Prove that
Question: If $f(x)=\frac{x-1}{x}$, find the value of $\left\{\frac{1}{x}\right\}$ Solution: Given, $f(x)=\frac{x-1}{x}$ $F(x)=1-1 / x$ To find $f(1 / x)$ replacing $x$ by $1 / x$ $F(1 / x)=1-1 /(1 / x)$ $F(1 / x)=1-x$...
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Question: Tick (✓) the correct answer: The ratio of cost price and selling price of an article is 20 : 21. What is the gain per cent on it? (a) $5 \%$ (b) $5 \frac{1}{2} \%$ (c) $6 \%$ (d) $6 \frac{1}{4} \%$ Solution: (a) $5 \%$ $\mathrm{CP}=$ Rs. $20 x$ $\mathrm{SP}=$ Rs. $21 x$ Gain = Rs. $(21-20)$ = Rs. X $\therefore$ Gain percentage $=\left(\frac{\text { gain }}{\mathrm{CP}} \times 100\right) \%$ $=\left(\frac{x}{20 x} \times 100\right) \%$ $=5 \%$...
Read More →Draw a ray diagram showing the path
Question: Draw a ray diagram showing the path of rays of light when it enters with oblique incidence from air into water, from water into air. Solution: When ray of light enters from air into water, it bends towards the normal as shown in figure I When ray of light enters from water into air, it bends away from the normal as shown in figure II....
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Question: Let $f=\{(1,6),(2,5),(4,3),(5,2),(8,-1),(10,-3)\}$ and $g=\{(2,0),(3,2),(5,6)$,(7,10),(8,12),(10,16)\}$ Find (i) dom (f $+g$ ) (ii) dom $\left(\frac{f}{g}\right)$. Solution: Given, $f=\{(1,6),(2,5),(4,3),(5,2),(8,-1),(10,-3)\}$ $g=\{(2,0),(3,2),(5,6),(7,10),(8,12),(10,16)\}$ (1) Domain of $f=\{1,2,4,5,8,10\}$ Domain of $\mathrm{g}=\{2,3,5,7,8,10\}$ Domain of $(f+g)=\{x: x \in D f \cap D g\}$ Where Df = Domain of function f, Dg = Domain of function g Domain of $(f+g)=\{2,5,8,10\}$. (2) ...
Read More →Under what condition in an arrangement of two plane mirrors,
Question: Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be the angle of incidence. Show the same with the help of diagram. Solution: Incident ray and reflected ray will always be parallel to each other if two plane mirrors are placed perpendicular to each other as shown in figure, i = r....
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\left(x^{x}\right) \sqrt{x}$ Solution: Let $y=(x)^{x} \sqrt{x}$ Taking log both the sides: $\Rightarrow \log y=\log (x)^{x} \sqrt{x}$ $\Rightarrow \log y=\log (x)^{x}+\log \sqrt{x}\{\log (a b)=\log a+\log b\}$ $\Rightarrow \log y=\log (x)^{x}+\log x^{\frac{1}{2}}$ $\Rightarrow \log y=x \log x+\frac{1}{2} \log x$ $\left\{\log x^{a}=a \log x\right\}$ $\Rightarrow \log y=\left(x+\frac{1}{2}\right) \log x$ Differentiating with re...
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Question: Tick (✓) the correct answer: On selling a chair for Rs 720, a man loses 25%. To gain 25% it must be sold for (a) Rs 900 (b) Rs 1200 (c) Rs 1080 (d) Rs 1440 Solution: (b) Rs 1200 $\mathrm{SP}=\mathrm{Rs} 720$ Loss percentage $=25 \%$ $\mathrm{CP}=\left\{\frac{100}{(100-\text { loss } \%)} \times \mathrm{SP}\right\}$ $=\operatorname{Rs}\left\{\frac{100}{(100-25)} \times \mathrm{SP}\right\}$ $=\operatorname{Rs}\left(\frac{100}{85} \times 720\right)$ $=\operatorname{Rs} 960$ $\therefore$ D...
Read More →How are power and focal length of a lens related ?
Question: How are power and focal length of a lens related ? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light ? (CBSE 2012) Solution: $\mathrm{P}=\frac{1}{f(\text { in metre })}=\frac{100}{f(\text { in } \mathrm{cm})}$ $f_{1}=20 \mathrm{~cm} \therefore \mathrm{P}_{1}=\frac{100}{20}=5.0 \mathrm{D}$ $f_{2}=40 \mathrm{~cm} \therefore \mathrm{P}_{2}=\frac{100}{40}=2.5 \mathrm{D}$ The lens of focal length 20 cm or p...
