In ∆ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O.
Question: In∆ABC,AB=ACand the bisectors of Band Cmeet at a pointO. Prove thatBO=COand the rayAOis the bisector of A. Solution: In triangle ABC, we have:AB = AC (Given) $\Rightarrow \angle B=\angle C$ $\Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle C$ $\Rightarrow \angle O B C=\angle O C B$ $\Rightarrow B O=C O$ Now, in $\triangle A O B$ and $\triangle A O C$, we have: $O B=O C \quad$ (Proved) $A B=A C \quad$ (Given) $A O=A O \quad$ (Common) $\therefore \triangle A O B \cong \triangle A O C ...
Read More →In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to
Question: In an AP. $S_{p}=q, S_{q}=p$ and $S_{r}$ denotes the sum of first $r$ terms. Then, $S_{p+q}$ is equal to (a) 0(b) (p+q)(c)p+q(d)pq Solution: In the given problem, we are given $S_{p}=q$ and $S_{q}=p$ We need to find $S_{p+q}$ Now, as we know, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ So, $S_{p}=\frac{p}{2}[2 a+(p-1) d]$ $q=\frac{p}{2}[2 a+(p-1) d]$ $2 q=2 a p+p(p-1) d$ .............(1) Similarly, $S_{q}=\frac{q}{2}[2 a+(q-1) d]$ $p=\frac{q}{2}[2 a+(q-1) d]$ $2 p=2 a q+q(q-1) d$...............(2...
Read More →In ∆ABC, D is the midpoint of BC. If DL ⊥ AB and DM ⊥ AC such that DL = DM, prove that AB = AC.
Question: In∆ABC,Dis the midpoint ofBC. IfDLABandDMACsuch thatDL=DM, prove thatAB=AC. Solution: In $\triangle \mathrm{BDL}$ and $\triangle \mathrm{CDM}$, we have: $\mathrm{BD}=\mathrm{CD} \quad$ (D is midpoint) $\mathrm{DL}=\mathrm{DM} \quad$ (Given) $\angle \mathrm{BLD}=\angle \mathrm{CMD} \quad\left(90^{\circ}\right.$ each $)$ Thus, $\triangle \mathrm{BDL} \cong \triangle \mathrm{CDM}$ (RHS criterion) $\Rightarrow \mathrm{BL}=\mathrm{MC}(\mathrm{CPCT})$ $\therefore \angle B=\angle C$ This make...
Read More →Solve the following
Question: If 215is divided by 13, the remainder is ___________. Solution: If 215is divided by 13 Since 215= (5 3)15 =10C0(5)15(3)0+ .......... Correction...
Read More →Which of the following corresponds to the principal value branch
Question: Which of the following corresponds to the principal value branch of $\tan ^{-1}$ ? (a) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)-\{0\}$ (d) $(0, \pi)$ Solution: The principal value branch of $\tan ^{-1} x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Hence, the correct answer is option (a)....
Read More →The position of the term independent of x in the expansion of
Question: The position of the term independent of $x$ in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is _____________ Solution: for $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ $T_{r+1}={ }^{10} C_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^{2}}\right)^{r}$ $={ }^{10} C_{r} \frac{1}{(\sqrt{3})^{10-r}} x^{\frac{10-r}{2}} x^{-2 r}\left(\frac{3}{2}\right)^{r}$ $T_{r+1}=\frac{{ }^{10} C_{r}}{(\sqrt{3})^{10-r}}\left(\frac{3}{2}\right)^{r} \qua...
Read More →In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ∆ABC such that AX = AY.
Question: In the adjoining figure,XandYare respectively two points on equal sidesABandACof∆ABCsuch thatAX=AY. Prove thatCX=BY. Solution: In $\triangle \mathrm{AXC}$ and $\triangle \mathrm{AYB}$, we have : $\mathrm{AC}=\mathrm{AB}$ (Given) $\mathrm{AX}=\mathrm{AY}$ (Given) $\angle \mathrm{BAC}=\angle \mathrm{CAB}$ (Angle common to $\triangle \mathrm{AXC}$ and $\triangle \mathrm{AYB}$ ) $\therefore \triangle \mathrm{AXC} \cong \triangle \mathrm{AYB} \quad$ (SAS criterion) So, $\mathrm{CX}=\mathrm{...
