The coefficient of
Question: The coefficient of $\frac{1}{x}$ in the expansion of $(1+x)^{n}\left(1+\frac{1}{x}\right)^{n}$ is (a) $\frac{n !}{[(n-1) !(n+1) !]}$ (b) $\frac{(2 n) !}{[(n-1) !(n+1) !]}$ (c) $\frac{(2 n) !}{(2 n-1) !(2 n+1) !}$ (d) none of these Solution: (b) $\frac{(2 n) !}{[(n-1) !(n+1) !]}$ Coefficient of $\frac{1}{x}$ in the given expansion $=$ Coefficient of 1 in $(1+x)^{n} \times$ Coefficient of $\frac{1}{x}$ in $\left(1+\frac{1}{x}\right)^{n}+$ Coefficient of $x$ in $(1+x)^{n} \times$ Coeffici...
Read More →The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively.
Question: The sidesBC,CAandABof ∆ABChave been produced toD,EandF, respectively. BAE+ CBF +ACD= ?(a) 240(b) 300(c) 320(d) 360 Solution: (d) 360 We have : $\angle 1+\angle B A E=180^{\circ} \ldots($ i $)$ $\angle 2+\angle C B F=180^{\circ} \quad \ldots($ ii $)$ and $\angle 3+\angle A C D=180^{\circ} \quad \ldots($ iii $)$ Adding $(i),(i i)$ and $(i i i)$, we get: $(\angle 1+\angle 2+\angle 3)+(\angle B A E+\angle C B F+\angle A C D)=540^{\circ}$ $\Rightarrow 180^{\circ}+\angle B A E+\angle C B F+\...
Read More →Solve the following
Question: If $T_{2} / T_{3}$ in the expansion of $(a+b)^{n}$ and $T_{3} / T_{4}$ in the expansion of $(a+b)^{n+3}$ are equal, then $n=$ (a) 3 (b) 4 (c) 5 (d) 6 Solution: (c) 5 In the expansion $(a+b)^{n}$, we have $\frac{T_{2}}{T_{3}}=\frac{{ }^{n} C_{1} a^{n-1} \times b^{1}}{{ }^{n} C_{2} a^{n-2} \times b^{2}}$ In the expansion $(a+b)^{n+3}$, we have $\frac{T_{3}}{T_{4}}=\frac{{ }^{n+3} C_{2} a^{n+1} b^{2}}{{ }^{n+3} C_{3} a^{n} b^{3}}$ Thus, we have $\frac{T_{2}}{T_{3}}=\frac{T_{3}}{T_{4}}$ $\...
Read More →The total number of terms in the expansion of
Question: The total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ after simplification is (a) 202 (b) 51 (c) 50 (d) none of these Solution: (b) 51 Here,n, i.e., 100, is even. $\therefore$ Total number of terms in the expansion $=\frac{n}{2}+1=\frac{100}{2}+1=51$...
Read More →In the given figure, the sides CB and BA of ∆ABC have been produced to D and E,
Question: In the given figure, the sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ABD = 110 and CAE = 135. Then ACB = ?(a) 65(b) 45(c) 55(d) 35 Solution: (a) 65 We have : $\angle A B D+\angle A B C=180^{\circ} \quad[\because C B D$ is a straight line $]$ $\Rightarrow 100^{\circ}+\angle A B C=180^{\circ}$ $\Rightarrow \angle A B C=70^{\circ}$ Side AB of triangle ABC is produced to E. $\therefore \angle C A E=\angle A B C+\angle A C B$ $\Rightarrow 135^{\circ}=70^{\...
Read More →The coefficient of x
Question: The coefficient of $x^{4}$ in $\left(\frac{x}{2}-\frac{3}{x^{2}}\right)^{10}$ is (a) $\frac{405}{256}$ (b) $\frac{504}{259}$ (c) $\frac{450}{263}$ (d) none of these Solution: (a) $\frac{405}{256}$ Suppose $x^{4}$ occurs at the $(r+1)$ th term in the given expansion. Then, we have $T_{r+1}={ }^{10} C_{r}\left(\frac{x}{2}\right)^{10-r}\left(\frac{-3}{2 x^{2}}\right)^{r}$ $=(-1)^{r}{ }^{10} C_{r} \frac{3^{r}}{2^{10-r}} x^{10-r-2 r}$ For this term to contain $x^{4}$, we must have : $10-3 r...
