A hemispherical bowl is made of steel 0.25 cm thick.
Question: A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl. Solution: Inner radius = 5 cm Outer radius = 5 + 0.25 = 5.25 Volume of steel used = Outer volume-Inner volume $=2 / 3 \times \pi \times\left((5.25)^{3}-(5)^{3}\right)$ $=2 / 3 \times 22 / 7 \times\left((5.25)^{3}-(5)^{3}\right)$ $=41.282 \mathrm{~cm}^{3}$...
Read More →A hemispherical tank has the inner radius of 2.8 m.
Question: A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters. Solution: Radius of the tank = 2.8 m Therefore Capacity $=2 / 3 \pi r^{3}$ $=2 / 3 \times 22 / 7 \times(2.8)^{3}=45.994 \mathrm{~m}^{3}$ $1 \mathrm{~m}^{3}=10001$ Therefore capacity in litres = 45994 litres...
Read More →Prove that:
Question: Prove that: (i) $\sin 65^{\circ}+\cos 65^{\circ}=\sqrt{2} \cos 20^{\circ}$ (ii) $\sin 47^{\circ}+\cos 77^{\circ}=\cos 17^{\circ}$ Solution: (i) Consider LHS : $\sin 65^{\circ}+\cos 65^{\circ}$ $=\sin 65^{\circ}+\cos \left(90^{\circ}-25^{\circ}\right)$ $=\sin 65^{\circ}+\sin 25^{\circ}$ $=2 \sin \left(\frac{65^{\circ}+25^{\circ}}{2}\right) \cos \left(\frac{65^{\circ}-25^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right...
Read More →The general solution of the differential equation
Question: The general solution of the differential equation $\frac{y d x-x d y}{y}=0$ is A. $x y=C$ B. $x=C y^{2}$ C. $y=C x$ D. $y=C x^{2}$ Solution: The given differential equation is: $\frac{y d x-x d y}{y}=0$ $\Rightarrow \frac{y d x-x d y}{x y}=0$ $\Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0$ Integrating both sides, we get: $\log |x|-\log |y|=\log k$ $\Rightarrow \log \left|\frac{x}{y}\right|=\log k$ $\Rightarrow \frac{x}{y}=k$ $\Rightarrow y=\frac{1}{k} x$ $\Rightarrow y=C x$ where $C=\f...
Read More →Find the volume of a sphere whose diameter is:
Question: Find the volume of a sphere whose diameter is: (i) 14 cm (ii) 3.5 dm (iii) 2.1 m Solution: (i) Diameter = 14 cm, Radius(r) = 14/2 = 7 cm Therefore volume $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(7)^{3}=1437.33 \mathrm{~cm}^{3}$ (ii) Diameter = 3.5 dm, Radius (r) = 3.52 = 1.75 dm Therefore volume $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(1.75)^{3}$ $=22.46 \mathrm{dm}^{3}$ (iii) Diameter = 2.1m, Radius(r) = 2.1/2 = 1.05 m Therefore volume $=4 / 3 \pi r^{3}$ $=4 / 3 \times 2...
Read More →The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.
Question: The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009? Solution: Let the population at any instant (t)bey. It is given that the rate of increase of population is proportional to the number of inhabitants at any instant. $\therefore \frac{d y}{d t} \propto y$ $\Rightarrow \frac{d y}{d...
Read More →Find the volume of a sphere whose radius is:
Question: Find the volume of a sphere whose radius is: (i) 2 cm (ii) 3.5 cm (iii) 10.5 cm. Solution: (i)Radius (r) = 2 cm Therefore volume $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(2)^{3}$ $=33.52 \mathrm{~cm}^{3}$ (ii) Radius (r) = 3.5 cm Therefore volume $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(3.5)^{3}=179.666 \mathrm{~cm}^{3}$ (iii) Radius (r) = 10.5 cm Therefore volume $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(10.5)^{3}=4851 \mathrm{~cm}^{3}$...
