Construct a quadratic in x such that A.M.
Question: Construct a quadratic inxsuch that A.M. of its roots isAand G.M. isG. Solution: Let the roots of the quadratic equation be $a$ and $b$. $A=\frac{a+b}{2}$ $\therefore a+b=2 A \quad \ldots \ldots \ldots(\mathrm{i})$ Also, $G^{2}=a b \quad \ldots \ldots$ (ii) The quadratic equation having roots a and b is given by $x^{2}-(a+b) x+a b=0$. $\therefore x^{2}-2 A x+G^{2}=0 \quad[$ Using (i) and (ii) $]$...
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Question: If $A=\left[\begin{array}{ll}a b \\ 0 1\end{array}\right]$, prove that $A^{n}=\left[\begin{array}{cc}a^{n} b\left(a^{n}-1\right) / a-1 \\ 0 1\end{array}\right]$ for every positive integer $n$. Solution: We shall prove the result by the principle of mathematical induction onn. Step 1: If $n=1$, by definition of integral power of a matrix, we have $A^{1}=\left[\begin{array}{cc}a^{1} b\left(a^{1}-1\right) / a-1 \\ 0 1\end{array}\right]=\left[\begin{array}{ll}a b \\ 0 1\end{array}\right]=A...
Read More →Find the two numbers whose A.M. is 25 and GM is 20.
Question: Find the two numbers whose A.M. is 25 and GM is 20. Solution: Let A.M. and G.M. between the two numbers $a$ and $b$ be $A$ and $G$, respectively. $A=25$ $\Rightarrow \frac{a+b}{2}=25$ $\Rightarrow a+b=50$ ...(i) Also $G=20$ $\Rightarrow \sqrt{a b}=20$ $\Rightarrow a b=400$ ....(ii) Now, putting the value of $a$ in $($ ii $)$ : $\Rightarrow(50-b) b=400$ $\Rightarrow b^{2}-50 b+400=0$ $\Rightarrow b^{2}-10 b-40 b+400=0$ $\Rightarrow b(b-10)-40(b-10)=0$ $\Rightarrow(b-10)(b-40)=0$ $\Right...
Read More →In the given figure, AB and CD are straight lines through the centre O of a circle.
Question: In the given figure,ABandCDare straight lines through the centreOof a circle. IfAOC= 80 and CDE= 40, find (i) DCE, (ii) ABC. Solution: (i)CED= 90(Angle in a semi circle)In ΔCED, we have:CED +EDC +DCE= 180 (Angle sum property of a triangle)⇒ 90+ 40+DCE= 180⇒DCE= (180 130) = 50 ...(i)DCE =50(ii)AsAOCandBOCare linear pair, we have:BOC =(180 80) = 100...(ii)In ΔBOC,we have:OBC+OCB+BOC= 180(Angle sum property of a triangle)⇒ABC +DCE +BOC= 180 [∵OBC =ABC andOCB =DCE]⇒ABC= 180 (BOC+DCE)⇒ABC =...
Read More →Find a point on the x-axis which is equidistant
Question: Find a point on thex-axis which is equidistant from the points (7, 6) and (3, 4). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ Here we are to find out a point on thexaxis which is equidistant from both the pointsA(7,6) andB(3,4). Let this point be denoted asC(x, y). Since the point lies on thex-axis the value of its ordinate will be 0...
Read More →In the adjoining figure, O is the centre of a circle.
Question: In the adjoining figure,Ois the centre of a circle. ChordCDis parallel to diameterAB. IfABC= 25, calculate CED. Solution: BCD =ABC= 25(Alternate angles)JoinCOandDO.We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by an arc at any point on the circumference.Thus,BOD= 2BCD⇒BOD= 2 25= 50Similarly,AOC= 2ABC⇒AOC= 2 25= 50ABis a straight line passing through the centre.i.e.,AOC +COD +BOD= 180⇒ 50+COD+ 50= 180⇒COD= (180 100)= 80 $\Rightarrow \...
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Question: If $A=\left[\begin{array}{ll}a b \\ 0 1\end{array}\right]$, prove that $A^{n}=\left[\begin{array}{cc}a^{n} b\left(a^{n}-1\right) / a-1 \\ 0 1\end{array}\right]$ for every positive integer $n$ Solution: We shall prove the result by the principle of mathematical induction onn. Step 1: If $n=1$, by definition of integral power of a matrix, we have $A^{1}=\left[\begin{array}{cc}a^{1} b\left(a^{1}-1\right) / a-1 \\ 0 1\end{array}\right]=\left[\begin{array}{ll}a b \\ 0 1\end{array}\right]=A$...
