Solve this following
Question: Show that $\mathrm{AB} \neq \mathrm{BA}$ in each of the following cases: $A=\left[\begin{array}{lll}1 2 3 \\ 0 1 0 \\ 1 1 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1 1 0 \\ 0 -1 1 \\ 2 3 4\end{array}\right]$ Solution: Given : $\mathrm{A}=\left[\begin{array}{lll}1 2 3 \\ 0 1 0 \\ 1 1 0\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}-1 1 0 \\ 0 -1 1 \\ 2 3 4\end{array}\right]$ Matrix A is of order $3 \times 3$, and Matrix $B$ is of order $3 \times 3$ To show : matr...
Read More →The value of
Question: The value of current in the $6 \Omega$ resistance is :4A8A10A6ACorrect Option: , 3 Solution: $\Rightarrow 10 \mathrm{~V}+12 \mathrm{~V}-1080+3 \mathrm{~V}-420=0$ $\Rightarrow \mathrm{V}=60$ $\therefore$ current in $6 \Omega=\frac{\mathrm{V}-0}{6}=10 \mathrm{~A}$ Hence option 3 ....
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{\sin x(2+3 \cos x)} d x$ Solution: To solve this type of solution, we are going to substitute the value of $\sin x$ and $\cos x$ in terms of $\tan (x / 2)$ $\sin x=\frac{2\left[\tan \frac{x}{2}\right]}{1+\tan ^{2} \frac{x}{2}}$ $\cos x=\frac{\left(1-\frac{\tan ^{2} x}{2}\right)}{1+\frac{\tan ^{2} x}{2}}$ $I=\int \frac{1}{\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\left(2+3 \cdot \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$ $I=\int ...
Read More →Solve this following
Question: Show that $A B \neq B A$ in each of the following cases: $A=\left[\begin{array}{cc}5 -1 \\ 6 7\end{array}\right]$ and $B=\left[\begin{array}{ll}2 1 \\ 3 4\end{array}\right]$ Solution: Given : $A=\left[\begin{array}{cc}5 -1 \\ 6 7\end{array}\right]$ and $B=\left[\begin{array}{cc}2 1 \\ 3 4\end{array}\right]$ Matrix A is of order $2 \times 2$ and Matrix B is of order $2 \times 2$ To show : matrix $\mathrm{AB} \neq \mathrm{BA}$ Formula used: Where $c_{i j}=a_{i 1} b_{1 j}+a_{i 2} b_{2 j}+...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x \sqrt{\frac{1-x}{1+x}} d x$ Solution: Let, $x=\sin t$ Differentiate both side with respect to t $\frac{d x}{d t}=\cos t \Rightarrow d x=\cos t \mathrm{dt}$ $y=\int \sin t \sqrt{\frac{1-\sin t}{1+\sin t}} \cos t d t$ $y=\int \sin t \sqrt{\frac{(1-\sin t)(1-\sin t)}{(1+\sin t)(1-\sin t)}} \cos t d t$ $y=\int \sin t(1-\sin t) d t$ $y=\int \sin t d t-\int \sin ^{2} t d t$ $y=-\cos t-\int \frac{1-\cos 2 t}{2} d t$ $y=-\cos t-\left(\frac{t}{2}-\frac{\sin 2 t}{4}\right)+c$ Ag...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1+x^{2}}{\sqrt{1+x^{2}}} d x$ Solution: $y=\int \sqrt{1+x^{2}} d x$ Use formula $\sqrt{a^{2}+x^{2}}=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}+a^{2}}\right)$ $y=\frac{x}{2} \sqrt{x^{2}+1}+\frac{1}{2} \ln \left(x+\sqrt{x^{2}+1}\right)+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{x^{5}}{\sqrt{1+x^{3}}} d x$ Solution: Let, $1+\mathrm{x}^{3}=\mathrm{t}$ Differentiate both side with respect to $t$ $3 x^{2} \frac{d x}{d t}=1 \Rightarrow x^{2} d x=\frac{d t}{3}$ $y=\frac{1}{3} \int \frac{(t-1)}{\sqrt{t}} d t$ $y=\frac{1}{3} \int \sqrt{t}-\frac{1}{\sqrt{t}} d t$ $y=\frac{1}{3}\left(\frac{2}{3} t^{\frac{3}{2}}-2 \sqrt{t}\right)+c$ Again, put $t=1+x^{3}$ $y=\frac{1}{3}\left(\frac{2}{3}\left(1+x^{3}\right)^{\frac{3}{2}}-2 \sqrt{1+x^{3}}\right)+c$ $y...
