if
Question: If $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$, then prove that $x=\frac{a-b}{1+a b}$ Solution: Let : $a=\tan m$ $b=\tan n$ $x=\tan y$ Now, $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$ $\Rightarrow \sin ^{-1} \frac{2 \tan m}{1+\tan ^{2} m}-\cos ^{-1} \frac{1-\tan ^{2} n}{1+\tan ^{2} n}=\tan ^{-1} \frac{2 \tan y}{1-\tan ^{2} y}$ $\Rightarrow \sin ^{-1}(\sin 2 m)-\cos ^{-1}(\cos 2...
Read More →In the adjoining figure, y = ?
Question: In the adjoining figure,y= ?(a) 36(b) 54(c) 63(d) 72 Solution: (b) 54 We have: $3 x+72=180^{\circ} \quad[\because A O B$ is a straight line $]$ $\Rightarrow 3 x=108$ $\Rightarrow x=36$ Also, $\angle A O C+\angle C O D+\angle B O D=180^{\circ} \quad[\because A O B$ is a straight line $]$ $\Rightarrow 36^{\circ}+90^{\circ}+y=180^{\circ}$ $\Rightarrow y=54^{\circ}$...
Read More →In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
Question: In the given figure, OAB= 75, OBA= 55 and OCD= 100. Then ODC= ?(a) 20(b) 25(c) 30(d) 35 Solution: (c) 30 In $\triangle O A B$, we have: $\angle O A B+\angle O B A+\angle A O B=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow 75^{\circ}+55^{\circ}+\angle A O B=180^{\circ}$ $\Rightarrow \angle A O B=50^{\circ}$ $\therefore \angle C O D=\angle A O B=50^{\circ} \quad$ [Vertically-Opposite Angles] In $\Delta O C D$, we have: $\angle C O D+\angle O C D+\angle O D C=180^{\cir...
Read More →Show that
Question: (i) $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$ (ii) $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1}\left(\frac{4}{5}\right)$ (iii) $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$ (iv) $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$ (v) $\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$ (vi) $2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$ (vii) $2 \tan ^{-1} \frac...
Read More →Find the middle terms in the expansion of:
Question: Find the middle terms in the expansion of: (i) $\left(3 x-\frac{x^{3}}{6}\right)^{9}$ (ii) $\left(2 x^{2}-\frac{1}{x}\right)^{7}$ (iii) $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ (iv) $\left(x^{4}-\frac{1}{x^{3}}\right)^{11}$ Solution: (i) Here,n, i.e. 9, is an odd number. Thus, the middle terms are $\left(\frac{n+1}{2}\right)$ th and $\left(\frac{n+1}{2}+1\right)$ th, i. e. 5 th and 6 th Now, $T_{5}=T_{4+1}={ }^{9} C_{4}(3 x)^{9-4}\left(\frac{-x^{3}}{6}\right)^{4}$ $=\frac{9 \times 8 \ti...
Read More →In the given figure, AB || CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB = ?
Question: In the given figure,AB||CD. IfEAB= 50 and ECD= 60, then AEB= ?(a) 50(b) 60(c) 70(d) 55 Solution: (c) 70 $A B \| C D$ and $B C$ is the transversal. $\therefore \angle A B E=\angle B C D=60^{\circ} \quad[$ Alternate Internal Angles $]$ In $\Delta A B E$, we have: $\angle E A B+\angle A B E+\angle A E B=180^{\circ} \quad[$ Sum of the angles of a triangle $]$ $\Rightarrow 50^{\circ}+60^{\circ}+\angle A E B=180^{\circ}$ $\Rightarrow \angle A E B=70^{\circ}$...
