The minimum value
Question: The minimum value of $x \log _{e} x$ is equal to (a) $\mathrm{e}$ (b) $1 / \mathrm{e}$ (c) $-1 / e$ (d) $2 / e$ (e) $-\mathrm{e}$ Solution: (c) $\frac{-1}{e}$ Here, $f(x)=x \log _{e} x$ $\Rightarrow f^{\prime}(x)=\log _{e} x+1$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \log _{e} x+1=0$ $\Rightarrow \log _{e} x=-1$ $\Rightarrow x=e^{-1}$ Now, $f^{\prime \prime}(x)=\frac{1}{x}$ $\Rightarrow f^{\prime \prime}\left(e^{-1}\right)=e0$ So, $x=e^{-1}$ is...
Read More →If the solve the problem
Question: Let $f(x)=2 x^{3}-3 x^{2}-12 x+5$ on $[-2,4]$. The relative maximum occurs at $x=$ (a) $-2$ (b) $-1$ (C) 2 (d) 4 Solution: (c) 2 Given : $f(x)=2 x^{3}-3 x^{2}-12 x+5$ $\Rightarrow f^{\prime}(x)=6 x^{2}-6 x-12$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 6 x^{2}-6 x-12=0$ $\Rightarrow x^{2}-x-2=0$ $\Rightarrow(x-2)(x+1)=0$ $\Rightarrow x=2,-1$ Now, $f^{\prime \prime}(x)=12 x-6$ $\Rightarrow f^{\prime \prime}(-1)=-12-6=-180$ So, $x=1$ is a local maxi...
Read More →(3x + 5) (1 + tan x)
Question: (3x + 5) (1 + tan x) Solution: Given $(3 x+5)(1+\tan x)$ Let $y=(3 x+5)(1+\tan x)$ Applying product rule of differentiation that is $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{t} \cdot \mathrm{y})=\mathrm{y} \cdot \frac{\mathrm{d} \mathrm{t}}{\mathrm{dx}}+\mathrm{t} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ $\Rightarrow y=(3 x+5)(1+\tan x)$ $\Rightarrow \frac{d y}{d x}=(1+\tan x) \frac{d}{d x}(3 x+5)+(3 x+5) \frac{d}{d x}(1+\tan x)$ $\Rightarrow \frac{d y}{d x}=3(1+\tan x)+(3 x+5)...
Read More →Show that the points P
Question: Show that the points P(2, 3, 5), Q(-4, 7, -7), R(-2, 1, -10) and S(4, -3, 2) are the vertices of a rectangle. Solution: To prove: Points P, Q, R, S forms rectangle. Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{Z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_...
Read More →The maximum value
Question: The maximum value of $\mathrm{f}(\mathrm{x})=\frac{x}{4+x+x^{2}}$ on $[-1,1]$ is (a) $-\frac{1}{4}$ (b) $-\frac{1}{3}$ (C) $\frac{1}{6}$ (d) $\frac{1}{5}$ Solution: (C) $\frac{1}{6}$ 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...
Read More →Show that the points A
Question: Show that the points A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram. Show that ABCD is not a rectangle. Solution: To prove: Points A, B, C, D form parallelogram Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$...
Read More →The function
Question: The function $\mathrm{f}(\mathrm{x})=2 x^{3}-15 x^{2}+36 x+4$ is maximum at $\mathrm{x}=$ (a) 3 (b) 0 (c) 4 (d) 2 Solution: (d) 2 Given : $f(x)=2 x^{3}-15 x^{2}+36 x+4$ $\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 6 x^{2}-30 x+36=0$ $\Rightarrow x^{2}-5 x+6=0$ $\Rightarrow(x-2)(x-3)=0$ $\Rightarrow x=2,3$ Now, $f^{\prime \prime}(x)=12 x-30$ $\Rightarrow f^{\prime \prime}(2)=24-30=-60$ So, $x=1$ is a local ...
Read More →Show that the points
Question: Show that the points A(1, 1, 1), B(-2, 4, 1), C(1, -5, 5) and D(2, 2, 5) are the vertices of a square. Solution: To prove: Points A, B, C, D form square. Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$ Here $\left(x_{1}, y_{1}, z_{1}\right)=(1,1,1)$ ...
Read More →If the solve the problem
Question: $f(x)=1+2 \sin x+3 \cos ^{2} x, 0 \leq x \leq \frac{2 \pi}{3}$ is (a) Minimum at $x=\pi / 2$ (b) Maximum at $x=\sin ^{-1}(1 / \sqrt{3})$ (c) Minimum at $x=\pi / 6$ (d) Maximum at $\sin ^{-1}(1 / 6)$ Solution: (a) Minimum at $x=\frac{\pi}{2}$ Given : $f(x)=1+2 \sin x+3 \cos ^{2} x$ $\Rightarrow f^{\prime}(x)=2 \cos x-6 \cos x \sin x$ $\Rightarrow f^{\prime}(x)=2 \cos x(1-3 \sin x)$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 2 \cos x(1-3 \sin x)=0$ ...
