Evaluate:
Question: Evaluate: (i) $\left(i^{41}+\frac{1}{i^{71}}\right)$ (ii) $\left(i^{53}+\frac{1}{i^{53}}\right)$ Solution: (i) $\left(i^{41}+\frac{1}{i^{71}}\right)=i^{41}+i^{-71}$ $\Rightarrow i^{4 \times 10+1}+i^{-4 \times 18+1}\left(\right.$ Since $\left.i^{4 n+1}=i\right)$ $\Rightarrow \mathrm{i}^{1}+\mathrm{i}^{1}$ $\Rightarrow 2 \mathrm{i}$ Hence, $\left(i^{41}+\frac{1}{i^{71}}\right)=2 i$ (ii) $\left(i^{53}+\frac{1}{i^{53}}\right)$ $\Rightarrow i^{53}+i^{-53}$ $\Rightarrow i^{4 \times 13+1}+i^{...
Read More →8 taps of the same size fill a tank in 27 minutes.
Question: 8 taps of the same size fill a tank in 27 minutes. If two taps go out of order, how long would the remaining taps take to fill the tank? Solution: Letxmin be the required number of time. Then, we have: No. of taps 8 6 Time (in min) 27 xx Clearly, less number of taps will take more time to fill the tank . So, it is a case of inverse proportion. Now, $8 \times 27=6 \times x$ $\Rightarrow x=\frac{8 \times 27}{6}$ $\Rightarrow x=36$ Therefore, it will take 36 min to fill the tank....
Read More →Solve this
Question: If $e^{x+y}-x=0$, prove that $\frac{d y}{d x}=\frac{1-x}{x}$ Solution: $e^{x+y}-x=0$ $e^{x+y}=x \ldots \ldots$ (i) Differentiating it with respect to $x$ using chain rule, $\frac{d}{d x}\left(e^{x+y}\right)=\frac{d}{d x}(x)$ $e^{x+y} \frac{d}{d x}(x+y)=1$ $1+\frac{d y}{d x}=\frac{1}{x}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}}-1$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{x}}{\mathrm{x}}$ Hence Proved....
Read More →7 teps of the same size fill a tank in 1 hour 36 minutes.
Question: 7 teps of the same size fill a tank in 1 hour 36 minutes. How long will 8 taps of the same size take to fill the tank? Solution: Letxbe the required number of taps. Then, we have: 1 h = 60 min i.e., 1 h 36 min = (60+36) min = 96 min No. of taps 7 8 Time (in min) 96 x Clearly, more number of taps will require less time to fill the tank. So, it is a case of inverse proportion. Now, $7 \times 96=8 \times x$ $\Rightarrow x=\frac{7 \times 96}{8}$ $\Rightarrow x=84$ Therefore, 8 taps of the ...
Read More →Solve this
Question: If $e^{y}=y^{x}$, prove that $\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}$ Solution: Here $e^{y}=y^{x}$ Taking log on both sides, $\log e^{y}=\log y^{x}$ $y \log e=x \log y$ [Using $\log a^{b}=b \log a$ ] $y=x \log y \ldots \ldots$ (i) Differentiating it with respect to $x$ using product rule, $\frac{d y}{d x}=\frac{d}{d x}(x \log y)$ $=x \frac{d y}{d x}(\log y)+\log y \frac{d}{d x}(x)$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\lo...
Read More →Evaluate:
Question: Evaluate: (i) $i^{-50}$ (ii) $\mathrm{i}^{-9}$ (ii) $\mathrm{i}^{-131}$. Solution: (i) L.H.S. $=\mathrm{i}^{-50}$ $\Rightarrow \mathrm{i}^{-4 \times 13+2}$ $\Rightarrow \mathrm{i}^{4 \mathrm{n}+2}$ $\Rightarrow-1$ Since it is of the form ${ }^{i^{4 n+2}}$ so the solution would be $-1$ (ii) L. H.S. $=\mathrm{i}^{-9}$ $\Rightarrow \mathrm{i}^{-4 \times 3+3}$ $\Rightarrow \mathrm{i}^{4 \mathrm{n}+3}$ $\Rightarrow \mathrm{i}^{3}=-\mathrm{i}$ Since it is of the form of $\mathrm{i}^{4 \mathr...