Read More →Find the middle terms in the expansion of:
Question: Find the middle terms in the expansion of: (i) $\left(3 x-\frac{x^{3}}{6}\right)^{9}$ (ii) $\left(2 x^{2}-\frac{1}{x}\right)^{7}$ (iii) $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ (iv) $\left(x^{4}-\frac{1}{x^{3}}\right)^{11}$ Solution: (i) Here,n, i.e. 9, is an odd number. Thus, the middle terms are $\left(\frac{n+1}{2}\right)$ th and $\left(\frac{n+1}{2}+1\right)$ th, i. e. 5 th and 6 th Now, $T_{5}=T_{4+1}={ }^{9} C_{4}(3 x)^{9-4}\left(\frac{-x^{3}}{6}\right)^{4}$ $=\frac{9 \times 8 \ti...
Read More →In the given figure, AB || CD. If ∠APQ = 70° and ∠PRD = 120°, then ∠QPR = ?
Question: In the given figure,AB||CD. IfAPQ= 70 and PRD= 120, then QPR= ?(a) 50(b) 60(c) 40(d) 35 Solution: (a) 50 $A B \| C D$ and $P Q$ is the transversal. $\therefore \angle P Q R=\angle A P Q=70^{\circ} \quad$ [Alternate Interior Angles] Side QR of traingle PQR is produced to D. $\therefore \angle P Q R+\angle Q P R=\angle P R D$ $\Rightarrow 70^{\circ}+\angle Q P R=120^{\circ}$ $\Rightarrow \angle Q P R=50^{\circ}$...
Read More →Find the sum of
Question: Find the sum of(i) the first 15 multiples of 8(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.(iii) all 3 digit natural numbers which are divisible by 13.(iv) all 3 - digit natural numbers, which are multiples of 11. Solution: In the given problem, we need to find the sum of terms for different arithmetic progressions. 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= comm...
Read More →In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, x = ?
Question: In the given figure, ifAB||CD,CD||EFandy:z= 3 : 7,x= ?(a) 108(b) 126(c) 162(d) 63 Solution: (b) 126 Let $y=(3 a)^{\circ}$ and $z=(7 a)^{\circ}$ Let the transversal intersect AB at P, CD at O and EF at Q. Then, we have: $\angle C O P=\angle D O F=y \quad$ [Vertically-Opposite Angles] $\therefore \angle O Q F+\angle D O Q=180^{\circ} \quad$ [Consecutive Interior Angles] $\Rightarrow 3 a+7 a=180^{\circ}$ $\Rightarrow 10 a=180^{\circ}$ $\Rightarrow a=18^{\circ}$ $\therefore y=3 \times 18^{...
Read More →Evaluate the following:
Question: Evaluate the following: (i) $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$ (ii) $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)$ (iii) $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ (iv) $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$ Solution: (i) $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(2 \tan ^{-1} \frac{1}{5}-\tan ^{-1} 1\right)$ $=\tan \left[\tan ^{-1}\left\{\frac{{ }^{2 \times \frac{1...
Read More →In the given figure, AB || CD. If ∠CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
Question: In the given figure,AB||CD. IfCAB= 80 and EFC= 25, then CEF= ?(a) 65(b) 55(c) 45(d) 75 Solution: (c) 45 $A B \| C D$ and $A F$ is the transversal. $\therefore \angle D C F=\angle C A B=80^{\circ} \quad$ [Corresponding Angles] Side EC of triangle EFC is produced to D. $\therefore \angle C E F+\angle E F C=\angle D C F$ $\Rightarrow \angle C E F+25^{\circ}=80^{\circ}$ $\Rightarrow \angle C E F=55^{\circ}$...
Read More →Find the middle term in the expansion of:
Question: Find the middle term in the expansion of: (i) $\left(\frac{2}{3} x-\frac{3}{2 x}\right)^{20}$ (ii) $\left(\frac{a}{x}+b x\right)^{12}$ (iii) $\left(x^{2}-\frac{2}{x}\right)^{10}$ (iv) $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$ Solution: (i) Here, n= 20 (Even number) Therefore, the middle term is the $\left(\frac{n}{2}+1\right)$ th term, i.e., the 11 th term. Now, $T_{11}=T_{10+1}$ $={ }^{20} C_{10}\left(\frac{2}{3} x\right)^{20-10}\left(\frac{3}{2 x}\right)^{10}$ $={ }^{20} C_{10} \fr...
