Fill in the blanks:
Question: Fill in the blanks: (i) $\sqrt[3]{125 \times 27}=3 \times \ldots$ (ii) $\sqrt[3]{8 \times \ldots}=8$ (iii) $\sqrt[3]{1728}=4 \times \ldots$ (iv) $\sqrt[3]{480}=\sqrt[3]{3} \times 2 \times \sqrt[3]{\ldots}$ (v) $\sqrt[3]{\cdots}=\sqrt[3]{7} \times \sqrt[3]{8}$ (vi) $\sqrt[3]{\cdots}=\sqrt[3]{4} \times \sqrt[3]{5} \times \sqrt[3]{6}$ (vii) $\sqrt[3]{\frac{27}{125}}=\frac{\ldots}{5}$ (viii) $\sqrt[3]{\frac{729}{1331}}=\frac{9}{\ldots}$ (ix) $\sqrt[3]{\frac{512}{\ldots}}=\frac{8}{13}$ Solu...
Read More →Fill in the blanks:
Question: Fill in the blanks: (i) $\sqrt[3]{125 \times 27}=3 \times \ldots$ (ii) $\sqrt[3]{8 \times \ldots}=8$ (iii) $\sqrt[3]{1728}=4 \times \ldots$ (iv) $\sqrt[3]{480}=\sqrt[3]{3} \times 2 \times \sqrt[3]{\ldots}$ (v) $\sqrt[3]{\cdots}=\sqrt[3]{7} \times \sqrt[3]{8}$ (vi) $\sqrt[3]{\cdots}=\sqrt[3]{4} \times \sqrt[3]{5} \times \sqrt[3]{6}$ (vii) $\sqrt[3]{\frac{27}{125}}=\frac{\ldots}{5}$ (viii) $\sqrt[3]{\frac{729}{1331}}=\frac{9}{\ldots}$ (ix) $\sqrt[3]{\frac{512}{\ldots}}=\frac{8}{13}$ Solu...
Read More →Prove the following
Question: $\sqrt{2}$ is a polynomial of degree (a) 2 (b) 0 (C) 1 (d) $1 / 2$ Solution: (b) $\sqrt{2}=-\sqrt{2} x^{\circ}$. Hence, $\sqrt{2}$ is a polynomial of degree 0 , because exponent of $x$ is 0 ....
Read More →Which one of the following is a polynomial?
Question: Which one of the following is a polynomial? (a) $\frac{x^{2}}{2}-\frac{2}{x^{2}}$ (b) $\sqrt{2 x}-1$ (c) $x^{2}+\frac{3 x^{3 / 2}}{\sqrt{x}}$ (d) $\frac{x-1}{x+1}$ Solution: (c) (a) Now, $\frac{x^{2}}{2}-\frac{2}{x^{2}}=\frac{x^{2}}{2}-2 x^{-2}$, it is not a polynomial, because exponent of $x$ is $-2$ which is not a whole number. (b) Now, $\sqrt{2 x}-1=\sqrt{2} x^{1 / 2}-1$, it is not a polynomial, because exponent of $x$ is $-\frac{1}{2}$ which is not a whole number. (c) Now, $x^{2}+\...
Read More →The angles of elevation of the top of a tower from two points at distances
Question: The angles of elevation of the top of a tower from two points at distances of 5 metres and 20 metres from the base of the tower and in the same straight line with it, are complementary. Find the height of the tower. Solution: Let the height of the tower be AB.We have, $\mathrm{AC}=5 \mathrm{~m}, \mathrm{AD}=20 \mathrm{~m}$ Let the angle of elevation of the top of the tower (i.e. $\angle \mathrm{ACB}$ ) from point $\mathrm{C}$ be $\theta$. Then, the angle of elevation of the top of the ...
