Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum.
Question: Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum. Solution: Let the two positive numbers be $x$ and y. Then, $x+y=15$ ....(1) Now, $z=x^{2}+y^{2}$ $\Rightarrow z=x^{2}+(15-x)^{2}$ [From eq. (1)] $\Rightarrow z=x^{2}+x^{2}+225-30 x$ $\Rightarrow z=2 x^{2}+225-30 x$ $\Rightarrow \frac{d z}{d x}=4 x-30$ For maximum or minimum values of $\mathrm{z}$, we must have $\frac{d z}{d x}=0$ $\Rightarrow 4 x-30=0$ $\Rightarrow x=\frac{15}{2}$ $\frac{d^{2} z}{d ...
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Question: $f(x)=x^{3}-1$ on $R$ Solution: We can observe thatf(x) increases when the values ofxincrease andf(x) decreases when the values ofxdecrease. Also,f(x) can be reduced by giving smaller values ofx.Similarly,f(x) can be enlarged by giving larger values ofx.So,f(x) does not have a minimum or maximum value....
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Question: $f(x)=16 x^{2}-16 x+28$ on $R$ Solution: Given: $f(x)=16 x^{2}-16 x+28$ $\Rightarrow f(x)=4\left(4 x^{2}-4 x+1\right)+24$ $\Rightarrow f(x)=4(2 x-1)^{2}+24$ Now, $4(2 x-1)^{2} \geq 0$ for all $x \in R$ $\Rightarrow f(x)=4(2 x-1)^{2}+24 \geq 24$ for all $x \in R$ $\Rightarrow f(x) \geq 24$ for all $x \in R$ The minimum value of $f$ is attained when $(2 x-1)=0$. $(2 x-1)=0$ $\Rightarrow x=\frac{1}{2}$ Therefore, the minimum value of $f$ at $x=\frac{1}{2}$ is 24 . Sincef(x) can be enlarge...
Read More →A variable line passes through a fixed point P.
Question: A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P. Solution: Let the variable line be $a x+b y=1$ We know that, length of the perpendicular from $(p, q)$ to the line $a x+b y+c=0$ is $\mathrm{d}=\left|\frac{\mathrm{ap}+\mathrm{bq}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|$ $\mathrm{d}_{1}=\left|\frac{2 \times \mathrm{a}+0 \times \...
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Question: $f(x)=-|x+1|+3$ on $R$ Solution: Given: $f(x)=-|x+1|+3$ Now, $-|x+1| \leq 0$ for all $x \in R$ $\Rightarrow f(x)=-|x+1|+3 \leq 3$ for all $x \in R$ $\Rightarrow f(x) \leq 3$ for all $x \in R$ The maximum value of $f$ is attained when $|x+1|=0$. $\Rightarrow x=-1$ Therefore, the maximum value of $f$ at $x=-1$ is 3 . Sincef(x) can be reduced, the minimum value does not exist, which is evident in the graph also.Hence, the functionfdoes not have a minimum value....
Read More →Find the equation of the parabola, which is symmetric about the y-axis and
Question: Find the equation of the parabola, which is symmetric about the y-axis and passes through the point P(2, -3). Solution: The equation of a parabola with vertex at the origin and symmetric about the y-axis is $x^{2}=4 a y$ Since point $P(2,-3)$ passes through above parabola we can write, $2^{2}=4 a(-3)$ $-4=-12 \mathrm{a}$ $\cdot a=-\frac{1}{3}$ Therefore, the equation of a parabola is $x^{2}=4 \cdot\left(-\frac{1}{3}\right) y$ $x^{2}=-\frac{4}{3} y$ - $3 x^{2}=-4 y$...
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Question: $f(x)=2 x^{3}+5$ on $R$ Solution: We can observe thatf(x) increases when the values ofxare increased andf(x) decreases when the values ofxare decreased. Also,f(x) can be reduced by giving small values ofx.Similarly,f(x) can be enlarged by giving large values ofx.So,f(x) does not have a minimum or maximum value....
Read More →If the equation of the base of an equilateral
Question: If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, 1), then find the length of the side of the triangle. Solution: Let $\triangle \mathrm{ABC}$ be an equilateral triangle. Given equation of the base $B C$ is $x+y=2$ We know that, in an equilateral triangle all angles are of $60^{\circ}$ So, in $\triangle \mathrm{ABD}$ $\sin 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{AB}}$\ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{\mathrm{AB}}\left[\because \s...
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Question: $f(x)=|\sin 4 x+3|$ on $R$ Solution: Given: $f(x)=|\sin 4 x+3|$ We know that $-1 \leq \sin 4 x \leq 1$. $\Rightarrow 2 \leq \sin 4 x+3 \leq 4$ $\Rightarrow 2 \leq|\sin 4 x+3| \leq 4$ $\Rightarrow 2 \leq f(x) \leq 4$ Hence, the maximum and minimum values of $f$ are 4 and 2 , respectively....
