The maximum value of sin
Question: The maximum value of $\sin ^{2}\left(\frac{2 \pi}{3}+x\right)+\sin ^{2}\left(\frac{2 \pi}{3}-x\right)$ is (a) 1/2 (b) 3/2 (c) 1/4 (d) 3/4 Solution: (b) 3/2 $\frac{2 \pi}{3}=120^{\circ}$ Let $f(x)=\sin ^{2}(90+30+x)+\sin ^{2}(90+30-x)$ $=[\cos (30+x)]^{2}+[\cos (30-x)]^{2} \quad[$ Using $\sin (90+A)=\cos A]$ $=\left[\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x\right]^{2}+\left[\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right]^{2}$ $=\frac{3}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x-\fra...
Read More →If A − B = π/4, then (1 + tan A) (1 − tan B) is equal to
Question: IfAB= /4, then (1 + tanA) (1 tanB) is equal to (a) 2 (b) 1 (c) 0 (d) 3 Solution: (a) 2 $\tan (A-B)=\tan \frac{\pi}{4}$ $\Rightarrow \frac{\tan A-\tan B}{1+\tan A \tan B}=1$ $\Rightarrow \tan A-\tan B=1+\tan A \tan B \quad \ldots(1)$ Now, $(1+\tan A)(1-\tan B)=1+\tan A-\tan B-\tan A \tan B$ $=1+1+\tan A \tan B-\tan A \tan B \quad($ Using eq $(1))$ $=2$...
Read More →AD is an altitude of an equilateral triangle ABC. On AD as base,
Question: AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed.Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4. Solution: We have an equilateral triangle $\triangle A B C$ in which $\mathrm{AD}$ is altitude. An equilateral triangle $\triangle A D E$ is drawn using $\mathrm{AD}$ as base. We have to prove that, $\frac{a r(\Delta A D E)}{a r(\Delta A B C)}=\frac{3}{4}$ Since the two triangles are equilateral, the two triangles will be similar...
Read More →If A − B = π/4, then (1 + tan A) (1 − tan B) is equal to
Question: IfAB= /4, then (1 + tanA) (1 tanB) is equal to (a) 2 (b) 1 (c) 0 (d) 3 Solution: (a) 2 $\tan (A-B)=\tan \frac{\pi}{4}$ $\Rightarrow \frac{\tan A-\tan B}{1+\tan A \tan B}=1$ $\Rightarrow \tan A-\tan B=1+\tan A \tan B \quad \ldots(1)$ Now, $(1+\tan A)(1-\tan B)=1+\tan A-\tan B-\tan A \tan B$ $=1+1+\tan A \tan B-\tan A \tan B \quad($ Using eq $(1))$ $=2$...
Read More →Solve the differential equation
Question: Solve the differential equation $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y(y \neq 0)$ Solution: $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y$ $\Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y}=x e^{\frac{x}{y}}+y^{2}$ $\Rightarrow e^{\frac{x}{y}}\left[y \cdot \frac{d x}{d y}-x\right]=y^{2}$ $\Rightarrow e^{\frac{x}{y}} \cdot \frac{\left[y \cdot \frac{d x}{d y}-x\right]}{y^{2}}=1$ ...(1) Let $e^{\frac{x}{y}}=z$. Differentiating it with respect toy, we...
Read More →If tan (A − B) = 1 and sec (A + B) =
Question: If $\tan (A-B)=1$ and $\sec (A+B)=\frac{2}{\sqrt{3}}$, the smallest positive value of $B$ is (a) $\frac{25 \pi}{24}$ (b) $\frac{19}{24}$ (c) $\frac{13 \pi}{24}$ (d) $\frac{11 \pi}{24}$ Solution: (b) $\frac{19}{24}$ Given: $\tan (A-B)=1$ and $\sec (A+B)=\frac{2}{\sqrt{3}}$ $\Rightarrow A-B=\frac{\pi}{4} \ldots(1)$ and $A+B=\frac{\pi}{6} \ldots(2)$ Adding these equations we get: $2 A=\frac{\pi}{4}+\frac{\pi}{6}$ $\Rightarrow A=\frac{5 \pi}{24}$ $\Rightarrow$ Smallest possible value of $B...
