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Question: Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x,[x]^{2}+2[x+2]-7=0$ has :no integral solution exactly four integral solutions exactly two solutions infinitely many solutionsCorrect Option: , 4 Solution: $[x]^{2}+2[x+2]-7=0$ $\Rightarrow[x]^{2}+2[x]+4-7=0$ $\Rightarrow[x]=1,-3$ $\Rightarrow x \in[1,2) \cup[-3,-2)$...
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Question: If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta \text { isin } \theta \\ i \sin \theta \cos \theta\end{array}\right], \quad\left(\theta=\frac{\pi}{24}\right)$ and $\mathrm{A}^{5}=\left[\begin{array}{ll}\mathrm{a} \mathrm{b} \\ \mathrm{c} \mathrm{d}\end{array}\right]$, where $\mathrm{i}=\sqrt{-1}$, then which one of the following is not true? $0 \leq a^{2}+b^{2} \leq 1$$\mathrm{a}^{2}-\mathrm{d}^{2}=0$$a^{2}-b^{2}=\frac{1}{2}$$a^{2}-c^{2}=1$Correct Option: , 3 Solution: $A^{2}=\left(\...
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Question: If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta \text { isin } \theta \\ i \sin \theta \cos \theta\end{array}\right], \quad\left(\theta=\frac{\pi}{24}\right)$ and $\mathrm{A}^{5}=\left[\begin{array}{ll}\mathrm{a} \mathrm{b} \\ \mathrm{c} \mathrm{d}\end{array}\right]$, where $\mathrm{i}=\sqrt{-1}$, then which one of the following is not true? 0 \leq a^{2}+b^{2} \leq 1$$\mathrm{a}^{2}-\mathrm{d}^{2}=0$$a^{2}-b^{2}=\frac{1}{2}$$a^{2}-c^{2}=1$Correct Option: , 3 Solution: $A^{2}=\left(\b...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sqrt{a^{2}-x^{2}} d x$ Solution: Let, $x=a \sin t$ Differentiate both side with respect to $t$ $\frac{d x}{d t}=a \cos t \Rightarrow \mathrm{dx}=\mathrm{a} \cos \mathrm{t} \mathrm{dt}$ $y=\int \sqrt{a^{2}-(a \sin t)^{2}} a \cos t d t$ $y=\int(a \cos t)(a \cos t) d t$ $y=\int a^{2} \cos ^{2} t d t$ $y=\int a^{2}\left(\frac{1+\cos 2 t}{2}\right) d t$ $y=\frac{a^{2}}{2} \int 1+\cos 2 t d t$ $y=\frac{a^{2}}{2}\left(t+\frac{\sin 2 t}{2}\right)+c$ Again, put $t=\sin ^{-1} \fr...
Read More →Evaluate the following integrals:
Question: $\int \tan ^{5} x d x$ Solution: $\int \tan ^{5} x d x$ We can write above integral as: $\int \tan ^{5} x d x=\int\left(\tan ^{3} x\right)\left(\tan ^{2} x\right) d x \cdots$ (Splitting $\tan ^{5} \mathrm{x}$ ) $=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$ (Using $\tan ^{2} x=\sec ^{2} x-1$ ) $=\int \sec ^{2} x\left(\tan ^{3} x\right) d x-\int\left(\tan ^{3} x\right) d x$ $=\int \sec ^{2} x\left(\tan ^{3} x\right) d x-\int\left(\tan ^{2} x\right)(\tan x) d x-\left(\right.$ Splittin...
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Question: Find the value of (x + y) from the following equation : $2\left[\begin{array}{ll}1 3 \\ 0 x\end{array}\right]+\left[\begin{array}{ll}y 0 \\ 1 2\end{array}\right]=\left[\begin{array}{ll}5 6 \\ 1 8\end{array}\right]$ Solution: Given $2\left[\begin{array}{ll}1 3 \\ 0 x\end{array}\right]+\left[\begin{array}{ll}y 0 \\ 1 2\end{array}\right]=\left[\begin{array}{ll}5 6 \\ 1 8\end{array}\right]$ ${\left[\begin{array}{cc}2 6 \\ 0 2 x\end{array}\right]+\left[\begin{array}{ll}y 0 \\ 1 2\end{array}...
