Solve the following
Question: If $r^{\text {th }}$ term is the middle term in the expansion of $\left(x^{2}-\frac{1}{2 x}\right)^{20}$, then $(r+3)^{t h}$ term is (a) ${ }^{20} C_{14}\left(\frac{x}{2^{14}}\right)$ (b) ${ }^{20} C_{12} x^{2} 2^{-12}$ (c) ${ }^{20} C_{7} x, 2^{-13}$ (d) none of these Solution: (c) ${ }^{20} C_{7} x, 2^{-13}$ Here $n$ is even So, The middle term in the given expansion is $\left(\frac{20}{2}+1\right)$ th $=11$ th term Therefore, $(r+3)$ th term is the 14 th term. $T_{14}={ }^{20} C_{13...
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Question: If $4 \cos ^{-1} x+\sin ^{-1} x=\pi$, then the value of $x$ is (a) $\frac{3}{2}$ (b) $\frac{1}{\sqrt{2}}$ (c) $\frac{\sqrt{3}}{2}$ (d) $\frac{2}{\sqrt{3}}$ Solution: (c) $\frac{\sqrt{3}}{2}$ We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$. $4 \cos ^{-1} x+\sin ^{-1} x=\pi$ $\Rightarrow 4 \cos ^{-1} x+\frac{\pi}{2}-\cos ^{-1} x=\pi$ $\Rightarrow 3 \cos ^{-1} x=\pi-\frac{\pi}{2}$ $\Rightarrow 3 \cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \cos ^{-1} x=\frac{\pi}{6}$ $\Rightarrow x=\co...
Read More →The middle term in the expansion of
Question: The middle term in the expansion of $\left(\frac{2 x}{3}-\frac{3}{2 x^{2}}\right)^{2 n}$ is (a) ${ }^{2 n} C_{n}$ (b) $(-1)^{n}{ }^{2 n} C_{n} x^{-n}$ (c) ${ }^{2 n} C_{n} x^{-n}$ (d) none of these Solution: (b) $(-1)^{n}{ }^{2 n} C_{n} x^{-n}$ Here, $n$ is even Middle term in the given expansion $=\left(\frac{2 n}{2}+1\right)$ th $=(n+1)$ th term $={ }^{2 n} C_{n}\left(\frac{2 x}{3}\right)^{2 n-n}\left(\frac{-3}{2 x^{2}}\right)^{n}$ $=(-1)^{n}{ }^{2 n} C_{n} x^{-n}$...
Read More →Write the first five terms of each of the following sequences whose nth terms are:
Question: Write the first five terms of each of the following sequences whosenth terms are: (a) $a_{n}=3 n+2$ (b) $a_{n}=\frac{n-3}{3}$ (c) $a_{n}=3^{n}$ (d) $a_{n}=\frac{3 n-2}{5}$ (e) $a_{n}=(-1)^{n} 2^{n}$ (f) $a_{n}=\frac{n(n-2)}{2}$ (g) an $=n^{2}-n+1$ (h) $a_{n}=2 n^{2}-3 n+1$ (i) $a_{n}=\frac{2 n-3}{6}$ Solution: Here, we are given thenthterm for various sequences. We need to find the first five terms of the sequence. (i) $a_{n}=3 n+2$ Here, thenthterm is given by the above expression. So...
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Question: If $3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$ is equal to (a) $\frac{1}{\sqrt{3}}$ (b) $-\frac{1}{\sqrt{3}}$ (c) $\sqrt{3}$ (d) $-\frac{\sqrt{3}}{4}$ Solution: (a) $\frac{1}{\sqrt{3}}$ Let $x=\tan y$ Then, $3 \sin ^{-1}\left(\frac{2 \tan y}{1+\tan ^{2} y}\right)-4\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)+2 \tan ^{-1}\left(\frac{2 \tan y}{1-\tan ^{2} y}\right)=\frac{\pi...
Read More →If the coefficients of 2nd, 3rd and 4th terms in the expansion of
Question: If the coefficients of 2 nd, 3 rd and 4 th terms in the expansion of $(1+x)^{n}, n \in N$ are in A.P., then $n=$ (a) 7 (b) 14 (c) 2 (d) none of these Solution: (a) 7 Coefficients of the 2 nd, 3 rd and 4 th terms in the given expansion are: ${ }^{n} C_{1},{ }^{n} C_{2}$ and ${ }^{n} C_{3}$ We have : $2 \times{ }^{n} C_{2}={ }^{n} C_{1}+{ }^{n} C_{3}$ Dividing both sides by ${ }^{n} C_{2}$, we get: $2=\frac{{ }^{n} C_{1}}{{ }^{n} C_{2}}+\frac{{ }^{n} C_{3}}{{ }^{n} C_{2}}$ $\Rightarrow 2...
