Construct each of the following angles, using ruler and compasses:
Question: Construct each of the following angles, using ruler and compasses:(i) 75(ii) 37.5(iii) 135(iv) 105(v) 22.5 Solution: (i)75Steps of construction1. Draw a line XY.2. Take a point O on XY.3.With Oas centre, draw a semi circle, cuttingXYatPandQ. 4. Construct $\angle Y O R=90^{\circ}$. 5. Draw the bisector of $\angle Y O R=90^{\circ}$ cutting the semi circle at point $S$. 6. With $S$ and $T$ as centres draw two arcs intersecting at point $A$. $\angle \mathrm{AOY}=75^{\circ}$ (ii) 37.5 Steps...
Read More →If the normal to the curve
Question: If the normal to the curve $y(x)=\int_{0}^{x}\left(2 t^{2}-15 t+10\right) d t$ at a point $(a, b)$ is parallel to the line $x+3 y=-5, a1$, then the value of $\mid \mathrm{a}+6 \mathrm{bl}$ is equal to Solution: $y(x)=\int_{0}^{x}\left(2 t^{2}-15 t+10\right) d t$ $\left.y^{\prime}(x)\right]_{x=a}=\left[2 x^{2}-15 x+10\right]_{a}=2 a^{2}-15 a+10$ Slope of normal $=-\frac{1}{3}$ $\Rightarrow \quad 2 a^{2}-15 a+10=3 \Rightarrow a=7$ $\ \quad a=\frac{1}{2}$ (rejected) $b=y(7)=\int_{0}^{7}\l...
Read More →Construct an angle of 90° using ruler and compasses and bisect it.
Question: Construct an angle of 90 using ruler and compasses and bisect it. Solution: Steps of construction:1. Draw a line segmentAB.2. WithAas the centre and and a small radius, draw an arc cuttingABat M.3. WithMas the centre and the same radius as above, draw an arc cutting the previously drawn arc atN.4. WithNas the centre and the same radius as above, draw an arc cutting the previously drawn arc atP.5. Again, withNas the centre and a radius more than half ofPN,draw an arc.6. WithPas the cent...
Read More →Draw an angle of 80° with the help of a protractor and bisect it.
Question: Draw an angle of 80 with the help of a protractor and bisect it. Measure each part of the bisected angle. Solution: Steps of construction 1. Draw $\angle \mathrm{AOB}=80^{\circ}$ using protractor. 2. With O as centre and a convenient radius, draw an arc cutting AO at N and OB at M.3. With N as centre and a convenient radius, draw an arc.4. Similarly, with M as centre and same radius, cut the previous drawn arc and name it as point C.5. Join OC.OC is the required angle bisector.On measu...
Read More →Solve the following
Question: Ifa= 1 +b+b2+b3+ ... to , then writebin terms ofa. Solution: Here, $a=1, b, b^{2}, b^{3}, \ldots \infty$ form an infinite G.P. $\therefore \mathrm{S}_{\infty}=\mathrm{a}=1+\mathrm{b}+\mathrm{b}^{2}+\mathrm{b}^{3}+\ldots \infty=\frac{1}{1-\mathrm{b}}$ $\Rightarrow a=\frac{1}{1-b}$ $\Rightarrow 1-b=\frac{1}{a}$ $\Rightarrow \mathrm{b}=1-\frac{1}{\mathrm{a}}$ $\therefore \mathrm{b}=\frac{\mathrm{a}-1}{\mathrm{a}}$...
Read More →Draw a line segment AB = 5.6 cm and draw its perpendicular bisector.
Question: Draw a line segmentAB= 5.6 cm and draw its perpendicular bisector. Measure the length of each part. Solution: Steps of Construction1. Draw a line AB = 5.6 cm.2. With A as centre and radius more than half of AB, draw one above and other below line AB.3. Similarly, with B as centre draw two arcs cutting the previous drawn arcs and name the points obtained as M and N respectively.4. Join MN intersecting AB at point O.MN is the required perpendicular bisector.AO = OB = 2.8 cm...
Read More →Write the product of n geometric means between two numbers a and b.