Read More →Solve this
Question: Prove that $\left|\begin{array}{ccc}a^{2} 2 a b b^{2} \\ b^{2} a^{2} 2 a b \\ 2 a b b^{2} a^{2}\end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$ Solution: Let $\mathrm{LHS}=\Delta=\mid \begin{array}{lll}a^{2} 2 a b b^{2}\end{array}$ $b^{2} \quad a^{2} \quad 2 a b$ $\begin{array}{lll}2 a b b^{2} a^{2} \mid\end{array}$ $=a^{2} \mid a^{2} \quad 2 a b$ $b^{2} \quad a^{2}|-(2 a b)| b^{2} \quad 2 a b$ $2 a b \quad a^{2}\left|+b^{2}\right| b^{2} \quad a^{2}$ $2 a b \quad b^{2} \mid \quad[$ Exp...
Read More →Show that:
Question: Show that: (i) $\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}=\sqrt[3]{\frac{729}{1000}}$ (ii) $\frac{\sqrt[3]{-512}}{\sqrt[3]{343}}=\sqrt[3]{\frac{-512}{343}}$ Solution: (i) $L H S=\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}=\frac{\sqrt[3]{9 \times 9 \times 9}}{\sqrt[3]{10 \times 10 \times 10}}=\frac{9}{10}$ $\mathrm{RHS}=\sqrt[3]{\frac{729}{1000}}=\sqrt[3]{\frac{9 \times 9 \times 9}{10 \times 10 \times 10}}=\sqrt[3]{\frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}}=\sqrt[3]{\left(\frac{9}{10}\ri...
Read More →Find the value of
Question: Find the value of $\frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$. Solution: We have $\frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$ $=\frac{4}{\left(6^{3}\right)^{-\frac{2}{3}}}+\frac{1}{\left(16^{2}\right)^{-\frac{3}{4}}}+\frac{2}{\left(3^{5}\right)^{-\frac{1}{5}}}$ $=\frac{4}{6^{3 \times\left(-\frac{2}{3}\right)}}+\frac{1}{16^{2 \times\left(-\frac{3}{4}\right)}}+\frac{2}{3^{5 \times\left(-\...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{ccc}a b-c c-b \\ a-c b c-a \\ a-b b-a c\end{array}\right|=(a+b-c)(b+c-a)(c+a-b)$ Solution: Let LHS $=\Delta=\mid \begin{array}{lll}a b-c c-b\end{array}$ $\begin{array}{clc}a-c b c-a \\ a-b b-a c \mid\end{array}$ $\Delta=\mid a \quad 0 \quad c-b+a$ $\begin{array}{ccc}a-c b+c-a 0 \\ a-b b+c-a c+a-b \mid\end{array}$ [Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{3}$ ] $=(b+c-a...
Read More →An observer 1.5 m tall is 30 m away from a chimney.
Question: An observer 1.5 m tall is 30 m away from a chimney. The angle ofelevation of the top of the chimney from his eye is 60.Find the height of the chimney. Solution: Let CE and AD be the heights of the observer and the chimney, respectively. We have, $\mathrm{BD}=\mathrm{CE}=1.5 \mathrm{~m}, \mathrm{BC}=\mathrm{DE}=30 \mathrm{~m}$ and $\angle \mathrm{ACB}=60^{\circ}$ In $\triangle \mathrm{ABC}$, $\tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$ $\Rightarrow \sqrt{3}=\frac{\mathrm{AD}-\mathr...
Read More →Simplify the following
Question: Simplify (256) $-\left(4^{-\frac{3}{2}}\right)$ Solution: $(256)^{-\left(4^{-\frac{3}{2}}\right)}=(256)^{-(4)^{-\frac{3}{2}}}=(256)^{-\left(2^{2}\right)^{-\frac{3}{2}}} .$ $=(256)^{-\left(2^{2 x-\frac{3}{2}}\right)}=(256)^{-\left(2^{-3}\right)}$ $\left[\because b^{\left(a^{m}\right)^{n}}=b^{a^{m n}}\right]$ $=\left(2^{8}\right)^{-\left(\frac{1}{2^{3}}\right)}=\left(2^{8}\right)^{-\frac{1}{8}}$ $=2^{8 x-\frac{1}{8}}=2^{-1}=\frac{1}{2}$...