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Question: Find the value of x and y, when i. $\left[\begin{array}{l}x+y \\ x-y\end{array}\right]=\left[\begin{array}{l}8 \\ 4\end{array}\right]$ Solution: If $\left[\begin{array}{ll}a b \\ c d\end{array}\right]=\left[\begin{array}{ll}e f \\ g h\end{array}\right]$ Then a=e, b=f, c=g, d=h Given $\left[\begin{array}{l}x+y \\ x-y\end{array}\right]=\left[\begin{array}{l}8 \\ 4\end{array}\right]$ So, $x+y=8$ and $x-y=4$ Adding these two gives $2 x=12$ $\Rightarrow x=6$ $y=2$ Conclusion : $x=6$ and $y=...
Read More →Different w.r.t tan-1 x when x ≠ 0.
Question: Different w.r.t tan-1x when x 0. Solution: Let $y=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $z=\tan ^{-1} x$ Now, put $x=\tan \theta$ $\therefore y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)$ and $z=\tan ^{-1}(\tan \theta)=\theta$ So, $\quad \tan \left(\frac{\sqrt{\sec \theta}-1}{\tan }\right)=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$ $\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)=\tan ^{-...
Read More →Evaluate the following integrals:
Question: $\int \tan ^{4} x d x$ Solution: $\int \tan ^{4} x d x$ We can write above integral as: $\int \tan ^{4} x d x=\int\left(\tan ^{2} x\right)\left(\tan ^{2} x\right) d x \cdots\left(\right.$ Splitting $\left.\tan ^{4} x\right)$ $=\int\left(\sec ^{2} x-1\right) \tan ^{2} x d x\left(U \operatorname{sing} \tan ^{2} x=\sec ^{2} x-1\right)$ Considering integral (1) Let $u=\tan x$ $d u=\sec ^{2} x d x$ Substituting values we get, $\int \sec ^{2} x\left(\tan ^{2} x\right) d x=\int u^{2} d u=\fra...
Read More →Differentiate x/sinx w.r.t sin x.
Question: Differentiate x/sinx w.r.t sin x. Solution: Let $y=\frac{x}{\sin x}$ and $z=\sin x$. Differentiating both the parametric functions w.r.t. $x$, $\frac{d y}{d x}=\frac{\sin x \cdot \frac{d}{d x}(x)-x \cdot \frac{d}{d x}(\sin x)}{(\sin x)^{2}}$ $=\frac{\sin x \cdot 1-x \cdot \cos x}{\sin ^{2} x}=\frac{\sin x-x \cos x}{\sin ^{2} x}$ $\frac{d z}{d x}=\cos x$ Now, $\frac{d y}{d z}=\frac{d y / d x}{d z / d x}=\frac{\frac{\sin x-x \cos x}{\sin ^{2} x}}{\cos x}=\frac{\sin x-x \cos x}{\sin ^{2} ...
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Question: If $A=\operatorname{diag}[2,-5,9], B=\operatorname{diag}[-3,7,14]$ and $C=\operatorname{diag}[4,-6,3]$, find: (i) $A+2 B$ (ii) $B+C-A$ Solution: If Z = diag[a,b,c], then we can write it as $Z=\left[\begin{array}{lll}a 0 0 \\ 0 b 0 \\ 0 0 c\end{array}\right]$ So, $A+2 B=\left[\begin{array}{ccc}2 0 0 \\ 0 -5 0 \\ 0 0 9\end{array}\right]+2\left(\left[\begin{array}{ccc}-3 0 0 \\ 0 7 0 \\ 0 0 14\end{array}\right]\right)$ $=\left[\begin{array}{ccc}2 0 0 \\ 0 -5 0 \\ 0 0 9\end{array}\right]+\...
