If 1x+2,1x+3,1x+5 are in A.P. Then, x =
Question: If $\frac{1}{x+2}, \frac{1}{x+3}, \frac{1}{x+5}$ are in A.P. Then, $x=$ (a) 5(b) 3(c) 1(d) 2 Solution: Here, we are given three terms, First term $\left(a_{1}\right)=\frac{1}{x+2}$ Second term $\left(a_{2}\right)=\frac{1}{x+3}$ Third term $\left(a_{3}\right)=\frac{1}{x+5}$ We need to find the value ofxfor 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=\left(\frac{1}{x+3}\right)-\left(\frac{1}{x+2}\rig...
Read More →if
Question: If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=$_________________________. Solution: Given: $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ ...(1) We know $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ ...(2) Adding $(1)$ and $(2)$, we get $2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}$ $\Rightarrow 2 \sin ^{-1} x=\frac{4 \pi}{6}$ $\Rightarrow \sin ^{-1} x=\frac{\pi}{3}$ $\Rightarrow x=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=\frac{\sqrt{3}}...
Read More →P is a point on the bisector of ∠ABC. If the line through P,
Question: Pis a point on the bisector of ABC. If the line throughP, parallel toBAmeetsBCatQ, prove that ΔBPQis an isosceles triangle. Solution: Given:BPis the bisector of ABC, andBA∥∥QP To prove: ΔBPQis an isosceles triangleProof: $\because \angle 1=\angle 2$(Given, $B P$ is the bisector of $\angle A B C$ ) And, $\angle 1=\angle 3$ (Alternate interior angles) $\therefore \angle 2=\angle 3$ So, $P Q=B Q$ (In a triangle, sides opposite to equal sides are equal.) But these are sides of $\Delta B P ...
Read More →The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
Question: The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is (a) 501th(b) 502th(c) 508th(d) none of these Solution: In the given problem, let us take the first term asaand the common difference asd. Here, we are given that, $a_{9}=449$...........(1) $a_{449}=9$............(2) We need to findn Also, we know, $a_{e}=a+(n-1) d$ For the $9^{\text {th }}$ term $(n=9)$, $a_{9}=a+(9-1) d$ $449=a+8 d$ (Using 1) $a=449-8 d$ ...........(3) Similarly, for the $449^{\text...
Read More →The value of
Question: The value of $\sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\right)$ is____________________. Solution: We know $\tan ^{-1} x=\sec ^{-1} \sqrt{1+x^{2}}$ and $\cot ^{-1} x=\operatorname{cosec}^{-1} \sqrt{1+x^{2}}$ So, $\tan ^{-1} 2=\sec ^{-1} \sqrt{1+2^{2}}=\sec ^{-1} \sqrt{5}$ $\cot ^{-1} 3=\operatorname{cosec}^{-1} \sqrt{1+3^{2}}=\operatorname{cosec}^{-1} \sqrt{10}$ $\therefore \sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\...
Read More →The value of
Question: The value of $\sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\right)$ is____________________. Solution: We know $\tan ^{-1} x=\sec ^{-1} \sqrt{1+x^{2}}$ and $\cot ^{-1} x=\operatorname{cosec}^{-1} \sqrt{1+x^{2}}$ So, $\tan ^{-1} 2=\sec ^{-1} \sqrt{1+2^{2}}=\sec ^{-1} \sqrt{5}$ $\cot ^{-1} 3=\operatorname{cosec}^{-1} \sqrt{1+3^{2}}=\operatorname{cosec}^{-1} \sqrt{10}$ $\therefore \sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\...
Read More →Solve the following
Question: If $13^{\text {th }}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is independent of $x$, then the value of $n$ is _______________ Solution: In $\left(x^{2}+\frac{2}{x}\right)^{n}$ $T_{r+1}={ }^{n} C_{r}\left(x^{2}\right)^{n-r}\left(\frac{2}{x}\right)^{r}$ $={ }^{n} C_{r} x^{2 n-2 r} \frac{(2)^{r}}{x^{r}}$ $={ }^{n} C_{r} x^{2 n-2 r} x^{-r}(2)^{r}$ $T_{r+1}={ }^{n} C_{r} 2^{r} x^{2 n-3 r}$ If forr= 12, i.e 13thterm is independent ofx 2n 3r= 0 ⇒ 2n= 3 12 i.e.n= 18...
Read More →The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O.
Question: The bisectors ofBandCof an isoscelesΔABCwithAB=ACintersect each other at a pointO. Show that the exterior angle adjacent toABCis equal toBOC. Solution: Given: Inan isoscelesΔABC,AB=AC,BOandCOare the bisectors ofABCandACB, respectively.To prove:ABD=BOCConstruction: ProduceCBto pointD.Proof:In ΔABC,∵∵AB=AC (Given)ACB=ABC (Angle opposite to equal sides are equal) $\Rightarrow \frac{1}{2} \angle A C B=\frac{1}{2} \angle A B C$ $\Rightarrow \angle O C B=\angle O B C \quad \ldots$. (i) (Give...