Read More →Solve this
Question: If $e^{x}+e^{y}=e^{x+y}$, prove that $\frac{d y}{d x}+e^{y-x}=0$ Solution: Here, $e^{x}+e^{y}=e^{x+y} \ldots \ldots$ (i) Differentiating both the sides using chain rule, $\frac{d}{d x} e^{x}+\frac{d}{d x} e^{y}=\frac{d}{d x}\left(e^{x+y}\right)$ $e^{x}+e^{y} \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y)$ $e^{x}+e^{y} \frac{d y}{d x}=e^{x+y}\left[1+\frac{d y}{d x}\right]$ $e^{y} \frac{d y}{d x}-e^{x+y} \frac{d y}{d x}=e^{x+y}-e^{x}$ $\frac{d y}{d x}\left(e^{y}-e^{x+y}\right)=e^{x+y}-e^{x}$...
Read More →Evaluate:
Question: Evaluate: (i) $(\sqrt{-1})^{192}$ (ii) $(\sqrt{-1})^{93}$ (iii) $(\sqrt{-1})^{30}$. Solution: Since $\mathrm{i}=\sqrt{-1}$ so (i) L.H.S. $=(\sqrt{-1})^{192}$ $\Rightarrow \mathrm{i}^{192}$ $\Rightarrow \mathrm{i}^{4 \times 48}=1$ Since it is of the form ${ }^{i^{4 n}}=1$ so the solution would be 1 (ii) L.H.S. $=(\sqrt{-1})^{93}$ $\Rightarrow \mathrm{i}^{4 \times 23+1}$ $\Rightarrow \mathrm{i}^{4 \mathrm{n}+1}$ $\Rightarrow \mathrm{i}^{1}=\mathrm{i}$ Since it is of the form of $\mathrm{...
Read More →A factory requires 42 machines to produce a given number of articles in 56 days.
Question: A factory requires 42 machines to produce a given number of articles in 56 days. How many machines would be required to produce the same number of articles in 48 days? Solution: Let x be the number of machines required to produce same number of articles in 48. Then, we have: No. of machines 42 x No. of days 56 48 Clearly, less number of days will require more number of machines. So, it is a case of inverse proportion. Now, $42 \times 56=x \times 48$ $\Rightarrow x=\frac{42 \times 56}{4...
Read More →Why are Arabari forests of Bengal
Question: Why are Arabari forests of Bengal known to be good example of conserved forests. Solution: Regeneration of Sal Forests An Example of Peoples Participation in the Management of Forests Despite best efforts, the West Bengal Forest Department could not revive the degraded Sal forests of Southwestern districts of the state. Excessive surveillance and policing of the degraded forests not only alienated the people but also resulted in frequent clashes between villagers and forest officials. ...
Read More →vA car takes 5 hours to reach a destination by travelling at the speed of 60 km/hr.
Question: A car takes 5 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 75 km/hr? Solution: Letx h be the required time taken. Then, we have: Speed (in km/h) 60 75 Time (in h) 5 x Clearly, the higher the speed, the lesser will be the the time taken. So, it is a case of inverse proportion. Now, $60 \times 5=75 \times x$ $\Rightarrow x=\frac{60 \times 5}{75}$ $\Rightarrow x=4$ Therefore, the car will reach its destinat...
