Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: The function $f(x)=2 \log (x-2)-x^{2}+4 x+1$ increases on the interval. A. $(1,2)$ B. $(2,3)$ C. $((1,3)$ D. $(2,4)$ Solution: Formula:- The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ Given:- $f(x)=2 \log (x-2)-x^{2}+4 x+1$ $\mathrm{d}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{dx}}\right)=\frac{2}{\mathrm{x}-2}-2 \m...
Read More →Transform the equation
Question: Transform the equation $2 x^{2}+y^{2}-4 x+4 y=0$ to parallel axes when the origin is shifted to the point (1, -2). Solution: Let the new origin be (h, k) = (1, -2) Then, the transformation formula become: $x=X+1$ and $y=Y+(-2)=Y-2$ Substituting the value of x and y in the given equation, we get $2 x^{2}+y^{2}-4 x+4 y=0$ Thus $2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0$ $\Rightarrow 2\left(X^{2}+1+2 X\right)+\left(Y^{2}+4-4 Y\right)-4 X-4+4 Y-8=0$ $\Rightarrow 2 X^{2}+2+4 X+Y^{2}+4-4 Y-4 X+4 ...
Read More →Find what the given equation becomes when the origin is shifted to the point
Question: Find what the given equation becomes when the origin is shifted to the point (1, 1). $x y-x-y+1=0$ Solution: Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: $x=X+1$ and $y=Y+1$ Substituting the value of x and y in the given equation, we get $x y-x-y+1=0$ Thus, $(X+1)(Y+1)-(X+1)-(Y+1)+1=0$ $\Rightarrow X Y+X+Y+1-X-1-Y-1+1=0$ $\Rightarrow X Y=0$ Hence, the transformed equation is XY = 0...
Read More →Find what the given equation becomes when the origin is shifted to the point
Question: Find what the given equation becomes when the origin is shifted to the point (1, 1). $x^{2}-y^{2}-2 x+2 y=0$ Solution: Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get $x^{2}-y^{2}-2 x+2 y=0$ Thus, $(X+1)^{2}-(Y+1)^{2}-2(X+1)+2(Y+1)=0$ $\Rightarrow\left(X^{2}+1+2 X\right)-\left(Y^{2}+1+2 Y\right)-2 X-2+2 Y+2=0$ $\Rightarrow X^{2}+1+2 X-Y^{2}-1-2 Y-2 X+2 Y=0$ $\Rightarro...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: The function $f(x)=x^{x}$ decreases on the interval. A. $(0, \mathrm{e})$ B. $(0,1)$ C. $(0,1 / \mathrm{e})$ D. $(1 / e, e)$ Solution: Formula:- The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ Given:- $f(x)=x^{x}$ $\mathrm{d}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{dx}}\right)=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm...
Read More →Find what the given equation becomes when the origin is shifted to the point
Question: Find what the given equation becomes when the origin is shifted to the point (1, 1). $x y-y^{2}-x+y=0$ Solution: Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: $x=X+1$ and $y=Y+1$ Substituting the value of x and y in the given equation, we get $x y-y^{2}-x+y=0$ Thus $(X+1)(Y+1)-(Y+1)^{2}-(X+1)+(Y+1)=0$ $\Rightarrow X Y+X+Y+1-\left(Y^{2}+1+2 Y\right)-X-1+Y+1=0$ $\Rightarrow X Y+X+Y+1-Y^{2}-1-2 Y-X+Y=0$ $\Rightarrow X Y-Y^{2}=0$ Hence, the transformed equa...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: The function $f(x)=\cos ^{-1} x+x$ increases in the interval. A. $(1, \infty)$ B. $(-1, \infty)$ C. $(-\infty, \infty)$ D. $(0, \infty)$ Solution: Formula:- The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ Given:- $f(x)=\cos ^{-1} x+x$ $\mathrm{d}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{dx}}\right)=\frac{\mathrm{x}^...
Read More →Find what the given equation becomes when the origin is shifted to the point
Question: Find what the given equation becomes when the origin is shifted to the point (1, 1). $x^{2}+x y-3 x-y+2=0$ Solution: Let the new origin be (h, k) = (1, 1) Then, the transformation formula become: x = X + 1 and y = Y + 1 Substituting the value of x and y in the given equation, we get $x^{2}+x y-3 x-y+2=0$ Thus $(X+1)^{2}+(X+1)(Y+1)-3(X+1)-(Y+1)+2=0$ $\Rightarrow\left(X^{2}+1+2 X\right)+X Y+X+Y+1-3 X-3-Y-1+2=0$ $\Rightarrow X^{2}+1+2 X+X Y-2 X-1=0$ $\Rightarrow X^{2}+X Y=0$ Hence, the tr...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: The interval of increase of the function $f(x)=x-e^{x}+\tan \left(\frac{2 \pi}{7}\right)$ is A. $(0, \infty)$ B. $(-\infty, 0)$ C. $(1, \infty)$ D. $(-\infty, 1)$ Solution: Formula:- The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ Given:- $f(x)=x-e^{x}+\tan \left(\frac{2 \pi}{7}\right)$ $\mathrm{d}\left(\frac{\mathrm{...