Read More →Sum of n terms of the series 2–√+8–√+18−−√+32−−√+..is
Question: Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots$ is (a) $\frac{n(n+1)}{2}$ (b) $2 n(n+1)$ (c) $\frac{n(n+1)}{\sqrt{2}}$ (d) 1 Solution: In the given problem, we need to find the sum of terms for a given arithmetic progression, $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$ So, 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...
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) $-\frac{7 \pi}{5}$ (c) $\frac{\pi}{10}$ (d) $-\frac{\pi}{10}$ Solution: $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)$ $=\sin ^{-1}\left[\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right]$ $=\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)$ $=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]$ $=\sin ^{-1}\left[\sin \left(-\frac{\pi}{10}\right)\right]$ $=-\frac{\pi}{10}$ Thus, the value ...
Read More →If A and B are the coefficient of
Question: If $A$ and $B$ are the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$ respectively, then $\frac{A}{B}=$ ________________ Solution: The coefficient ofxnis (1 +x)2nis? SinceTr+1=2nCrxr Forxncoefficient putr = n i.e coefficient ofxnis2nCn i.e. A =2nCn and for coefficient ofxnin (1 +x)2n1 Tr+1=2n1Crxr Putr = n i.e. coefficient ofxnin (1 +x)2n1is2n1Cn i.e B =2n1Cn $\therefore \frac{A}{B}=\frac{{ }^{2 n} C_{n}}{{ }^{2 n-1} C_{n}}=\frac{(2 n) ! n ! n !}{n ! n !(2...
Read More →The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
Question: The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is(a) 5(b) 10(c) 12(d) 14(e) 20 Solution: In the given problem, we have an A.P. Here, we need to find the number of termsnsuch that the sum ofnterms is 406. So here, we will use the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ Where;a= first term for the given A.P. d= common difference of the given A.P. n= number of terms The first term (a) = 3 The sum ofnterms (Sn) = 406 Common difference of the A.P. $(d...
Read More →if the
Question: If $3 \tan ^{-1} x+\cot ^{-1} x=\pi$, then $x$ equals (a) 0 (b) 1 (c) $-1$ (d) $\frac{1}{2}$ Solution: $3 \tan ^{-1} x+\cot ^{-1} x=\pi$ $\Rightarrow 2 \tan ^{-1} x+\tan ^{-1} x+\cot ^{-1} x=\pi$ $\Rightarrow 2 \tan ^{-1} x+\frac{\pi}{2}=\pi$ $\left[\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]$ $\Rightarrow 2 \tan ^{-1} x=\pi-\frac{\pi}{2}=\frac{\pi}{2}$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{4}$ $\Rightarrow x=\tan \frac{\pi}{4}=1$ Thus, the value of $x$ is 1 . Hence, the correct answ...
Read More →Solve the following
Question: If $x^{4}$ occurs in the $r^{\text {th }}$ terms in the expansion of $\left(x^{4}+\frac{1}{x^{3}}\right)^{15}$, then $r=$ ____________ Solution: for $\left(x^{4}+\frac{1}{x^{3}}\right)^{15}$ n= 15 $T_{r+1}={ }^{15} C_{4}\left(x^{4 r}\right)\left(\frac{1}{x^{3}}\right)^{15-r}$ $={ }^{15} C_{4} x^{4 r} x^{-(45-3 r)}$ $={ }^{15} C_{4} x^{4 r} x^{3 r-45}$ $={ }^{15} C_{4} x^{7 r-45}$ To getx4, 7r 45 = 4 i.e. 7r= 49 i.er= 7...
Read More →The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O.
Question: The bisectors ofBandCof an isosceles triangle withAB=ACintersect each other at a pointO.BOis produced to meetACat a pointM. Prove thatMOC=ABC. Solution: Given:In isoscelesΔ∆ABC,AB=AC;OBandOCare bisectors ofBandC, respectively.To prove:MOC=ABCProof:In ∆ABC,∵∵AB=AC (Given)ABC= ACB (Angles opposite to equal sides are equal) $\Rightarrow \frac{1}{2} \angle A B C=\frac{1}{2} \angle A C B$ ⇒⇒OBC= OCB (Given,OBandOCare the bisectors of Band C, respectively) .....(i)Now, in ∆OBC, MOCis an exte...
Read More →The total number of terms in the expansion of
Question: The total number of terms in the expansion of (1 +x)2n (1 x)2n___________. Solution: In $(1+x)^{2 n}-(1-x)^{2 n}$ Since $(1+x)^{2 n}={ }^{2 n} C_{0}+{ }^{2 n} C_{1} x+\ldots \ldots+{ }^{2 n} C_{2 n} x^{2 n}$ $(1-x)^{2 n}={ }^{2 n} C_{0}-{ }^{2 n} C_{1} x+\ldots \ldots+{ }^{2 n} C_{2 n} x^{2 n}$ Subtracting above two, i. e. $(1+x)^{2 n}-(1-x)^{2 n}=2\left[{ }^{2 n} C_{1} x+{ }^{2 n} C_{3} x^{3}+\ldots{ }^{2 n} C_{2 n-1} x^{2 n-1}\right]$ i. e. number of terms here $=2 \times \frac{n}{2}...
