Solve the following
Question: If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A.P. and G.P. are both $a, b$ and $c$ respectively, show that $a^{b-c} b^{c-a} c^{a-b}=1$. Solution: LetAbe the first term andD be the common difference of the AP. Therefore, $a_{p}=A+(p-1) D=a \quad \ldots(1)$ $a_{q}=A+(q-1) D=b \quad \ldots(2)$ $a_{r}=A+(r-1) D=c \quad \ldots(3)$ Also, supposeA' be the first term andRbe the common ratio of the GP. Therefore, $a_{p}=A^{\prime} R^{p-1}=a \quad \ldots \ldots(4)$ $a_...
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Question: If $A=\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]$, then find $A^{2}-5 A-14 l$. Hence, obtain $A^{3}$. Solution: Given: $A=\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]$ $A^{2}=\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]$ $=\left[\begin{array}{cc}9+20 -15-10 \\ -12-8 20+4\end{array}\right]$ $=\left[\begin{array}{cc}29 -25 \\ -20 24\end{array}\right]$ $A^{2}-5 A-14 I=\left[\begin{array}{cc}29 -25 \\ -20 24\...
Read More →Find the angle subtended at the origin by the line segment
Question: Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0). 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 right angled triangle the angle opposite the hypotenuse subtends an angle of. Here let the given points beA(0,100), B(10,0). Let the origin be denoted byO(0,0). Let us find the distanc...
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Question: If $A=\left[\begin{array}{ccc}1 0 -3 \\ 2 1 3 \\ 0 1 1\end{array}\right]$, then verify that $A^{2}+A=A(A+l)$, where $/$ is the identity matrix. Solution: To verify: $A^{2}+A=A(A+I)$, Given: $A=\left[\begin{array}{ccc}1 0 -3 \\ 2 1 3 \\ 0 1 1\end{array}\right]$ $A^{2}=\left[\begin{array}{ccc}1 0 -3 \\ 2 1 3 \\ 0 1 1\end{array}\right]\left[\begin{array}{ccc}1 0 -3 \\ 2 1 3 \\ 0 1 1\end{array}\right]$ $=\left[\begin{array}{ccc}1+0+0 0+0-3 -3+0-3 \\ 2+2+0 0+1+3 -6+3+3 \\ 0+2+0 0+1+1 0+3+1\...
Read More →In the given figure, O is the canter of the circle and ∠AOB = 70°.
Question: In the given figure,Ois the canter of the circle andAOB= 70.Calculate the values of (i) OCA, (ii) OAC. Solution: (i) 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. Thus,AOB= 2OCA $\Rightarrow \angle O C A=\left(\frac{\angle A O B}{2}\right)=\left(\frac{70^{\circ}}{2}\right)=35^{\circ}$ (ii) OA = OC (Radii of a circle)OAC=OCA [Base angles of an isosceles triangle are equal] $=35^{\circ}$...
Read More →If a, b, c are three distinct real numbers in G.P. and a + b + c = xb,
Question: Ifa,b,care three distinct real numbers in G.P. anda+b+c=xb, then prove that eitherx 1 orx 3. Solution: Let $r$ be the common ratio of the given G.P. $\therefore b=a r$ and $c=a r^{2}$ Now, $a+b+c=b x$ $\Rightarrow a+a r+a r^{2}=a r x$ $\Rightarrow r^{2}+(1-x) r+1=0$ $r$ is always a real number. $\therefore \mathrm{D} \geq 0$ $\Rightarrow(1-x)^{2}-4 \geq 0$ $\Rightarrow x^{2}-2 x-3 \geq 0$ $\Rightarrow(x-3)(x+1) \geq 0$ $\Rightarrow x3$ or $x-1$ and $x \neq 3$ or $-1 \quad[\because a, b...
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Question: If $A=\left[\begin{array}{cc}0 -x \\ x 0\end{array}\right], B=\left[\begin{array}{ll}0 1 \\ 1 0\end{array}\right]$ and $x^{2}=-1$, then show that $(A+B)^{2}=A^{2}+B^{2}$. Solution: Given: $A=\left[\begin{array}{cc}0 -x \\ x 0\end{array}\right], B=\left[\begin{array}{ll}0 1 \\ 1 0\end{array}\right]$ and $x^{2}=-1$ To show: $(A+B)^{2}=A^{2}+B^{2}$ LHS: $A+B=\left[\begin{array}{cc}0 -x \\ x 0\end{array}\right]+\left[\begin{array}{ll}0 1 \\ 1 0\end{array}\right]$ $=\left[\begin{array}{cc}0...