Read More →If x = 3sint – sin 3t, y = 3cost – cos 3t, find
Question: Ifx= 3sint sin 3t,y= 3cost cos 3t, find $\frac{d y}{d x}$ at $t=\frac{\pi}{3}$ Solution: Given, x= 3sint sin 3t,y= 3cost cos 3t Now, differentiating both the parametric functions w.r.t t, we have $\frac{d x}{d t}=3 \cos t-\cos 3 t .3=3(\cos t-\cos 3 t)$ $\frac{d y}{d t}=-3 \sin t+\sin 3 t .3=3(-\sin t+\sin 3 t)$ So, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3(-\sin t+\sin 3 t)}{3(\cos t-\cos 3 t)}=\frac{-\sin t+\sin 3 t}{\cos t-\cos 3 t}$ Putting, $t=\frac{\pi}{3}$ $\frac{d y}...
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Question: If $\mathrm{A}=\left[\begin{array}{rrr}1 -3 2 \\ 2 0 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}2 -1 -1 \\ 1 0 -1\end{array}\right]$, find a matrix $\mathrm{C}$ such that $(\mathrm{A}+\mathrm{B}+\mathrm{C})$ is a zero matrix. Solution: Given A+B+C is zero matrix i.e A+B+C = 0 $\left[\begin{array}{ccc}1 -3 2 \\ 2 0 2\end{array}\right]+\left[\begin{array}{ccc}2 -1 -1 \\ 1 0 -1\end{array}\right]+C=0$ $C=-\left[\begin{array}{ccc}1 -3 2 \\ 2 0 2\end{array}\right]-\left[\be...
Read More →Find the matrix X such that
Question: Find the matrix X such that 2A B + X = O, where $\mathrm{A}=\left[\begin{array}{ll}3 1 \\ 0 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}-2 1 \\ 0 3\end{array}\right]$ Solution: Given 2A B + X = 0 $2\left(\left[\begin{array}{ll}3 1 \\ 0 2\end{array}\right]\right)-\left[\begin{array}{cc}-2 1 \\ 0 3\end{array}\right]+X=0$ $X=\left[\begin{array}{cc}-2 1 \\ 0 3\end{array}\right]-2\left(\left[\begin{array}{ll}3 1 \\ 0 2\end{array}\right]\right)$ $=\left[\begin{array}{cc}-2 1 ...
Read More →If x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that
Question: Ifx=asin2t(1 + cos2t)andy=bcos2t(1cos2t), show that $\left(\frac{d y}{d x}\right)_{\text {at } t=\frac{\pi}{4}}=\frac{b}{a}$ Solution: Given, x=asin2t(1 + cos2t)andy=bcos2t(1cos2t) Differentiating both the parametric equations w.r.t t, we have $\frac{d x}{d t}=a\left[\sin 2 t \cdot \frac{d}{d t}(1+\cos 2 t)+(1+\cos 2 t) \cdot \frac{d}{d t} \sin 2 t\right]$ $=a[\sin 2 t \cdot(-\sin 2 t) \cdot 2+(1+\cos 2 t)(\cos 2 t) \cdot 2]$ $=a\left[-2 \sin ^{2} 2 t+2 \cos 2 t+2 \cos ^{2} 2 t\right]$...
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Question: If $A=\left[\begin{array}{cc}-2 3 \\ 4 5 \\ 1 -6\end{array}\right]$ and $B=\left[\begin{array}{cc}5 2 \\ -7 3 \\ 6 4\end{array}\right]$, find a matrix $C$ such that $A+B-C=0$ Solution: Given A + B C = 0 $\left[\begin{array}{cc}-2 3 \\ 4 5 \\ 1 -6\end{array}\right]+\left[\begin{array}{cc}5 2 \\ -7 3 \\ 6 4\end{array}\right]-C=0$ $\begin{aligned} C =\left[\begin{array}{cc}-2 3 \\ 4 5 \\ 1 -6\end{array}\right]+\left[\begin{array}{cc}5 2 \\ -7 3 \\ 6 4\end{array}\right] \\ C =\left[\begin{...