Read More →If the solve the problem
Question: Let $x, y$ be two variables and $x0, x y=1$, then minimum value of $x+y$ is (a) 1 (b) 2 (c) $2 \frac{1}{2}$ (d) $3 \frac{1}{3}$ Solution: (b) 2 Given : $x y=1$ $\Rightarrow y=\frac{1}{x}$ $f(x)=x+\frac{1}{x}$ $\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 1-\frac{1}{x^{2}}=0$ $\Rightarrow x^{2}-1=0$ $\Rightarrow x^{2}=1$ $\Rightarrow x=\pm 1$ $\Rightarrow x=1$ (Given : $x1$ ) $\Rightarrow y=1$ Now, $f^{\pr...
Read More →Show that the points A
Question: Show that the points A(0, 1, 2), B(2, -1, 3) and C(1, -3, 1) are the vertices of an isosceles right-angled triangle. Solution: To prove: Points A, B, C form isosceles triangle. Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$ Here, $\left(x_{1}, y_{1}...
Read More →If the solve the problem
Question: If $(\mathrm{x})=\frac{1}{4 x 2+2 x+1}$, then its maximum value is (a) $\frac{4}{3}$ (b) $\frac{2}{3}$ (c) 1 (d) $\frac{3}{4}$ Solution: (a) $\frac{4}{3}$ Maximum value of $\frac{1}{4 x^{2}+2 x+1}=$ Minimum value of $4 x^{2}+2 x+1$ Now, $f(x)=4 x^{2}+2 x+1$ $\Rightarrow f^{\prime}(x)=8 x+2$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 8 x+2=0$ $\Rightarrow 8 x=-2$ $\Rightarrow x=\frac{-1}{4}$ Now, $f^{\prime \prime}(x)=8$ $\Rightarrow f^{\prime \pri...
Read More →Show that the points
Question: Show that the points A(4, 6, -5), B(0, 2, 3) and C(-4, -4, -1) from the vertices of an isosceles triangle. Solution: To prove: Points A, B, C form isosceles triangle. Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$ Here, $\left(x_{1}, y_{1}, z_{1}\ri...
Read More →If the solve the problem
Question: If $(x)=x+\frac{1}{x}, x0$, then its greatest value is (a) $-2$ (b) 0 (c) 3 (d) none of these 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 $f^{\prime}(x)=0$ $\Rightarrow 1-\frac{1}{x^{2}}=0$ $\Rightarrow x^{2}-1=0$ $\Rightarrow x^{2}=1$ $\Rightarrow x=\pm 1$ $\Rightarrow x=1$ (Given : $x0$ ) Now, $f^{\prime \prime}(x)=\frac{2}{x^{3}}$ $\Rightarrow f^{\prime \prime}(1)=20$ So, $x=1$ is a local min...
Read More →Show that the points
Question: Show that the points $A(1,-1,-5), b(3,1,3)$ and $C(9,1,-3)$ are the vertices of an equilateral triangle. Solution: To prove: Points A, B, C form equilateral triangle. Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z...
Read More →The minimum value
Question: The minimum value of $\left(x^{2}+\frac{250}{x}\right)$ is (a) 75 (b) 50 (c) 25 (d) 55 Solution: (a) 75 Given : $f(x)=x^{2}+\frac{250}{x}$ $\Rightarrow f^{\prime}(x)=2 x-\frac{250}{x^{2}}$ For a local maxima or a local minima, we must have' $f^{\prime}(x)=0$ $\Rightarrow 2 x-\frac{250}{x^{2}}=0$ $\Rightarrow 2 x^{3}-250=0$ $\Rightarrow x^{3}=125$ $\Rightarrow x=5$ Now, $f^{\prime \prime}(x)=2+\frac{500}{x^{3}}$ $\Rightarrow f^{\prime \prime}(5)=2+\frac{500}{5^{3}}=\frac{750}{125}=60$ S...
Read More →Find the distance between the points :
Question: Find the distance between the points : (i) $A(5,1,2)$ and $B(4,6,-1)$ (ii) $\mathbf{P}(1,-1,3)$ and $\mathbf{Q}(2,3,-5)$ (iii) $R(1,-3,4)$ and $S(4,-2,-3)$ (iv) $C(9,-12,-8)$ and the origin Solution: Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$ (i) $A(5,1,2)$ and $B(4,6,-1)$ Here, $\left(x_{1}, y_{1}, z_{1}\r...
