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Question: If $y=\log \frac{x^{2}+x+1}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$, find $\frac{d y}{d x}$ Solution: Here, $y=\log \frac{x^{2}+x+1}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$ Differentiating it with respect to $x$ using chain and quotient rule, $\frac{d y}{d x}=\frac{d}{d x} \log \frac{x^{2}+x+1}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \frac{d}{d x} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$ $\frac{d y}{d x}=...
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Question: Prove that $\frac{1}{i}-\frac{1}{i^{2}}+\frac{1}{i^{3}}-\frac{1}{i^{4}}=0$ Solution: Given: $\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{i}^{2}}+\frac{1}{\mathrm{i}^{3}}-\frac{1}{\mathrm{i}^{4}}$ To prove : $\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{i}^{2}}+\frac{1}{\mathrm{i}^{3}}-\frac{1}{\mathrm{i}^{4}}=0$. $\Rightarrow$ L.H.S. $=\mathrm{i}^{-1}-\mathrm{i}^{-2}+\mathrm{i}^{-3}-\mathrm{i}^{-4}$ $\Rightarrow \mathrm{i}^{-4 \times 1+3}-\mathrm{i}^{-4 \times 1+2}+\mathrm{i}^{-4 \times 1+3}-\mathr...
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Question: Tick (✓) the correct answer: A pole 14 m high casts a shadow of 10 m. At the same time, what will be the height of a tree, the length of whose shadow is 7 metres? (a) 20 m (b) 9.8 m (c) 5 m (d) none of these Solution: (b) 9.8 m Letxm be the height of the tree. Height of object 14 xx Length of shadow 10 7 The more the length of the shadow, the more will be the height of the tree. Now, $\frac{14}{10}=\frac{x}{7}$ $\Rightarrow x=\frac{14 \times 7}{10}$ $\Rightarrow x=9.8$ Therefore, a 9.8...
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Question: Tick (✓) the correct answer: The weight of 12 sheets of a thick paper is 40 grams. How many sheets would weight 1 kg? (a) 480 (b) 360 (c) 300 (d) none of these Solution: (c) 300 Let x sheets weigh 1 kg. Now, 1 kg = 1000 g No. of sheets 12 xx Weight (in g) 40 1000 Now,$\frac{12}{40}=\frac{x}{1000}$ $\Rightarrow x=\frac{12 \times 1000}{40}$ $\Rightarrow x=300$...
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Question: Tick (✓) the correct answer: A car is travelling at a uniform speed of 75 km/hr. How much distance will it cover in 20 minutes? (a) 25 km (b) 15 km (c) 30 km (d) 20 km Solution: (a) 25 km Letxkm be the required distance. Now, 1 h = 60 min Distance (in km) 75 xx Time (in min) 60 20 Less distance will be covered in less time Now, $\frac{75}{60}=\frac{x}{20}$ $\Rightarrow x=\frac{75 \times 20}{60}$ $\Rightarrow x=25 \mathrm{~km}$...
Read More →Find the derivative of the function f(x) given by
Question: Find the derivative of the function $f(x)$ given by $f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$ and hence find $f^{\prime}(1)$ Solution: Here, $f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$ $f(1)=(2)(2)(2)(2)=16$ Taking log on both sides we get, $\log (f(x))=\log (1+x)+\log \left(1+x^{2}\right)+\log \left(1+x^{4}\right)+\log \left(1+x^{8}\right)$ Differentiating it with respect to $x$ we get, $\frac{1}{f(x)} \frac{d(f(x))}{d x}=\fr...
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Question: Prove that $6 i^{50}+5 i^{33}-2 i^{15}+6 i^{48}=7 i$ Solution: Given: $6 i^{50}+5 i^{33}-2 i^{15}+6 i^{48}$ To prove: $6 i^{50}+5 i^{33}-2 i^{15}+6 i^{48}=7 i$ $\Rightarrow 6 i^{4 \times 12+2}+5 i^{4 \times 8+1}-2 i^{4 \times 3+3}+6 i^{4 \times 12}$ $\Rightarrow 6 i^{2}+5 i^{1}-2 i^{3}+6 i^{0}$ $\Rightarrow-6+5 i+2 i+6$ $\Rightarrow 7 \mathrm{i}$ $\Rightarrow$ L.H.S $=$ R.H.S Hence proved....
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Question: Tick (✓) the correct answer: A machine fills 420 bottles in 3 hours. How many bottles will it fill in 5 hours? (a) 252 (b) 700 (c) 504 (d) 300 Solution: (b) 700 Letxbe the number of bottles filled in 5 hours. No. of bottles 420 xx Time (h) 3 5 More number of bottles will be filled in more time. Now, $\frac{420}{3}=\frac{x}{5}$ $\Rightarrow x=\frac{420 \times 5}{3}$ $\Rightarrow x=700$ Therefore, 700 bottles would be filled in 5 h....