Read More →The principal value branch
Question: The principal value branch of $\sec ^{-1}$ is (a) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ (b) $[0, \pi]-\left\{\frac{\pi}{2}\right\}$ (c) $(0, \pi)$ (d) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Solution: The principal value branch of $\sec ^{-1} x$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$. Hence, the correct answer is option (b)....
Read More →In the expansion of
Question: In the expansion of $\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$, the value of the constant term is ________________ Solution: $\operatorname{In}\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$ $T_{r+1}={ }^{16} C_{r}\left(x^{2}\right)^{16-r}\left(\frac{-1}{x^{2}}\right)^{r}$ $={ }^{16} C_{r} x^{32-2 r}(-1)^{r} x^{-2 r}$ i.e. $T_{r+1}={ }^{16} C_{r}(-1)^{r} x^{32-4 r}$ for constant term, $x^{32-4 r}=x^{0}$ i.e. $32-4 r=0$ i.e. $r=8$ $\therefore T_{8+1}={ }^{16} C_{8}$ i.e value of constant term i...
Read More →In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle.
Question: In the given figure,Ois a point in the interior of squareABCDsuch that ΔOABis an equilateral triangle. Show that ΔOCDis an isosceles triangle. Solution: Given:InsquareABCD,ΔOABis an equilateral triangle.To prove:ΔOCDis an isosceles triangle.Proof: $\because \angle D A B=\angle C B A=90^{\circ} \quad$ (Angles of square $A B C D$ ) And, $\angle O A B=O B A=60^{\circ}$ (Angles of equilateral $\triangle O A B$ ) $\therefore \angle D A B-\angle O A B=\angle C B A-\angle O B A=90^{\circ}-60^...
Read More →If Sn denote the sum of the first n terms of an A.P.
Question: If $S_{n}$ denote the sum of the first $n$ terms of an A.P. If $S_{2 n}=3 S_{n}$, then $S_{3 n}: S_{n}$ is equal to (a) 4(b) 6(c) 8(d) 10 Solution: Here, we are given an A.P. whose sum of $n$ terms is $S_{n}$ and $S_{2 n}=3 S_{n}$. We need to find $\frac{S_{3 n}}{S_{n}}$. Here we use the following formula for the sum ofnterms of an A.P., $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ Where;a= first term for the given A.P. d= common difference of the given A.P. n= number of terms So, first we findS3n...
Read More →The coefficient of
Question: The coefficient of $a^{-6} b^{4}$ in the expansion $\left(\frac{1}{a}-\frac{2}{3} b\right)^{10}$ is Solution: $\operatorname{In}\left(\frac{1}{a}-\frac{2}{3} b\right)^{10}$' $T_{r+1}={ }^{10} C_{r}\left(\frac{1}{a}\right)^{10-r}\left(\frac{-2 b}{3}\right)^{r}$ for coefficient of $a^{-6} b^{4}$ put $10-r=6$ i. e. $r=4$ We get coefficient of $a^{-6} b^{4}={ }^{10} C_{4}\left(\frac{-2}{3}\right)^{4}$ $=\frac{10 !}{4 ! 6 !} \times \frac{2^{4}}{3^{4}}$ $=\frac{10 \times 9 \times 8 \times 7}...
Read More →One branch of cos-1 other than the principal value branch corresponds to
Question: One branch of $\cos ^{-1}$ other than the principal value branch corresponds to (a) $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$ (b) $[\pi, 2 \pi]-\left\{\frac{3 \pi}{2}\right\}$ (c) $(0, \pi)$ (d) $[2 \pi, 3 \pi]$ Solution: The domain of the function $f(x)=\cos ^{-1} x$ is $[-1,1]$. The range of $\cos ^{-1} x$ in one of the intervals $\ldots,[-\pi, 0],[0, \pi],[\pi, 2 \pi],[2 \pi, 3 \pi], \ldots$ is one-one and onto with the range $[-1,1]$. Thus, one branch of $\cos ^{-1} x$ other th...