Read More →If ∆ABC and ∆BDE are equilateral triangles,
Question: If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE. Solution: Given: In ΔABC and ΔBDE are equilateral triangles. D is the midpoint of BC. To find:Ar∆ABCAr∆BDE In ΔABC and ΔBDE $\triangle \mathrm{ABC} \sim \triangle \mathrm{BDE}(\mathrm{AAAcrietria}$ of similarity, all angles of equilateral triangle are equal $)$ Since D is the midpoint of BC, BD : DC = 1. We know that the ratio of areas of two similar triangles is equal ...
Read More →If tan (π/4 + x) + tan (π/4 − x) = a,
Question: If tan (/4 +x) + tan (/4 x) =a, then tan2(/4 +x) + tan2(/4 x) = (a)a2+ 1 (b)a2+ 2 (c)a2 2 (d) None of these Solution: (c)a2 2 Given: $\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)=a$ $\Rightarrow\left[\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)\right]^{2}=a^{2}$ $\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)+2 \tan \left(\frac{\pi}{4}-x\right) \tan \left(\frac{\pi}{4}+x\right)=a^{2}$ $\Rightarrow \t...
Read More →If tan (π/4 + x) + tan (π/4 − x) = a,
Question: If tan (/4 +x) + tan (/4 x) =a, then tan2(/4 +x) + tan2(/4 x) = (a)a2+ 1 (b)a2+ 2 (c)a2 2 (d) None of these Solution: (c)a2 2 Given: $\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)=a$ $\Rightarrow\left[\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)\right]^{2}=a^{2}$ $\Rightarrow \tan ^{2}\left(\frac{\pi}{4}+x\right)+\tan ^{2}\left(\frac{\pi}{4}-x\right)+2 \tan \left(\frac{\pi}{4}-x\right) \tan \left(\frac{\pi}{4}+x\right)=a^{2}$ $\Rightarrow \t...
Read More →Find the particular solution of the differential equation
Question: Find the particular solution of the differential equation $\left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0$, given that $y=1$ when $x=0$ Solution: $\left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0$ $\Rightarrow \frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}=0$ Integrating both sides, we get: $\tan ^{-1} y+\int \frac{e^{x} d x}{1+e^{2 x}}=\mathrm{C}$ ...(1) Let $e^{x}=t \Rightarrow e^{2 x}=t^{2} .$ $\Rightarrow \frac{d}{d x}\left(e^{x}\right)=\frac{d t}{d x}$ $\R...
Read More →If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC.
Question: If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE. Solution: Given: In ΔABC, D is a point on side AB such that AD : DB= 3 : 2. E is a point on side BC such that DE || AC. To find: $\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{BDE}}$ In ΔABC, $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$ Since DE\|AC, $\frac{E C}{E B}=\frac{3}{2}$ According to converse of basic proportionality theorem if a line...
Read More →The value of cos
Question: The value of cos (36 A) cos (36 +A) + cos (54 +A) cos (54 A) is (a) sin 2A (b) cos 2A (c) cos 3A (d) sin 3A Solution: (b) cos 2A $\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\cos \left(54^{\circ}+A\right) \cos \left(54^{\circ}-A\right)$ $=\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\sin \left[90^{\circ}-\left(54^{\circ}+A\right)\right] \sin \left[90^{\circ}-\left(54^{\circ}-A\right)\right] \quad\left[\right.$ Since $\left.\sin \left(90^{\circ}-\theta...
Read More →ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m,
Question: ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that the area of ∆APQ is one-sixteenth of the area of ∆ABC. Solution: Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q. AP = 1cm, PB = 3cm, AQ = 1.5cm and QC = 4.5cm. To find: $\operatorname{Ar}(\triangle \mathrm{APQ})=\frac{1}{16}(\triangle \mathrm{ABC})$ In ΔABC, $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$ $\frac{...
