Solve the following
Question: If (1 x+x2)n=a0+a1x+a2x2+...+a2nx2n, thena0+a2+a4+...+a2nequals (a) $\frac{3^{n}+1}{2}$ (b) $\frac{3^{n}-1}{2}$ (c) $\frac{1-3^{n}}{2}$ (d) $3^{n}+\frac{1}{2}$ Solution: Given; $\left(1-x+x^{2}\right) n=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{2 n} x^{2 n} \quad \ldots(1)$ In order to find $a_{0}+a_{2}+a_{4}+\ldots+a_{2 n}$ i.e. all even terms are involved replacexby 1 in equation (1) we get $(1-1+1)^{n}=a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{2 n}$ i.e. $1=a_{0}+a_{1}+a_{2}+\ldots+a_{2 n} \quad...
Read More →The value of
Question: The value of $\sin \left(2 \sin ^{-1}(.6)\right)$ is (a) 0.48 (b) 0.96 (c) 1.2 (d) sin 1.2 Solution: $\sin \left(2 \sin ^{-1} 0.6\right)$ $=\sin \left[\sin ^{-1}\left(2 \times 0.6 \times \sqrt{1-(0.6)^{2}}\right)\right]$ $\left[2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\right]$ $=\sin \left[\sin ^{-1}(2 \times 0.6 \times \sqrt{1-0.36})\right]$ $=\sin \left[\sin ^{-1}(2 \times 0.6 \times \sqrt{0.64})\right]$ $=\sin \left[\sin ^{-1}(2 \times 0.6 \times 0.8)\right]$ $=\sin \...
Read More →If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
Question: If the first, second and last term of an A.P. area,band 2arespectively, its sum is (a) $\frac{a b}{2(b-a)}$ (b) $\frac{a b}{(b-a)}$ (c) $\frac{3 a b}{2(b-a)}$ (d) none of these Solution: In the given problem, we are given first, second and last term of an A.P. We need to find its sum. So, here First term =a Second term (a2) = b Last term (l)= 2a Now, using the formula $a_{n}=a+(n-1) d$ $a_{2}=a+(2-1) d$ $b=a+d$ $d=b-a$ ...........(1) Also, $a_{n}=a+(n-1) d$ $2 a=a+n d-d$ $2 a-a=n d-d$ ...
Read More →Solve the following
Question: If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}$, then (a) Re (z) = 0 (b) Im (z) = 0 (c) Re (z) 0, Im (z) 0 (d) Re (z) 0, Im (z) 0 Solution: for $\frac{\sqrt{3}}{2}+\frac{i}{2}$ $r=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$ and $\theta=\tan ^{-1} \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\tan ^{-1} \frac{1}{\sqrt{3}}$ i. e. $\theta=\frac{\pi}{6}$ i.e. $\frac{\sqrt{3}}{2}+\frac{i}{2}=r(\cos \theta+i \sin \theta)$ we have $\theta=\frac{\pi}{...
Read More →if the
Question: If $\alpha \leq 2 \sin ^{-1} x+\cos ^{-1} x \leq \beta$, then (a) $\alpha=-\frac{\pi}{2}, \beta=\frac{\pi}{2}$ (b) $\alpha=0, \beta=\pi$ (c) $\alpha=-\frac{\pi}{2}, \beta=\frac{3 \pi}{2}$ (d) $\alpha=0, \beta=2 \pi$ Solution: $2 \sin ^{-1} x+\cos ^{-1} x$ $=\sin ^{-1} x+\sin ^{-1} x+\cos ^{-1} x$ $=\sin ^{-1} x+\frac{\pi}{2}$ $\left(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$ Now, $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$ $\Rightarrow-\frac{\pi}{2}+\frac{\pi}{2} \leq \s...
Read More →If the sum of first n even natural numbers is equal to k times
Question: If the sum of firstneven natural numbers is equal toktimes the sum of firstnodd natural numbers, thenk= (a) $\frac{1}{n}$ (b) $\frac{n-1}{n}$ (C) $\frac{n+1}{2 n}$ (d) $\frac{n+1}{n}$ Solution: In the given problem, we are given that the sum of the firstneven natural numbers is equal toktimes the sum of firstnodd natural numbers. We need to find the value ofk Now, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2. ...