Read More →Solve this
Question: If $(\cos x)^{y}=(\tan y)^{x}$, prove that $\frac{d y}{d x}=\frac{\log \tan y+y \tan x}{\log \cos x-x \sec y \operatorname{cosec} y}$ Solution: Here, $(\cos x)^{y}=(\tan y)^{x}$ Taking log on both sides, $\log (\cos x)^{y}=\log (\tan y)^{x}$ $y \log (\cos x)=x \log (\tan y)\left[U \sin g \log a^{b}=b \log a\right]$ Differentiating it with respect to $x$ using product rule and chain rule, $\frac{d}{d x}[y \log \cos x]=\frac{d}{d x}[x \log \tan y]$ $y \frac{d}{d x}(\log \cos x)+\log \cos...
Read More →What is the importance of forests
Question: What is the importance of forests as a resource ? Solution: Economic Reasons: Food: Tribals obtain most of their food requirements from the forests, e.g., fruits, tubers, fleshy roots, leaves. Nuts: Pine Nut (Chilgoza), Almond, Walnut and Cashewnut are obtained from forests trees. Spices: Cardamom, Cinnamon, Nutmeg and Cloves are spices obtained from forest plants. Commercial Products: A number of forest products are of commercial importance, g., rubber, resin, tannins, tendu, lac, cor...
Read More →Evaluate:
Question: Evaluate: (i) $i^{19}$ (ii) $i^{62}$ (ii) $\mathrm{i}^{373} .$ Solution: We all know that $i=\sqrt{(}-1)$. And ${ }^{i^{4 n}}=1$ $\mathrm{i}^{4 \mathrm{n}+1}=\mathrm{i}$ (where $\mathrm{n}$ is any positive integer) $\mathrm{i}^{4 \mathrm{n}+2}=-1$ $\mathrm{i}^{4 \mathrm{n}+3}=-1$ So, (i) L.H.S $=\mathrm{i}^{19}$ $=i^{4 \times 4+3}$ $=i^{4 n+3}$ Since it is of the form $\mathrm{i}^{4 \mathrm{n}+3}$ so the solution would be simply $-\mathrm{i}$ Hence the value of $\mathrm{i}^{19}$ is $-\...
Read More →6 cows can graze a field in 28 days.
Question: 6 cows can graze a field in 28 days. How long would 14 cows take to graze the same field? Solution: Letxbe the number of days. Then, we have No. of days 28 x No. of cows 6 14 Clearly, more number of cows will take less number of days to graze the field. So, it is a case of inverse proportion. Now, $28 \times 6=x \times 14$ $\Rightarrow x=\frac{28 \times 6}{14}$ $\Rightarrow x=12$ Therefore, 14 cows will take 12 days to graze the field....
Read More →Solve this
Question: If $(\sin x)^{y}=(\cos y)^{x}$, prove that $\frac{d y}{d x}=\frac{\log \cos y-y \cot x}{\log \sin x+x \tan y}$ Solution: Here, $(\sin x)^{y}=(\cos y)^{x}$ Taking log on both sides, $\log (\sin x)^{y}=\log (\cos y)^{x}$ $y \log (\sin x)=x \log (\cos y)\left[U \operatorname{sing} \log a^{b}=b \log a\right]$ Differentiating it with respect to $x$ using product rule and chain rule, $\frac{d}{d x}[y \log \sin x]=\frac{d}{d x}[x \log \cos y]$ $y \frac{d}{d x}(\log \sin x)+\log \sin x \frac{d...
Read More →Suggest a few useful ways of utilising
Question: Suggest a few useful ways of utilising waste water. Solution: Waste or used water can also become a resource. Treated municipal water can be poured in irrigation channels for supply to crop fields, Treated waste water can be used in urban areas for watering gardens, lawns and washing vehicles, Industries can treat their waste water and recycle the same, Waste water passed into ponds recharges the ground water, Sewage sludge, separated from waste water is a source of manure, compost and...
Read More →Is water conservation necessary ?