Read More →In the given figure, ABCD is a square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.
Question: In the given figure,ABCDis a square andPis a point inside it such thatPB=PD. Prove thatCPAis a straight line. Solution: In $\triangle \mathrm{PAD}$ and $\triangle \mathrm{PAB}$, we have : $\mathrm{AD}=\mathrm{AB} \quad$ (Side of a square) $\mathrm{AP}=\mathrm{AP} \quad($ Common $)$ $\mathrm{PD}=\mathrm{PB} \quad$ (Given) $\triangle \mathrm{PAD} \cong \mathrm{PAB} \quad$ (SSS criterion) $\Rightarrow \angle \mathrm{APD}=\angle \mathrm{APB}$ In $\triangle \mathrm{CPD}$ and $\triangle \mat...
Read More →The ratio of the coefficients of x
Question: The ratio of the coefficients ofxmandxnin the expansion of (1 +x)m+nis ___________. Solution: For (1 +x)m+n Tr+1=m+nCrxr for coefficient ofxm,r=m i.e coefficient ism+nCm and for coefficient ofxn,r = n i.e. coefficient ism+nCn $\Rightarrow$ Ratio of coefficient of $x^{m}$ to $x^{n}$ is $\frac{m+n}{m+n} C_{m}=\frac{{ }^{m+n} C_{m+n-m}}{m+n C_{n}}$ $=\frac{{ }^{m+n} C_{n}}{{ }^{m+n} C_{n}}=1$...
Read More →if the
Question: If $\tan ^{-1} x=\frac{x}{10}$ for some $x \in \mathrm{R}$, then the value of $\cot ^{-1} x$ is (a) $\frac{\pi}{5}$ (b) $\frac{2 \pi}{5}$ (c) $\frac{3 \pi}{5}$ (d) $\frac{4 \pi}{5}$ Solution: Disclaimer:The solution has been provided for the following question. If $\tan ^{-1} x=\frac{\pi}{10}$ for some $x \in \mathrm{R}$, then the value of $\cot ^{-1} x$ is (a) $\frac{\pi}{5}$ (b) $\frac{2 \pi}{5}$ (c) $\frac{3 \pi}{5}$ (d) $\frac{4 \pi}{5}$ Solution: We know $\tan ^{-1} x+\cot ^{-1} x...
Read More →Middle term in the expansion of
Question: Middle term in the expansion of (a3+ba)28is ___________. Solution: In (a3+ba)28 Here n = 28 So, there is one middle term i.e. $\left(\frac{28}{2}+1\right)$ th term which is 15 th term $\therefore$ Middle term is $T_{15}$ i. e. $T_{14+1}={ }^{28} C_{14}\left(a^{3}\right)^{28-14}(b a)^{14} \quad$ (using $T_{r+1}={ }^{n} C_{r} x^{r} y^{n-r}$ for $\left.(x+y)^{n}\right)$ $={ }^{28} C_{14}\left(a^{3}\right)^{14} b^{14} a^{14}$ $={ }^{28} C_{14} a^{56} a^{14}$...
Read More →If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
Question: If in an A.P. $S_{n}=n^{2} p$ and $S_{m}=m^{2} p$, where $S_{r}$ denotes the sum of $r$ terms of the A.P., then $S_{p}$ is equal to (a) $\frac{1}{2} p^{3}$ (b) $m n p$ (c) $p^{3}$ (d) $(m+n) p^{2}$ Solution: In the given problem, we are given an A.P whose $S_{n}=n^{2} \mathrm{p}$ and $S_{n}=m^{2} \mathrm{p}$ We need to find $S_{p}$ Now, as we know, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ Where, first term =a Common difference =d Number of terms =n So, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $n^{2} p...
Read More →In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint of BC.