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Question: Find matrix $X$, if $\left[\begin{array}{rrr}3 5 -9 \\ -1 4 -7\end{array}\right]+X=\left[\begin{array}{lll}6 2 3 \\ 4 8 6\end{array}\right]$. Solution: Given $\left[\begin{array}{ccc}3 5 -9 \\ -1 4 -7\end{array}\right]+x=\left[\begin{array}{lll}6 2 3 \\ 4 8 6\end{array}\right]$ $x=\left[\begin{array}{lll}6 2 3 \\ 4 8 6\end{array}\right]-\left[\begin{array}{ccc}3 5 -9 \\ -1 4 -7\end{array}\right]$ $=\left[\begin{array}{ccc}3 -3 12 \\ 5 4 13\end{array}\right]$ Conclusion : $x=\left[\begi...
Read More →If x = ecos2t and y = esin2t, prove that dy/ dx = – y log x/ x log y.
Question: Ifx=ecos2tandy=esin2t, prove thatdy/ dx =ylogx/ xlogy. Solution: Given, x=ecos2tandy=esin2t So, cos 2t = log x and sin 2t = log y Now, differentiating both the parameter functions w.r.t t, we have $\frac{d x}{d t}=e^{\cos 2 t} \cdot \frac{d}{d t}(\cos 2 t)=e^{\cos 2 t}(-\sin 2 t) \cdot \frac{d}{d t}(2 t)$ $=-e^{\cos 2 t} \cdot \sin 2 t \cdot 2=-2 e^{\cos 2 t} \cdot \sin 2 t$ Now, $\quad y=e^{\sin 2 t}$ $\frac{d y}{d t}=e^{\sin 2 t} \cdot \frac{d}{d t}(\sin 2 t)=e^{\sin 2 t} \cdot \cos ...
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Question: Find matrices $A$ and $B$, if $2 A-B=\left[\begin{array}{rrr}6 -6 0 \\ -4 2 1\end{array}\right]$ and $2 B+A=\left[\begin{array}{ccc}3 2 5 \\ -2 1 -7\end{array}\right]$ Solution: Add 2(2A-B) and (2B+A) $2(2 A-B)+(2 B+A)=2\left(\left[\begin{array}{ccc}6 -6 0 \\ -4 2 1\end{array}\right]\right)+\left[\begin{array}{ccc}3 2 5 \\ -2 1 -7\end{array}\right]$ $5 A=\left(\left[\begin{array}{ccc}12 -12 0 \\ -8 4 2\end{array}\right]\right)+\left[\begin{array}{ccc}3 2 5 \\ -2 1 -7\end{array}\right]$...
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Question: Find matrices $A$ and $B$, if $A+B=\left[\begin{array}{ccc}1 0 2 \\ 5 4 -6 \\ 7 3 8\end{array}\right]$ and $A-B=\left[\begin{array}{ccc}-5 -4 8 \\ 11 2 0 \\ -1 7 4\end{array}\right]$ Solution: Add (A+B) and (A-B) We get $(A+B)+(A-B)=\left[\begin{array}{ccc}1 0 2 \\ 5 4 -6 \\ 7 3 8\end{array}\right]+\left[\begin{array}{ccc}-5 -4 8 \\ 11 2 0 \\ -1 7 4\end{array}\right]$ $2 \mathrm{~A}=\left[\begin{array}{ccc}-4 -4 10 \\ 16 6 -6 \\ 6 10 12\end{array}\right]$ $\mathrm{A}=\left[\begin{array...