Read More →Evaluate each of the following
Question: Evaluate each of the following (i) $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$ (ii) $\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$ (iii) $\sqrt[3]{\frac{729}{216}} \times \frac{6}{9}$ (iv) $\sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}}-1$ (v) $\sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13}$ Solution: (i) To evaluate the value of the given expression, we need to proceed as follows: $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}=\sqrt[3]{3 \times 3 \t...
Read More →Evaluate each of the following
Question: Evaluate each of the following (i) $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$ (ii) $\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$ (iii) $\sqrt[3]{\frac{729}{216}} \times \frac{6}{9}$ (iv) $\sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}}-1$ (v) $\sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13}$ Solution: (i) To evaluate the value of the given expression, we need to proceed as follows: $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}=\sqrt[3]{3 \times 3 \t...
Read More →Evaluate each of the following
Question: Evaluate each of the following (i) $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$ (ii) $\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$ (iii) $\sqrt[3]{\frac{729}{216}} \times \frac{6}{9}$ (iv) $\sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}}-1$ (v) $\sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13}$ Solution: (i) To evaluate the value of the given expression, we need to proceed as follows: $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}=\sqrt[3]{3 \times 3 \t...
Read More →A kite is flying at a height of 75 m from the level ground,
Question: A kite is flying at a height of 75 m from the level ground, attached to a string inclined at a 60 to the horizontal. Find the length of the string assuming that there is no slack in it. Solution: Let $O X$ be the horizontal ground and $A$ be the position of the kite. Also, let $O$ be the position of the observer and $O A$ be the thread. Now, draw $A B \perp O X$. We have: $\angle B O A=60^{\circ}, O A=75 \mathrm{~m}$ and $\angle O B A=90^{\circ}$ Height of the kite from the ground $=A ...
Read More →Evaluate each of the following
Question: Evaluate each of the following (i) $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$ (ii) $\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$ (iii) $\sqrt[3]{\frac{729}{216}} \times \frac{6}{9}$ (iv) $\sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}}-1$ (v) $\sqrt[3]{0.1 \times 0.1 \times 0.1 \times 13 \times 13 \times 13}$ Solution: (i) To evaluate the value of the given expression, we need to proceed as follows: $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}=\sqrt[3]{3 \times 3 \t...
Read More →Prove the following
Question: If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, then find the value of $x^{2}+y^{2} ?$ Solution: Now, $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ [multiplying numerator and denominator by $\sqrt{3}+\sqrt{2}$ ] $=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$ [using identity, $(a+b)(a-b)=a^{2}-b^{2}$ ] $=\frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2 \cdot \sqrt{3} \cdot...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{ccc}1 1+p 1+p+q \\ 2 3+2 p 4+3 p+2 q \\ 3 6+3 p 10+6 p+3 q\end{array}\right|=1$ Solution: Let LHS $=\Delta=\mid \begin{array}{lll}1 1+p 1+p+q\end{array}$ $\begin{array}{lll}2 3+2 p 4+3 p+2 q \\ 3 6+3 p 10+6 p+3 q \mid\end{array}$ $=\mid \begin{array}{lll}1 1 1+\mathrm{p}\end{array}$ $\begin{array}{llll}3 6 10+6 p|+| 1 p q \\ 2 2 p 2 q \\ 3 3 p 3 q \mid \end{array}$ $\begin{array}{lllll}3 6 10+6 p |+| 1 p q \\ 2 2 p 2 q \\ 3 3 p 3 q \mid \end{array}$ $=\...
Read More →A tower stands vertically on the ground. From a point on the ground which is 20 m
Question: A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60. Find the height of the tower. Solution: LetABbe the tower standing vertically on the groundandObe the position of the observer.We now have: $O A=20 m, \angle O A B=90^{\circ}$ and $\angle A O B=60^{\circ}$ Let: $A B=h \mathrm{~m}$ Now, in the right $\Delta O A B$, we have: $\frac{A B}{O A}=\tan 60^{\circ}=\sqrt{3}$ $\Rig...