Read More →If the coefficients of the (n + 1)
Question: If the coefficients of the (n+ 1)thterm and the (n+ 3)thterm in the expansion of (1 +x)20are equal, then the value ofnis (a) 10 (b) 8 (c) 9 (d) none of these Solution: (c) 9 Coefficient of $(n+1)$ th term $=$ Coefficient of $(n+3)$ th We have : ${ }^{20} C_{n}={ }^{20} C_{n+2}$ $\Rightarrow 2 n+2=20 \quad\left[\because\right.$ if ${ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x=y$ or $\left.x+y=n\right]$ $\Rightarrow n=9$...
Read More →if the
Question: If $\theta=\sin ^{-1}\left\{\sin \left(-600^{\circ}\right)\right\}$, then one of the possible values of $\theta$ is (a) $\frac{\pi}{3}$ (b) $\frac{\pi}{2}$ (c) $\frac{2 \pi}{3}$ (d) $-\frac{2 \pi}{3}$ Solution: (a) $\frac{\pi}{3}$ We know $\sin ^{-1}(\sin x)=x$ Now, $\theta=\sin ^{-1}\left\{\sin \left(-600^{\circ}\right)\right\}$ $=\sin ^{-1}\left\{\sin \left(720^{\circ}-600^{\circ}\right)\right\}$ $=\sin ^{-1}\left\{\sin \left(120^{\circ}\right)\right\}$ $[\because \sin x=\sin (\pi-x)...
Read More →The coefficient of x
Question: The coefficient ofx8y10in the expansion of (x+y)18is (a)18C8 (b)18p10 (c) 218 (d) none of these Solution: (a) ${ }^{18} C_{8}$ Suppose the $(r+1)$ th term in the given expansion contains $x^{8} y^{10}$. Then, we have $T_{r+1}={ }^{18} C_{r} x^{18-r} y^{r}$ For the coefficient of $x^{8} y^{10}$ We have $r=10$ Hence, the required coefficient is ${ }^{18} C_{10}$ or ${ }^{18} C_{8}$...
Read More →The coefficient of x
Question: The coefficient of $x^{5}$ in the expansion of $(1+x)^{21}+(1+x)^{22}+\ldots+(1+x)^{30}$ (a)51C5 (b)9C5 (c)31C621C6 (d)30C5+20C5 Solution: (c) ${ }^{31} C_{6}-{ }^{21} C_{6}$ We have $(1+x)^{21}+(1+x)^{22}+\ldots(1+x)^{30}$ $=(1+x)^{21}\left[\frac{(1+x)^{10}-1}{(1+x)-1}\right]$ $=\frac{1}{x}\left[(1+x)^{31}-(1+x)^{21}\right]$ Coefficient of $x^{5}$ in the given expansion $=$ Coefficient of $x^{5}$ in $\frac{1}{x}\left[(1+x)^{31}-(1+x)^{21}\right]$ $=$ Coefficient of $x^{6}$ in $\left[(...
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Question: $\sin \left\{2 \cos ^{-1}\left(\frac{-3}{5}\right)\right\}$ is equal to (a) $\frac{6}{25}$ (b) $\frac{24}{25}$ (c) $\frac{4}{5}$ (d) $-\frac{24}{25}$ Solution: (d) $-\frac{24}{25}$ Let $\cos ^{-1}\left(-\frac{3}{5}\right)=x, 0 \leq x \leq \pi$ Then, $\cos x=-\frac{3}{5}$ $\therefore \sin x=\sqrt{1-\cos ^{2} x}=\sqrt{1-\left(-\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$ Now, $\sin \left\{2 \cos ^{-1}\left(-\frac{3}{5}\right)\right\}=\sin (2 x)$ $=2 \sin x \cos x$ $=2 \times...
Read More →The coefficient of the term independent of x in the expansion of
Question: The coefficient of the term independent of $x$ in the expansion of $\left(a x+\frac{b}{x}\right)^{14}$ is (a) $14 ! a^{7} b^{7}$ (b) $\frac{14 !}{7 !} a^{7} b^{7}$ (c) $\frac{14 !}{(7 !)^{2}} a^{7} b^{7}$ (d) $\frac{14 !}{(7 !)^{3}} a^{7} b^{7}$ Solution: (c) $\frac{14 !}{(7 !)^{2}} a^{7} b^{7}$ Suppose $(r+1)$ th term in the given expansion is independent of $x$. Then, we have $T_{r+1}={ }^{14} C_{r}(a x)^{14-r}\left(\frac{b}{x}\right)^{r}$ $={ }^{14} C_{r} a^{14-r} b^{r} x^{14-2 r}$ ...
