Using square root table, find the square root
Question: Using square root table, find the square root1312 Solution: Using the table to find $\sqrt{2}$ and $\sqrt{41}$ $\sqrt{1312}=\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 41}$ $=2 \times 2 \sqrt{2} \times \sqrt{41}$ $=2 \times 2 \times 1.414 \times 6.4031$ $=36.222$...
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Question: Using square root table, find the square root25725 Solution: Using the table to find $\sqrt{3}$ and $\sqrt{7}$ $\sqrt{25725}=\sqrt{3 \times 5 \times 5 \times 7 \times 7 \times 7}$ $=\sqrt{3} \times 5 \times 7 \times \sqrt{7}$ $=1.732 \times 5 \times 7 \times 2.646$ $=160.41$...
Read More →If x = a cos3θ and y = b sin3θ, prove that
Question: If $x=a \cos ^{3} \theta$ and $y=b \sin ^{3} \theta$, prove that $\left(\frac{x}{a}\right)^{2 / 3}+\left(\frac{y}{b}\right)^{2 / 3}=1$ Solution: We have $x=a \cos ^{3} \theta$ $=\frac{x}{a}=\cos ^{3} \theta \quad \ldots$ (i) Again, $y=b \sin ^{3} \theta$ $=\frac{y}{b}=\sin ^{3} \theta \quad \ldots$ (ii) Now, LHS $=\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}$ $=\left(\cos ^{3} \theta\right)^{\frac{2}{3}}+\left(\sin ^{3} \theta\right)^{\frac{2}{3}} \quad...
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Question: Using square root table, find the square root6929 Solution: Using the table to find $\sqrt{41}$ $\sqrt{6929}=\sqrt{169} \times \sqrt{41}$ $=13 \times 6.4031$ $=83.239$...
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Question: Using square root table, find the square root3509 Solution: Using the table to find $\sqrt{29}$ $\sqrt{3509}=\sqrt{121} \times \sqrt{29}$ $=11 \times 5.3851$ $=59.235$...
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Question: Using square root table, find the square root8700 Solution: Using the table to find $\sqrt{3}$ and $\sqrt{29}$ $\sqrt{8700}=\sqrt{3} \times \sqrt{29} \times \sqrt{100}$ $=1.7321 \times 5.385 \times 10$ $=93.27$...
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Question: If $\left(\frac{x}{a} \sin \theta-\frac{y}{b} \cos \theta\right)=1$ and $\left(\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta\right)=1$, prove that $\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=2$ Solution: We have $\left(\frac{x}{a} \sin \theta-\frac{y}{b} \cos \theta\right)=1$ Squaring both side, we have: $\left(\frac{x}{a} \sin \theta-\frac{y}{b} \cos \theta\right)^{2}=(1)^{2}$ $\Rightarrow\left(\frac{x^{2}}{a^{2}} \sin ^{2} \theta+\frac{y^{2}}{b^{2}} \cos ^{2} \theta-2 \fr...
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Question: Using square root table, find the square root540 Solution: Using the table to find $\sqrt{3}$ and $\sqrt{5}$ $\sqrt{540}=\sqrt{54} \times \sqrt{10}$ $=2 \times 3 \sqrt{3} \times \sqrt{5}$ $=2 \times 3 \times 1.732 \times 2.2361$ $=23.24$...
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Question: Using square root table, find the square root198 Solution: Using the table to find $\sqrt{2}$ and $\sqrt{11}$ $\sqrt{198}=\sqrt{2} \times \sqrt{9} \times \sqrt{11}$ $=1.414 \times 3 \times 3.317$ $=14.070$...
Read More →The value of
Question: The value of $(256)^{0.16} \times(256)^{0.09}$ is (a) 4 (b) 16 (c) 64 (d) $256.25$ Solution: (a) (a) $(256)^{0.16} \times(256)^{0.09}=(256)^{\frac{16}{100}} \times(256)^{\frac{9}{100}}$ $=(256)^{\frac{16}{100}}+\frac{9}{100}$ $\left[\because x^{a} \cdot x^{b}=x^{a+b}\right]$ $=(256)^{\frac{25}{100}}=(256)^{\frac{1}{4}}$ $=\left(4^{4}\right)^{\frac{1}{4}}=4$ $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$...
