Solve this
Question: If $f(x)=\left\{\begin{array}{ll}\frac{x^{2}}{2}, \text { if } 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1x \leq 2\end{array}\right.$. Show that $f$ is continuous at $x=1$. Solution: Given: $f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \quad \text { if } 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1\mathrm{x} \leq 2\end{array}\right.$ We have $(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarro...
Read More →A rectangular sheet of paper 30 cm × 18 cm
Question: A rectangular sheet of paper 30 cm 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed. Solution: Lethcm be the length of the paper andr cm be the radius of the paper. We know that the rectangular sheet of paper 30 cm x 18 cm can be transformed into two types of cylinder. Type 1: Length = 30 cm Diamet...
Read More →State whether any given set is finite or infinite:
Question: State whether any given set is finite or infinite: $H=\{x \in Z:-15x15]$ Solution: Integers = -3, -2, -1, 0, 1, 2, 3, The integers lies between -15 and 15 are finite. the given set is finite...
Read More →An AP consists of 37 terms.
Question: An AP consists of 37 terms. The sum of the three middle most terms and the sum of the last three terms is 429. Find the AP. Solution: Since, total number of terms [Odd] $\therefore$ Middle term $=\left(\frac{37+1}{2}\right)$ th term $=19$ th term So, the three middle most terms $=18$ th 19 th and 20 th. By given condition, Sum of the three middle most terms $=225$ $a_{18}+a_{19}+a_{20}=225$ $\Rightarrow \quad(a+17 d)+(a+18 d)+(a+19 d)=225$ $\Rightarrow \quad 3 a+54 d=225$ $\Rightarrow ...
Read More →The eighth term of an AP is half
Question: The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term. Solution: Let a and d be the first term and common difference of an AP respectively. Now, by given condition, $a_{8}=\frac{1}{2} a_{2}$ $\Rightarrow$$a+7 d=\frac{1}{2}(a+d)$$\left[\because a_{n}=a+(n-1) d\right]$ $\Rightarrow$ $2 a+14 d=a+d$ $\Rightarrow$ $a+13 d=0$ $\ldots($ i) and $a_{11}=\frac{1}{3} a_{4}+1$ $\Rightarrow$ $a+10 d=\frac{1}{3}[a+3 d]+1$...
Read More →Solve this
Question: If $f(x)=\left\{\begin{array}{ll}\frac{x^{2}}{2}, \text { if } 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1x \leq 2\end{array} .\right.$ Show that $f$ is continuous at $x=1 .$ Solution: Given: $f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \text { if } 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1\mathrm{x} \leq 2\end{array}\right.$ We have $(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}...
Read More →A cylindrical water tank of diameter 1.4 m and height 2.1 m
Question: A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled? Solution: Radius of the cylindrical tank $=0.7 \mathrm{~m}$ Height of the cylindrical tank $=2.1 \mathrm{~m}$ Volume of the cylindrical tank $=\pi(0.7)^{2}(2.1) \mathrm{m}^{3}$ Length of the water column flown from the pipe in $1 \mathrm{~s}=2 \mathrm{~m}$ Let the time taken to comple...
Read More →Solve this
Question: If $f(x)= \begin{cases}\frac{x^{2}}{2}, \text { if } 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1x \leq 2\end{cases}$ Show that $f$ is continuous at $x=1$. Solution: Given: $f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \text { if } 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1\mathrm{x} \leq 2\end{array}\right.$ We have $(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}...
Read More →From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second.
Question: From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour. Solution: Radius of the water tap $=0.75 \mathrm{~cm}=0.0075 \mathrm{~m}$ Length of the water flowing in $1 \mathrm{~s}=7 \mathrm{~m}=700 \mathrm{~cm}$ Volume of water flowing in $1 \mathrm{~s}=\pi(0.0075)^{2} \times 700$ Volume of the water flowing in 1 hour $=\pi(0.0075)^{2} \times 700 \times 60 \times 60$ Volume of the water flowing in ...