Read More →In Figure (1), O is the centre of the circle. If ∠OAB = 40° and ∠OCB = 30°, find ∠AOC.
Question: (i) In Figure (1),Ois the centre of the circle. IfOAB= 40 and OCB= 30, find AOC.(ii) In Figure (2),A,BandCare three points on the circle with centreOsuch that AOB= 90 and AOC= 110. Find BAC. Solution: (i) JoinBO. In ΔBOC, we have:OC = OB(Radii of a circle)⇒OBC=OCBOBC= 30 ...(i)In ΔBOA, we have:OB = OA (Radii of a circle)⇒OBA=OAB[∵OAB= 40]⇒OBA= 40 ...(ii)Now, we have:ABC=OBC+OBA= 30+ 40 [From (i) and (ii)]ABC= 70The angle subtended by an arc of a circle at the centre is double the angle...
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Question: If $A=\left[\begin{array}{ll}0 0 \\ 4 0\end{array}\right]$, find $A^{16}$ Solution: Given : $A=\left[\begin{array}{ll}0 0 \\ 4 0\end{array}\right]$ Here, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}0 0 \\ 4 0\end{array}\right]\left[\begin{array}{ll}0 0 \\ 4 0\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}0+0 0+0 \\ 0+0 0+0\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ $A^{4}=A^{2} A^{2}$ $\Rightarrow A^{8}=\left[\beg...
Read More →Two circles with centres O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through A or B, intersecting the circles at P and Q.
Question: Two circles with centresOandO' intersect at two pointsAandB. A linePQis drawn parallel toOO' throughAorB, intersecting the circles atPandQ. Prove thatPQ= 2OO'. Solution: Given: Two circles with centresOandO' intersect at two pointsAandB. Draw a linePQparallel toOO' throughB,OXperpendicular toPQ,O'Yperpendicular toPQ, join all.We know that perpendicular drawn from the centre to the chord, bisects the chord.PX=XBandYQ=BYPX+YQ=XB+BYOn addingXB+BYon both sides, we getPX+YQ+XB+BY= 2(XB+BY)⇒...
Read More →If a, b, c are in A.P. and a, b, d are in G.P.,
Question: Ifa,b,care in A.P. anda,b,dare in G.P., show thata, (ab), (dc) are in G.P. Solution: $a, b$ and $c$ are in A.P. $\therefore 2 b=a+c \quad \ldots \ldots(\mathrm{i})$ Also, $a, b$ and $d$ are in G.P. $\therefore b^{2}=a d \quad \ldots \ldots$ (ii) Now, $(a-b)^{2}=a^{2}-2 a b+b^{2}$ $\Rightarrow(a-b)^{2}=a^{2}-a(a+c)+a d \quad$ [Using (i) and (ii)] $\Rightarrow(a-b)^{2}=a^{2}-a^{2}-a c+a d$ $\Rightarrow(a-b)^{2}=a d-a c$ $\Rightarrow(a-b)^{2}=a(d-c)$ Therefore, $a,(a-b)$ and $(d-c)$ are i...
Read More →Find a 2 × 2 matrix A such that
Question: Find a 2 2 matrixAsuch that $A\left[\begin{array}{rr}1 -2 \\ 1 4\end{array}\right]=6 I_{2}$ Solution: Let $\mathrm{A}=\left[\begin{array}{ll}w x \\ y z\end{array}\right]$ Now, $\left[\begin{array}{cc}w x \\ y z\end{array}\right]\left[\begin{array}{cc}1 -2 \\ 1 4\end{array}\right]=6 I_{2}$ $\Rightarrow\left[\begin{array}{ll}w+x -2 w+4 x \\ y+z -2 y+4 z\end{array}\right]=6\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$ $\Rightarrow\left[\begin{array}{ll}w+x -2 w+4 x \\ y+z -2 y+4 z\...