Read More →The ratio of the coefficient of x
Question: The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in $\left(x^{2}+\frac{2}{x}\right)^{15}$, is (a) 12 : 32 (b) 1 : 32 (c) 32 : 12 (d) 32 : 1 Solution: $\ln \left(x^{2}+\frac{2}{x}\right)^{15}$ we have $T_{r+1}={ }^{15} C_{r}\left(x^{2}\right)^{15-r}\left(\frac{2}{x}\right)^{r}$ i.e. General term is $T_{r+1}={ }^{15} C_{r} x^{30-3 r} 2^{r}$ Hence for the term independent ofx, 30 3r= 0 i.e.r= 10 henceT11has coefficient15C10210 ...(1) and term withx15will have 30 3r=...
Read More →In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.
Question: In the given figure,OA=OBandOP=OQ. Prove that (i)PX=QX, (ii)AX=BX. Solution: Proof: In $\triangle \mathrm{OQA}$ and $\triangle \mathrm{OPB}$, we have : $\mathrm{OQ}=\mathrm{OP} \quad$ (Given) $\mathrm{OA}=\mathrm{OB} \quad$ (Given) $\angle \mathrm{AOQ}=\angle \mathrm{BOP} \quad$ (Common) $\triangle \mathrm{OQA} \cong \mathrm{OPB} \quad$ (SAS criterion) $\angle O A Q=\angle O B P \quad$ (Corresponding angles of congruent triangles) Now, consider triangles BQX and APX. Given : $\mathrm{O...
Read More →The value of tan
Question: The value of $\tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right)$ (a) $\frac{19}{8}$ (b) $\frac{8}{19}$ (C) $\frac{19}{12}$ (d) $\frac{3}{4}$ Solution: $\tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right)=\tan \left(\tan ^{-1} \frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+\tan ^{-1} \frac{1}{4}\right)$ $=\tan \left(\tan ^{-1} \frac{\frac{4}{5}}{\frac{3}{5}}+\tan ^{-1} \frac{1}{4}\right)$ $=\tan \left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{4}\right)$ $=\tan \lef...
Read More →The first and last term of an A.P. are a and l respectively.
Question: The first and last term of an A.P. are $a$ and $/$ respectively. If $S$ is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^{2}-a^{2}}{k-(l+a)}$, then $k=$ Solution: In the given problem, we are given the first, last term, sum and the common difference of an A.P. We need to find the value ofk Here, First term =a Last term =l Sum of all the terms =S Common difference $(d)=\frac{l^{2}-a^{2}}{k-(l+a)}$ Now, as we know, $l=a+(n-1) d$ .........(1) Further,...
Read More →If the coefficients of x
Question: If the coefficients ofx7andx8in(2+x3)n2+x3nare equal, thennis (a) 56 (b) 55 (c) 45 (d) 15 Solution: In $(a+x)^{n}$ $T_{r+1}={ }^{n} C_{r} a^{n-r} x^{r}$ $\therefore \operatorname{In}\left(2+\frac{x}{3}\right)^{n}$ Coefficient ofx7is $T_{8}={ }^{n} C_{7}(2)^{n-7}\left(\frac{x}{3}\right)^{7}$ $T_{8}={ }^{n} C_{7} \frac{2^{\text {n- }-} x^{7}}{3^{7}}$ and coefficient ofx8is $T_{9}={ }^{n} C_{8}(2)^{n-8}\left(\frac{x}{3}\right)^{8}$ $T_{9}={ }^{n} C_{8} \frac{2^{n-8} x^{8}}{3^{8}}$ Since $...
Read More →The domain of
Question: The domain of $\cos ^{-1}\left(x^{2}-4\right)$ is (a) $[3,5]$ (b) $[-1,1]$ (c) $[-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]$ (d) $[-\sqrt{5},-\sqrt{3}] \cap[\sqrt{3}, \sqrt{5}]$ Solution: The domain of $\cos ^{-1}(x)$ is $[-1,1]$ $\therefore-1 \leq x^{2}-4 \leq 1$ $\Rightarrow-1+4 \leq x^{2}-4+4 \leq 1+4$ $\Rightarrow 3 \leq x^{2} \leq 5$ $\Rightarrow \pm \sqrt{3} \leq x \leq \pm \sqrt{5}$ $\Rightarrow x \in[-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]$ Hence, the correct answer is...
Read More →ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersects the side AB at D.