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Question: Prove that $1+i^{2}+i^{4}+i^{6}=0$ Solution: L.H.S. $=1+i^{2}+i^{4}+i^{6}$ To Prove: $1+i^{2}+i^{4}+i^{6}=0$ $\Rightarrow 1+(-1)+1+i^{2}$ Since, $i^{4 n}=1$ (Where n is any positive integer) $\Rightarrow i^{4 n+2}$ $\Rightarrow i^{2}=-1$ $\Rightarrow 1+-1+1+-1=0$ Hence proved....
Read More →If 8 oranges cost ₹ 52, how many oranges can be bought for ₹ 169?
Question: If 8 oranges cost₹ 52, how many oranges can be bought for₹ 169? (a) 13 (b) 18 (c) 26 (d) 24 Solution: Let the number of oranges that can be bought for ₹169 bex. Quantity 8 x Price(in ₹) 52 169 As the quantity increases the price also increases. So, this is a case of direct proportion. $\therefore \frac{\circ}{52}=\frac{x}{169}$ $\Rightarrow x=\frac{8 \times 169}{52}=26$ Thus, 26 oranges can be bought for ₹169. Hence, the correct answer is option (c)....
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Question: If $y=x \sin y$, prove that $\frac{d y}{d x}=\frac{y}{x(1-x \cos y)}$ Solution: Here, $y=x \sin y$ siny $=\frac{y}{x}$ ....(1) Differentiating it with respect to $x$ using product rule, $\frac{d y}{d x}=\frac{d}{d x}(x$ siny $)$ $\frac{d y}{d x}=x \frac{d}{d x}(\sin y)+\sin y \frac{d}{d x}(x)$ $\frac{d y}{d x}=x \cos y \frac{d y}{d x}+\sin y(1)$ $\frac{d y}{d x}-x \cos y \frac{d y}{d x}=\sin y$ $\frac{d y}{d x}(1-x \cos y)=\sin y$ $\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$ $\frac{d y}...
Read More →If 14 kg of pulses cost ₹ 882, what is the cost of 22 kg of pulses?
Question: If 14 kg of pulses cost₹ 882, what is the cost of 22 kg of pulses? (a)₹ 1254 (b)₹ 1298 (c)₹ 1342 (d) ₹ 1386 Solution: Let 22 kg of pulses cost ₹x. Quantity(in kg) 14 22 Price(in ₹) 882 x As the quantity increases, the price also increases. So, it is a case of direct proportion. $\therefore \frac{14}{882}=\frac{22}{x}$ $\Rightarrow x=\frac{22 \times 882}{14}=1386$ Thus, the cost of 22 kg of pulses is ₹1,386. Hence, the correct answer is option (d)....
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Question: If $y=x \sin y$, prove that $\frac{d y}{d x}=\frac{y}{x(1-x \cos y)}$ Solution: Here, $y=x \sin y$ siny $=\frac{y}{x}$ ....(1) Differentiating it with respect to $x$ using product rule, $\frac{d y}{d x}=\frac{d}{d x}(x$ siny $)$ $\frac{d y}{d x}=x \frac{d}{d x}(\sin y)+\sin y \frac{d}{d x}(x)$ $\frac{d y}{d x}=x \cos y \frac{d y}{d x}+\sin y(1)$ $\frac{d y}{d x}-x \cos y \frac{d y}{d x}=\sin y$ $\frac{d y}{d x}(1-x \cos y)=\sin y$ $\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$ $\frac{d y}...
Read More →If x and y vary inversely and x = 18 when y = 8,
Question: Ifxandyvary inversely andx= 18 wheny= 8, findxwheny= 16. Solution: $x$ and $y$ var $y$ inversely. i.e. $x y=$ constant Now, $18 \times 8=x_{1} \times 16$ $\Rightarrow \frac{18 \times 8}{16}=x_{1}$ $\Rightarrow 9=x_{1}$ $\therefore$ Value of $x=9$...
Read More →If x and y vary inversely and x = 15 when y = 6,
Question: Ifxandyvary inversely andx= 15 wheny= 6, findywhenx= 9. Solution: $x$ and $y$ var $y$ inversely. i.e. $x y=$ constant Now, $15 \times 6=9 \times y_{1}$ $\Rightarrow y_{1}=\frac{15 \times 6}{9}$ $\Rightarrow y_{1}=10$ $\therefore$ Value of $y=10$, when $x=9$...
Read More →If x and y vary inversely and x = 15 when y = 6,
Question: Ifxandyvary inversely andx= 15 wheny= 6, findywhenx= 9. Solution: $x$ and $y$ var $y$ inversely. i.e. $x y=$ constant Now, $15 \times 6=9 \times y_{1}$ $\Rightarrow y_{1}=\frac{15 \times 6}{9}$ $\Rightarrow y_{1}=10$ $\therefore$ Value of $y=10$, when $x=9$...
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Question: If $x y \log (x+y)=1$, prove that $\frac{d y}{d x}=\frac{y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}$ Solution: Here, $x y \log (x+y)=1 \ldots \ldots(i)$ Differentiating it with respect to $x$ using the chain rule and product rule, $\frac{d y}{d x}(x y \log (x+y))=\frac{d}{d x}(1)$ $x y \frac{d}{d x} \log (x+y)+x \log (x+y) \frac{d y}{d x}+y \log (x+y) \frac{d}{d x}(x)=0$ $\left(\frac{x y}{(x+y)}\right)(1+)+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0$ $\left(\frac{x y}{(x+y)}...