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Question: $x=\frac{1+\log t}{t^{2}}, \quad y=\frac{3+2 \log t}{t}$ Solution: On differentiating both the given parametric functions w.r.t. t, we have $\frac{d x}{d t}=\frac{t^{2} \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t}\left(t^{2}\right)}{t^{4}}$ $=\frac{t^{2} \cdot\left(\frac{1}{t}\right)-(1+\log t) \cdot 2 t}{t^{4}}=\frac{t-(1+\log t) \cdot 2 t}{t^{4}}$ $=\frac{t[1-2-2 \log t]}{t^{4}}=\frac{-(1+2 \log t)}{t^{3}}$ $y=\frac{3+2 \log t}{t}$ Next, $\frac{d y}{d t}=$$\frac{t \cd...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \tan ^{3} \mathrm{x} \mathrm{dx}$ Solution: $\int \tan ^{3} x d x$ We can write above integral as: $\int \tan ^{3} x d x=\int\left(\tan ^{2} x\right)(\tan x) d x \cdots\left(\right.$ Splitting $\left.\tan ^{3} x\right)$ $=\int\left(\sec ^{2} x-1\right)(\tan x) d x\left(U \operatorname{sing} \tan ^{2} x=\sec ^{2} x-1\right)$ Considering integral (1) Let $u=\tan x$ $d u=\sec ^{2} x d x$ Substituting values we get, $\int \sec ^{2} x(\tan x) d x=\int u d u=\frac{u^{2}}{2}+C$...
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Question: If $5 \mathrm{~A}=\left[\begin{array}{ccc}5 10 -15 \\ 2 3 4 \\ 1 0 -5\end{array}\right]$, find $\mathrm{A}$. Solution: $5 A=\left[\begin{array}{ccc}5 10 -15 \\ 2 3 4 \\ 1 0 -5\end{array}\right]$ $A=\left[\begin{array}{ccc}\frac{5}{5} \frac{10}{5} \frac{-15}{5} \\ \frac{2}{5} \frac{3}{5} \frac{4}{5} \\ \frac{1}{5} \frac{0}{5} \frac{-5}{5}\end{array}\right]$ $A=\left[\begin{array}{ccc}1 2 -3 \\ \frac{2}{5} \frac{3}{5} \frac{4}{5} \\ \frac{1}{5} 0 -1\end{array}\right]$ Conclusion: $A=\lef...
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Question: Let $\mathrm{A}=\left[\begin{array}{ccc}0 1 -2 \\ 5 -1 -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ccc}1 -3 -1 \\ 0 -2 5\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ccc}2 -5 1 \\ -4 0 6\end{array}\right]$. Compute $5 \mathrm{~A}-3 \mathrm{~B}+4 \mathrm{C}$. Solution: $5 A-3 B+4 C=5\left(\left[\begin{array}{ccc}0 1 -2 \\ 5 -1 -4\end{array}\right]\right)-3\left(\left[\begin{array}{ccc}1 -3 -1 \\ 0 -2 5\end{array}\right]\right)+4\left(\left[\begin{array}{ccc}2 -5 1 \\ ...
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Question: Let $A=\left[\begin{array}{ll}2 4 \\ 3 2\end{array}\right], B=\left[\begin{array}{cc}1 3 \\ -2 5\end{array}\right]$ and $C=\left[\begin{array}{cc}-2 5 \\ 3 4\end{array}\right]$. Find: i. $A+2 B$ ii. $B-4 c$ iii. $A-2 B+3 C$ Solution: $A+2 B=\left[\begin{array}{ll}2 4 \\ 3 2\end{array}\right]+2\left(\left[\begin{array}{cc}1 3 \\ -2 5\end{array}\right]\right)$ $=\left[\begin{array}{ll}2 4 \\ 3 2\end{array}\right]+\left[\begin{array}{cc}2 6 \\ -4 10\end{array}\right]$ $=\left[\begin{array...
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