Read More →Prove that:
Question: Prove that: (i) $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$ (ii) $\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x$ Solution: (i) Consider LHS : $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$ $=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}$...
Read More →Prove that:
Question: Prove that: (i) $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$ (ii) $\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x$ Solution: (i) Consider LHS : $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$ $=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}$...
Read More →Find a particular solution of the differential equation
Question: Find a particular solution of the differential equation $(x+1) \frac{d y}{d x}=2 e^{-y}-1$, given that $y=0$ when $x=0$ Solution: $(x+1) \frac{d y}{d x}=2 e^{-y}-1$ $\Rightarrow \frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1}$ $\Rightarrow \frac{e^{y} d y}{2-e^{y}}=\frac{d x}{x+1}$ Integrating both sides, we get: $\int \frac{e^{y} d y}{2-e^{y}}=\log |x+1|+\log \mathrm{C}$ ...(1) Let $2-e^{y}=t$. $\therefore \frac{d}{d y}\left(2-e^{y}\right)=\frac{d t}{d y}$ $\Rightarrow-e^{y}=\frac{d t}{d y}$ $...
Read More →Prove that:
Question: Prove that: (i) cos 55 + cos 65 + cos 175 = 0 (ii) sin 50 sin 70 + sin 10 = 0 (iii) cos 80 + cos 40 cos 20 = 0 (iv) cos 20 + cos 100 + cos 140 = 0 (v) $\sin \frac{5 \pi}{18}-\cos \frac{4 \pi}{9}=\sqrt{3} \sin \frac{\pi}{9}$ (vi) $\cos \frac{\pi}{12}-\sin \frac{\pi}{12}=\frac{1}{\sqrt{2}}$ (vii) sin 80 cos 70 = cos 50 (viii) sin 51 + cos 81 = cos 21 Solution: (i) Consider LHS : $\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}$ $=2 \cos \left(\frac{55^{\circ}+65^{\circ}}{2}\right) \cos ...
Read More →Prove that:
Question: Prove that: (i) cos 55 + cos 65 + cos 175 = 0 (ii) sin 50 sin 70 + sin 10 = 0 (iii) cos 80 + cos 40 cos 20 = 0 (iv) cos 20 + cos 100 + cos 140 = 0 (v) $\sin \frac{5 \pi}{18}-\cos \frac{4 \pi}{9}=\sqrt{3} \sin \frac{\pi}{9}$ (vi) $\cos \frac{\pi}{12}-\sin \frac{\pi}{12}=\frac{1}{\sqrt{2}}$ (vii) sin 80 cos 70 = cos 50 (viii) sin 51 + cos 81 = cos 21 Solution: (i) Consider LHS : $\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}$ $=2 \cos \left(\frac{55^{\circ}+65^{\circ}}{2}\right) \cos ...
Read More →Prove that:
Question: Prove that: (i) sin 38 + sin 22 = sin 82 (ii) cos 100 + cos 20 = cos 40 (iii) sin 50 + sin 10 = cos 20 (iv) sin 23 + sin 37 = cos 7 (v) sin 105 + cos 105 = cos 45 (vi) sin 40 + sin 20 = cos 10 Solution: (i) Consider LHS : $s$ in $38^{\circ}+\sin 22^{\circ}$ $=2 \sin \left(\frac{38^{\circ}+22^{\circ}}{2}\right) \cos \left(\frac{38^{\circ}-22^{\circ}}{2}\right)$ $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \sin 30^{\circ} \...
Read More →Express each of the following as the product of sines and cosines:
Question: Express each of the following as the product of sines and cosines: (i) sin 12x+ sin 4x (ii) sin 5x sinx (iii) cos 12x+ cos 8x (iv) cos 12x cos 4x (v) sin 2x+ cos 4x Solution: (i) sin 12x+ sin 4x $=2 \sin \left(\frac{12 x+4 x}{2}\right) \cos \left(\frac{12 x-4 x}{2}\right)$ $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \sin 8 x \cos 4 x$ (ii) sin 5x sinx $=2 \sin \left(\frac{5 x-x}{2}\right) \cos \left(\frac{5 x+x}{2}\right...