Read More →Prove that the curves
Question: Prove that the curvesy2= 4xandx2+y2 6x+ 1 = 0 touch each other at the point (1, 2). Solution: Given curve equations are:y2= 4x .(1) andx2+y2 6x+ 1 = 0 .. (i) Now, differentiating (i) w.r.t. x, we get 2y.(dy/dx) = 4 ⇒ dy/dx = 2/y Slope of tangent at (1, 2), m1= 2/2 = 1 Differentiating (ii) w.r.t. x, we get 2x + 2y.(dy/dx) 6 = 0 2y. dy/dx = 6 2x ⇒ dy/dx = (6 2x)/ 2y Hence, the slope of the tangent at the same point (1, 2) ⇒ m2= (6 2 x 1)/ (2 x 2) = 4/4 = 1 Its seen that m1= m2= 1 at the ...
Read More →Find the angle of intersection of the curves
Question: Find the angle of intersection of the curvesy= 4 x2andy=x2. Solution: The given curves arey= 4 x2. (i) andy=x2 (ii) And, we know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection. Now, differentiating equations (i) and (ii) w.r.t x, we have dy/dx = -2x ⇒ m1= -2x m1is the slope of the tangent to the curve (i). and dy/dx = 2x ⇒ m2= 2x m2is the slope of the tangent to the curve (ii). So, m1= -2x and m...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{x^{2}}{\sqrt{1-x}} d x$ Solution: Let, $x=\sin ^{2} t$ Differentiate both side with respect to t $\frac{d x}{d t}=2 \sin t \cos t d t \Rightarrow \mathrm{dx}=2 \sin \mathrm{t} \cos \mathrm{tdt}$ $y=\int \frac{\sin ^{4} t}{\cos t} 2 \sin t \cos t d t$ $y=2 \int \sin ^{5} t d t$ $y=2 \int\left(1-\cos ^{2} t\right)^{2} \sin t d t$ Let, $\cos t=z$ Differentiate both side with respect to $z$ $-\sin t \frac{d t}{d z}=1 \Rightarrow \sin \mathrm{tdt}=-\mathrm{dz}$ $y=-2 \i...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{x \sqrt{1+x^{n}}} d x$ Solution: Let, $\sqrt{1+x^{n}}=t$ Differentiate both side with respect to $\mathrm{t}$ $\frac{n x^{n-1}}{2 \sqrt{1+x^{n}}} \frac{d x}{d t}=1 \Rightarrow \frac{d x}{x \sqrt{1+x^{n}}}=\frac{2 d t}{n\left(t^{2}-1\right)}$ $y=\int \frac{2}{n\left(t^{2}-1\right)} d t$ Use formula $\int \frac{1}{t^{2}-a^{2}} d t=\frac{1}{2 a} \ln \left(\frac{t-a}{t+a}\right)$ $y=\frac{1}{n} \ln \left(\frac{t-1}{t+1}\right)+c$ Again put $t=\sqrt{1+x^{n}}$ $y=\fra...
Read More →Solve this following
Question: Compute $\mathrm{AB}$ and $\mathrm{BA}$, which ever exists when $A=\left[\begin{array}{cc}2 1 \\ 3 2 \\ -1 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 0 1 \\ -1 2 1\end{array}\right]$ Solution: Given : $A=\left[\begin{array}{cc}2 1 \\ 3 2 \\ -1 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 0 1 \\ -1 2 1\end{array}\right]$ Matrix $A$ is of order $3 \times 2$ and Matrix $B$ is of order $2 \times 3$ To find : matrices $A B$ and $B A$ Formula used : Where $c_{i j}=a_{i 1}...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x^{3}(\log x)^{2} d x$ Solution: Use method of integration by parts $y=\log ^{2} x \int x^{3} d x-\int \frac{d}{d x} \log ^{2} x\left(\int x^{3} d x\right) d x$ $y=\log ^{2} x \frac{x^{4}}{4}-\int \frac{2 \log x}{x} \frac{x^{4}}{4} d x$ $y=\frac{x^{4}}{4} \log ^{2} x-\frac{1}{2}\left(\log x \int x^{3} d x-\int \frac{d}{d x} \log x\left(\int x^{3} d x\right) d x\right.$ $y=\frac{x^{4}}{4} \log ^{2} x-\frac{1}{2}\left(\log x \frac{x^{4}}{4}-\int \frac{1}{x} \frac{x^{4}}{4}...