Read More →In the given figure, side BC of ∆ABC has been produced to a point D.
Question: In the given figure, sideBCof ∆ABChas been produced to a pointD. If A= 3y, B=x, C= 5y and CBD= 7y. Then, the value ofxis (a)60(b) 50(c) 45(d) 35 Solution: Disclaimer: In the questionACDshould be 7y.In the given figure,ACB +ACD = 180 (Linear pair of angles)5y+7y =180⇒12y=180⇒y=15 .....(1)In ∆ABC,A +B +ACB = 180 (Angle sum property)3y +x +5y =180⇒x +8y=180⇒x+8 15=180 [Using (1)]⇒x+120=180⇒x =180120 =60Thus, the value ofxis 60.Hence, the correct answer is option (a)....
Read More →The coefficient of x
Question: The coefficient of $x^{-3}$ in the expansion of $\left(x-\frac{m}{x}\right)^{11}$ is (a) $-924 m^{7}$ (b) $-792 m^{5}$ (c) $-792 m^{6}$ (d) $-330 m^{7}$ Solution: (d) $-330 m^{7}$ Let $x^{-3}$ occur at $(r+1)$ th term in the given expansion. Then, we have $T_{r+1}={ }^{11} C_{r} x^{11-r}\left(\frac{-m}{x}\right)^{r}$ $=(-1)^{r} \times{ }^{11} C_{r} m^{r} x^{11-r-r}$ For this term to contain $x^{-3}$, we must have $11-2 r=-3$ $\Rightarrow r=7$ $\therefore$ Required coefficient $=(-1)^{7...
Read More →The coefficient of x
Question: The coefficient of $x^{-3}$ in the expansion of $\left(x-\frac{m}{x}\right)^{11}$ is (a) $-924 m^{7}$ (b) $-792 m^{5}$ (c) $-792 m^{6}$ (d) $-330 m^{7}$ Solution: (d) $-330 m^{7}$ Let $x^{-3}$ occur at $(r+1)$ th term in the given expansion. Then, we have $T_{r+1}={ }^{11} C_{r} x^{11-r}\left(\frac{-m}{x}\right)^{r}$ $=(-1)^{r} \times{ }^{11} C_{r} m^{r} x^{11-r-r}$ For this term to contain $x^{-3}$, we must have $11-2 r=-3$ $\Rightarrow r=7$ $\therefore$ Required coefficient $=(-1)^{7...
Read More →In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively.
Question: In the given figure,BOandCOare the bisectors ofBand Crespectively. If A= 50 then BOC= ?(a) 130(b) 100(c) 115(d) 120 Solution: (c) 115 In $\Delta A B C$, we have: $\angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow 50^{\circ}+\angle B+\angle C=180^{\circ}$ $\Rightarrow \angle B+\angle C=130^{\circ}$ $\Rightarrow \frac{1}{2} \angle B+\frac{1}{2} \angle C=65^{\circ} \quad \ldots(i)$ $\ln \Delta O B C$, we have: $\angle O B C+\angle O C B+\angle B ...
Read More →If the fifth term of the expansion
Question: If the fifth term of the expansion $\left(a^{2 / 3}+a^{-1}\right)^{n}$ does not contain ' $a$ '. Then $n$ is equal to (a) 2 (b) 5 (c) 10 (d) none of these Solution: (c) 10 $T_{5}=T_{4+1}$ $={ }^{n} C_{4}\left(a^{2 / 3}\right)^{n-4}\left(a^{-1}\right)^{4}$ $={ }^{n} C_{4} a^{\left(\frac{2 n-8}{3}-4\right)}$ For this term to be independent of a, we must have $\frac{2 n-8}{3}-4=0$ $\Rightarrow 2 n-20=0$ $\Rightarrow n=10$...
Read More →In a ∆ABC, it is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and ∠ACD = 90°. If BC is produced to E then ∠ECD = ?
Question: In a ∆ABC,it is given that A: B: C= 3 : 2 : 1 and ACD =90.IfBCis produced toEthen ECD =?(a) 60(b) 50(c) 40(d) 25 Solution: Let $\angle A=(3 x)^{\circ}, \angle B=(2 x)^{\circ}$ and $\angle C=x^{\circ}$ Then, $3 x+2 x+x=180^{\circ} \quad[$ Sum of the angles of a triangle $]$ $\Rightarrow 6 x=180^{\circ}$ $\Rightarrow x=30^{\circ}$ Hence, the angles are $\angle A=3 \times 30^{\circ}=90^{\circ}, \angle B=2 \times 30^{\circ}=60^{\circ}$ and $\angle C=30^{\circ}$ Side BC of triangle ABC is p...
