Solve the following
Question: If thenthtermanof a sequence is given byan=n2n+ 1, write down its first five terms. Solution: Given : $a_{n}=n^{2}-n+1$ For $n=1, a_{1}=1^{2}-1+1$ = 1 For $n=2, a_{2}=2^{2}-2+1$ = 3 For $n=3, a_{3}=3^{2}-3+1$ = 7 For $n=4, a_{4}=4^{2}-4+1$ = 13 For $n=5, a_{5}=5^{2}-5+1$ = 21 Thus, the first five terms of the sequence are 1, 3, 7, 13, 21....
Read More →Find the principal values of each of the following:
Question: Find the principal values of each of the following: (i) $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ (ii) $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ (iii) $\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$ (iv) $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$ Solution: (i) Let $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$ Then, $\tan y=\frac{1}{\sqrt{3}}$ We know that the range of the principal value branch is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.\ Thus, $\tan y=\frac{1}{\sqrt{3}}=\ta...
Read More →Draw a pair of tangents to a circle of radius 4.5 cm,
Question: Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45. Solution: Steps of ConstructionStep 1. Draw a circle with centre O and radius 4.5 cm.Step 2. Draw any diameter AOB of the circle.Step 3. ConstructBOC = 45 such that radius OC cuts the circle at C.Step 4. Draw AM AB and CN OC. Suppose AM and CN intersect each other at P. Here, AP and CP are the pair of tangents to the circle inclined to each other at an angle of 45....
Read More →Draw a circle of radius 3 cm. Take two points P and Q on one of its extended
Question: Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. Solution: Given that Construct a circle of radius, andextended diameter each at distance of 7cm from its centre. Construct the pair of tangents to the circle from these two points. We follow the following steps to construct the given Step of construction Step: I- First of all we draw a circle of radi...
Read More →The sums of first n terms of two A.P.'s are in the ratio (7n + 2) :
Question: The sums of firstnterms of two A.P.'s are in the ratio (7n+ 2) : (n+ 4). Find the ratio of their 5thterms. Solution: Let the first term, the common difference and the sum of the first $n$ terms of the first A .P. be $a_{1}, d_{1}$ and $\mathrm{S}_{1}$, respectively, and those of the second A.P. be $a_{2}, d_{2}$ and $\mathrm{S}_{2}$, respectively. Then, we have, $S_{1}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]$ And, $S_{2}=\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]$ Given: $\frac{S_...
Read More →Two parallel lines l and m are intersected by a transversal t.
Question: Two parallel lineslandmare intersected by a transversalt. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. Solution: Given:l||mand thebisectorsofinterioranglesintersectatBandD.To prove:ABCDis a rectangle.Proof:Since,l||m (Given) So, $\angle P A C=\angle A C R$ (Alternate interior angles) $\Rightarrow \frac{1}{2} \angle P A C=\frac{1}{2} \angle A C R$ $\Rightarrow \angle B A C=\angle A C D$ but, these are a pair of alternate interior angles forABand...
Read More →Draw a circle of radius 6 cm. From a point 10 cm away from its centre,
Question: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Solution: Given that Construct a circle of radius, andform its centre, construct the pair of tangents to the circle. Find the length of tangents. We follow the following steps to construct the given tep of construction Step: I- First of all we draw a circle of radius. Step: II- Make a pointPat a distance of, and join. Step: III -Draw a right bis...
Read More →The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6.
Question: The sums ofnterms of two arithmetic progressions are in the ratio 5n+ 4 : 9n+ 6. Find the ratio of their 18th terms. Solution: Let there be two A.P.s. Let their first terms be $a_{1}$ and $a_{2}$ and their common differences be $d_{1}$ and $d_{2}$. Given : $\frac{5 n+4}{9 n+6}=\frac{\text { Sum of } n \text { terms in the first A.P. }}{\text { Sum of } n \text { terms in the second A.P. }}$ $\Rightarrow \frac{5 n+4}{9 n+6}=\frac{2 a_{1}+\left[(n-1) d_{1}\right]}{2 a_{2}+\left[(n-1) d_{...
Read More →In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC.
Question: In the adjoining figure,ABCDis a parallelogram andEis the midpoint of sideBC. IfDEandABwhen produced meet atF, prove thatAF= 2AB. Solution: Given:ABCD is a parallelogram.BE = CE (E is the mid point of BC)DE and AB when produced meet at F.To prove:AF = 2ABProof:In parallelogram ABCD, we have: AB|| DC DCE = EBF (Alternate interior angles)In∆DCE and ∆BFE, we have:DCE=EBF (Proved above) DEC =BEF (Vertically opposite angles)Also, BE =CE (Given)∆DCE ∆BFE (By ASA congruence rule) DC = BF (...