Read More →The value of cos
Question: The value of cos (36 A) cos (36 +A) + cos (54 +A) cos (54 A) is (a) sin 2A (b) cos 2A (c) cos 3A (d) sin 3A Solution: (b) cos 2A $\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\cos \left(54^{\circ}+A\right) \cos \left(54^{\circ}-A\right)$ $=\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\sin \left[90^{\circ}-\left(54^{\circ}+A\right)\right] \sin \left[90^{\circ}-\left(54^{\circ}-A\right)\right] \quad\left[\right.$ Since $\left.\sin \left(90^{\circ}-\theta...
Read More →If tanθ
Question: If $\tan \theta=\frac{1}{2}$ and $\tan \phi=\frac{1}{3}$, then the value of $\theta+\phi$ is (a) $\frac{\pi}{6}$ (b) $\pi$ (c) 0 (d) $\frac{\pi}{4}$ Solution: It is given that $\tan \theta=\frac{1}{2}$ and $\tan \phi=\frac{1}{3}$. Now, $\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$ $=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}$ $=\frac{\frac{5}{6}}{\frac{5}{6}}$ = 1 $\therefore \theta+\phi=\frac{\pi}{4} \quad\left(\tan \frac{\pi}{4}=1\ri...
Read More →Find the equation of the curve passing through the point
Question: Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is, $\sin x \cos y d x+\cos x \sin y d y=0$ Solution: The differential equation of the given curve is: $\sin x \cos y d x+\cos x \sin y d y=0$ $\Rightarrow \frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0$ $\Rightarrow \tan x d x+\tan y d y=0$ Integrating both sides, we get: $\log (\sec x)+\log (\sec y)=\log \mathrm{C}$ $\log (\sec x \cdot \sec y)=\log \mathrm...
Read More →The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.
Question: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. Solution: Given: Theareas of two similar triangles ABC and PQR are in the ratio 9 : 16.BC = 4.5cm. To find: length of QR We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.ar∆ABCar∆PQR=BCQR2 $\frac{9}{16}=\left(\frac{4.5}{Q R}\right)^{2}$ $\frac{3}{4}=\frac{4.5}{Q R}$ $\mathrm{QR}=\frac{4 \times 4.5}{3}$ $Q R=...
Read More →If sin (π cos x) = cos (π sin x),
Question: If sin ( cosx) = cos ( sinx), then sin 2x= (a) $\pm \frac{3}{4}$ (b) $\pm \frac{4}{3}$ (c) $\pm \frac{1}{3}$ (d) none of these Solution: (c) $\pm \frac{1}{3}$ $\sin (\pi \cos x)=\cos (\pi \sin x)$ As we know that $\sin x=-\cos \left(\frac{\pi}{2}+x\right)$ $\Rightarrow-\cos \left(\frac{\pi}{2}+\pi \cos x\right)=\cos (\pi \sin x)$ $\Rightarrow \frac{-\pi}{2}-\pi \cos x=\pi \sin x$ $\Rightarrow \pi \sin x-\pi \cos x=\frac{\pi}{2}$ $\Rightarrow \sin x-\cos x=\frac{1}{2}$ Squaring both sid...
Read More →If sin (π cos x) = cos (π sin x),
Question: If sin ( cosx) = cos ( sinx), then sin 2x= (a) $\pm \frac{3}{4}$ (b) $\pm \frac{4}{3}$ (c) $\pm \frac{1}{3}$ (d) none of these Solution: (c) $\pm \frac{1}{3}$ $\sin (\pi \cos x)=\cos (\pi \sin x)$ As we know that $\sin x=-\cos \left(\frac{\pi}{2}+x\right)$ $\Rightarrow-\cos \left(\frac{\pi}{2}+\pi \cos x\right)=\cos (\pi \sin x)$ $\Rightarrow \frac{-\pi}{2}-\pi \cos x=\pi \sin x$ $\Rightarrow \pi \sin x-\pi \cos x=\frac{\pi}{2}$ $\Rightarrow \sin x-\cos x=\frac{1}{2}$ Squaring both sid...