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Question: $f(x)=\sin 2 x+5$ on $R$ Solution: Given: $f(x)=\sin 2 x+5$ We know that $-1 \leq \sin 2 x \leq 1$ $\Rightarrow-1+5 \leq \sin 2 x+5 \leq 1+5$ $\Rightarrow 4 \leq \sin 2 x+5 \leq 6$ $\Rightarrow 4 \leq f(x) \leq 6$ Hence, the maximum and minimum values of $f$ are 6 and 4 , respectively....
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Question: $f(x)=|x+2|$ on $R$ Solution: Given: $f(x)=|x+2|$ Now, $|x+2| \geq 0$ for all $x \in \mathrm{R}$ Thus, $f(x) \geq 0$ for all $x \in \mathbf{R}$ Therefore, the minimum value of $f$ at $x=-2$ is 0 . Since $f(x)$ can be enlarged, the maximum value does not exist, which is evident in the graph also. Hence, the function $f$ does not have a maximum value....
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Question: $f(x)=(x-1)^{2}+2$ on $R$ Solution: Given: $f(x)=-(x-1)^{2}+2$ Now, $(x-1)^{2} \geq 0$ for all $x \in \mathbf{R}$ $\Rightarrow f(x)=-(x-1)^{2}+2 \leq 2$ for all $x \in \mathbf{R}$ The maximum value of $f(x)$ is attained when $(x-1)=0$. $(x-1)=0$ $\Rightarrow x=1$ Therefore, the maximum value of $f(x)=2$ Since $f(x)$ can be reduced, the minimum value does not exist, which is evident in the graph also. Hence, function $f$ does not have a minimum value....
Read More →Find the equation of the parabola with vertex at the origin, passing through
Question: Find the equation of the parabola with vertex at the origin, passing through the point P(5, 2) and symmetric with respect to the y-axis. Solution: The equation of a parabola with vertex at the origin and symmetric about the y-axis is $x^{2}=4 a y$ Since point P(5,2) passes through above parabola we can write, $5^{2}=4 a(2)$ $\cdot 25=8 a$ $a=\frac{25}{8}$ Therefore, the equation of a parabola is $x^{2}=4 \cdot \frac{25}{8} y$ $x^{2}=\frac{25}{2} y$ - $2 x^{2}=25 y$...
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Question: $f(x)=4 x^{2}-4 x+4$ on $R$ Solution: Given: $f(x)=4 x^{2}-4 x+4$ $\Rightarrow f(x)=\left(4 x^{2}-4 x+1\right)+3$ $\Rightarrow f(x)=(2 x-1)^{2}+3$ Now, $(2 x-1)^{2} \geq 0$ for all $x \in R$ $\Rightarrow f(x)=(2 x-1)^{2}+3 \geq 3$ for all $x \in R$ $\Rightarrow f(x) \geq 3$ for all $x \in R$ The minimum value of $f$ is attained when $(x-1)=0$. $(2 x-1)=0$ $\Rightarrow x=\frac{1}{2}$ Thus, the minimum value of $f(x)$ at $x=\frac{1}{2}$ is 3 . Since $f(x)$ can be enlarged, the maximum va...
Read More →For what values of a and b the intercepts
Question: For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x 3y + 6 = 0 on the axes. Solution: Given equation is $a x+b y+8=0$ It can also be re-written as $a x+b y=-8$ Now, dividing by $-8$ to both the sides, we get $\frac{a}{-8} x+\frac{b}{-8} y=1$ $\Rightarrow \frac{x}{-\frac{8}{a}}+\frac{y}{-\frac{8}{b}}=1$ So, the intercepts on the axes are $-\frac{8}{a}$ and $-\frac{8...
Read More →$f(x)=4 x^{2}-4 x+4$ on $R$
Question: $f(x)=4 x^{2}-4 x+4$ on $R$ Solution: Given: $f(x)=4 x^{2}-4 x+4$ $\Rightarrow f(x)=\left(4 x^{2}-4 x+1\right)+3$ $\Rightarrow f(x)=(2 x-1)^{2}+3$ Now, $(2 x-1)^{2} \geq 0$ for all $x \in R$ $\Rightarrow f(x)=(2 x-1)^{2}+3 \geq 3$ for all $x \in R$ $\Rightarrow f(x) \geq 3$ for all $x \in R$ The minimum value of $f$ is attained when $(x-1)=0$. $(2 x-1)=0$ $\Rightarrow x=\frac{1}{2}$ Thus, the minimum value of $f(x)$ at $x=\frac{1}{2}$ is 3 . Since $f(x)$ can be enlarged, the maximum va...
Read More →Find the equation of the parabola with vertex at the origin and focus F(0, 5)
Question: Find the equation of the parabola with vertex at the origin and focus F(0, 5) Solution: Vertex : A (0,0) Given focus $F(0,5)$ is of the form $F(0, a)$ For Vertex $A(0,0)$ and Focus $F(0, a)$, equation of parabola is $x^{2}=4 a y$ Here, a = 5 Therefore, equation of parabola, $x^{2}=20 y$...