Read More →If the coefficient of x in
Question: If the coefficient of $x$ in $\left(x^{2}+\frac{\lambda}{x}\right)^{5}$ is 270, then $\lambda=$ (a) 3 (b) 4 (c) 5 (d) none of these Solution: (a) 3 The coefficient of $x$ in the given expansion where $x$ occurs at the $(r+1)$ th term. We have: ${ }^{5} C_{r}\left(x^{2}\right)^{5-r}\left(\frac{\lambda}{x}\right)^{r}$ $={ }^{5} C_{r} \lambda^{r} x^{10-2 r-r}$ For it to contain $x$, we must have : $10-3 r=1$ $\Rightarrow r=3$ $\therefore$ Coefficient of $x$ in the given expansion: ${ }^{5...
Read More →If the sum of odd numbered terms and the sum of even numbered terms in the expansion of
Question: If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $(x+a)^{n}$ are $A$ and $B$ respectively, then the value of $\left(x^{2}-a^{2}\right)^{n}$ is (a) $A^{2}-B^{2}$ (b) $A^{2}+B^{2}$ (c) $4 A B$ (d) none of these Solution: (a) $A^{2}-B^{2}$ If $A$ and $B$ denote respectively the sums of odd terms and even terms in the expansion $(x+a)^{n}$ Then, $(x+a)^{n}=A+B \quad \ldots(1)$ $(x-a)^{n}=A-B \quad \ldots(2)$ Multplying both the equations we get $(x+a)...
Read More →Side BC of ∆ABC has been produced to D on left and to E on right hand side of BC such that ∠ABD = 125° and ∠ACE = 130°.
Question: Side BC of ∆ABC has been produced to D on left and to E on right hand side of BC such that ABD = 125 and ACE = 130. Then A = ?(a) 50(b) 55(c) 65(d) 75 Solution: (d) 75 We have : $\therefore \angle A B D+\angle A B C=180^{\circ} \quad[\because D E$ is a straight line $]$ $\Rightarrow 125^{\circ}+\angle A B C=180^{\circ}$ $\Rightarrow \angle A B C=55^{\circ}$ Also, $\angle A C E+\angle A C B=180^{\circ}$ $\Rightarrow 130^{\circ}+\angle A C B=180^{\circ}$ $\Rightarrow \angle A C B=50^{\ci...
Read More →In the expansion of
Question: In the expansion of $\left(\frac{1}{2} x^{1 / 3}+x^{-1 / 5}\right)^{8}$, the term independent of $x$ is (a)T5 (b)T6 (c)T7 (d)T8 Solution: (b) $T_{6}$ Suppose the (r+ 1)th term in the given expansion is independent ofx. Thus, we have: $T_{r+1}={ }^{8} C_{r}\left(\frac{1}{2} x^{1 / 3}\right)^{8-r}\left(x^{-1 / 5}\right)^{r}$ $={ }^{8} C_{r} \frac{1}{2^{8-r}} x^{\frac{8-r}{3}-\frac{r}{5}}$ For this term to be independent of $x$, we must have $\frac{8-r}{3}-\frac{r}{5}=0$ $\Rightarrow 40-5...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following:If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is(a) 87(b) 88(c) 89(d) 90 Solution: In the given problem, we are given 7thand 13thterm of an A.P. We need to find the 26thterm Here, $a_{7}=34$ $a_{13}=64$ Now, we will find $a_{7}$ and $a_{13}$ using the formula $a_{n}=a+(n-1) d$ So, $a_{7}=a+(7-1) d$ $34=a+6 d$ .......(1) Also, $a_{13}=a+(13-1) d$ $64=a+12 d$.......(2) Further, to solve foraandd On sub...
Read More →In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?
Question: In ∆ABC, sideBC is produced to D. IfABC = 50 and ACD = 110 thenA = ?(a) 160(b) 60(c) 80(d) 30 Solution: $\therefore \angle A+\angle B=\angle A C D$ $\Rightarrow \angle A+50^{\circ}=110^{\circ}$ $\Rightarrow \angle A=60^{\circ}$ Hence, the correct answer is option (b)....