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Question: Using square root table, find the square root82 Solution: Using the table to find $\sqrt{2}$ and $\sqrt{41}$ $\sqrt{82}=\sqrt{2} \times \sqrt{41}$ $=1.414 \times 6.403$ $=9.055$...
Read More →The value of
Question: The value of $\sqrt[4]{(81)^{-2}}$ is (a) $\frac{1}{9}$ (b) $\frac{1}{3}$ (c) 9 (d) $\frac{1}{81}$ Solution: (a) $\sqrt[4]{(81)^{-2}}=\sqrt[4]{\frac{1}{(81)^{2}}}=\frac{1}{(81)^{2 / 4}}=\frac{1}{(81)^{1 / 2}}$ $\left[\because \sqrt[m]{a^{n}}=a^{n / m}\right]$ $=\frac{1}{\left(9^{2}\right)^{1 / 2}}=\frac{1}{9^{2 / 2}}=\frac{1}{9}$ $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$...
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Question: Using square root table, find the square root74 Solution: Using the table to find $\sqrt{2}$ and $\sqrt{37}$ $\sqrt{74}=\sqrt{2} \times \sqrt{37}$ $=1.414 \times 6.083$ $=8.602$...
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Question: Using square root table, find the square root15 Solution: Using the table to find $\sqrt{3}$ and $\sqrt{5}$ $\sqrt{15}=\sqrt{3} \times \sqrt{5}$ $=1.732 \times 2.236$ $=3.873$...
Read More →If x = a sec θ + b tan θ and y = a tan θ + b sec θ,
Question: Ifx=asec +btan andy=atan +bsec , prove that (x2y2) = (a2b2). Solution: We have $x^{2}-y^{2}=\left[(a \sec \theta+b \tan \theta)^{2}-(a \tan \theta+b \sec \theta)^{2}\right]$ $=\left(a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a b \sec \theta \tan \theta\right)-\left(a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta+2 a b \tan \theta \sec \theta\right)$ $=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta$ $=\left(a^{2} \sec ^{2} \theta-a^{2} \ta...
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Question: The product $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$ equals to (a) $\sqrt{2}$ (b) 2 (c) $\sqrt[12]{2}$ (d) $\sqrt[12]{32}$ Thinking Process Take the LCM of indices of three irrational numbers. Then, convert all individually in the form whose index will be equal to LCM. Solution: (b) LCM of 3,4 and $12=12$ $\sqrt[3]{2}=\sqrt[12]{2^{4}}$ $\left[\because \sqrt[m]{a}=\sqrt[m n]{a^{n}}\right]$ $\sqrt[4]{2}=\sqrt[12]{2^{3}}$ and $\sqrt[12]{32}=\sqrt[12]{2^{5}}$ $\therefore$ Produc...
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Question: The product $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$ equals to (a) $\sqrt{2}$ (b) 2 (c) $\sqrt[12]{2}$ (d) $\sqrt[12]{32}$ Thinking Process Take the LCM of indices of three irrational numbers. Then, convert all individually in the form whose index will be equal to LCM. Solution: (b) LCM of 3,4 and $12=12$ $\sqrt[3]{2}=\sqrt[12]{2^{4}}$ $\left[\because \sqrt[m]{a}=\sqrt[m n]{a^{n}}\right]$ $\sqrt[4]{2}=\sqrt[12]{2^{3}}$ and $\sqrt[12]{32}=\sqrt[12]{2^{5}}$ $\therefore$ Produc...
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Question: Using square root table, find the square root7 Solution: From the table, we directly find that the square root of 7 is 2.646....