Read More →Find the circumcentre of the triangle whose
Question: Find the circumcentre of the triangle whose vertices are (2, 3), (1, 0), (7, 6). 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 given to beA(2.3),B(1,0) andC(7,6). Let the circumce...
Read More →If a, b, c are in A.P. and a, x, b and b, y, c are in G.P., show that
Question: If $a, b, c$ are in A.P. and $a, x, b$ and $b, y, c$ are in G.P., show that $x^{2}, b^{2}, y^{2}$ are in A.P. Solution: $a, b$ and $c$ are in A.P. $\begin{array}{ll}\therefore 2 b=a+c \ldots \ldots . \text { (i) }\end{array}$ $a, x$ and $b$ are in G.P. $\therefore x^{2}=a b \quad \ldots \ldots$ (ii) And, $b, y$ and $c$ are also in G.P. $\therefore y^{2}=b c$ Now, putting the values of $a$ and $c$ : $\Rightarrow 2 b=\frac{x^{2}}{b}+\frac{y^{2}}{b}$ $\Rightarrow 2 b^{2}=x^{2}+y^{2}$ Ther...
Read More →In the adjoining figure, OPQR is a square. A circle drawn with centre O cuts the square at X and Y.
Question: In the adjoining figure,OPQRis a square. A circle drawn with centreOcuts the square atXandY. Prove thatQX=QY. Solution: Given:OPQRis a square. A circle with centreOcuts the square atXandY.To prove:QX = QYConstruction:JoinOXandOY. Proof:In ΔOXPand ΔOYR, we have:OPX = ORY (90 each)OX = OY (Radii of a circle)OP = OR (Sides of a square) ΔOXPΔOYR(BY RHS congruency rule)⇒PX = RY (By CPCT)⇒PQ - PX=QR - RY (PQandQRare sides of a square)⇒QX = QYHence, proved....
Read More →If a, b, c are in A.P., b,c,d are in G.P. and
Question: If $a, b, c$ are in A.P., $b, c, d$ are in G.P. and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in A.P., prove that $a, c, e$ are in G.P. Solution: $a, b$ and $c$ are in A.P. $\therefore 2 b=a+c \quad \ldots \ldots .(\mathrm{i})$ Also, $b, c$ and $d$ are in G.P. $\therefore c^{2}=b d \quad \ldots \ldots$ (ii) And $\frac{1}{c}, \frac{1}{d}$ and $\frac{1}{e}$ are in A.P. $\therefore \frac{2}{d}=\frac{1}{c}+\frac{1}{e}$ $\Rightarrow d=\frac{2 c e}{c+e} \quad \ldots \ldots$ (iii) $\because...
Read More →In the adjoining figure, AB and AC are two equal chords of a circle with centre O.
Question: In the adjoining figure,ABandACare two equal chords of a circle with centreO. Show thatOlies on the bisector of BAC. Solution: Given:ABandACare two equal chords of a circle with centreO. To prove:OAB = OACConstruction:JoinOA, OBandOC.Proof:In ΔOABand ΔOAC, we have:AB = AC (Given)OA = OA (Common)OB = OC(Radii of a circle) ΔOABΔ OAC(By SSS congruency rule)⇒OAB = OAC (CPCT)Hence, pointOlies on the bisector ofBAC....
Read More →The three vertices of a parallelogram are (3, 4) (3, 8)
Question: The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex. Solution: We are given three vertices of aparallelogramA(3, 4), B(3, 8) and C(9, 8). We know that diagonals of a parallelogram bisect each other. Let the fourth vertexbe D(x, y). Mid point of $B D=\left(\frac{x+3}{2}, \frac{y+8}{2}\right)$ Mid point of $A C=\left(\frac{9+3}{9}, \frac{4+8}{9}\right)=(6,6)$ Since the mid point of BD = mid point of AC $\mathrm{So},\left(\frac{x+3}{2}, \frac{y+8}{2}...