Read More →If the sum of n terms of an A.P.
Question: If the sum of $n$ terms of an A.P. is $n P+\frac{1}{2} n(n-1) Q$, where $P$ and $Q$ are constants, find the common difference. Solution: We have: $S_{n}=n P+\frac{1}{2} n(n-1) Q$ For $n=1, S_{1}=P+0=P$ For $n=2, S_{2}=2 P+Q$ Also, $a_{1}=S_{1}=P$ $a_{2}=S_{2}-S_{1}$ $=2 P+Q-P=P+Q$ $\therefore d=a_{2}-a_{1}=P+Q-P=Q$...
Read More →In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB.
Question: In the adjoining figure,ABCDis a parallelogram in whichABis produced toEso thatBE=AB. Prove thatEDbisectsBC. Solution: In∆ODCand∆OEB, we have:DC = BE (∵DC = AB)COD= BOE (Vertically opposite angles)OCD= OBE ( Alternate interior angles)i.e., ∆ODC∆OEB⇒OC = OB (CPCT)We know thatBC = OC + OB. ED bisects BC....
Read More →Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.
Question: Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms. Solution: Given: $S_{n}=3 n^{2}$ For $n=1, S_{1}=3 \times 1^{2}=3$ For $n=2, S_{2}=3 \times 2^{2}=12$ For $n=3, S_{3}=3 \times 3^{2}=27$ and so on $\therefore S_{1}=a_{1}=3$ $a_{2}=S_{2}-S_{1}=12-3=9$ $a_{3}=S_{3}-S_{2}=27-12=15$ and so on Thus, the A.P. is $3,9,15 \ldots$...
Read More →Find the value
Question: Find the value of $\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$. Solution: $\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$ $=\tan ^{-1}\left\{\tan \left(\frac{\pi}{3}\right)\right\}-\cot ^{-1}\left(\cot \frac{5 \pi}{6}\right)$ $\left[\because\right.$ Range of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ; \frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and range of $\cot ^{-1}$ is $\left.[0, \pi] ; \frac{5 \pi}{6} \in[0, \pi]\right]$ $=\frac{\pi}{3}-\frac{5 \pi}{...
Read More →ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.
Question: ABCDis a rectangle in which diagonalACbisects Aas well as C. Show that (i)ABCDis a square, (ii) diagonalBDbisects Bas well as D. Solution: Given: In rectangleABCD,ACbisectsA,i.e.1=2andACbisectsC,i.e.3=4.Toprove:(i)ABCDis a square,(ii) diagonalBDbisects Bas well as D.Proof: (i) Since, $A D \| B C$(Opposite sides of a rectangle are parallel.) So, $\angle 1=\angle 4$ (Alternate interior angles) But, $\angle 1=\angle 2$ (Given) So, $\angle 2=\angle 4$ In $\Delta A B C$, Since, $\angle 2=\a...
Read More →Solve the following
Question: IfS1be the sum of (2n+ 1) terms of an A.P. andS2be the sum of its odd terms, the prove that: S1:S2= (2n+ 1) : (n+ 1) Solution: Let the A.P. be $a, a+\mathrm{d}, a+2 d \ldots$ $\therefore S_{1}=\frac{2 n+1}{2}[2 a+(2 n+1-1) d]$ $\Rightarrow S_{1}=\frac{2 n+1}{2}[2 a+(2 n) d]$ $\Rightarrow S_{1}=(2 n+1)(a+n d) \quad \ldots(i)$ $S_{2}=\frac{n+1}{2}[2 a+(n+1-1) \times 2 d]$ $\Rightarrow S_{2}=\frac{n+1}{2}[2 a+2 n d]$ $\Rightarrow S_{2}=(n+1)[a+n d]$ ....(ii) From (i) and (ii), we get: $\f...
Read More →Find the value
Question: Find the value of $\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)$ Solution: $\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]$ $=\tan ^{-1}\left[\tan \left(\frac{\pi}{8}\right)\right]$ $=\frac{\pi}{8}$...
Read More →In the given figure, a circle with centre O is inscribed in a quadrilateral ABCD such that,
Question: In the given figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, B = 90 and DS = 5 cm, then the radius of the circle (in cm) is (a) 11 (b) 18 (c) 6 (d) 15 [CBSE 2013] Solution: It is given that the sides BC, AB, AD and CD touches a circle with centre O at points P, Q, R and S, respectively.AB = 29 cm, AD = 23 cm, B = 90 and DS = 5 cm.Now,DR = DS = 5 cm (Lengths...