Question: ΔABCis a right triangle right angled atAsuch thatAB=ACand bisector of Cintersects the sideABatD. Prove thatAC+AD=BC. Solution: Given: In right triangle ΔABC,BAC= 90,AB=ACand ACD=BCD.To prove:AC+AD=BCProof:LetAB=AC=xandAD=y.InΔ∆ABC, $B C^{2}=A B^{2}+A C^{2}$ $\Rightarrow B C^{2}=x^{2}+x^{2}$ $\Rightarrow B C^{2}=2 x^{2}$ $\Rightarrow B C=\sqrt{2 x^{2}}$ $\Rightarrow B C=x \sqrt{2}$ Now, $\frac{B D}{A D}=\frac{B C}{A C}$ (An angle bisector of an angle of a triangle divides the opposite s...
Read More →Let Sn denote the sum of n terms of an A.P. whose first term is a.
Question: Let $S_{n}$ denote the sum of $n$ terms of an A.P. whose first term is $a$. If the common difference $d$ is given by $d=S_{n}-k S_{n-1}+S_{n-2}$, then $k=$ (a) 1(b) 2(c) 3(d) none of these. Solution: In the given problem, we are given $d=S_{n}-k S_{n-1}+S_{n-2}$ We need to find the value ofk So here, First term =a Common difference =d Sum ofnterms =Sn Now, as we know, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ .........(1) Also, forn-1terms, $S_{n-1}=\frac{n-1}{2}[2 a+[(n-1)-1] d]$ $=\frac{n-1}{...
Read More →The total number of terms in the expansion of
Question: The total number of terms in the expansion of (x+a)51 (xa)51after simplification is (a) 102 (b) 25 (c) 26 (d) none of these Solution: for $(x+a)^{51}-(x-a)^{51}$ Since $(x+a)^{51}={ }^{51} C_{0} x^{51}+{ }^{51} C_{1} x^{50} a+{ }^{51} C_{2} x^{49} a^{2}+\ldots+{ }^{51} C_{51} a^{51}$ and $(x-a)^{51}={ }^{51} C_{0} x^{51}-{ }^{51} C_{1} x^{50} a+{ }^{51} C_{2} x^{49} a^{2} \ldots{ }^{51} C_{51} a^{51}$ Subtracting above values, $(x+a)^{51}-(x-a)^{51}=2\left({ }^{51} C_{1} x^{50} a+{ }^{...
Read More →if the
Question: If $x1$, then $2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ is equal to (a) $4 \tan ^{-1} x$ (b) 0 (c) $\frac{\pi}{2}$ (d) $\pi$ Solution: $2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2 \tan ^{-1} x+2 \tan ^{-1} x \quad\left[\because \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2 \tan ^{-1} x\right]$ $=4 \tan ^{-1} x$ Hence, the correct answer is option (a)...
Read More →The value of sin
Question: The value of $\sin \left(2\left(\tan ^{-1} 0.75\right)\right)$ is equal to (a) $0.75$ (b) $1.5$ (c) $0.96$ (d) $\sin ^{-1} 1.5$ Solution: $\sin \left(2\left(\tan ^{-1} 0.75\right)\right)=\sin \left(2 \tan ^{-1} 0.75\right)$ $=\sin \left(\sin ^{-1} \frac{2 \times 0.75}{1+(0.75)^{2}}\right)$ $=\sin \left(\sin ^{-1} 0.96\right)$ $=0.96$ Hence, the correct answer is option (c)....
Read More →If the middle term of
Question: If the middle term of $\left(\frac{1}{x}+x \sin x\right)^{10}$ is equal to $7 \frac{7}{8}$, then the value of $x$ is (a) $2 n \pi+\frac{\pi}{6}$ (b) $n \pi+\frac{\pi}{6}$ (c) $n \pi+(-1)^{n} \frac{\pi}{6}$ (d) $n \pi+(-1)^{n} \frac{\pi}{3}$ Solution: Given middle term of $\left(\frac{1}{x}+x \sin x\right)^{10}$ is $7 \frac{7}{8}$ Sincen= 10 i.e. middle term is $\left(\frac{n}{2}+1\right)^{\text {th }}$ term is 6th term. $\therefore T_{6}=T_{5+1}={ }^{10} C_{5}\left(\frac{1}{x}\right)^{...
Read More →ABCD is a quadrilateral such that diagonal AC bisects the angles ∠A and ∠C.
Question: ABCDis a quadrilateral such that diagonalACbisects the angles Aand C. Prove thatAB=ADandCB=CD. Solution: Given: InquadrilateralABCD,ACbisects the angles Aand C.To prove:AB=ADandCB=CDProof:InΔ∆ABCandΔ∆ADC,BAC=DAC (Given,ACbisects the angles A)AC=AC (Common side)BCA=DCA (Given,ACbisects the angles C)By ASA congruence criteria,Δ∆ABCΔ∆ADCHence,AB=ADandCB=CD. (CPCT)...