Read More →Solve the following
Question: If $x^{a}=x^{b / 2} z^{b / 2}=z^{c}$, then prove that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. Solution: Here, $x^{a}=(x z)^{\frac{b}{2}}=z^{c}$ Now, taking log on both the sides: $\log (\mathrm{x})^{\mathrm{a}}=\log (\mathrm{xz})^{\frac{b}{2}}=\log (\mathrm{z})^{\mathrm{c}}$ $\Rightarrow \operatorname{alog} \mathrm{x}=\frac{b}{2} \log (\mathrm{xz})=\mathrm{c} \log \mathrm{z}$ $\Rightarrow \operatorname{alog} \mathrm{x}=\frac{b}{2} \log \mathrm{x}+\frac{b}{2} \log \mathrm{z}...
Read More →An equilateral triangle of side 9 cm is inscribed in a circle.
Question: An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle. Solution: Let ΔABCbe an equilateral triangle of side 9 cm.LetADbe one of its median. Then,ADBC (ΔABCis an equilateral triangle) Also, $B D=\left(\frac{B C}{2}\right)=\left(\frac{9}{2}\right)=4.5 \mathrm{~cm}$ In right angled ΔADB, we have: $A B^{2}=A D^{2}+B D^{2}$ $\Rightarrow A D^{2}=A B^{2}-B D^{2}$ $\Rightarrow A D=\sqrt{A B^{2}-B D^{2}}$ $=\sqrt{(9)^{2}-\left(\frac{9}{2}\right)^{2}} \math...
Read More →Which point on y-axis is equidistant from (2, 3)
Question: Which point ony-axis is equidistant from (2, 3) and (4, 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 $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ Here we are to find out a point on they-axis which is equidistant from both the pointsA(2,3) andB(4,1). Let this point be denoted asC(x, y). Since the point lies on they-axis the value of its ordinate will be 0. Or in other words w...
Read More →Solve the following
Question: If $\frac{1}{a+b}, \frac{1}{2 b}, \frac{1}{b+c}$ are three consecutive terms of an A.P., prove that $a, b, c$ are the three consecutive terms of a G.P. Solution: Here, $\frac{1}{a+b}, \frac{1}{2 b}$ and $\frac{1}{b+c}$ are in A.P. $\therefore 2 \times \frac{1}{2 b}=\frac{1}{a+b}+\frac{1}{b+c}$ $\Rightarrow \frac{1}{b}=\frac{b+c+a+b}{(a+b)(b+c)}$ $\Rightarrow(a+b)(b+c)=b(2 b+a+c)$ $\Rightarrow a b+a c+b^{2}+b c=2 b^{2}+a b+b c$ $\Rightarrow 2 b^{2}-b^{2}=a c$ $\Rightarrow b^{2}=a c$ Thu...
Read More →The coordinates of the point P are (−3, 2).
Question: The coordinates of the pointPare (3, 2). Find the coordinates of the pointQwhich lies on the line joiningPand origin such that OP = OQ. Solution: If $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ are given as two points, then the co-ordinates of the midpoint of the line joining these two points is given as $\left(x_{m}, y_{m}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$ It is given that the point P has co-ordinates (3,2) Here we are asked to find out t...
Read More →If pth, qth, rth and sth terms of an A.P. be in G.P.,
Question: Ifpth,qth,rthandsthterms of an A.P. be in G.P., then prove thatpq,qr,rsare in G.P. Solution: Here, $a_{p}=a+(p-1) d$ $a_{q}=a+(q-1) d$ $a_{r}=a+(r-1) d$ $a_{s}=a+(s-1) d$ It is given that $a_{p}, a_{q}, a_{r}$ and $a_{s}$ are in G.P. $\therefore \frac{a_{q}}{a_{p}}=\frac{a_{r}}{a_{q}}=\frac{a_{q}-a_{r}}{a_{p}-a_{q}}=\frac{q-r}{p-q} \quad \ldots \ldots(\mathrm{i})$ Similarly, $\frac{a_{r}}{a_{q}}=\frac{a_{s}}{a_{r}}=\frac{a_{r}-a_{s}}{a_{q}-a_{r}}=\frac{r-s}{q-r} \quad \ldots \ldots$ (i...
Read More →Prove that the points (−2, 5),
Question: Prove that the points (2, 5), (0, 1) and (2, 3) are collinear. 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 $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}^{2}\right)}$ For three points to be collinear the sum of distances between two pairs of points should be equal to the third pair of points. The given points areA(2,5),B(0,1) andC(2,3) Let us find the distances between the possible pairs of p...
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