Read More →In the given figure, DE and DF are tangents from an external point D to a
Question: In the given figure, DE and DF are tangents from an external point D to a circle with centre A.If DE = 5 cm and DE DF, then the radius of the circle is (a) 3 cm (b) 5 cm (c) 4 cm (d) 6 cm [CBSE 2013] Solution: It is given that DE and DF are tangents from an external point D to a circle with centre A.DE = 5 cm and DE DF. Join AE and AF.Now, DE is a tangent at E and AE is the radius through the point of contact E. $\therefore \angle \mathrm{AED}=90^{\circ} \quad$ (Tangent at any point of...
Read More →In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms.
Question: In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms. Show that 20th term is 112. Solution: Given: $a=2, S_{5}=\frac{1}{4}\left(S_{10}-S_{5}\right)$ We have: $S_{5}=\frac{5}{2}[2 \times 2+(5-1) d]$ $\Rightarrow S_{5}=5[2+2 d] \ldots(i)$ Also, $S_{10}=\frac{10}{2}[2 \times 2+(10-1) d]$ $\Rightarrow S_{10}=5[4+9 d] \ldots . .(i i)$ $\because S_{5}=\frac{1}{4}\left(S_{10}-S_{5}\right)$ From (i) and (ii), we have: $\Rightarrow 5[2+2 \mathr...
Read More →Find the value
Question: Find the value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ Solution: $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]$ $=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]$ $=\frac{\pi}{6}$...
Read More →In the given figure, a circle touches the side DF of ΔEDF at H and touches ED
Question: In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ΔEDF is (a) 18 cm (b) 13.5 cm (c) 12 cm (d) 9 cm [CBSE 2012] Solution: In the given figure, DH and DK are tangents drawn to the circle from an external point D. DH = DK (Lengths of tangents drawn from an external point to a circle are equal)FH and FM are tangents drawn to the circle from an external point F. FH = FM (Lengths of tange...
Read More →if
Question: If $\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0$, find the value of $x$ Solution: $\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0$ $\Rightarrow \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=\cos \left(\frac{\pi}{2}\right)$ $\Rightarrow \sin ^{-1} \frac{2}{5}+\cos ^{-1} x=\frac{\pi}{2}$ $\therefore x=\frac{2}{5} \quad\left[\because \sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}\right]$...
Read More →How many terms of the A.P.
Question: How many terms of the A.P. $-6,-\frac{11}{2},-5, \ldots$ are needed to give the sum $-25 ?$ Solution: Given: An A.P. with $a=-6$ and $d=-\frac{11}{2}-(-6)=\frac{1}{2}$ $S_{n}=-25$ $\therefore-25=\frac{n}{2}\left[2 \times(-6)+(n-1) \frac{1}{2}\right]$ $\Rightarrow-25=\frac{n}{2}\left[-12+\frac{n}{2}-\frac{1}{2}\right]$ $\Rightarrow-50=n\left[\frac{n}{2}-\frac{25}{2}\right]$ $\Rightarrow-100=n(n-25)$ $\Rightarrow n^{2}-25 n+100=0$ $\Rightarrow(n-20)(n-5)=0$ $\Rightarrow n=20$ or $n=5$...
Read More →In the given figure, the sides AB ,
Question: In the given figure, the sides AB , BC and CA of triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is (a) 11 cm (b) 10 cm (c) 14 cm (d) 15 cm [CBSE 2012] Solution: It is given that the sides AB , BC and CA of ∆ABC touch a circle at P, Q and R, respectively.Also, PA = 4 cm, PB = 3 cm and AC = 11 cmWe know that, the lengths of tangents drawn from an external point to a circle are equal. AR = AP = 4 cmBQ = BP = 3 cmNow, CR =...
Read More →The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°.
Question: The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60. Find the angles of the parallelogram. Solution: Given: InparallelogramABCD,DPAB,AQBCandPDQ=60InquadrilateralDPBQ, by angle sum property, we have $\angle P D Q+\angle D P B+\angle B+\angle B Q D=360^{\circ}$ $\Rightarrow 60^{\circ}+90^{\circ}+\angle B+90^{\circ}=360^{\circ}$ $\Rightarrow \angle B=360^{\circ}-240^{\circ}$ $\Rightarrow \angle B=120^{\circ}$ Therefore, $\an...
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