Read More →Find the equation of the line passing
Question: Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7. Solution: Given lines are 2x + y = 5 1 x + 3y = -8 2 Firstly, we find the point of intersection of equation 1 and equation 2 Multiply the equation 2 by 2, we get 2x + 6y = -16 .3 On subtracting equation 3 from 1, we get 2x + y 2x 6y = 5 (-16) On simplifying we get ⇒-5y = 5 + 16 ⇒ -5y = 21 $\Rightarrow y=-\frac{21}{5}$ Putting the value of $y$ in...
Read More →Find the equation of the parabola with focus
Question: Find the equation of the parabola with focus $F(0,-3)$ and directrix $y=3$. Solution: Given equation of directrix : y = 3 - $y-3=0$ Above equation is of the form, y - a = 0 Focus of the parabola $F(0,-3)$ is of the form $F(0,-a)$ Therefore, a = 3 For directrix with equation $y-a=0$ and focus $(0,-a)$, equation of the parabola is, $x^{2}=-4 a y$ $\cdot x^{2}=-12 y$...
Read More →Find the absolute maximum and minimum values of a function $f$ given by
Question: Find the absolute maximum and minimum values of a function $f$ given by $f(x)=2 x^{3}-15 x^{2}+36 x+1$ on the interval $[1,5]$ Solution: Given : $f(x)=2 x^{3}-15 x^{2}+36 x+1$ $\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$ For a local maximum or a local minimum, we have $f^{\prime}(x)=0$ $\Rightarrow 6 x^{2}-30 x+36=0$ $\Rightarrow x^{2}-5 x+6=0$ $\Rightarrow(x-3)(x-2)=0$ $\Rightarrow x=2$ and $x=3$ Thus, the critical points of $f$ are $1,2,3$ and 5 . Now, $f(1)=2(1)^{3}-15(1)^{2}+36(1)+1...
Read More →Find the equation of the parabola with focus
Question: Find the equation of the parabola with focus $F(4,0)$ and directrix $x=-4$. Solution: Given equation of directrix : $x=-4$ x + 4 = 0 Above equation is of the form, $x+a=0$ Focus of the parabola $F(4,0)$ is of the form $F(a, 0)$ Therefore, $\mathrm{a}=4$ For directrix with equation $\mathrm{x}+\mathrm{a}=0$ and focus $(\mathrm{a}, 0)$, equation of the parabola is, $y^{2}=4 a x$ $\cdot y^{2}=16 x$...
Read More →Show that the tangent of an angle between the lines
Question: Show that the tangent of an angle between the lines $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is $\frac{2 a b}{a^{2}-b^{2}}$ Solution: Given $\frac{x}{a}+\frac{y}{b}=1$$\ldots \ldots 1$ $\frac{x}{a}-\frac{y}{b}=1$$\ldots \ldots .2$ Firstly, we find the slope of the given lines $\frac{x}{a}+\frac{y}{b}=1$ Above equation can be written as $\Rightarrow \frac{y}{b}=1-\frac{x}{a}$ On rearranging we get $\Rightarrow y=b-\frac{b}{a} x$ $\R...
Read More →Find the equation of the parabola with vertex at the origin and focus at
Question: Find the equation of the parabola with vertex at the origin and focus at F(-2, 0). Solution: Vertex : A (0,0) Given focus $F(-2,0)$ is of the form $F(-a, 0)$ For Vertex $A(0,0)$ and Focus $F(-a, 0)$, equation of parabola is $y^{2}=-4 a x$ Here, a = 2 Therefore, equation of parabola, $y^{2}=-8 x$...
Read More →Find the absolute maximum and minimum values of a function $f$ given by
Question: Find the absolute maximum and minimum values of a function $f$ given by $f(x)=12 x^{4 / 3}-6 x^{1 / 3}, x \in[-1,1]$ Solution: Given : $f(x)=12 x^{\frac{4}{3}}-6 x^{\frac{1}{3}}$ $\Rightarrow f^{\prime}(x)=16 x^{\frac{1}{3}}-2 x^{\frac{-2}{3}}=\frac{2(8 x-1)}{x^{\frac{2}{3}}}$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{2(8 x-1)}{x^{\frac{2}{3}}}=0$ $\Rightarrow 8 x-1=0$ $\Rightarrow x=\frac{1}{8}$ Thus, the critical points of $f$ are $-1, ...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $3 x^{2}=-16 y$ Solution: Given equation : $3 x^{2}=-16 y$ $x^{2}=-\frac{16}{3} y$ Comparing the given equation with parabola having an equation, $x^{2}=4 a y$ $4 a=\frac{16}{3}$ $a=\frac{4}{3}$ Focus: $F(0,-a)=F\left(0,-\frac{4}{3}\right)$ Vertex: $A(0,0)=A(0,0)$ Equation of the directrix: $y-a=0$ $y-\frac{4}{3}=0$ $y=\frac{4}{3}$ Lenth of lat...
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