Question: Write the product ofngeometric means between two numbersaandb. Solution: Let $G_{1}, G_{2}, \ldots, G_{n}$ be $n$ geometric means between two quantities $a$ and $b$. Thus, $a, G_{1}, G_{2}, \ldots, G_{n}, b$ is a G.P. Let $r$ be the common ratio of this G.P. $\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ And, $G_{1}=a r, G_{2}=a r^{2}, G_{3}=a r^{3}, \ldots, G_{n}=a r^{n}$ Now, product of $n$ geometric means $=G_{1} \cdot G_{2} \cdot G_{3} \cdot \ldots \cdot G_{n}=(a r)\left(a...
Read More →The range of a $in mathbb{R}$ for which the function
Question: The range of $a \in \mathbb{R}$ for which the function $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$ $\mathrm{x}+2 \mathrm{n} \pi, \mathrm{n} \in \mathbb{N}$, has critical points, is(1) $(-3,1)$(2) $\left[-\frac{4}{3}, 2\right]$(3) $[1, \infty)$(4) $(-\infty,-1]$Correct Option: , 2 Solution: $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+(a-7) \sin x$ $f(x)=(4 a-3)(1)+(a-7) \cos x=0$ $\cos x=\frac{3-4 a}{a-7}$ $\frac{3 a+4}{a-7...
Read More →if
Question: If $A=\left[\begin{array}{rr}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$, then verify that $A^{\top} A=I_{2}$ Solution: Given : $A=\left[\begin{array}{cc}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$ $A^{T}=\left[\begin{array}{cc}\cos \alpha -\sin \alpha \\ \sin \alpha \cos \alpha\end{array}\right]$ Now, $A^{T} A=I_{2}$ Consider : LHS $=A^{T} A$ $=\left[\begin{array}{cc}\cos \alpha -\sin \alpha \\ \sin \alpha \cos \alpha\end{array}\ri...
Read More →if
Question: If $A^{T}=\left[\begin{array}{cc}3 4 \\ -1 2 \\ 0 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1 2 1 \\ 1 2 3\end{array}\right]$, find $A^{T}-B^{T}$. Solution: Given: $A^{T}=\left[\begin{array}{cc}3 4 \\ -1 2 \\ 0 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1 2 1 \\ 1 2 3\end{array}\right]$ $B^{T}=\left[\begin{array}{cc}-1 1 \\ 2 2 \\ 1 3\end{array}\right]$ Now, $A^{T}-B^{T}=\left[\begin{array}{cc}3 4 \\ -1 2 \\ 0 1\end{array}\right]-\left[\begin{array}{cc}-1 1 \\ 2 2 ...
Read More →Write the quadratic equation the arithmetic and geometric
Question: Write the quadratic equation the arithmetic and geometric means of whose roots areAandGrespectively. Solution: Let the roots of the required quadratic equation be $a$ and $b$. $\therefore A=\frac{a+b}{2}$ and $G=\sqrt{a b}$ The equation having $a$ and $b$ as its roots is $x^{2}-x(a+b)+a b=0$ $\Rightarrow x^{2}-2 A x+G^{2}=0 \quad\left[\because A=\frac{a+b}{2}\right.$ and $\left.G=\sqrt{a b}\right]$...
Read More →If second, third and sixth terms of an A.P. are consecutive terms of a G.P.,
Question: If second, third and sixth terms of an A.P. are consecutive terms of a G.P., write the common ratio of the G.P. Solution: Here, second term, $a_{2}=a+d$ Third term, $a_{3}=a+2 d$ Sixth term, $a_{6}=a+5 d$ As, $a_{2}, a_{3}$ and $a_{6}$ are in G.P. $\therefore$ First term of G.P. $=a_{2}=A=a+d$ Second term of G.P. $=A r=a+2 d$ Third term of G.P. $=A r^{2}=a+5 d$ $\therefore(a+2 d)^{2}=(a+d) \times(a+5 d)$ $\Rightarrow a^{2}+4 a d+4 d^{2}=a^{2}+6 a d+5 d^{2}$ $\Rightarrow 2 a d+d^{2}=0$ ...
