The sum of the digits in unit place
Question: The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time isA. 432B. 108C. 36D. 18 Solution: B. 108 Explanation: The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time =(3+4+5+6)3! =108 Hence, Option (B) 108 is the correct answer....
Read More →The number of different four digit numbers
Question: The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once isA. 120B. 96C. 24D. 100 Solution: C. 24 Explanation: We know that, nPr= $\frac{n !}{(n-r) !}$ Four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once =4P4= 4! = 24 Hence, Option (C) 24 is the correct answer....
Read More →The number of possible outcomes
Question: The number of possible outcomes when a coin is tossed 6 times isA. 36B. 64C. 12D. 32 Solution: B. 64 Explanation: The coin is tossed 6 times. Number ofpossible outcomes = 2 The number of possible outcomes when a coin is tossed 6 times is 26= 64 Hence, Option (B) 64 is the correct answer....
Read More →solve this
Question: $f(x)=x^{3}(2 x-1)^{3}$ Solution: Given : $f(x)=x^{3}(2 x-1)^{3}$ $\Rightarrow f^{\prime}(x)=3 x^{2}(2 x-1)^{3}+6 x^{3}(2 x-1)^{2}$ For the local maxima or minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}(2 x-1)^{3}+6 x^{3}(2 x-1)^{2}=0$ $\Rightarrow 3 x^{2}(2 x-1)^{2}(2 x-1+2 x)=0$ $\Rightarrow x^{2}(2 x-1)^{2}(4 x-1)=0$ $\Rightarrow x=0, \frac{1}{2}$ and $\frac{1}{4}$ Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $\frac{1}{4}, x=\frac{1...
Read More →If nC12 = nC8, then n is equal to
Question: IfnC12=nC8, then n is equal toA. 20B. 12C. 6D. 30 Solution: A. 20 Explanation: According to the question, nC12=nC8 We know that, nCr n 8 = 12 n = 20 Hence, Option (A) 20 is the correct answer....
Read More →A group consists of 4 girls and 7 boys.
Question: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girls (ii) at least one boy and one girl (iii) at least three girls. Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ (i) No girls Total number of ways the team can have no girls =4C0.7C5=21 (ii) at least one boy and one girl Case(A)1 boy and 4 girls =7C1.4C4 =7 Case(B)2 boys and3 girls =7C2.4C3 =84 Case(C) 3boys and 2girls =7C3.4C2 =210 Case(D)4 boys and 1 girls =7C...
Read More →If the solve the problem
Question: $\mathrm{f}(\mathrm{x})=x \sqrt{1-x}, x0$ Solution: Given : $f(x)=x \sqrt{1-x}$ $\Rightarrow f^{\prime}(x)=\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}}=\frac{2-3 x}{2 \sqrt{1-x}}$ For the local maxima or minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0$ $\Rightarrow x=\frac{2}{3}$ Since, $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{2}{3}, x=\frac{2}{3}$ is a point of maxima. The local maximum value of $f(x)$ at $x=\frac{2}{3}$...
Read More →Find the equation of the circle whose centre is
Question: Find the equation of the circle whose centre is (2, - 5) and which passes through the point (3, 2). Solution: The general form of the equation of a circle is: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Where, (h, k) is the centre of the circle. r is the radius of the circle. In this question we know that $(h, k)=(2,-5)$, so for determining the equation of the circle we need to determine the radius of the circle. Since the circle passes through $(3,2)$, that pair of values for $x$ and $y$ must satisfy...
Read More →If the solve the problem
Question: $f(x)=2 \sin x-x,-\frac{\pi}{2}\leq x \leq \frac{\pi}{2}$ Solution: Given : $f(x)=2 \sin x-x$ $\Rightarrow f^{\prime}(x)=2 \cos x-1$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 2 \cos x-1=0$ $\Rightarrow \cos x=\frac{1}{2}$ $\Rightarrow x=\frac{\pi}{3}$ or $\frac{-\pi}{3}$ Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{\pi}{3}, x=\frac{\pi}{3}$ is the point of local maxima. The local maximum value of $f(...
Read More →Find the centre and radius of each of the following circles :
Question: Find the centre and radius of each of the following circles : $x^{2}+(y-1)^{2}=2$ Solution: The general form of the equation of a circle is: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Where, $(\mathrm{h}, \mathrm{k})$ is the centre of the circle. $r$ is the radius of the circle. Comparing the given equation of circle with general form we get: $h=0, k=1, r^{2}=2$ $\Rightarrow$ centre $=(0,1)$ and radius $=\sqrt{2}$ units. Ans: centre $=(0,1)$ and radius $=\sqrt{2}$ units ....
Read More →Find the centre and radius of each of the following circles :
Question: Find the centre and radius of each of the following circles : $(x+5)^{2}+(y-3)^{2}=20$ Solution: The general form of the equation of a circle is: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Where, (h, k) is the centre of the circle. r is the radius of the circle Comparing the given equation of circle with general form we get: $h=-5, k=3, r^{2}=20$ $\Rightarrow$ centre $=(-5,3)$ and radius $=\sqrt{2} 0=2 \sqrt{5}$ units. Ans: centre $=(-5,3)$ and radius $=2 \sqrt{5}$ units....