Read More →Find the values of a and b so that the function f given by
Question: Find the values of $a$ and $b$ so that the function $f$ given by $f(x)=\left\{\begin{aligned} 1, \text { if } x \leq 3 \\ a x+b, \text { if } 3x5 \text { is continuous at } x=3 \text { and } x=5 \\ 7, \text { if } x \geq 5 \end{aligned}\right.$ Solution: Given: $f(x)=\left\{\begin{aligned} 1, \text { if } \mathrm{x} \leq 3 \\ a x+b, \text { if } 3x5 \\ 7, \text { if } \mathrm{x} \geq 5 \end{aligned}\right.$ We have $(\mathrm{LHL}$ at $x=3)=\lim _{x \rightarrow 3} f(x)=\lim _{h \rightar...
Read More →A cylindrical tube, open at both ends, is made of metal.
Question: A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal. Solution: Here,r= Inner radius = 5.2 cm R= Outer radius t= Thickness = 0.8 cm h = Length = 25 cm R=r + t= 5.2 cm + 0.8 cm = 6 cm Volume of the metal =h(R2- r2) $=\frac{22}{7}$$\times(25) \times\left((6)^{2}-(5.2)^{2}\right)$ = 704 cm3 Thus, the volume of the metal is 704 cm3....
Read More →Prove the following
Question: Find the (i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5. (ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5. (iii) sum of those integers from 1 to 500 which are multiples of 2 or 5. Solution: (i) Since, multiples of 2 as well as of 5 = LCM of (2, 5) = 10 Multiples of 2 as well as of 5 between 1 and 500 is 10, 20, 30. 490 which form an AP with first term (a) = 10 and common difference (d) = 20 -10 = 10 nth term an=Last...
Read More →Water flows out through a circular pipe whose internal diameter is 2 cm,
Question: Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes? Solution: Radius of the circular pipe $=0.01 \mathrm{~m}$ Length of the water column in $1 \mathrm{sec}=6 \mathrm{~m}$ Volume of the water flowing in $1 s=\pi r^{2} h=\pi(0.01)^{2}(6) \mathrm{m}^{3}$ Volume of the water flowing in $30 \mathrm{mins}=\pi(0.01)^{2}(6) \time...
Read More →In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
Question: In each of the following, find the value of the constantk so that the given function is continuous at the indicated point; (i) $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 2 k x}{x^{2}}, \text { if } \quad x \neq 0 \\ 8 \text {, if } x=0\end{array}\right.$ at $x=0$ (ii) $f(x)=\left\{\begin{aligned}(x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\ k, \text { if } x=1 \end{aligned}\right.$ at $x=1$ (iii) $f(x)=\left\{\begin{array}{c}k\left(x^{2}-2 x\right), \text { if } x0 \\ \cos x, \t...
Read More →The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq.
Question: The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube. Solution: r= Inner radii of the tube R= Outer radii of the tube h = Length of the tube 2h(R-r) = 88 ... (1) h(R2-r2) = 176 ... (2) Substituting h = 14 cm in equation (1) and (2): (R-r) = 88/28 ... (1) (R-r)(R+r)= 176/14 ... (2) Simplifying the second equation by substituting it with the first equation:...
Read More →In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
Question: In each of the following, find the value of the constantk so that the given function is continuous at the indicated point; (i) $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 2 k x}{x^{2}}, \text { if } \quad x \neq 0 \\ 8 \text {, if } x=0\end{array}\right.$ at $x=0$ (ii) $f(x)=\left\{\begin{aligned}(x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\ k, \text { if } x=1 \end{aligned}\right.$ at $x=1$ (iii) $f(x)=\left\{\begin{array}{c}k\left(x^{2}-2 x\right), \text { if } x0 \\ \cos x, \t...
Read More →In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
Question: In each of the following, find the value of the constantk so that the given function is continuous at the indicated point; (i) $f(x)=\left\{\begin{array}{c}\frac{1-\cos 2 k x}{x^{2}}, \text { if } \\ 8\end{array}\right.$,$x \neq 0$ if $x=0$ at $x=0$ (ii) $f(x)=\left\{\begin{aligned}(x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\ k, \text { if } x=1 \end{aligned}\right.$ at $x=1$ (iii) $f(x)=\left\{\begin{array}{c}k\left(x^{2}-2 x\right), \text { if } x0 \\ \cos x, \text { if } x \...
Read More →2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter.