Read More →Solve the following
Question: If $A_{1}, A_{2}$ be two AM's and $G_{1}, G_{2}$ be two GM's between a and $b$, then find the value of $\frac{A_{1}+A_{2}}{G_{1} G_{2}}$. Solution: It is given that $A_{1}$ and $A_{2}$ are the A.M.s between $a$ and $b$. Thus, $a, A_{1}, A_{2}$ and $b$ are in A.P. with common difference d. Here, $d=\frac{b-a}{3}$ $\therefore A_{1}=a+\frac{b-a}{3}=\frac{2 a+b}{3}$ and $A_{2}=a+\frac{2(b-a)}{3}=\frac{a+2 b}{3}$ It is also given that $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$ are the G.M.s betw...
Read More →In the given figure, ∠AOB = 90° and ∠ABC = 30°.
Question: In the given figure,AOB= 90 and ABC= 30. Then, CAO= ?(a) 30(b) 45(c) 60(d) 90 Solution: (c) 60We have:AOB = 2ACB $\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\left(\frac{1}{2} \times 90^{\circ}\right)=45^{\circ}$ COA = 2CBA = (2 30) = 60COD = 180 -COA = (180 - 60) = 120 $\Rightarrow \angle \mathrm{CAO}=\frac{1}{2} \angle \mathrm{COD}=\left(\frac{1}{2} \times 120^{\circ}\right)=60^{\circ}$...
Read More →Solve the following
Question: Ifpth,qthandrthterms of a G.P. arex,y,zrespectively, then write the value ofxqryrpzpq. Solution: Let us take a G.P. whose first term isAand common ratio isR. According to the question, we have: $A R^{p-1}=x$ $A R^{q-1}=y$ $A R^{r-1}=z$ $\therefore x^{q-r} y^{r-p} z^{p-q}$ $=A^{q-r} \times R^{(p-1)(q-r)} \times A^{r-p} \times R^{(q-1)(r-p)} \times A^{p-q} \times R^{(r-1)(p-q)}$ $=A^{q-r+r-p+p-q} \times R^{(p r-p r-q+r)+(r q-r+p-p q)+(p r-p-q r+q)}$ $=A^{0} \times R^{0}$ $=1$ $\therefore...
Read More →Find the length of the medians of a ΔABC having
Question: Find the length of the medians of a ΔABC having vertices at A(0,1), B(2, 1) and C(0, 3). Solution: We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,1); B (2, 1) and C (0, 3). So we should find the mid-points of the sides of the triangle. In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as, $\mathrm{P}(x, y)=\left(\fr...
Read More →Solve the following
Question: Ifpth,qthandrthterms of a G.P. arex,y,zrespectively, then write the value ofxqryrpzpq. Solution: Let us take a G.P. whose first term isAand common ratio isR....
Read More →If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is
Question: If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is9292, then write its first term and common difference. Solution: Let us take a G.P. whose first term isaand common difference isr. $\therefore S_{\infty}=\frac{a}{1-r}$ $\Rightarrow \frac{a}{1-r}=3$ ...(i) And, sum of the terms of the G.P. $a^{2},(a r)^{2},\left(a r^{2}\right)^{2}, \ldots \infty:$ $S_{\infty}^{\prime}=\frac{a^{2}}{1-r^{2}}$ $\Rightarrow \frac{a^{2}}{1-r^{2}}=\frac{9}{2}$ ....(ii) $\...
Read More →In the given figure, A and B are the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at points P and Q respectively.
Question: In the given figure,AandBare the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at pointsPandQrespectively. IfAB= 4 cm, then the length of common chordPQis(a) 3 cm(b) 6 cm(c) 7.5 cm(d) 9 cm Solution: (b) 6 cmWe know that the line joining their centres is the perpendicular bisector of the common chord.Join AP.Then AP = 5 cm; AB = 4 cmAlso, AP2= BP2+ AB2Or BP2= AP2- AB2Or BP2= 52- 42Or BP = 3 cm ΔABP is a right angled and PQ = 2 BP = (2 3) cm = 6 cm...