Read More →If a cone of maximum volume is inscribed in a given sphere,
Question: If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is (a) $\frac{3}{4}$ (b) $\frac{1}{3}$ (C) $\frac{1}{4}$ (d) $\frac{2}{3}$ Solution: (d) $\frac{2}{3}$ Let $h, r, V$ and $R$ be the height, radius of the base, volume of the cone and the radius of the sphere, respectively. Given : $h=R+\sqrt{R^{2}-r^{2}}$ $\Rightarrow h-R=\sqrt{R^{2}-r^{2}}$ Squaring both side, we get $h^{2}+R^{2}-2 h R=R^{2}-r^{2}$ $\Right...
Read More →In which octant does each of the given points lie?
Question: In which octant does each of the given points lie? (i) $(-4,-1,-6)$ (ii) $(2,3,-4)$ (iii) $(-6,5,-1)$ (iv) $(4,-3,-2)$ (v) $(-1,-6,5)$ (vi) $(4,6,8)$ Solution: The position of a point in a octant is signified by the signs of the x, y, z coordinates. Here is a table showing signs of the x, y, z coordinates in all the octants. According to the table (i) $(-4,-1,-6)$ lies in octant VII (ii) $(2,3,-4)$ lies in octant $\mathrm{V}$ (iii) $(-6,5,-1)$ lies in octant VI (iv) $(4,-3,-2)$ lies in...
Read More →If the solve the problem
Question: $f(x)=\sin +\sqrt{3} \cos x$ is maximum when $x=$ (a) $\frac{\pi}{3}$ (b) $\frac{\pi}{4}$ (C) $\frac{\pi}{6}$ (d) 0 Solution: (c) $\frac{\pi}{6}$ Given: $f(x)=\sin x+\sqrt{3} \cos x$ $\Rightarrow f^{\prime}(x)=\cos x-\sqrt{3} \sin x$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \cos x-\sqrt{3} \sin x=0$ $\Rightarrow \cos x=\sqrt{3} \sin x$ $\Rightarrow \tan x=\frac{1}{\sqrt{3}}$ $\Rightarrow x=\frac{\pi}{6}$ Now, $f^{\prime \prime}(x)=-\sin x-\sqrt{...
Read More →In which plane does the point (4, -3, 0) lie?
Question: In which plane does the point (4, -3, 0) lie? Solution: Here the x, y, z coordinates of the point are 4, -3, 0. As the distance of point along the z-axis is 0, the plane in which the point lies is the xy-plane....
Read More →If a point lies on yz-plane then what is its x-coordinate?
Question: If a point lies on yz-plane then what is its x-coordinate? Solution: x-coordinate is the distance of a point from the origin parallel or along the xaxis. To measure the x coordinate, you must move either to the left of the origin or to its right. In case of a point lying on the yz-plane, you do not move to the right or to the left of the origin. Hence x coordinate is 0 for a point on the yz-plane....
Read More →If a point lies on the z-axis, then find its x-coordinate and y-coordinate.
Question: If a point lies on the z-axis, then find its x-coordinate and y-coordinate. Solution: X and y coordinates of a point are its distance from the origin along or parallel to the horizontal x-axis and y-axis. To measure the x and y coordinates, you must move either to the left of the origin or to its right. In case of a point on the z-axis, you do not move to the right or to the left of the origin. Hence x and y coordinates are 0 for a point on the z-axis...
Read More →The least and greatest values
Question: The least and greatest values of $f(x)=x^{3}-6 x^{2}+9 x$ in $[0,6]$, are (a) 3,4 (b) 0,6 (c) 0,3 (d) 3,6 Solution: Given : $f(x)=x^{3}-6 x^{2}+9 x$ $\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+9$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}-12 x+9=0$ $\Rightarrow x^{2}-4 x+3=0$ $\Rightarrow(x-1)(x-3)=0$ $\Rightarrow x=1,3$ Now, $f(0)=0^{3}-6(0)^{2}+9(0)=0$ $f(1)=1^{3}-6(1)^{2}+9(1)=1-6+9=4$ $f(3)=3^{3}-6(3)^{2}+9(3)=27-54+27=0$ $f(6)=6^{3}-6(6)^...
Read More →If the solve the problem
Question: If $x+y=8$, then the maximum value of $x y$ is (a) 8 (b) 16 (c) 20 (d) 24 Solution: (b) 16 Given : $x+y=8$ $\Rightarrow y=8-x$ ....(1) Let $f(x)$ be $x y$. $\Rightarrow f(x)=x(8-x)$ $[$ From eq. $(1)]$ $\Rightarrow f^{\prime}(x)=8-2 x$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 8-2 x=0$ $\Rightarrow 8=2 x$ $\Rightarrow x=4$ $\Rightarrow y=8-4=4$ $[$ From eq. $(1)]$ Now, $f^{\prime \prime}(x)=-2$ $\Rightarrow f^{\prime \prime}(4)=-20$ So, $x=4$ is ...
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