 6 (b) 4 (c) 8 (d) none of these Solution: (b) 4 Given : $f(x)=x^{4}-x^{2}-2 x+6$ $\Rightarrow f^{\prime}(x)=4 x^{3}-2 x-2$ $\Rightarrow f^{\prime}(x)=(x-1)\left(4 x^{2}+4 x+2\right)$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow(x-1)\left(4 x^{2}+4 x+2\right)=0$ $\Rightarrow(x-1)=0$ $\Rightarrow x=1$ Now, $f^{\prime \prime}(x)=12 x^{2}-2$ $\Rightarrow f^{\prime \prime}(1)=12-2=100$ So...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are (0, 10) and the length of whose latus rectum is 9 units. Solution: Given: Foci are $(0, \pm 10)$ and the length of latus rectum is 9 units Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ The length of the latus rectum is 9 units. Therefore, $\frac{2 b^{2}}{a}=9 \Rightarrow b^{2}=\frac{9}{2} a \cdots$ (1) The foci are given at (0, 10) That means, ae = 10, whe...
Read More →If the solve the problem
Question: Let $f(x)=x^{3}+3 x^{2}-9 x+2$. Then, $f(x)$ has (a) a maximum at $x=1$ (b) a minimum at $x=1$ (c) neither a maximum nor a minimum at $x=-3$ (d) none of these Solution: (b) a minimum at $x=1$ Given: $f(x)=x^{3}+3 x^{2}-9 x+2$ $\Rightarrow f^{\prime}(x)=3 x^{2}+6 x-9$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}+6 x-9=0$ $\Rightarrow x^{2}+2 x-3=0$ $\Rightarrow(x+3)(x-1)=0$ $\Rightarrow x=-3,1$ Now, $f^{\prime \prime}(x)=6 x+6$ $\Rightarrow f...
Read More →For the function
Question: For the function $\mathrm{f}(\mathrm{x})=x+\frac{1}{x}$ (a) $x=1$ is a point of maximum (b) $x=-1$ is a point of minimum (c) maximum value $$ minimum value (d) maximum value $$ minimum value Solution: (d) maximum value $$ minimum value 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 $f^{\prime}(x)=0$ $\Rightarrow 1-\frac{1}{x^{2}}=0$ $\Rightarrow x^{2}-1=0$ $\Rightarrow x^{2}=1$ $\Rightarrow x=\pm 1$ Now, $f^...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are (0, 13) and the length of whose conjugate axis is 24. Solution: Given: Foci are (0, 13), the conjugate axis is of the length 24 Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ The conjugate axis is of the length 24 , i.e., $2 \mathrm{~b}=24$ Therefore, $b=12$ The foci are given at $(0, \pm 13)$ $\mathrm{A}$ and $\mathrm{B}$ are the foci. That means, ae $=13$...
Read More →The minimum value
Question: The minimum value of $\frac{x}{\log _{e} x}$ is (a) $\mathrm{e}$ (b) $1 / e$ (c) 1 (d) none of these Solution: $(a) e$ Given: $f(x)=\frac{x}{\log _{e} x}$ $\Rightarrow f^{\prime}(x)=\frac{\log _{e} x-1}{\left(\log _{e} x\right)^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{\log _{e} x-1}{\left(\log _{e} x\right)^{2}}=0$ $\Rightarrow \log _{e} x-1=0$ $\Rightarrow \log _{e} x=1$ $\Rightarrow x=e$ Now, $f^{\prime \prime}(x)=\frac{-1}{x\left(...
Read More →Find the equation of the hyperbola whose foci are
Question: Find the equation of the hyperbola whose foci are (0, 13) and the length of whose conjugate axis is 24. Solution: Given: Foci are (0, 13), the conjugate axis is of the length 24 Need to find: The equation of the hyperbola. Let, the equation of the hyperbola be: $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ Vertices are $(\pm 3,0)$, that means, $a=3$ And also given, the eccentricity, $e=\frac{4}{3}$ We know that, $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$ Therefore $\Rightarrow \sqrt{1+\frac{b^{2}}...
Read More →If the solve the problem
Question: If $a x+\frac{b}{x} \geq c$ for all positive $x$ where $a, b,0$, then (a) $a b\frac{c^{2}}{4}$ (b) $a b \geq \frac{c^{2}}{4}$ (c) $a b \geq \frac{c}{4}$ Solution: $(\mathrm{b}) a b \geq \frac{c^{2}}{4}$ Given : $a x+\frac{b}{x} \geq c$ Minimum value of $a x+\frac{b}{x}=c$ Now, $f(x)=a x+\frac{b}{x}$ $\Rightarrow f^{\prime}(x)=a-\frac{b}{x^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow a-\frac{b}{x^{2}}=0$ $\Rightarrow a x^{2}-b=0$ $\Rightarrow a...
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