Read More →Find the co-ordinates of the point on the curve
Question: Find the co-ordinates of the point on the curve x+ y= 4 at which tangent is equally inclined to the axes. Solution: Equation of the curve is given by, x+ y= 4 Now, let (x1, y1) be he required point on the curve So, x1+ y1= 4 On differentiating on both the sides w.r.t. x1, we get $\frac{d}{d x_{1}} \sqrt{x_{1}}+\frac{d}{d x_{1}} \sqrt{y_{1}}=\frac{d}{d x_{1}}(4)$ $\frac{1}{2 \sqrt{x_{1}}}+\frac{1}{2 \sqrt{y_{1}}} \cdot \frac{d y_{1}}{d x_{1}}=0$ $\Rightarrow \frac{1}{\sqrt{x_{1}}}+\frac...
Read More →Find the co-ordinates of the point on the curve
Question: Find the co-ordinates of the point on the curve x+ y= 4 at which tangent is equally inclined to the axes. Solution: Equation of the curve is given by, x+ y= 4 Now, let (x1, y1) be he required point on the curve So, x1+ y1= 4 On differentiating on both the sides w.r.t. x1, we get $\frac{d}{d x_{1}} \sqrt{x_{1}}+\frac{d}{d x_{1}} \sqrt{y_{1}}=\frac{d}{d x_{1}}(4)$ $\frac{1}{2 \sqrt{x_{1}}}+\frac{1}{2 \sqrt{y_{1}}} \cdot \frac{d y_{1}}{d x_{1}}=0$ $\Rightarrow \frac{1}{\sqrt{x_{1}}}+\frac...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\log (1-x)}{x^{2}} d x$ Solution: Use method of integration by parts $y=\log (1-x) \int \frac{1}{x^{2}} d x-\int \frac{d}{d x} \log (1-x)\left(\int \frac{1}{x^{2}} d x\right) d x$ $y=-\log (1-x) \frac{1}{x}-\int \frac{1}{(1-x) x} d x$ $y=-\frac{1}{x} \log (1-x)-\int \frac{x+(1-x)}{(1-x) x} d x$ $y=-\frac{1}{x} \log (1-x)-\int \frac{1}{(1-x)}+\frac{1}{x} d x$ $y=-\frac{1}{x} \log (1-x)+\log (1-x)-\log x+c$ $y=\left(1-\frac{1}{x}\right) \log (1-x)-\log x+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\log x}{x^{3}} d x$ Solution: Use method of integration by parts $y=\log x \int \frac{1}{x^{3}} d x-\int \frac{d}{d x} \log x\left(\int \frac{1}{x^{3}} d x\right) d x$ $y=-\log x \frac{1}{2 x^{2}}+\int \frac{1}{2 x^{3}} d x$ $y=-\frac{1}{2 x^{2}} \log x-\frac{1}{4 x^{2}}+c$ $y=-\frac{1}{4 x^{2}}(2 \log x+1)+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \log \left(x+\sqrt{x^{2}+a^{2}}\right) d x$ Solution: Use method of integration by parts $y=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \int d x-\int \frac{d}{d x} \log \left(x+\sqrt{x^{2}+a^{2}}\right)\left(\int d x\right) d x$ $y=x \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\int \frac{1+\frac{2 x}{2 \sqrt{x^{2}+a^{2}}}}{x+\sqrt{x^{2}+a^{2}}} x d x$ $y=x \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\int \frac{x}{\sqrt{x^{2}+a^{2}}} d x$ Let, $\mathrm{x}^{2}+\mathrm{a}^{2}=\mathrm{t}$ ...