Read More →At what point must the origin be shifted
Question: At what point must the origin be shifted, if the coordinates of a point (-4,2) become (3, -2)? Solution: Let (h, k) be the point to which the origin is shifted. Then, $x=-4, y=2, X=3$ and $Y=-2$ $\therefore \mathrm{x}=\mathrm{X}+\mathrm{h}$ and $\mathrm{y}=\mathrm{Y}+\mathrm{k}$ $\Rightarrow-4=3+h$ and $2=-2+k$ $\Rightarrow h=-7$ and $k=4$ Hence, the origin must be shifted to (-7, 4)...
Read More →If the origin is shifted to the point
Question: If the origin is shifted to the point (2, -1) by a translation of the axes, the coordinates of a point become (-3, 5). Find the origin coordinates of the point. Solution: Let the new origin be (h, k) = (2, -1) and (x, y) = (-3, 5) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: x = X + h and y = Y + k $\Rightarrow-3=X+2$ and $5=Y+(-1)$ $\Rightarrow X=-5$ and $Y=6$ Thus, the new coordinates are (-5, 6)...
Read More →Find the intervals in which $f(x)$ is increasing or decreasing:
Question: Find the intervals in which $f(x)$ is increasing or decreasing: i. $f(x)=x|x|, x \in R$ ii. $f(x)=\sin x+|\sin x|, 0x \leq 2 \pi$ iii. $f(x)=\sin x(1+\cos x), 0x\pi / 2$ Solution: (i): Consider the given function, $f(x)=x|x|, x \in R$ $\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}-\mathrm{x}^{2}, \mathrm{x}0 \\ \mathrm{x}^{2}, \mathrm{x}0\end{array}\right.$ $\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}-2 \mathrm{x}, \mathrm{x}0 \\ 2 \mathrm{x}, \mathrm{x}0\en...
Read More →If the origin is shifted to the point
Question: If the origin is shifted to the point (0, -2) by a translation of the axes, the coordinates of a point become (3, 2). Find the original coordinates of the point. Solution: Let the new origin be (h, k) = (0, -2) and (x, y) = (3, 2) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: $x=X+h$ and $y=Y+k$ $\Rightarrow 3=X+0$ and $2=Y+(-2)$ $\Rightarrow X=3$ and $Y=4$ Thus, the new coordinates are (3, 4)...
Read More →Prove the following
Question: If $\frac{(1+i)^{2}}{2-i}=x+i y$, , then find the value of $x+y$ Solution: According to the question, We have, $x+i y=\frac{(1+i)^{2}}{2-i}$ $=\frac{1+2 i+i^{2}}{2-i}$ $=\frac{2 \mathrm{i}}{2-\mathrm{i}}$ Rationalizing the denominator, $=\frac{2 \mathrm{i}(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})}$ $=\frac{4 \mathrm{i}+2 \mathrm{i}^{2}}{4-\mathrm{i}^{2}}$ $=\frac{4 \mathrm{i}-2}{4+1}$ $=\frac{-2}{5}+\frac{4 i}{5}$ Thus, $\mathrm{X}=-\frac{2}{5}, \mathrm{y}=\frac{4}{5}$ Hence, $x+y=-...
Read More →If the origin is shifted to the point
Question: If the origin is shifted to the point (-3, -2) by a translation of the axes, find the new coordinates of the point (3, -5). Solution: Let the new origin be (h, k) = (-3, -2) and (x, y) = (3, -5) be the given point. Let the new coordinates be $(X, Y)$ We use the transformation formula: $x=X+h$ and $y=Y+k$ $\Rightarrow 3=X-3$ and $-5=Y-2$ $\Rightarrow X=6$ and $Y=-3$ Thus, the new coordinates are (6, -3)...
Read More →If the origin is shifted to the point
Question: If the origin is shifted to the point (1, 2) by a translation of the axes, find the new coordinates of the point (3, -4). Solution: Let the new origin be (h, k) = (1, 2) and (x, y) = (3, -4) be the given point. Let the new coordinates be (X, Y) We use the transformation formula: $x=X+h$ and $y=Y+k$ $\Rightarrow 3=X+1$ and $-4=Y+2$ $\Rightarrow X=2$ and $Y=-6$ Thus, the new coordinates are $(2,-6)$...
Read More →Solve the following
Question: if $\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$, , then find $(x, y)$. Solution: According to the question, We have, $x+i y=\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}$ $=\left(\frac{(1+i)^{2}}{1-i^{2}}\right)^{3}-\left(\frac{(1-i)^{2}}{1-i^{2}}\right)^{3}$ $=\left(\frac{1+2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\right)^{3}-\left(\frac{1-2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\right)^{3}$ $=\left(\frac{2 \mathrm{i}}{2}\right)^{3}-\left(\frac{-2 \ma...