Read More →The value of the expression
Question: The value of the expression $\tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right)$ is (a) $2+\sqrt{5}$ (b) $\sqrt{5}-2$ (c) $\frac{\sqrt{5}+2}{2}$ (d) $5+\sqrt{2}$ Solution: Let $\cos ^{-1} \frac{2}{\sqrt{5}}=\theta$. Then, $\cos \theta=\frac{2}{\sqrt{5}}$ Now, $\tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right)$ $=\tan \left(\frac{\theta}{2}\right)$ $=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$ $=\sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}}$ $=\sqrt{\frac{\...
Read More →If the first term of an A.P. is 2 and common difference is 4,
Question: If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is(a) 3200 (b) 1600 (c) 200 (d) 2800 Solution: In the given problem, we need to find the sum of 40 terms of an arithmetic progression, where we are given the first term and the common difference. So, 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 Gi...
Read More →The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC.
Question: The line segments joining the midpointsMandNof parallel sidesABandDCrespectively of a trapeziumABCDis perpendicular to both the sidesABandDC. Prove thatAD = BC. Solution: Given:In trapeziumABCD,MandNare mid-pointsofABandDC,MNABandMNDC.To prove:AD = BCConstruction: JoinCMandDM.Proof:InΔCMNandΔDMN,MN=MN (Common sides)CNM=DNM= 90 (Given,MNDC)CN=DN (Given,Nis the mid-pointDC)By SAS congruence criteria,ΔCMNΔDMNSo,CM=DM (CPCT) .....(i)And,CMN=DMN (CPCT)But,AMN=BMN= 90 (Given,MNAB)⇒⇒AMN-CMN=B...
Read More →The coefficient of the middle term in the expansion of
Question: The coefficient of the middle term in the expansion of (1 +x)10is ___________. Solution: In (1 +x)10 Tr+1=10Crxr Middle term is obtained whenr= 5 i.e.Tr+1=10C5x5 i.e. coefficient of middle term is10C5...
Read More →The middle term in the expansion of
Question: The middle term in the expansion of $\left(x-\frac{1}{x}\right)^{18}$ is ______________ Solution: $\left(x-\frac{1}{x}\right)^{18}$ Since 18 is even there is only one middle term i.e. $\left(\frac{2 n}{2}+1\right)^{\text {th }}$ term i.e. (n+ 1)thterm i.e. $T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(\frac{-1}{x}\right)^{n}$ $={ }^{2 n} C_{n}(-1)^{n} x^{2 n-n} x^{-n}$ $={ }^{2 n} C_{n}(-1)^{n} x^{0} \quad$ i. e. ${ }^{18} C_{9}(-1)^{18}={ }^{18} C_{9}$...
Read More →If Sr denotes the sum of the first r terms of an A.P.
Question: If $S_{r}$ denotes the sum of the first $r$ terms of an A.P. Then, $S_{3 n}$ : $\left(S_{2 n}-S_{n}\right)$ is (a)n(b) 3n(c) 3(d) none of these Solution: Here, we are given an A.P. whose sum of $r$ terms is $S_{r}$. We need to find $\frac{S_{3 n}}{S_{2 m}-S_{n}}$. 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, $S_{3 ...
Read More →The number of terms in the expansion of
Question: The number of terms in the expansion of $\left\{\left(2 x+y^{3}\right)^{4}\right\}^{7}$ is ______________ Solution: In {(2x+y3)4}7 n=?? Since In (a + b)n; number of terms isn+ 1 {(2x+y3)}28 Number of terms is 28 + 1 = 29...
Read More →Which of the following is the principal value branch
Question: Which of the following is the principal value branch of $\operatorname{cosec}^{-1}$ ? (a) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (b) $[0, \pi]-\left\{\frac{\pi}{2}\right\}$ (c) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (d) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ Solution: The principal value branch of $\operatorname{cosec}^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$. Hence, the correct answer is option (d)....
Read More →The sum of the series
Question: The sum of the series $\sum_{r=0}^{10}{ }^{20} \mathrm{C}_{f}$ is ____________ Solution: for $\sum_{r=0}^{10}{ }^{20} C_{r}$ for $n=2$ consider, $(1+x)^{20}={ }^{20} C_{0}+{ }^{20} C_{1} x+\ldots \ldots+{ }^{20} C_{20} x^{20}$ i. e. $\sum_{r=0}^{10}{ }^{20} C_{r}={ }^{20} C_{0}+{ }^{20} C_{1}+{ }^{20} C_{2}+\ldots \ldots+{ }^{20} C_{10}$ Since ${ }^{20} C_{1}={ }^{20} C_{19}$ ${ }^{20} C_{2}={ }^{20} C_{18}$ ${ }^{20} C_{11}={ }^{20} C_{9}$ i. e. $(1+x)^{20}=2\left({ }^{20} C_{0}+{ }^{...
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