Read More →Prove that the curves xy = 4
Question: Prove that the curvesxy= 4 andx2+y2= 8 touch each other. Solution: Given curves are equations of two circles, xy= 4 .. (i) and x2+y2= 8 . (ii) Different equation (i) w.r.t., x $x \cdot \frac{d y}{d x}+y \cdot 1=0$ $\Rightarrow \frac{d y}{d x}=-\frac{y}{x} \Rightarrow m_{1}=-\frac{y}{x}$ $\ldots($ iii $)$ where, $m_{1}$ is the slope of the tangent to the curve. Differentiating eq. (ii) w.r.t. $x$ $2 x+2 y \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y} \Rightarrow m_{2}...
Read More →Find the condition that the curves
Question: Find the condition that the curves 2x=y2and 2xy=kintersect orthogonally. Solution: Its seen that the given curves are equation of two circles. 2x=y2.. (1) and 2xy=k ..(2) We know that, two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90o. Now, differentiating equations (1) and (2) w.r.t. t, we get $2.1=2 y \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{y} \Rightarrow m_{1}=\frac{1}{y}$ $\left...
Read More →Evaluate the following integrals:
Question: Evaluate $\int(x+1)^{2} e^{x} d x$ Solution: $y=\int\left(x^{2}+2 x+1\right) e^{x} d x$ $y=\int\left(x^{2}+2 x\right) e^{x} d x+\int e^{x} d x$ We know that $\int\left(f(x)+f^{\prime}(x)\right) e^{x} d x=f(x) e^{x}$ Here, $f(x)=x^{2}$ then $f^{\prime}(x)=2 x$ $y=x^{2} e^{x}+e^{x}+c$ $y=\left(x^{2}+1\right) e^{x}+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x \sin ^{3} x d x$ Solution: We know that $\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$ $y=\int x\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x$ $y=\frac{3}{4} \int x \sin x d x-\frac{1}{4} \int x \sin 3 x d x$ Use method of integration by parts $y=\frac{3}{4}\left(x \int \sin x d x-\int \frac{d}{d x} x\left(\int \sin x d x\right) d x\right)$ $-\frac{1}{4}\left(x \int \sin 3 x d x-\int \frac{d}{d x} x\left(\int \sin 3 x d x\right) d x\right) y$ $=\frac{3}{4}\left(-x \cos x+\in...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x \sec ^{2} 2 x d x$ Solution: Use method of integration by parts $y=x \int \sec ^{2} 2 x d x-\int \frac{d}{d x} x\left(\int \sec ^{2} 2 x d x\right) d x$ $y=x \frac{\tan 2 x}{2}-\int \frac{\tan 2 x}{2} d x$ Use formula $\int \tan x d x=\log \sec x$ $y=\frac{x}{2} \tan 2 x-\frac{\log (\sec 2 x)}{4}+c$...
Read More →x and y are the sides of two squares such that
Question: xandyare the sides of two squares such thaty=xx2. Find the rate of change of the area of second square with respect to the area of first square. Solution: Lets consider the area of the first square A1= x2 And, area of the second square be A2= y2 Now, A1= x2and A2= y2= (x x2)2 Differentiating both A1and A2w.r.t. t, we get $\frac{d \mathrm{~A}_{1}}{d t}=2 x \cdot \frac{d x}{d t}$ and $\frac{d \mathrm{~A}_{2}}{d t}=2\left(x-x^{2}\right)(1-2 x) \cdot \frac{d x}{d t}$ Thus, $\frac{d \mathrm...
Read More →The volume of a cube increases at a constant rate.
Question: The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side. Solution: Lets assume x to be the length of the cube. So, the volume of the cube V = x3. (1) Given that, dV/dt = K Now, on differentiating the equation (1) w.r.t. t, we get dV/dt = 3x2. dx/dt = K (constant) So, dx/dt = K/3x2 Now, Surface area of the cube, S = 6x2 Differentiating both sides w.r.t. t, we get $\frac{\overline{d s}}{d t}=6 \cdot 2 \cdot...
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