Read More →Let F defined on [0,1] be twice differentiable
Question: Let $F$ defined on $[0,1]$ be twice differentiable such that $\mid f^{\prime \prime}(x) \leq 1$ for all $x \in[0,1] .$ If $f(0)=f(1)$, then show that $\left|f^{\prime}(x)\right|1$ for all $x \in[0,1]$ ? Solution: As $f(0)=f(1)$ and $f$ is differentiable, hence by Rolles theorem: $f^{\prime}(c)=0$ for some $c \in[0,1]$ let us now apply LMVT (as function is twice differentiable) for point $c$ and $x \in[0,1]$, hence, $\frac{\left|f^{\prime}(x)-f(c)\right|}{x-c}=f^{\prime \prime}(d)$ $\Ri...
Read More →Find the equations of the medians of a triangle whose sides are given
Question: Find the equations of the medians of a triangle whose sides are given by the equations 3x + 2y + 6 = 0, 2x 5y + 4 = 0 and x -3y 6 = 0. Solution: The given equations are $3 x+2 y+6=0 \ldots$ (i) $2 x-5 y+4=0 \ldots$ (ii) and $x-3 y-6=0 \ldots$ (iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC Firstly, we solve the equation (i) and (ii) $3 x+2 y+6=0 \ldots(i)$ $2 x-5 y+4=0 \ldots(i i)$ Multiplying the eq. (i) by 2 and (ii) by 3, we get $6 x+4 y+12=...
Read More →Show that
Question: Show that $f(x)=x+\cos x-a$ is an increasing function on $R$ for all values of a ? Solution: We have, $f(x)=x+\cos x-a$ $f^{\prime}(x)=1-\sin x=\frac{2 \cos ^{2} x}{2}$ Now, $x \in R$ $\Rightarrow \frac{\cos ^{2} x}{2}0$ $\Rightarrow \frac{2 \cos ^{2} x}{2}0$ $\Rightarrow f^{\prime}(x)0$ Hence, $f(x)$ is an increasing function for $x \in R$...
Read More →Prove the following
Question: Evaluate $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)$, where $n \in N$ Solution: According to the question, We have, $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)=\sum_{n=1}^{13}(1+i) i^{n}$ $=(1+i)\left(1+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}+i^{12}+i^{13}\right)$ $=(1+i) \frac{i\left(i^{13}-1\right)}{i-1}$ $=(1+i) \frac{i(i-1)}{i-1}$ $=(1+i) i$ $=i+i^{2}$ $=i-1$ $\therefore \sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)=1-1$...
Read More →Find the values of b for which the function
Question: Find the values of $b$ for which the function $f(x)=\sin x-b x+c$ is a decreasing function on $R$ ? Solution: We have, $f(x)=\sin x-b x+c$ $f^{\prime}(x)=\cos x-b$ Given that $f(x)$ is on decreasing function on $R$ $\therefore \mathrm{f}^{\prime}(\mathrm{x})0$ for all $\mathrm{x} \in \mathrm{R}$ $\Rightarrow \cos x-b0$ for $a l l x \in R$ $\Rightarrow b\cos x$ for all $x \in R$ But the last value of $\cos x$ in 1 $\therefore \mathrm{b} \geq 1$...
Read More →For a positive integer n,
Question: For a positive integer n, find the value of (1 i)n(1 1/i)n. Solution: According to the question, We have, $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}=(1-i)^{n}\left(1-\frac{1}{i^{2}}\right)^{n}$ = (1 i)n(1 + i)n = (1 i2)n = 2n Therefore, (1 i)n(1 1/i)n= 2n...
Read More →Find the value(s) of a for which
Question: Find the value(s) of a for which $f(x)=x^{3}-a x$ is an increasing function on $R$ ? Solution: We have, $f(x)=x^{3-a x}$ $f^{\prime}(x)=3 x^{2}-a$ Given that $f(x)$ is on increasing function $\therefore \mathrm{f}^{\prime}(\mathrm{x}) 0$ for all $x \in \mathrm{R}$ $\Rightarrow 3 x^{2}-a0$ for all $x \in R$ $\Rightarrow \mathrm{a}3 \mathrm{x}^{2}$ for all $\mathrm{x} \in \mathrm{R}$ But the last value of $3 x^{2}=0$ for $x=0$ $\therefore a \leq 0$...
Read More →Show that
Question: Show that $f(x)=x^{2}-x \sin x$ is an increasing function on $(0, \pi / 2) ?$ Solution: We have, $f(x)=x^{2}-x \sin x$ $f^{\prime}(x)=2 x-\sin x-x \cos x$ Now, $X \in\left(0, \frac{\pi}{2}\right)$ $\Rightarrow 0 \leq \sin x \leq 1,0 \leq \cos x \leq 1$, $\Rightarrow 2 x-\sin x-x \cos x0$ $\Rightarrow f^{\prime}(x) \geq 0$ Hence, $f(x)$ is an increasing function on $\left(0, \frac{\pi}{2}\right)$....
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