Read More →Express each of the following as the product of sines and cosines:
Question: Express each of the following as the product of sines and cosines: (i) sin 12x+ sin 4x (ii) sin 5x sinx (iii) cos 12x+ cos 8x (iv) cos 12x cos 4x (v) sin 2x+ cos 4x Solution: (i) sin 12x+ sin 4x $=2 \sin \left(\frac{12 x+4 x}{2}\right) \cos \left(\frac{12 x-4 x}{2}\right)$ $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \sin 8 x \cos 4 x$ (ii) sin 5x sinx $=2 \sin \left(\frac{5 x-x}{2}\right) \cos \left(\frac{5 x+x}{2}\right...
Read More →If tan α
Question: If $\tan \alpha=\frac{1}{1+2^{-x}}$ and $\tan \beta=\frac{1}{1+2^{x+1}}$, then write the value of $\alpha+\beta$ lying in the interval $\left(0, \frac{\pi}{2}\right)$. Solution: $\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$ $=\frac{\frac{1}{1+2^{-x}}+\frac{1}{1+2^{x+1}}}{1-\frac{1}{\left(1+2^{-x}\right)\left(1+2^{x+1}\right)}}$ $=\frac{1+2^{x+1}+1+2^{-x}}{1+2^{x+1}+2^{-x}+2^{-x+x+1}-1}$ $=\frac{2+2^{x+1}+2^{-x}}{2+2^{x+1}+2^{-x}}$ $=1$ Therefore, $\alpha...
Read More →If sin α − sin β = a and cos α + cos β = b,
Question: If sin sin =aand cos + cos =b, then write the value of cos ( + ). Solution: $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$ $=\frac{2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta+2-2}{2}$ $=\frac{\sin ^{2} \alpha+\cos ^{2} \alpha+\sin ^{2} \beta+\cos ^{2} \beta+2 \cos \alpha \cos \beta-2 \sin \alpha \sin \beta-2}{2}$ $=\frac{(\sin \alpha-s \text { in } \beta)^{2}+(\cos \alpha+\cos \beta)^{2}-2}{2}$ $=\frac{a^{2}+b^{2}-2}{2}$...
Read More →If A + B = C
Question: IfA+B=C, then write the value of tanAtanBtanC. Solution: $\tan A \tan B \tan C=\tan A \tan B \tan (A+B)$ $[$ Using $A+B=C]$ $=\tan A \tan B \times \frac{\tan A+\tan B}{1-\tan A \tan B}$ $=\frac{\tan ^{2} A \tan B+\tan A \tan ^{2} B}{1-\tan A \tan B}$ $=\frac{\tan ^{2} A \tan B+\tan A \tan ^{2} B+\tan A+\tan B-\tan A-\tan B}{1-\tan A \tan B}$ $=\frac{-\tan A(1-\tan A \tan B)-\tan B(1-\tan A \tan B)+\tan A+\tan B}{1-\tan A \tan B}$ $=\frac{-(1-\tan A \tan B)(\tan A+\tan B)+\tan A+\tan B}...
Read More →If a = b cos
Question: If $a=b \cos \frac{2 \pi}{3}=c \cos \frac{4 \pi}{3}$, then write the value of $a b+b c+c a$. Solution: $a=b \cos 120^{\circ}=c \cos 240^{\circ}$ $\Rightarrow a=-\frac{1}{2} b=-\frac{1}{2} c$ Therefore, $a b+b c+c a=\frac{-1}{2} b \times b+b \times b+b \times \frac{-1}{2} b$ $=-b^{2}+b^{2}$ $=0$...