Read More →Find the cube root of each of the following rational numbers:
Question: Find the cube root of each of the following rational numbers: (i) 0.001728 (ii) 0.003375 (iii) 0.001 (iv) 1.331 Solution: (i)We have: $0.001728=\frac{1728}{1000000}$ $\therefore \sqrt[3]{0.001728}=\sqrt[3]{\frac{1728}{1000000}}=\frac{\sqrt[3]{1728}}{\sqrt[3]{1000000}}$ Now On factorising 1728 into prime factors, we get: $1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$ On grouping the factors in triples of equal factors, we get: $1728=\{2 \times 2 \times ...
Read More →Prove the following
Question: If $a=\frac{3+\sqrt{5}}{2}$, then find the value of $a^{2}+\frac{1}{a^{2}}$ Solution: Given, $a=\frac{3+\sqrt{5}}{2}$ $\ldots$ (i) Now, $\frac{1}{a}=\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}$ [multiplying numerator and denominator by $3-\sqrt{5}$ ] $=\frac{6-2 \sqrt{5}}{3^{2}-(\sqrt{5})^{2}}$ [using identity, $(a-b)(a+b)=a^{2}-b^{2}$ ] $=\frac{6-2 \sqrt{5}}{9-5}=\frac{6-2 \sqrt{5}}{4}$ $\Rightarrow$ $\frac{1}{a}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{ccc}x+4 x x \\ x x+4 x \\ x x x+4\end{array}\right|=16(3 x+4)$ Solution: Let LHS $=\Delta=\mid x+4 \quad x \quad x$ $\begin{array}{ccc}x x+4 x \\ x x x+4 \mid\end{array}$ $=\mid 3 x+4 \quad 3 x+4 \quad 3 x+4$ $\begin{array}{cccc}\mathrm{x} \mathrm{x}+4 \mathrm{x} \\ \mathrm{x} \mathrm{x} \mathrm{x}+4 \text { [Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \text { ] }\end{array}$ $=(3 x+4) \mid 1 \quad 1 \quad 1$ $\beg...
Read More →Find the cube root of each of the following rational numbers:
Question: Find the cube root of each of the following rational numbers: (i) $\frac{-125}{729}$ (ii) $\frac{10648}{12167}$ (iii) $\frac{-19683}{24389}$ (iv) $\frac{686}{-3456}$ (v) $\frac{-39304}{-42875}$ Solution: (i) Let us consider the following rational number: $\frac{-125}{729}$ Now, $\sqrt[3]{\frac{-125}{729}}$ $=\frac{\sqrt[3]{-125}}{\sqrt[3]{729}} \quad\left(\because \sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\right)$ $=\frac{-\sqrt[3]{125}}{\sqrt[3]{729}} \quad(\because \sqrt[3...
Read More →Prove that
Question: If $\tan \theta=\sqrt{3}$, then $\sec \theta=?$ (a) $\frac{2}{\sqrt{3}}$ (b) $\frac{\sqrt{3}}{2}$ (c) $\frac{1}{2}$ (d) 2 Solution: (d) 2 Let us first draw a right $\triangle \mathrm{ABC}$ right angled at $\mathrm{B}$ and $\angle \mathrm{A}=\theta$. Given: $\tan \theta=\sqrt{3}$ But $\tan \theta=\frac{B C}{A B}$ So, $\frac{B C}{A B}=\frac{\sqrt{3}}{1}$ Thus, $B C=\sqrt{3} k$ and $A B=k$ Using Pythagoras theorem, we get:AC2= AB2+ BC2 $\Rightarrow \mathrm{AC}^{2}=(\sqrt{3} k)^{2}+(k)^{2}...
Read More →Prove the following
Question: If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, then find the value of $\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}$ Solution: We have, $\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}$ $=\frac{4(3 \sqrt{3}+2 \sqrt{2})+3(3 \sqrt{3}-2 \sqrt{2})}{(3 \sqrt{3}-2 \sqrt{2})(3 \sqrt{3}+2 \sqrt{2})}$ $=\frac{12 \sqrt{3}+8 \sqrt{2}+9 \sqrt{3}-6 \sqrt{2}}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}}$ [using identity, $(a+b)(a-b)=a^{2}-b^{2}$ ] $=\frac{21 \sqrt{3}+2 \sqrt{2}}{...
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