Read More →In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such
Question: In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. FindBPAB. Solution: Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q such that PQ || BC and PQ divides ΔABC in two parts equal in area. To find: $\frac{\mathrm{BP}}{\mathrm{AB}}$ We have PQ $\|$ BC And $A r(\triangle \mathrm{APQ})=A r($ quad $\mathrm{BPQC})$ $\operatorname{Ar}(\triangle \mathrm{APQ})+\operatorname{Ar}(\triangle \mathr...
Read More →Show that the general solution of the differential equation
Question: Show that the general solution of the differential equation $\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$ is given by $(x+y+1)=A(1-x-y-2 x y)$, where $A$ is parameter Solution: $\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0$ $\Rightarrow \frac{d y}{d x}=-\frac{\left(y^{2}+y+1\right)}{x^{2}+x+1}$ $\Rightarrow \frac{d y}{y^{2}+y+1}=\frac{-d x}{x^{2}+x+1}$ $\Rightarrow \frac{d y}{y^{2}+y+1}+\frac{d x}{x^{2}+x+1}=0$ Integrating both sides, we get: $\int \frac{d y}{y^{2}+y+1}+\int \frac{d...
Read More →If tan θ
Question: If $\tan \theta_{1} \tan \theta_{2}=k$, then $\frac{\cos \left(\theta_{1}-\theta_{2}\right)}{\cos \left(\theta_{1}+\theta_{2}\right)}=$ (a) $\frac{1+k}{1-k}$ (b) $\frac{1-k}{1+k}$ (c) $\frac{k+1}{k-1}$ (d) $\frac{k-1}{k+1}$ Solution: (a) $\frac{1+k}{1-k}$ $\frac{\cos \left(\theta_{1}-\theta_{2}\right)}{\cos \left(\theta_{1}+\theta_{2}\right)}$ $=\frac{\cos \theta_{1} \cos \theta_{2}+\sin \theta_{1} \sin \theta_{2}}{\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}}$ Divid...
Read More →If ∆ABC ∼ ∆DEF such that AB = 5 cm,
Question: If $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ such that $\mathrm{AB}=5 \mathrm{~cm}$, area $(\Delta \mathrm{ABC})=20 \mathrm{~cm}^{2}$ and area $(\Delta \mathrm{DEF})=45 \mathrm{~cm}^{2}$, determine $\mathrm{DE}$. Solution: Given: The area of two similar $\triangle \mathrm{ABC}=20 \mathrm{~cm}^{2}, \triangle \mathrm{DEF}=45 \mathrm{~cm}^{2}$ respectively and $\mathrm{AB}=5 \mathrm{~cm}$. To find: measure of DE We know that the ratio of areas of two similar triangles is equal to ...
Read More →The value of cos
Question: The value of $\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)$ is (a) $\frac{1}{2} \cos 2 x$ (b) 0 (c) $-\frac{1}{2} \cos 2 x$ (d) $\frac{1}{2}$ Solution: (a) $\frac{1}{2} \cos 2 x$ $\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)$ $=\cos \left(\frac{\pi}{6}+x+\frac{\pi}{6}-x\right) \cos \left(\frac{\pi}{6}+x-\frac{\pi}{6}+x\right) \quad\left[\right.$ Using $\left.\cos (A+B) \cos (A-B)=\cos ^{2} A-\sin ^{2} B\right]$ $=\cos \frac{...
Read More →The areas of two similar triangles are 121 cm2 and 64 cm2 respectively.
Question: The areas of two similar triangles are $121 \mathrm{~cm}^{2}$ and $64 \mathrm{~cm}^{2}$ respectively. If the median of the first triangle is $12.1 \mathrm{~cm}$, find the corresponding median of the other. Solution: Given: The area of two similar triangles is $121 \mathrm{~cm}^{2}$ and $64 \mathrm{~cm}^{2}$ respectively. IF the median of the first triangle is $12.1 \mathrm{~cm}$ To find: corresponding medians of the other triangle We know that the ratio of areas of two similar triangle...
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