Question: 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire. Solution: Diameter of the cylindrical wire = 0.25 cm Radius of the cylindrical wire = 0.125 cm Volume of the brass = 2.2 dm3= 2200cm3 Volume of the brass = Volume of the cylindrical wire Length of wire $=\frac{2200 \mathrm{~cm}^{3}}{\frac{22}{7} \times(0.125 \mathrm{~cm})^{2}}=44800 \mathrm{~cm}=448 \mathrm{~m}$ Thus, length of the wire is 448 m....
Read More →2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter.
Question: 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire. Solution: Diameter of the cylindrical wire = 0.25 cm Radius of the cylindrical wire = 0.125 cm Volume of the brass = 2.2 dm3= 2200cm3 Volume of the brass = Volume of the cylindrical wire Length of wire $=\frac{2200 \mathrm{~cm}^{3}}{\frac{22}{7} \times(0.125 \mathrm{~cm})^{2}}=44800 \mathrm{~cm}=448 \mathrm{~m}$ Thus, length of the wire is 448 m....
Read More →Find the length of 13.2 kg of copper wire of diameter 4 mm,
Question: Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm. Solution: Density of copper = Weight/Volume = 8.4 gram/1 cm3= 8.4 gram/cm3 Volume $=$ Weight $/$ Density $=13.2 \mathrm{~kg} \times 1000 \mathrm{gram} / \mathrm{kg} / 8.4 \mathrm{gram} / \mathrm{cm}^{3}=1571.43 \mathrm{~cm}^{3}$ $\mathrm{L}=\frac{V}{\pi r^{2}}=\frac{1571.43 \mathrm{~cm}^{3}}{\frac{22}{7} \times(0.2 \mathrm{~cm})^{2}}=12500.01 \mathrm{~cm}=125 \mathrm{~m}$ Thus, length o...
Read More →A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm.
Question: A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawnout into a wire of diameter 1 mm. What will be the length of the wire so formed? Solution: Diameter of the ductile metal = 1 cm Radius of the ductile metal = 0.5 cm Volume of the ductile metal = r2(length) =(0.5 cm)2(5 cm) = 1.25 cm3 Ductile metal is drawn into a wire of diameter 1 mm. Radius of the wire = 0.5 mm = 0.05 cm Length of wire $=\frac{1.25 \pi \mathrm{cm}^{3}}{\pi(0.05 \mathrm{...
Read More →The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter.
Question: The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what is the rise of water level in the tank due to it? Solution: Length of the water on a roof = 18 m Breadth of the water on a roof = 16.5 m Height of the water on a roof = 10 cm=0.1 m Volume of the water on a roof $=$ Length $\times$ Breadth $\times$ Height $=18 \mathrm{~m} \times 16.5 \mathrm{~m} \times 0.1 \mathrm{~m}=29.7 \mathrm{~m}^{...
Read More →A rectangular sheet of paper 30 cm × 18 cm
Question: A rectangular sheet of paper 30 cm 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed. Solution: Case 1: Height =hcm = 30 cm Diameter =dcm = 18 cm Radius =rcm = 9 cm Volume = (r2h) or ((9 )2(30 )) = 2430 cm3 Case 2: Height=hcm = 18 cm Diameter =dcm = 30 cm, Radius =rcm = 15 cm Volume = (r2h) or ((15 ...
Read More →A cylindrical container with diameter of base 56 cm
Question: A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm 22 cm 14 cm. Find the rise in the level of the water when the solid is completely submerged. Solution: Diameter of the cylindrical container =dcm = 56 cm Radius of the cylindrical container =rcm = 28 cm Volume of cylindrical container = Volume of the rectangular solid Length of the rectangular solid = 32 cm Breadth of the rectangular solid = 22 cm ...
Read More →A well with 14 m diameter is dug 8 m deep.
Question: A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment. Solution: Diameter of the well =dm = 14 m Height of the well =hm = 8 m Radius of the well =rm = 7 m Volume of the well = r2h= (7 m)2(8 m) = 1232 m3 Volume of the well = Volume of the embankment An embankment is a hollow cylinder with thickness. Its inner radius would be equal to the radius of the well, i.e...
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