Read More →If the sum of the binomial coefficients of the expansion
Question: If the sum of the binomial coefficients of the expansion $\left(2 x+\frac{1}{x}\right)^{n}$ is equal to 256, then the term independent of $x$ is (a) 1120 (b) 1020 (c) 512 (d) none of these Solution: (a) 1120 Suppose $(r+1)$ th tem in the given expansion is independent of $x$. Then, we have $T_{r+1}={ }^{n} C_{r}(2 x)^{n-r}\left(\frac{1}{x}\right)^{r}$ $={ }^{n} C_{r} 2^{n-r} x^{n-2 r}$ For this term to be independent of $x$, we must have $n-2 r=0$ $\Rightarrow r=n / 2$ $\therefore$ Req...
Read More →If the sum of the binomial coefficients of the expansion
Question: If the sum of the binomial coefficients of the expansion $\left(2 x+\frac{1}{x}\right)^{n}$ is equal to 256, then the term independent of $x$ is (a) 1120 (b) 1020 (c) 512 (d) none of these Solution: (a) 1120 Suppose $(r+1)$ th tem in the given expansion is independent of $x$. Then, we have $T_{r+1}={ }^{n} C_{r}(2 x)^{n-r}\left(\frac{1}{x}\right)^{r}$ $={ }^{n} C_{r} 2^{n-r} x^{n-2 r}$ For this term to be independent of $x$, we must have $n-2 r=0$ $\Rightarrow r=n / 2$ $\therefore$ Req...
Read More →The value of
Question: The value of $\cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ is (a) $\frac{\pi}{2}$ (b) $\frac{5 \pi}{3}$ (C) $\frac{10 \pi}{3}$ (d) 0 Solution: (d) 0 We have $\cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{\pi}{3}\right)\right\}+\sin ^{-1}\left\{\sin \left(2 \pi-\frac{\pi}{3}\right)\right\}$ $=\cos ^{-1}\left\{\cos \left(\frac{\pi}{3}\right)\right\}+\sin ^{-1}\lef...
Read More →In the given figure, lines AB and CD intersect at a point O.
Question: In the given figure, lines AB and CD intersect at a pointO. The sidesCAandOBhave been produced toEandFrespectively such that OAE=x and DBF=y. If OCA= 80, COA= 40 and BDO= 70 thenx +y = ?(a)190(b) 230(c) 210(d) 270 Solution: In the given figure,BOD =COA (Vertically opposite angles)BOD = 40 .....(1)In∆ACO,OAE =OCA +COA (Exterior angle of a triangle is equal to the sum of two opposite interior angles)⇒x = 80 +40 =120 .....(2)In∆BDO,DBF =BDO +BOD (Exterior angle of a triangle is equal to t...
Read More →In the given figure, two rays BD and CE intersect at a point A.
Question: In the given figure, two raysBDandCEintersect at a pointA. The sideBCof ∆ABChave been produced on both sides to pointsFandGrespectively. IfABF=x, ACG=y and DAE=zthenz= ?(a)x + y 180(b)x + y +180(c) 180 (x + y)(d)x + y +360 Solution: In the given figure,ABF +ABC = 180 (Linear pair of angles)x +ABC = 180⇒ABC = 180x .....(1)Also,ACG +ACB = 180 (Linear pair of angles)y+ACB = 180⇒ACB = 180y .....(2)Also,BAC =DAE =z .....(3) (Vertically opposite angles)In∆ABC,BAC +ABC +ACB =180 (Angle sum pr...
Read More →The the given figure, EAD ⊥ BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also,
Question: The the given figure,EADBCD. RayFACcuts rayEADat a pointAsuch that EAF= 30. Also, in ∆BAC,BAC=x and ABC= (x+ 10). Then, the value ofxis(a) 20(b) 25(c) 30(d) 35 Solution: In the given figure,CAD =EAF (Vertically opposite angles)CAD =30 In∆ABD,ABD +BAD +ADB=180 (Angle sum property)⇒(x+ 10) + (x +30) + 90 =180⇒ 2x + 130 =180⇒2x=180 130 = 50⇒x= 25Thus, the value ofxis 25.Hence, the correct answer is option (b)....
Read More →The value of
Question: The value of $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)$ is (a) $\frac{3 \pi}{5}$ (b) $-\frac{\pi}{10}$ (c) $\frac{\pi}{10}$ (d) $\frac{7 \pi}{5}$ Solution: (b) $-\frac{\pi}{10}$ $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)=\sin ^{-1}\left\{\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right\}$ $=\sin ^{-1}\left\{\cos \left(\frac{3 \pi}{5}\right)\right\}$ $=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right\}$ $=\frac{\pi}{2}-\frac{3 \pi}{5}$ $=-\frac{\pi}{10}$...
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