Read More →(i) For two matrices A and B,
Question: (i) For two matrices $A$ and $B, A=\left[\begin{array}{lll}2 1 3 \\ 4 1 0\end{array}\right], B=\left[\begin{array}{rr}1 -1 \\ 0 2 \\ 5 0\end{array}\right]$ verify that $(A B)^{T}=B^{T} A^{T}$ (ii) For the matrices $A$ and $B$, verify that $(A B)^{T}=B^{T} A^{T}$, where $A=\left[\begin{array}{ll}1 3 \\ 2 4\end{array}\right], B=\left[\begin{array}{ll}1 4 \\ 2 5\end{array}\right]$ Solution: (i) Given : $A=\left[\begin{array}{lll}2 1 3 \\ 4 1 0\end{array}\right]$ $A^{T}=\left[\begin{array}...
Read More →Solve the following
Question: If logxa,ax/2and logbxare in G.P., then write the value ofx. Solution: $\log _{x} a, a^{\frac{x}{2}}$ and $\log _{b} x$ are in G.P. $\therefore\left(\mathrm{a}^{\frac{x}{2}}\right)^{2}=\log _{x} \mathrm{a} \times \log _{\mathrm{b}} x$ $\Rightarrow a^{x}=\frac{\log _{b} a}{\log _{b} x} \times \log _{b} x$ $\Rightarrow a^{x}=\log _{b} a$ Now, by taking $\log _{a}$ on both the sides: $\Rightarrow \mathrm{x}=\log _{\mathrm{a}}\left(\log _{\mathrm{b}} \mathrm{a}\right)$...
Read More →Two chords AB and CD of a circle intersect each other at a point E outside the circle.
Question: Two chordsABandCDof a circle intersect each other at a pointEoutside the circle. IfAB= 11 cm,BE= 3 cm andDE= 3.5 cm, thenCD= ?(a) 10.5 cm(b) 9.5 cm(c) 8.5 cm(d) 7.5 cm Solution: (c) 8.5 cmJoin AC. Then AE : CE = DE : BE (Intersecting secant theorem) AE BE = DE CELet CD =xcmThen AE = (AB + BE) = (11 + 3) cm = 14 cm; BE = 3cm; CE = (x+ 3.5) cm; DE = 3.5 cm 14 3 = (x+ 3.5) 3.5 $\Rightarrow x+3.5=\frac{14 \times 3}{3.5}=\frac{42}{3.5}=12$ ⇒x= (12 - 3.5) cm = 8.5 cmHence, CD = 8.5 cm...
Read More →If (p + q)th and (p − q)th terms of a G.P. are m and n respectively,
Question: If (p+q)thand (pq)thterms of a G.P. aremandnrespectively, then write ispth term. Solution: Here, $(p+q)^{\text {th }}$ term $=\mathrm{m}$ $\Rightarrow a r^{(p+q)-1}=m \quad \ldots \ldots$ (i) And, $(\mathrm{p}-\mathrm{q})^{\mathrm{th}}$ term $=\mathrm{n}$ $\Rightarrow a r^{(p-q)-1}=n \quad \ldots \ldots$ (ii) Dividing (i) by (ii): $\frac{a r(p+q)-1}{a r(p-q)-1}=\frac{m}{n}$ $\Rightarrow r^{2 q}=\frac{m}{n}$ $\Rightarrow r^{q}=\sqrt{\frac{m}{n}}$ Now, from (i): $a\left(r^{p-1} \times r^...
Read More →If the coordinates of the mid-points of the sides of a triangle be
Question: If the coordinates of the mid-points of the sides of a triangle be (3, 2), (3, 1) and (4, 3), then find the coordinates of its vertices. Solution: The co-ordinates of the midpoint $\left(x_{m}, y_{m}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by, $\left(x_{m}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$ Let the three vertices of the triangle be $A\left(x_{A}, y_{A}\right), B\left(x...
Read More →In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ∠ADC = 95° and ∠ECF = 20°.
Question: In the given figure,ABCDis a cyclic quadrilateral in whichDCis produced toEandCFis drawn parallel toABsuch thatADC= 95 and ECF= 20. Then, BAD= ?(a) 95(b) 85(c) 105(d) 75 Solution: (c) 105We have:ABC + ADC = 180⇒ABC + 95 = 180⇒ABC= (180 - 95) = 85Now,CF || AB and CB is the transversal.BCF =ABC = 85 (Alternate interior angles)⇒BCE = (85 + 20) = 105⇒DCB = (180 - 105) = 75Now, BAD + BCD = 180⇒BAD + 75 = 180⇒BAD = (180 - 75) = 105...
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