Read More →In the given figure, AB || CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
Question: In the given figure,AB||CD. IfAOC= 30 and OAB= 100, then OCD= ?(a) 130(b) 150(c) 80(d) 100 Solution: (a) 130 Draw $O E\|A B\| C D$ Now, $O E^{\prime} \mid A B$ and $O A$ is the transversal. $\therefore \angle O A B+\angle A O E=180^{\circ} \quad$ [Angles on the same side of a transversal line are supplementary] $\Rightarrow \angle O A B+\angle A O C+\angle C O E=180^{\circ}$ $\Rightarrow 100^{\circ}+30^{\circ}+\angle C O E=180^{\circ}$ $\Rightarrow \angle C O E=50^{\circ}$ Also, $O E \...
Read More →Show that the expansion
Question: Show that the expansion of $\left(x^{2}+\frac{1}{x}\right)^{12}$ does not contain any term involving $x^{-1}$. Solution: Suppose $x^{-1}$ occurs at the $(r+1)$ th term in the given expression. Then, $T_{r+1}={ }^{12} C_{r}\left(x^{2}\right)^{12-r}\left(\frac{1}{x}\right)^{r}$ $={ }^{12} C_{r} x^{24-2 r-r}$ For this term to contain $x^{-1}$, we must have $24-3 r=-1$ $\Rightarrow 3 r=25$ $\Rightarrow r=\frac{25}{3}$ It is not possible, as $r$ is not an integer. Hence, the expansion of $\...
Read More →In the given figure, AB || CD. If ∠BAO = 60° and ∠OCD = 110°, then ∠AOC = ?
Question: In the given figure,AB||CD. IfBAO= 60 and OCD= 110, then AOC= ?(a) 70(b) 60(c) 50(d) 40 Solution: (c) 50 Draw $E O F\|A B\| C D$. Now, $E O \| A B$ and $O A$ is the transversal. $\therefore \angle E O A=\angle O A B=60^{\circ} \quad$ [Alternate Interior Angles] Also, $O F \| C D$ and $O C$ is the transversal. $\therefore \angle C O F+\angle O C D=180^{\circ} \quad$ [Angles on the same side of a transversal line are supplementary] $\Rightarrow \angle C O F+110^{\circ}=180^{\circ}$ $\Rig...
Read More →For the principal values, evaluate each of the following:
Question: For the principal values, evaluate each of the following: (i) $\cos ^{-1} \frac{1}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)$ (ii) (iii) $\sin ^{-1}\left(-\frac{1}{2}\right)+2 \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ (iv) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ Solution: (i) $\cos ^{-1}(\cos x)=x$ $\sin ^{-1}(\sin x)=x$ $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$ $=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 ...
Read More →Does the expansion of
Question: Does the expansion of $\left(2 x^{2}-\frac{1}{x}\right)$ contain any term involving $x^{9} ?$ Solution: Suppose $x^{9}$ occurs in the given expression at the $(r+1)$ th term. Then, we have: $T_{r+1}={ }^{20} C_{r}\left(2 x^{2}\right)^{20-r}\left(\frac{-1}{x}\right)^{r}$ $=(-1)^{r}{ }^{20} C_{r}(2)^{20-r}(x)^{40-2 r-r}$ For this term to contain $x^{9}$, we must have $40-3 r=9$ $\Rightarrow 3 r=31$ $\Rightarrow r=\frac{31}{3}$ It is not possible, as $r$ is not an integer. Hence, there is...
Read More →Does the expansion of
Question: Does the expansion of $\left(2 x^{2}-\frac{1}{x}\right)$ contain any term involving $x^{9} ?$ Solution: Suppose $x^{9}$ occurs in the given expression at the $(r+1)$ th term. Then, we have: $T_{r+1}={ }^{20} C_{r}\left(2 x^{2}\right)^{20-r}\left(\frac{-1}{x}\right)^{r}$ $=(-1)^{r}{ }^{20} C_{r}(2)^{20-r}(x)^{40-2 r-r}$ For this term to contain $x^{9}$, we must have $40-3 r=9$ $\Rightarrow 3 r=31$ $\Rightarrow r=\frac{31}{3}$ It is not possible, as $r$ is not an integer. Hence, there is...