Read More →If a is the G.M. of 2 and
Question: If $a$ is the $G . M$. of 2 and $\frac{1}{4}$, find $a$. Solution: $a$ is the G.M. between 2 and $\frac{1}{4}$. $\therefore a^{2}=2 \times \frac{1}{4}$ $\Rightarrow a^{2}=\frac{1}{2}$ $\Rightarrow a=\sqrt{\frac{1}{2}}$ $\Rightarrow a=\frac{1}{\sqrt{2}}$...
Read More →Find the coordinates of the circumcentre of the triangle whose
Question: Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (1, 6) and (4, 1). Also, find its circumradius. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle. Here the three vertices of the triangle are g...
Read More →Find the geometric means of the following pairs of numbers:
Question: Find the geometric means of the following pairs of numbers: (i) 2 and 8 (ii)a3bandab3 (iii) 8 and 2 Solution: (i) Let the G.M. between 2 and 8 be $G$. Then, $2, G$ and 8 are in G.P. $\therefore G^{2}=2 \times 8$ $\Rightarrow G^{2}=16$ $\Rightarrow G=\pm \sqrt{16}$ $\Rightarrow G=\pm 4$ (ii) Let the G.M. between $a^{3} b$ and $a b^{3}$ be $G .$ Then, $a^{3} b, G$ and $a b^{3}$ are in G.P. $\therefore G^{2}=a^{3} b \times a b^{3}$ $\Rightarrow G^{2}=a^{4} b^{4}$ $\Rightarrow G=\sqrt{a^{4...
Read More →In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O
Question: In the adjoining figure,DEis a chord parallel to diameterACof the circle with centreO. IfCBD= 60, calculate CDE. Solution: Angles in the same segment of a circle are equal.i.e.,CAD =CBD= 60We know that an angle in a semicircle is a right angle.i.e.,ADC= 90In ΔADC,we have:ACD+ADC+CAD= 180 (Angle sum property of a triangle)⇒ACD+ 90+ 60= 180⇒ACD= 180 (90+ 60) = (180 150)= 30⇒CDE=ACD= 30 (Alternate angles asACparallel toDE)Hence,CDE= 30...
Read More →In the given figure, ∠ABD = 54° and ∠BCD = 43°, calculate
Question: In the given figure,ABD= 54 and BCD= 43, calculate (i) ACD (ii) BAD (iii) BDA. Solution: (i)We know that the angles in the same segment of a circle are equal.i.e.,ABD =ACD= 54(ii)We know that the angles in the same segment of a circle are equal.i.e.,BAD =BCD= 43(iii)In ΔABD,we have:BAD+ADB+DBA =180 (Angle sum property of a triangle)⇒ 43+ADB+ 54= 180⇒ADB= (180 97) = 83⇒BDA= 83...
Read More →In the given figure, O is the centre of the circle.
Question: In the given figure,Ois the centre of the circle. IfACB= 50, find OAB. Solution: We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.AOB= 2ACB = 2 50 [Given] AOB= 100 ...(i)Let us consider the triangle ΔOAB.OA = OB(Radii of a circle)Thus,OAB=OBAIn ΔOAB, we have:AOB+OAB+OBA= 180⇒ 100+OAB+OAB= 180⇒ 100+ 2OAB= 180⇒ 2OAB= 180 100= 80⇒OAB= 40Hence,OAB=40...
Read More →Show that the points (−3, 2), (−5,−5), (2, −3) and
Question: Show that the points (3, 2), (5,5), (2, 3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ In a rhombus all the sides are equal in length. And the area A of a rhombus is given as $A=\frac{1}{2}$ (Product of both diagonals) Here the four points areA(3,2), B(5,5),C(2,...
Read More →In the given figure, O is the centre of the circle.
Question: In the given figure,Ois the centre of the circle. IfABD= 35 and BAC= 70, find ACB. Solution: It is clear thatBDis the diameter of the circle.Also, we know that the angle in a semicircle is a right angle.i.e.,BAD= 90Now, considering the ΔBAD, we have:ADB+BAD+ABD= 180 (Angle sum property of a triangle)⇒ADB+ 90+ 35= 180⇒ADB= (180- 125) = 55Angles in the same segment of a circle are equal.Hence,ACB=ADB= 55...
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Question: If $A=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]$, prove that $A^{n}=\left[\begin{array}{ll}1 n \\ 0 1\end{array}\right]$ for all positive integers $n$. Solution: We shall prove the result by the principle of mathematical induction onn.Step 1: Ifn= 1, by definition of integral powers of matrix, we have $A^{1}=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]=A$ So, the result is true for $n=1$. Step 2: Let the result be true for $n=m$. Then, $A^{m}=\left[\begin{array}{cc}1 m...