Read More →In which direction will she move the screen
Question: In which direction will she move the screen to obtain a sharp image of the building ? What is the approximate focal length of this lens ? Solution: A real image can be obtained on the screen. Therefore, the lens used is convex lens as it forms real as well as virtual image. The distance of the real image formed by a convex lens from the lens decreases as the object distance from the lens increases. Hence, the screen has to be moved towards the lens to obtain the sharp image of the buil...
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Question: $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-5}{5-\mathrm{x}}$ then find dom (f) and range (f). Solution: Given, $f(x)=\frac{x-5}{5-x}$ (i) dom(f) Here f(x) is a polynomial function whose domain is R except for points at which denominator becomes zero. Hence x 5 The domain is $(-\infty, \infty)-\{5\}$ (ii) range(f) Let $y=\frac{x-5}{5-x}$ For the specified domain $y=-1$ Range is $\{-1\}$....
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Question: Tick (✓) the correct answer The selling price of an article is $\frac{6}{5}$ of the cost price. The gain per cent is (a) 20% (b) 25% (c) 30% (d) 120% Solution: (a) 20% Let Rs $x$ be the CP of each article. SP of an article $=$ Rs $\frac{6}{5} \mathrm{x}$ Now, gain $=(\mathrm{SP}-\mathrm{CP})$ $=\operatorname{Rs}\left(\frac{6}{5} x-x\right)$ $=\operatorname{Rs} \frac{x}{5}$ $\therefore$ Gain percentage $=\left(\frac{\text { gain }}{\mathrm{CP}} \times 100\right) \%$ $=\left(\frac{\frac{...
Read More →A convex lens of focal length 20 cm
Question: A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement ? If yes, where shall the object be placed in each case for obtaining these images ? Solution: The statement is correct. A convex lens of focal length 20 cm will produce a magnified Virtual image if object is placed at a distance less than 20 cm from the lens. A convex lens of focal length 20 cm will produce a magnified real image if object is placed at a distance grea...
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Question: Tick (✓) the correct answer: By selling a radio for Rs 950, a man loses 5%. What per cent shall he gain by selling it for Rs 1040? (a) 4% (b) 4.5% (c) 5% (d) 9% Solution: (a) 4% SP of the radio $=$ Rs 950 Loss $=5 \%$ $\mathrm{CP}=\left\{\frac{100}{(100-\text { loss })} \times \mathrm{SP}\right\}$ $=$ Rs $\left\{\frac{100}{(100-5)} \times 950\right\}$ $=$ Rs $\left(\frac{100}{95} \times 950\right)$ $=$ Rs 1000 Now, gain $=\mathrm{Rs}(1040-1000)$ = Rs 40 $\therefore$ Gain percentage $=\...
Read More →Write the domain and the range of the function,
Question: Write the domain and the range of the function, $f(x)=\sqrt{x-[x]}$. Solution: Given, $\mathrm{f}(\mathrm{x})=\sqrt{x-[x]}$ Where [x] is the Greatest Integer Function of x. $f(x)=\sqrt{\{x\}}$ Where {x} is fractional part of x. The graph of $f(x)$ is (i) dom(f) Domain of{x} is R. The value of the fractional part of x is always either positive or zero. Hence domain of f(x) is R. (ii) range(f) Range of $\{x\}$ is $[0,1)$. As the root value $[0,1)$ between interval lies between $[0,1)$. H...
Read More →Refractive index of diamond with respect
Question: Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond. Solution: $n_{d g}=1.6$ and $n_{g u}=1.5$ $n_{\alpha}^{\circ}=?$ Now $n_{d g}=\frac{n_{d a}}{n_{g a}}$ $\therefore n_{d u}=n_{d_{d}} \times n_{s e}=1.6 \times 1.5=2.4$...
Read More →Write the domain and the range of the function,
Question: Write the domain and the range of the function, $f(x)=\sqrt{x-[x]}$. Solution: Given, $\mathrm{f}(\mathrm{x})=\sqrt{x-[x]}$ Where [x] is the Greatest Integer Function of x. $f(x)=\sqrt{\{x\}}$ Where {x} is fractional part of x. The graph of $f(x)$ is (i) dom(f) Domain of{x} is R. The value of the fractional part of x is always either positive or zero. Hence domain of f(x) is R....