Read More →Find the principal values of each of the following:
Question: Find the principal values of each of the following: (i) $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ (ii) $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ (iii) $\cos ^{-1}\left(\sin \frac{4 \pi}{3}\right)$ (iv) $\cos ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ Solution: (i) Let $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=y$ Then, $\cos y=-\frac{\sqrt{3}}{2}$ We know that the range of the principal value branch is $[0, \pi]$. Thus, $\cos y=-\frac{\sqrt{3}}{2}=\cos \left(\frac{5 \pi}{6}\right)...
Read More →In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
Question: In the given figure,ABis a mirror,PQis the incident ray andQRis the reflected ray. If PQR= 108, find AQP.(a) 72(b) 18(c) 36(d) 54 Solution: (c) 36 We know that angle of incidence = angle of reflection. Then, let $\angle A Q P=\angle B Q R=x^{\circ}$ Now, $\angle A Q P+\angle P Q R+\angle B Q R=180^{\circ} \quad[\because A Q B$ is a straight line $]$ $\Rightarrow x+108+x=180^{\circ}$ $\Rightarrow 2 x=72^{\circ}$ $\Rightarrow x=36^{\circ}$ $\therefore \angle A Q P=36^{\circ}$...
Read More →Which term in the expansion of
Question: Which term in the expansion of $\left\{\left(\frac{x}{\sqrt{y}}\right)^{1 / 3}+\left(\frac{y}{x^{1 / 3}}\right)^{1 / 2}\right\}^{21}$ contains $x$ and $y$ to one and the same power? Solution: Suppose $T_{r+1}$ th term in the given expression contains $x$ and $y$ to one and the same power. Then, $T_{r+1}$ th term is ${ }^{21} C_{r}\left[\left(\frac{x}{\sqrt{y}}\right)^{1 / 3}\right]^{21-r}\left[\left(\frac{y}{x^{1 / 3}}\right)^{1 / 2}\right]^{r}$ $={ }^{21} C_{r}\left(\frac{x^{(21-r) / ...
Read More →In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD = ?
Question: In the given figure, straight linesABandCDintersect atO. IfAOC+ BOD= 130, then AOD= ?(a) 65(b) 115(c) 110(d) 125 Solution: (b) 115 We have: $\angle A O C=\angle B O D \quad[$ Vertically-Opposite Angles $]$ $\therefore \angle A O C+\angle B O D=130^{\circ}$ $\Rightarrow \angle A O C+\angle A O C=130^{\circ} \quad[\because \angle A O C=\angle B O D]$ $\Rightarrow 2 \angle A O C=130^{\circ}$ $\Rightarrow \angle A O C=65^{\circ}$ Now, $\angle A O C+\angle A O D=180^{\circ}[\because C O D$ ...
Read More →Find the sum of the first
Question: Find the sum of the first (i) 11 terms of the A.P.2, 6, 10. 14(ii) 13 terms of the A.P. 6, 0, 6, 12, ...(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8. Solution: In the given problem, we need to find the sum of terms for different arithmetic progressions. 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 (i) $2,...
Read More →In the given figure, straight lines AB and CD intersect at O.
Question: In the given figure, straight lines $A B$ and $C D$ intersect at $O$. If $\angle A O C=\Phi, \angle B O C=\theta$ and $\theta=3 \phi$, then $\phi=$ ? (a) 30(b) 40(c) 45(d) 60 Solution: (c) 45 We have: $\theta+\phi=180^{\circ} \quad[\because A O D$ is a straight line $]$ $\Rightarrow 3 \phi+\phi=180^{\circ} \quad[\because \theta=3 \phi]$ $\Rightarrow 4 \phi=180^{\circ}$ $\Rightarrow \phi=45^{\circ}$...
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