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Question: Prove that: $\left|\begin{array}{ccc}b+c a-b a \\ c+a b-c b \\ a+b c-a c\end{array}\right|=3 a b c-a^{3}-b-c^{3}$ Solution: Let $\mathrm{LHS}=\Delta=\mid b+c a-b \quad a$ $c+a b-c b$ $a+b c-a c$ $\Delta=(b+c) \mid b-c \quad b$ $c-a c|-(a-b)| c+a b$ $a+b c|+a| c+a b-c$ $a+b c-a \mid \quad[$ Expanding $]$ $=(b+c)\left\{b c-c^{2}-b c+a b\right\}-(a-b)\left\{c^{2}+a c-a b-b^{2}\right\}+a\left\{c^{2}-a^{2}-a b+a c-b^{2}+b c\right\}$ $=b c^{2}-c^{3}+a b c-a c^{2}-a^{2} c+a^{2} b+a b^{2}+b c^...
Read More →If a cos θ + b sin θ = m and a sin θ − b cos θ
Question: Ifacos +bsin =mandasin bcos =n, prove that (m2+n2) = (a2+b2). Solution: We have $m^{2}+n^{2}=\left[(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}\right]$ $=\left(a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \cos \theta \sin \theta\right)+\left(a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta\right)$ $=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta$ $=\left(a^{2} \cos ^{2} \theta+a^{2} \s...
Read More →Show that none of the following is an identity:
Question: Show that none of the following is an identity:(i) cos2 + cos = 1(ii) sin2 + sin = 2(iii) tan2 + sin = cos2 Solution: (i) $\cos ^{2} \theta+\cos \theta=1$ $\mathrm{LHS}=\cos ^{2} \theta+\cos \theta$ $=1-\sin ^{2} \theta+\cos \theta$ $=1-\left(\sin ^{2} \theta-\cos \theta\right)$ Since $\mathrm{LHS} \neq \mathrm{RHS}$, this is not an identity. (ii) $\sin ^{2} \theta+\sin \theta=1$ $\mathrm{LHS}=\sin ^{2} \theta+\sin \theta$ $=1-\cos ^{2} \theta+\sin \theta$+ $=1-\left(\cos ^{2} \theta-\...
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Question: $\sqrt[4]{\sqrt[3]{2^{2}}}$ equals to (a) $2^{-\frac{1}{6}}$ (b) $2^{-6}$ (c) $2^{\frac{1}{6}}$ (d) $2^{6}$ Solution: (c) $\sqrt[4]{\sqrt[3]{2^{2}}}=\left[\sqrt[3]{2^{2}}\right]^{1 / 4}=\left[\left(2^{2}\right)^{1 / 3}\right]^{1 / 4}$ $\left[\because \sqrt[n]{a}=a^{1 / n}\right]$ $=\left[2^{2 / 3}\right]^{1 / 4}=2^{\frac{2}{3} \cdot \frac{1}{4}}=2^{\frac{1}{6}}$ $\left[\because\left(a^{m}\right)^{n}=a^{m n}\right]$...
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Question: If $\sqrt{2}=1.4142 \ldots$, then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to (a) $2.4142 \ldots$ (b) $5.8282 \ldots$ (c) $0.4142 \ldots$ (d) $0.1718 \ldots$ Solution: $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1}}$ [inside the root, multiplying numerator and denominator by $(\sqrt{2}-1)]$ $=\sqrt{\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2})^{2}-(1)^{2}}}$ [using identity $(a-b)(a+b)=a^{2}-b^{2}$ ] $=\sqrt{\frac{(\sqrt{2}-1)^...
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Question: Prove that $\frac{\cot ^{2} \theta(\sec \theta-1)}{(1+\sin \theta)}+\frac{\sec ^{2} \theta(\sin \theta-1)}{(1+\sec \theta)}=0$ Solution: $\mathrm{LHS}=\frac{\cot ^{2} \theta(\sec \theta-1)}{(1+\sin \theta)}+\frac{\sec ^{2} \theta(\sin \theta-1)}{(1+\sec \theta)}$ $=\frac{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\left(\frac{1}{\cos \theta}-1\right)}{(1+\sin \theta)}+\frac{\frac{1}{\cos ^{2} \theta}(\sin \theta-1)}{\left(1+\frac{1}{\cos \theta}\right)}$ $=\frac{\frac{\cos ^{2} \theta}{\s...
Read More →The value of
Question: The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is (a) $\sqrt{2}$ (b) 2 (c) 4 (d) 8 Solution: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}}$ $=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}=2$...
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