Read More →If
Question: If $\frac{\cos (x-y)}{\cos (x+y)}=\frac{m}{n}$, then write the value of $\tan x \tan y$ Solution: $\frac{\cos (x-y)}{\cos (x+y)}=\frac{m}{n}$ $\Rightarrow \frac{\cos x \cos y+\sin x \sin y}{\cos x \cos y-\sin x \sin y}=\frac{m}{n}$ $\Rightarrow \frac{1+\tan x \tan y}{1-\tan x \tan y}=\frac{m}{n}$ [ Dividing numerator and denominator of LHS by $\cos x \cos y$ ] $\Rightarrow n+n \tan x \tan y=m-m \tan x \tan y$ $\Rightarrow \tan x \tan y(m+n)=m-n$ $\Rightarrow \tan x \tan y=\frac{m-n}{m+...
Read More →If tan (A + B) = p and tan (A − B) = q
Question: If tan (A+B) =pand tan (AB) =q, then write the value of tan 2B. Solution: $\tan 2 B=\tan (B+B)$ $=\tan (A+B-(A-B))$ $=\frac{\tan (A+B)-\tan (A-B)}{1+\tan (A+B) \tan (A-B)}$ $=\frac{p-q}{1+p q} \quad[\because \tan (A+B)=p$ and $\tan (A-B)=q]$...
Read More →Write the interval in which the value of
Question: Write the interval in which the value of $5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+3$ lies. Solution: Let $f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+3$ $=5 \cos x+3\left(\cos x \cos 60^{\circ}-\sin x \sin 60^{\circ}\right)+3$ $=5 \cos x+\frac{3}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+3$ $=\frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+3$ We know that, $-\sqrt{\left(\frac{13}{2}\right)^{2}+\left(\frac{3 \sqrt{3}}{2}\right)^{2}} \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \...
Read More →If 12 sin x − 9sin
Question: If 12 sinx 9sin2xattains its maximum value atx= , then write the value of sin . Solution: Let $f(x)=12 \sin x-9 \sin ^{2} x$ $=-\left(9 \sin ^{2} x-12 \sin x\right)$ $=-\left[(3 \sin x)^{2}-2.3 \sin x .2+2^{2}-4\right]$ $=-\left[(3 \sin x-2)^{2}-4\right]$ $=4-(3 \sin x-2)^{2}$ Minimum value of $(3 \sin x-2)^{2}$ is 0 . Therefore, maximum value of $f(x)=4-(3 \sin x-2)^{2}$ is 4 . We are given that $12 \sin x-9 \sin ^{2} x$ will attain its maximum value at $x=\alpha$. $\therefore 12 \sin...
Read More →Write the maximum value of
Question: Write the maximum value of 12 sinx 9 sin2x. Solution: Let $f(x)=12 \sin x-9 \sin ^{2} x$ $=-\left(9 \sin ^{2} x-12 \sin x\right)$ $=-\left[(3 \sin x)^{2}-2.3 \sin x .2+2^{2}-4\right]$ $=-\left[(3 \sin x-2)^{2}-4\right]$ $=4-(3 \sin x-2)^{2}$ Minimum value of $(3 \sin x-2)^{2}$ is 0 . Therefore, maximum value of $4-(3 \sin x-2)^{2}$ would be 4 ....
Read More →Write the maximum and minimum values of
Question: Write the maximum and minimum values of 3 cosx+ 4 sinx+ 5. Solution: Let $f(x)=3 \cos x+4 \sin x+5$ We know that, $-\sqrt{3^{2}+4^{2}} \leq 3 \cos x+4 \sin x \leq \sqrt{3^{2}+4^{2}}$ $\Rightarrow-5 \leq 3 \cos x+4 \sin x \leq 5$ $\Rightarrow-5+5 \leq 3 \cos x+4 \sin x+5 \leq 5+5$ $\Rightarrow 0 \leq f(x) \leq 10$ Hence, maximum and minimum vales of $f(x)$ are 0 and 10 respectively....
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