Question: In the given figure,ABCDis a quadrilateral in whichAB||DCandPis the midpoint ofBC. On producing,APandDCmeet atQ. Prove that (i)AB=CQ, (ii)DQ=DC+AB. Solution: In $\triangle \mathrm{ABP}$ and $\triangle \mathrm{QCP}$, we have: $\angle \mathrm{BPA}=\angle \mathrm{CPQ} \quad($ Vertically opposite angle $)$ $\angle \mathrm{PAB}=\angle \mathrm{PQC} \quad$ (Alternate angles) $\mathrm{PB}=\mathrm{PC} \quad(\mathrm{P}$ is the midpoint) $\triangle \mathrm{ABP} \cong \triangle \mathrm{QCP} \quad$...
Read More →The number of terms in the expansion of (x + y + z)
Question: The number of terms in the expansion of (x+y+z)nis ___________. Solution: $(x+y+z)^{n}$ $=[x+(y+z)]^{n}$ $={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}(y+z)+{ }^{n} C_{2} x^{n-2}(y+z)^{2}+\ldots+{ }^{n} C_{n}(y+z)^{n}$ Number of terms in the expansion $=1+2+\ldots+n+(n+1)$ $=\frac{(n+1)(n+2)}{2}={ }^{n+2} C_{2}$...
Read More →The largest coefficient in
Question: The largest coefficient in (1 +x)41is ___________. Solution: In (1 +x)41 Sincenis odd i.e 41 The largest coefficient is41C21or41C20 i. e. $\left(\frac{n+1}{2}\right.$ or $\left.\frac{n+1}{2}+1\right)$ th term gives largest coefficient...
Read More →The value of
Question: The value of $\cot \left(\sin ^{-1} x\right)$ is $(a) \frac{\sqrt{1+x^{2}}}{x}$ $(\mathrm{b}) \frac{x}{\sqrt{1+x^{2}}}$ $(\mathrm{c}) \frac{1}{x}$ $(\mathrm{d}) \frac{\sqrt{1-x^{2}}}{x}$ Solution: We know $\sin ^{-1} x=\cot ^{-1} \frac{\sqrt{1-x^{2}}}{x}$ $\therefore \cot \left(\sin ^{-1} x\right)=\cot \left(\cot ^{-1} \frac{\sqrt{1-x^{2}}}{x}\right)$ $\Rightarrow \cot \left(\sin ^{-1} x\right)=\frac{\sqrt{1-x^{2}}}{x}$ $\left[\cot \left(\cot ^{-1} x\right)=x, \forall x \in \mathrm{R}\...
Read More →In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP.
Question: In the given figure,ABCis an equilateral triangle;PQ||ACandACis produced toRsuch thatCR= BP. Prove thatQRbisectsPC. Solution: Since $\Delta \mathrm{ABC}$ is an equilateral $\Delta$, then $\angle \mathrm{ABC}=\angle \mathrm{BCA}=\angle \mathrm{CAB}=60^{\circ}$ Since PQ $\| \mathrm{CA}$ and PC is a transversal, then $\angle \mathrm{BPQ}=\angle \mathrm{BCA}=60^{\circ} \quad$ (Corresponding angles) Since PQ $\| \mathrm{CA}$ and QA is a transversal, then $\angle \mathrm{BQP}=\angle \mathrm{...
Read More →The largest coefficient in (1 + x)
Question: The largest coefficient in (1 +x)30is ___________. Solution: for $(1+x)^{30}$ $T_{r+1}={ }^{n} C_{r}(1)^{30-r} x^{r}$ Since $n=30 \quad$ i.e. $\quad$ even The largest coefficient is fo $r r=\frac{n}{2}=15$ i.e for middle term which is30C15....
Read More →If S1 is the sum of an arithmetic progression of 'n' odd number of terms and
Question: If $\mathrm{S}_{1}$ is the sum of an arithmetic progression of ' $n$ ' odd number of terms and $\mathrm{S}_{2}$ the sum of the terms of the series in odd places, then $\frac{S_{1}}{S_{2}}=$ (a) $\frac{2 n}{n+1}$ (b) $\frac{n}{n+1}$ (c) $\frac{n+1}{2 n}$ (d) $\frac{n+1}{n}$ Solution: In the given problem, we are givenas the sum of an A.P of n odd number of terms andthe sum of the terms of the series in odd places. We need to find $\frac{S_{1}}{S_{2}}$ Now, leta1,a2.anbe thenterms of A.P...
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