Read More →If four numbers in A.P. are such that their sum is 50
Question: If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are(a) 5, 10, 15, 20(b) 4, 101, 16, 22(c) 3, 7, 11, 15(d) none of these Solution: Here, we are given that four numbers are in A.P., such that their sum is 50 and the greatest number is 4 times the smallest. So, let us take the four terms as $a-d, a, a+d, a+2 d$. Now, we are given that sum of these numbers is 50, so we get, $(a-d)+(a)+(a+d)+(a+2 d)=50$ $a-d+a+a+d+a+2 d=5...
Read More →if the
Question: If $\sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$, where $a, x \in(0,1)$, then, the value of $x$ is (a) 0 (b) $\frac{a}{2}$ (c) $a$ (d) $\frac{2 a}{1-a^{2}}$ Solution: $\sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ $\Rightarrow 2 \tan ^{-1} a+2 \tan ^{-1} a=2 \tan ^{-1} x$ $\Rightarrow 4 \tan ^{-1} a=2 \tan ^{-1}...
Read More →If the coefficients of 2nd, 3rd and the 4th terms in the expansion of
Question: If the coefficients of 2nd, 3rdand the 4thterms in the expansion of (1 +x)nare in A.P., then the value ofnis (a) 2 (b) 7 (c) 11 (d) 14 Solution: Given coefficient of 2nd, 3rdand 4thterms in the expansion (1 +x)nare A.P Since coefficient of (r+ 1)thterms isnCr 2nd,3rdand 4thcoefficient are such that 2nCr=nC1+nC3 i. e. $\quad 2 \times \frac{n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{6}$ i. e. $n(n-1)=n\left[1+\frac{(n-1)(n-2)}{6}\right]$ i.e. $\quad n-1=1+\frac{(n-1)(n-2)}{6}$ i.e. $6(n-1)=6+n^{2}-...
Read More →In the given figure, line l is the bisector of an angle ∠A and B is any point on l.
Question: In the given figure, linelis the bisector of an angle AandBis any point onl. IfBPandBQare perpendiculars fromBto the arms of A, show that(i) ΔAPB ΔAQB(ii)BP=BQ, i.e.,Bis equidistant from the arms ofA. Solution: Given:In the given figure, BAQ= BAP,BPAPandBQAQ.To prove:(i) ΔAPB ΔAQB(ii)BP=BQ, i.e.,Bis equidistant from the arms ofA.Proof:(i) In ΔAPBand ΔAQB,BAQ=BAP, (Given)APB=AQB= 90 (Given,BPAPandBQAQ)AB=AB (Common side)By AAS congruence criteria,ΔAPB ΔAQB(ii)∵∵ΔAPB ΔAQB [From (i)]BP=BQ...
Read More →The two successive terms in the expansion of
Question: The two successive terms in the expansion of (1 +x)24whose coefficients are in the ratio 1 : 4 are (a) 3rdand 4th (b) 4thand 5th (c) 5thand 6th (d) 6thand 7th Solution: For (1 +x)24two successive terms have coefficients in ration 1 : 4 Let the two successive terms be (r+ 1)thand (r+ 2) terms i. e. $T_{r+1}={ }^{24} C_{r} x^{r}$ and $T_{r+2}={ }^{24} C_{r+1} x^{r+1}$ Given that $\frac{{ }^{24} C_{r}}{{ }^{24} C_{r+1}}=\frac{1}{4}$ i. e. $\frac{(24) !}{r !(24-r) !} \times \frac{(r+1) !(2...
Read More →if the
Question: If $\tan ^{-1}(\cot \theta)=2 \theta$, then $\theta=$ (a) $\pm \frac{\pi}{3}$ (b) $\pm \frac{\pi}{4}$ (c) $\pm \frac{\pi}{6}$ (d) none of these Solution: (c) $\pm \frac{\pi}{6}$ We have, $\tan ^{-1}(\cot \theta)=2 \theta$ $\Rightarrow \tan 2 \theta=\cot \theta$ $\Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{1}{\tan \theta}$ $\Rightarrow 2 \tan ^{2} \theta=1-\tan ^{2} \theta$ $\Rightarrow 3 \tan ^{2} \theta=1$ $\Rightarrow \tan ^{2} \theta=\frac{1}{3}$ $\Rightarrow \tan \th...
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