Read More →If in the expansion of
Question: If in the expansion of (1 +y)n, the coefficients of 5th, 6th and 7th terms are in A.P., thennis equal to (a) 7, 11 (b) 7, 14 (c) 8, 16 (d) none of these Solution: (b) 7, 14 Coefficients of the 5 th, 6 th and 7 th terms in the given expansion are ${ }^{n} C_{4},{ }^{n} C_{5}$ and ${ }^{n} C_{6}$ These coefficients are in $A P$. Thus, we have $2^{n} C_{5}={ }^{n} C_{4}+{ }^{n} C_{6}$ On dividing both sides by ${ }^{n} C_{5}$, we get: $2=\frac{{ }^{n} C_{4}}{{ }^{n} C_{5}}+\frac{{ }^{n} C...
Read More →In the expansion of
Question: In the expansion of $\left(x-\frac{1}{3 x^{2}}\right)^{9}$, the term independent of $x$ is (a)T3 (b)T4 (c)T5 (d) none of these Solution: (b) $T_{4}$ Suppose $T_{\mathrm{r}+1}$ is the term in the given expression that is independent of $x$. Thus, we have : $T_{r+1}={ }^{9} C_{r} x^{9-r}\left(\frac{-1}{3 x^{2}}\right)^{r}$ $=(-1)^{r}{ }^{9} C_{r} \frac{1}{3^{r}} x^{9-r-2 r}$ For this term to be independent of $x$, we must have $9-3 r=0$ $\Rightarrow r=3$ Hence, the required term is the 4...
Read More →In a ∆ABC, if ∠A − ∠B = 42° and ∠B − ∠C = 21° then ∠B = ?
Question: Ina ∆ABC, if A B = 42 and B C = 21 then B = ? (a) 32(b) 63(c) 53(d) 95 Figure Solution: (c) 53 Let $\angle A-\angle B=42^{\circ} \quad \ldots(i)$ and $\angle B-\angle C=21^{\circ} \quad \ldots($ ii $)$ Adding $(i)$ and $(i i)$, we get: $\begin{array}{ll}\angle A-\angle C=63^{\circ} \\ \angle B=\angle A-42^{\circ} {[\text { Using }(i)]} \\ \angle C=\angle A-63^{\circ} {[\text { Using }(i i i)]}\end{array}$ $\therefore \angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a...
Read More →If the sum of first p term of an A.P. is ap2 + bp,
Question: If the sum of first $\rho$ term of an A.P. is $a p^{2}+b p$, find its common difference. Solution: Here, we are given, $S_{p}=a p^{2}+b p$ Let us take the first term asaand the common difference asd. Now, as we know, $a_{p}=S_{p}-S_{p-1}$ So, we get, $a_{p}=\left(a p^{2}+b p\right)-\left[a(p-1)^{2}+b(p-1)\right]$ $=a p^{2}+b p-\left[a\left(p^{2}+1-2 p\right)+b p-b\right]$ $\left[\right.$ Using $\left.(a-b)^{2}=a^{2}+b^{2}-a b\right]$ $=a p^{2}+b p-\left(a p^{2}+a-2 a p+b p-b\right)$ $=...
Read More →If in the expansion of
Question: If in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}, x^{-17}$ occurs in $r$ th term, then (a)r= 10 (b)r= 11 (c)r= 12 (d)r= 13 Solution: (c)r = 12 Here, $T_{r}={ }^{15} C_{r-1}\left(x^{4}\right)^{15-r+1}\left(\frac{-1}{x^{3}}\right)^{r-1}$ $=(-1)^{r} \times{ }^{15} C_{r-1} x^{64-4 r-3 r+3}$ For this term to contain $x^{-17}$, we must have : $67-7 r=-17$ $\Rightarrow r=12$...
Read More →In ∆ABC, if 3∠A = 4∠B = 6∠C then A : B : C = ?
Question: In ∆ABC, if 3A = 4B = 6C then A : B : C = ?(a) 3 : 4 : 6(b) 4 : 3 : 2(c) 2 : 3 : 4(d) 6 : 4 : 3 Solution: LCM of 3, 4 and 6 = 123A = 4B = 6C (Given)Dividing throughout by 12, we get $\frac{3 \angle \mathrm{A}}{12}=\frac{4 \angle \mathrm{B}}{12}=\frac{6 \angle \mathrm{C}}{12}$ $\Rightarrow \frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}$ Let $\frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}=k$, where $k$ is some cons...