Question: Is water conservation necessary ? Give reasons. Solution: Distribution of fresh water is highly uneven. Large tracts are deficient in rain as well as ground water, At most places more water is withdrawn from reservoir and underground source than their recharging Requirement in urban and industrial areas is nearly always higher than the availability, Further demand for water is rising by 4 8% annually in all fields, whether agriculture, industry or domestic use. Therefore, water conserv...
Read More →12 men can dig a pond in 8 days.
Question: 12 men can dig a pond in 8 days. How many men can dig it in 6 days? Solution: Letxbe the required number of men. Then, we have: No. of days 8 6 No. of men 12 x Clearly, more men will require less number of days to dig the pond. So, it is a case of inverse proportion. Now, $8 \times 12=6 \times x$ $\Rightarrow x=\frac{8 \times 12}{6}$ $\Rightarrow x=16$ Therefore, 16 men can dig the pond in 6 days....
Read More →Prepare a list of five activities that you perform daily
Question: Prepare a list of five activities that you perform daily in which natural resources can be conserved or energy utilisation can be minimised. Solution: Judicious use of electricity by switching off lights and electrical appliances not required, Replacement of incandescent bulbs with fluorescent, compact fluorescent ones and LED bulbs. Replacement of electricity or gas operated geysers with solar water heaters, Replacement of electricity generating sets with solar light, Having more natu...
Read More →In the context of conservation of natural resources,
Question: In the context of conservation of natural resources, explain the terms reduce, recycle and reuse. From among the materials that we use in daily life, identify two materials for each category. Solution: Three Rs reduce, recycle and reuse. Reduce:It is to reduce consumption by preventing wastage. Switching off unnecessary lights, fans and other electrical appliances, Repair of leaky taps. Reducing food wastage, Walking down to nearby market instead of using vehicle. Recycle:Separation of...
Read More →Solve this
Question: If $y^{x}=e^{y-x}$ prove that $\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}$ Solution: Here, $y^{x}=e^{y-x}$ Taking log on both sides, $\log y^{x}=\log e^{y-x}$ $x \log y=(y-x) \log e\left[\right.$ Since, $\left.\log (A B)=\log A+\log B ; \log a^{b}=b \log a\right]$ $x \log y=(y-x) \ldots \ldots .(i)$ Differentiating with respect to $x$ using product rule, $\frac{d}{d x}(x \log y)=\frac{d}{d x}(y-x)$ $\left[\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})+\log \mathrm{y} \fr...
Read More →(a) Locate and name the water reservoirs
Question: (a) Locate and name the water reservoirs in figures (i) and (ii). (b) Which has advantage over the other and why ? Solution: (a) Water reservoir in figure (i) is pond while it is underground water body (ground water) in figure (ii). (b) Ground water is more advantageous than pond water. For Benefits: Prevents flooding, Checks soil erosion. Retains water underground and prevents drought, Increases life of downstream reservoirs and dams, Higher biomass production and income of water shed...
Read More →If 35 men can reap a field in 8 days,
Question: If 35 men can reap a field in 8 days, in how many days can 20 men reap the same field? Solution: Letxbe the required number of days. Then, we have: No. of days 8 x No. of men 35 20 Clearly, less men will take more days to reap the field. So, it is a case of inverse proportion. Now, $8 \times 35=x \times 20$ $\Rightarrow \frac{8 \times 35}{20}=x$ $\Rightarrow 14=x$ Therefore, 20 men can reap the same field in 14 days....
Read More →Suggest a few measures for controlling
Question: Suggest a few measures for controlling carbon dioxide levels in the atmosphere. Solution: Increasing Vegetation Cover. It will increase utilisation of atmospheric CO2in photosynthesis. Seeding of Oceans With Phytoplankton. Increased photosynthetic activity of oceans will result in decreasing CO2concentration. Carbonation. CO2released during combustion should not be allowed to pass into atmosphere. Instead, it can be changed into carbonates. Alternate Sources of Energy. Instead of fossi...
Read More →