Read More →Insert 5 geometric means between
Question: Insert 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}$. Solution: Let the $5 G .$ M.s between $\frac{32}{9}$ and $\frac{81}{2}$ be $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}$ and $\mathrm{G}_{5}$. $\frac{32}{9}, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}, \frac{81}{2}$ $\Rightarrow a=\frac{32}{9}, n=7$ and $a_{7}=\frac{81}{2}$ $\because a_{7}=\frac{81}{2}$ $\Rightarrow a r^{6}=\frac{81}{2}$ $\Rightarrow r^{6}=\frac{81...
Read More →In the given figure, O is the centre of the circle. if ∠PBC = 25° and ∠APB = 110°,
Question: In the given figure,Ois the centre of the circle. ifPBC= 25 and APB= 110, find the value of ADB. Solution: From the given diagram, we have: ACB=PCBBPC= (180- 110) = 70 (Linear pair)Considering ΔPCB, we have:PCB+BPC+PBC= 180 (Angle sum property)⇒PCB+ 70+ 25= 180⇒PCB= (180 95) = 85⇒ACB=PCB= 85We know that the angles in the same segment of a circle are equal.ADB=ACB= 85...
Read More →If two opposite vertices of a square are (5, 4)
Question: If two opposite vertices of a square are (5, 4) and (1, 6), find the coordinates of its remaining two vertices. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal totimes the side of the square. Here let the two ...
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Question: If $A=\left[\begin{array}{ccc}2 0 1 \\ 2 1 3 \\ 1 -1 0\end{array}\right]$, find $A^{2}-5 A+4 /$ and hence find a matrix $X$ such that $A^{2}-5 A+4 /+X=0$ Solution: Given: $A=\left[\begin{array}{ccc}2 0 1 \\ 2 1 3 \\ 1 -1 0\end{array}\right]$ $A^{2}=\left[\begin{array}{ccc}2 0 1 \\ 2 1 3 \\ 1 -1 0\end{array}\right]\left[\begin{array}{ccc}2 0 1 \\ 2 1 3 \\ 1 -1 0\end{array}\right]$ $=\left[\begin{array}{lll}4+0+1 0+0-1 2+0+0 \\ 4+2+3 0+1-3 2+3+0 \\ 2-2+0 0-1-0 1-3+0\end{array}\right]$ $=...
Read More →Insert 5 geometric means between 16 and
Question: Insert 5 geometric means between 16 and $\frac{1}{4}$. Solution: Let the $5 \mathrm{G} . \mathrm{M} . \mathrm{s}$ betweem 16 and $\frac{1}{4}$ be $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}$ and $\mathrm{G}_{5}$. $16, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}, \frac{1}{4}$ $\Rightarrow a=16, n=7$ and $a_{7}=\frac{1}{4}$ $\because a_{7}=\frac{1}{4}$ $\Rightarrow a r^{6}=\frac{1}{4}$ $\Rightarrow r^{6}=\frac{1}{4 \times 16}$ $\Righ...
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Question: (i) If $P(x)=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$, then show that $P(x) P(y)=P(x+y)=P(y) P(x)$. (ii) If $P=\left[\begin{array}{lll}x 0 0 \\ 0 y 0 \\ 0 0 z\end{array}\right]$ and $Q=\left[\begin{array}{lll}a 0 0 \\ 0 b 0 \\ 0 0 c\end{array}\right]$, prove that $P Q=\left[\begin{array}{ccc}x a 0 0 \\ 0 y b 0 \\ 0 0 z c\end{array}\right]=Q P$ Solution: (i) Given: $P(x)=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ then, $P(y...
Read More →Insert 6 geometric means between 27 and
Question: Insert 6 geometric means between 27 and $\frac{1}{81}$. Solution: Let the 6 G.M.s between 27 and $\frac{1}{81}$ be $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}$ and $\mathrm{G}_{6}$. Thus, $27, \mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}, \mathrm{G}_{6}$ and $\frac{1}{81}$ are in G.P. $\therefore a=27, n=8$ and $a_{8}=\frac{1}{81}$ $\because a_{8}=\frac{1}{81}$ $\Rightarrow \mathrm{ar}^{7}=\frac{1}{81}$ $\Rightarrow ...
Read More →Find the value of k, if the point P (0, 2)
Question: Find the value ofk, if the pointP(0, 2) is equidistant from (3,k) and (k, 5). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ It is said thatP(0,2) is equidistant from bothA(3,k) andB(k,5). So, using the distance formula for both these pairs of points we have $A P=\sqrt{(3)^{2}+(k-2)^{2}}$ $B P=\sqrt{(k)^{2}+(3)^{2}}$ Now since both thes...
Read More →Find the centre of the circle passing through (5, −8),
Question: Find the centre of the circle passing through (5, 8), (2, 9) and (2, 1). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The centre of a circle is at equal distance from all the points on its circumference. Here it is given that the circle passes through the pointsA(5,8),B(2,9) andC(2,1). Let the centre of the circle be represented by th...
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