Read More →The middle term in the expansion of
Question: The middle term in the expansion of $\left(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}}\right)^{10}$ is (a) 251 (b) 252 (c) 250 (d) none of these Solution: (b) 252 Here, $n$, i. e., 10 , is an even number. $\therefore$ Middle term $=\left(\frac{10}{2}+1\right)$ th term $=6$ th term Thus, we have $T_{6}=T_{5+1}$ $={ }^{10} C_{5}\left(\frac{2 x^{2}}{3}\right)^{10-5}\left(\frac{3}{2 x^{2}}\right)^{5}$ $=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times \frac{2^{5}}...
Read More →If 4/5, a, 2 are three consecutive terms of an A.P., then find the value of a.
Question: If $\frac{4}{5}, a, 2$ are three consecutive terms of an A.P., then find the value of $a$. Solution: Here, we are given three consecutive terms of an A.P. First term $\left(a_{1}\right)=\frac{4}{5}$ Second term $\left(a_{2}\right)=a$ Third term $\left(a_{3}\right)=2$ We need to find the value ofa.So, in an A.P. the difference of two adjacent terms is always constant. So, we get, $d=a_{2}-a_{1}$ $d=a-\frac{4}{5}$$\ldots(1)$ Also, $d=a_{3}-a_{2}$ $d=2-a$$\cdots(2)$ Now, on equating (1) a...
Read More →In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.
Question: In the given figure, AB || CD and EF AB. If EG is the transversal such that GED = 130, find EGF. Solution: In the given figure,AB || CD and GE is the transversal.GED +EGF = 180 (Sum of adjacent interior angles on the same side of the transversal is supplementary)⇒130+EGF= 180⇒EGF= 180130 =50Thus, the measure of EGF is 50...
Read More →If an the expansion of
Question: If an the expansion of $(1+x)^{15}$, the coefficients of $(2 r+3)^{\text {th }}$ and $(r-1)^{\text {th }}$ terms are equal, then the value of $r$ is (a) 5 (b) 6 (c) 4 (d) 3 Solution: (a) 5 Coefficients of $(2 r+3)$ th and $(r-1)$ th terms in the given expansion are ${ }^{15} C_{2 r+2}$ and ${ }^{15} C_{r-2}$. Thus, we have ${ }^{15} C_{2 r+2}={ }^{15} C_{r-2}$ $\Rightarrow 2 r+2=r-2 \quad$ or $2 r+2+r-2=15 \quad\left[\because\right.$ if ${ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x=y$ or ...
Read More →In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively.
Question: In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If EGB = 35 and QP EF, find the measure of PQH. Solution: In the given figure,AB || CD and EF is a transversal.PHQ =EGB (Pair of alternate exterior angles)⇒PHQ =35In∆PHQ,PHQ +QPH +PQH=180 (Angle sum property)⇒35 +90 +x=180⇒125 +x=180⇒x=180 125 = 55Thus, the measure of PQH is 55....
Read More →For what value of p are 2p + 1, 13, 5p − 3
Question: For what value of $p$ are $2 p+1,13,5 p-3$ are three consecutive terms of an A.P.? Solution: Here, we are given three terms, First term (a1) = Second term (a2) = Third term (a3) = We need to find the value ofpfor which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get, $d=a_{2}-a_{1}$ $d=13-(2 p+1)$ $d=13-2 p-1$ $d=12-2 p$ $\ldots(1)$ Also, $d=a_{3}-a_{2}$ $d=(5 p-3)-13$ $d=5 p-3-13$ $d=5 p-16$...........(2) Now, on equating (1) ...
Read More →In the expansion of
Question: In the expansion of $\left(x^{2}-\frac{1}{3 x}\right)^{9}$, the term without $x$ is equal to (a) $\frac{28}{81}$ (b) $\frac{-28}{243}$ (c) $\frac{28}{243}$ (d) none of these Solution: (c) $\frac{28}{243}$ Suppose the (r+ 1)th term in the given expansion is independent ofx. Then , we have: $T_{r+1}={ }^{9} C_{r}\left(x^{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^{r}$ $=(-1)^{r}{ }^{9} C_{r} \frac{1}{3^{r}} x^{18-2 r-r}$ For this term to be independent of $x$, we must have: $18-3 r=0$ $\...
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