Read More →How is the refractive index of a medium
Question: How is the refractive index of a medium related to the speed of light ? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media. Solution: Refractive index of a medium $=\frac{\text { speed of light in air }}{\text { speed of light in medium }}$ i.e. $n=\frac{c}{v}$ For medium $\mathrm{I}, n_{1}=\frac{c}{v_{1}}$ For medium II, $n_{2}=\frac{c}{v_{2}}$ $\therefore n_{21}=\frac{n_{2}}{n_{1}}=\frac{c}{v_{2}} \times \frac{v...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $x^{\tan ^{-1} x}$ Solution: Lety $=x^{\tan ^{-1} x}$ Taking log both the sides: $\Rightarrow \log y=\log x^{\tan ^{-1} x}$ $\Rightarrow \log y=\tan ^{-1} \times \log \times\left\{\log x^{2}=\operatorname{alog} x\right\}$ Differentiating with respect to $x$ : $\Rightarrow \frac{d(\log y)}{d x}=\frac{d\left(\tan ^{-1} x \log x\right)}{d x}$ $\Rightarrow \frac{d(\log y)}{d x}=\tan ^{-1} x \times \frac{d(\log x)}{d x}+\log x \tim...
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Question: Tick (✓) the correct answer: Oranges are bought at 5 for Rs 10 and sold at 6 for Rs 15. His gain per cent is (a) 50% (b) 40% (c) 35% (d) 25% Solution: (d) 25% L. C.M of 5 and $6=(5 \times 1 \times 6)=30$ Let 30 be the number of oranges bought. $\mathrm{CP}$ of 5 oranges $=\mathrm{Rs} 10$ CP of 1 oranges $=$ Rs $\left(\frac{10}{5}\right)$ = Rs 2 $\therefore \mathrm{CP}$ of 30 oranges $=\mathrm{Rs}(2 \times 30)$ = Rs 60 SP of 6 oranges $=$ Rs 15 SP of 1 oranges $=\operatorname{Rs}\left(\...
Read More →A pencil when dipped in water in
Question: A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water, we use liquids like, kerosene or turpentine. Support your answer with reason. Solution: A pencil dipped in water appears to be bent at the interface of air and water due to the refraction of light. The refraction of light occurs because the speed of light changes when light travels from one medium to another. T...
Read More →Find the range of the function f
Question: Find the range of the function f(x) = |x|. Solution: |x| is defined as |x|= x; x=0 $-x ; x0$ The value of |x| is never a negative value. Hence range of $|x|$ is $[0, \infty)$....
Read More →Why does a light ray incident on a rectangular
Question: Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram. Solution: Perform an experiment to demonstrate that light bends from its path, when it falls obliquely on the surface of a glass slab. Also show that angle of incidence is about equal to the emergent angle. Take a glass slab and place it on a white sheet of paper fixed on a drawing board. Mark the boundary ABCD of the glass slab. Fix two pins P1and P2, v...
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Question: Tick (✓) the correct answer: Ravi buys some toffees at 5 for a rupee and sells them at 2 for a rupee. His gain per cent is (a) 30% (b) 40% (c) 50% (d) 150% Solution: (d) 150% L. C. M of 5 and $2=(5 \times 1 \times 2)=10$ Let 10 be the number of toffees bought. CP of 5 toffees $=$ Rs 1 $\mathrm{CP}$ of 1 toffee $=\mathrm{Rs}\left(\frac{1}{5}\right)$ $\therefore \mathrm{CP}$ of 10 toffees $=\mathrm{Rs}\left(\frac{1}{5} \times 10\right)$ = Rs 2 SP of 2 toffees $=$ Rs 1 SP of 1 toffee $=$ ...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $(\tan x)^{1 / x}$ Solution: Let $y=(\tan x)^{\frac{1}{x}}$ Taking log both the sides: $\Rightarrow \log y=\log (\tan x)^{\frac{1}{x}}$ $\Rightarrow \log y=\frac{1}{x} \log \tan x\left\{\log x^{a}=\operatorname{alog} x\right\}$ Differentiating with respect to $x$ : $\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\frac{1}{\mathrm{x}} \log \tan \mathrm{x}\right)}{\mathrm{dx}}$ $\Rightarrow \fra...
Read More →Identify the device used as a spherical mirror or lens in the following cases,
Question: Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case. (a) Object is placed between device and its focus, image formed is enlarged and behind it. (b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of an object. (c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of th...
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