Read More →A school has 9 periods a day each of 40 minutes duration.
Question: A school has 9 periods a day each of 40 minutes duration. How long would each period be, if the school has 8 periods a day, assuming the number of school hours to be the same? Solution: Letxmin be the duration of each period when the school has 8 periods a day. No. of periods 9 8 Time (in min) 40 x Clearly, if the number of periods reduces, the duration of each period will increase. So, it is a case of inverse proportion. Now, $9 \times 40=8 \times x$ $\Rightarrow x=\frac{9 \times 40}{...
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Question: If $(\sin x)^{y}=x+y$, prove that $\frac{d y}{d x}=\frac{1-(x+y) y \cot x}{(x+y) \log \sin x-1}$ Solution: Here $(\sin x)^{y}=x+y$ Taking log both sides, $\log (\sin x)^{y}=\log (x+y)$ $y \log (\sin x)=\log (x+y)\left[\right.$ Using $\log a^{b}=b \log a$ ] Differentiating it with respect to $x$ using the chain rule and product rule, $\frac{d}{d x}(y \log (\sin x))=\frac{d}{d x} \log (x+y)$ $y \frac{d}{d x} \log (\sin x)+\log \sin x \frac{d y}{d x}=\frac{1}{(x+y)} \frac{d}{d x}(x+y)$ $\...
Read More →In a hostel, 75 students had food provision for 24 days.
Question: In a hostel, 75 students had food provision for 24 days. If 15 students leave the hostel, for how many days would the food provision last? Solution: Letxbe the required number of days. Then, we have: No. of students 75 60 No. of days 24 x Clearly, less number of students will take more days to finish the food. So, it is a case of inverse proportion. Now, $75 \times 24=60 \times x$ $\Rightarrow x=\frac{75 \times 24}{60}$ $\Rightarrow x=30$ Therefore, the food will now last for 30 days....
Read More →In a hostel, 75 students had food provision for 24 days.
Question: In a hostel, 75 students had food provision for 24 days. If 15 students leave the hostel, for how many days would the food provision last? Solution: Letxbe the required number of days. Then, we have: No. of students 75 60 No. of days 24 x Clearly, less number of students will take more days to finish the food. So, it is a case of inverse proportion. Now, $75 \times 24=60 \times x$ $\Rightarrow x=\frac{75 \times 24}{60}$ $\Rightarrow x=30$ Therefore, the food will now last for 30 days....
Read More →A garrison of 900 men had provisions for 42 days.
Question: A garrison of 900 men had provisions for 42 days. However, a reinforcement of 500 men arrived. For how many days will the food last now? Solution: Letxbe the required number of days. Then, we have: No. of men 900 1400 No. of days 42 x Clearly, more men will take less number of days to finish the food. So, it is a case of inverse proportion. Now, $900 \times 42=1400 \times x$ $\Rightarrow x=\frac{900 \times 42}{1400}$ $\Rightarrow x=27$ Therefore, the food will now last for 27 days....
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Question: If $x \sin (a+y)+\sin a \cos (a+y)=0$, prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$ Solution: Here, $x \sin (a+y)+\sin a \cos (a+y)=0$ $x=\frac{-\sin a \cos (a+y)}{x \sin (a+y)}$ Differentiating it with respect to $x$ using the chain rule and product rule, $\frac{d}{d x}[x \sin (a+y)+\sin a \cos (a+y)]=0$ $x \frac{d}{d x} \sin (a+y)+\sin (a+y) \frac{d x}{d x}+\sin a \frac{d}{d x} \cos (a+y)+\cos (a+y) \frac{d}{d x} \sin a=0$ $x \cos (a+y)\left(0+\frac{d y}{d x}\right)+\si...
Read More →A farmer has enough food to feed 28 animals in his cattle for 9 days.
Question: A farmer has enough food to feed 28 animals in his cattle for 9 days. How long would the food last, if there were 8 more animals in his cattle? Solution: Let x be the required number of days. Then, we have: No. of days 9 x No. of animals 28 36 Clearly, more number of animals will take less number of days to finish the food. So, it is a case of inverse proportion. Now, $9 \times 28=x \times 36$ $\Rightarrow x=\frac{9 \times 28}{36}$ $\Rightarrow x=7$ Therefore, the food will last for 7 ...
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Question: If $y=x \sin (a+y)$, prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$ Solution: Here $y=x \sin (a+y)$ Differentiating it with respect to $x$ using the chain rule and product rule, $\frac{d y}{d x}=x \frac{d}{d x} \sin (a+y)+\sin (a+y) \frac{d x}{d x}$ $\frac{d y}{d x}=x \cos (a+y) \frac{d y}{d x}+\sin (a+y)$ $(1-x \cos (a+y)) \frac{d y}{d x}=\sin (a+y)$ $\frac{d y}{d x}=\frac{\sin (a+y)}{(1-x \cos (a+y))}$ $\frac{d y}{d x}=\frac{\sin (a+y)}{\left(1 \frac{y}{\...
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