Read More →If the solve the problem
Question: $f(x)=\sin 2 x-x,-\frac{\pi}{2}\leq x \leq \frac{\pi}{2}$ Solution: Given : $f(x)=\sin 2 x-x$ $\Rightarrow f^{\prime}(x)=2 \cos 2 x-1$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 2 \cos 2 x-1=0$ $\Rightarrow \cos 2 x=\frac{1}{2}$ $\Rightarrow x=\frac{-\pi}{6}$ or $\frac{\pi}{6}$ Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{\pi}{6}, x=\frac{\pi}{6}$ is the point of local maxima. The local maximum value ...
Read More →A sports team of 11 students is to be constituted,
Question: A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and atleast 5 from Class X II. If there are 20 students in each of these classes, in how many ways can the team be constituted? Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ A team of 11 students can be constituted in the following two ways (i) 5 students from class XI and 6 students from XII =20C5.20C6 (ii) 6 students from class XI and 5 students from XII =20C6.20C5 The number ways the team ca...
Read More →In how many ways can a football team of 11 players
Question: In how many ways can a football team of 11 players be selected from 16 players? How many of them will (i) include 2 particular players? (ii) exclude 2 particular players? Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ According to the question, 11 players can be selected out of 16 =16C11 (i) include 2 particular players =14C9 (ii) exclude 2 particular players =14C11...
Read More →Find the centre and radius of each of the following circles :
Question: Find the centre and radius of each of the following circles : $\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{3}\right)^{2}=\frac{1}{16}$ Solution: The general form of the equation of a circle is: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Where, $(h, k)$ is the centre of the circle. $r$ is the radius of the circle. Comparing the given equation of circle with general form we get: $\mathrm{h}=1 / 2, \mathrm{k}=-1 / 3, \mathrm{r}^{2}=1 / 16$ $\Rightarrow$ centre $=(1 / 2,-1 / 3)$ and radius $=1 / 4$ u...
Read More →A bag contains six white marbles and five red marbles.
Question: A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if (a) they can be of any colour (b) two must be white and two red and (c) they must all be of the same colour. Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ According to the question, Number of white marbles = 6, Number of red marbles = 5 Total number of marbles = 6 white + 5 red = 11 marbles (a)If they can be of any colour Then, any 4 marbles out of 1...
Read More →18 mice were placed in two experimental groups
Question: 18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups? Solution: According to the question, Number of mice =18, Number of groups = 3 Since the groups are equally large, The number of mice in each group can be = 6 mice The number of ways of placement of mice =18! For each group the placement of mice = 6! Hence, the required number of ways = 18!/(6!6!6!) = 18!/(6!)3...
Read More →Find the centre and radius of each of the following circles :
Question: Find the centre and radius of each of the following circles : $(x-3)^{2}+(y-1)^{2}=9$ Solution: The general form of the equation of a circle is: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Where, (h, k) is the centre of the circle. r is the radius of the circle. Comparing the given equation of circle with general form we get: $h=3, k=1, r^{2}=9$ $\Rightarrow$ centre $=(3,1)$ and radius $=3$ units. Ans: centre = (3, 1) and radius = 3 units....
Read More →A convex polygon has 44 diagonals.
Question: A convex polygon has 44 diagonals. Find the number of its sides. [Hint: Polygon of n sides has (nC2 n) number of diagonals.] Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ Let the number of sides the given polygon have = n Now, The number of line segments obtained by joining n vertices =nC2 So, number of diagonals of the polygon =nC2 n = 44 $\frac{n(n-1)}{2}-n=44$ n2 3n 88 = 0 (n 11) (n + 8) = 0 n = 11 or n = 8 The polygon has 11sides....
Read More →Find the equation of a circle with
Question: Find the equation of a circle with Centre at the origin and radius 4 Solution: The general form of the equation of a circle is: $(x-h)^{2}+(y-k)^{2}=r^{2}$ Where, (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle in he general form: $\Rightarrow(x-0)^{2}+(y-0)^{2}=4^{2}$ $\Rightarrow x^{2}+y^{2}=16$ Ans; equation of a circle with. Centre at the origin and radius 4 is: $x^{2}+y^{2}=16$...
Read More →In an examination, a student has to answer 4 questions
Question: In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice. Solution: We know that, nCr $=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$ According to the question, Total number of questions =5 Number of questions to be answered =4 Compulsory questions are question number 1 and 2 Hence, the number of ways in which the student can make the choice =3...
Read More →If the solve the problem
Question: $f(x)=\cos x, 0x\pi$ Solution: Given : $f(x)=\cos x$ $\Rightarrow f^{\prime}(x)=-\sin x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow-\sin x=0$ $\Rightarrow \sin x=0$ $\Rightarrow x=0$ or $\pi$ Since $0x\pi$, none is in the interval $(0, \pi)$....
Read More →In a certain city, all telephone numbers have six digits,
Question: In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all si x digits distinct? Solution: According to the question, All telephone numbers have six digits Given that, The first two digits = 41 or 42 or 46 or 62 or 64 Hence, the number of 2 digits that the telephone number begins with =5 First two digits can be filled in 5 ways, The remaining four-digits can be filled in8P4ways, 8P4= 8!/(8 ...
Read More →If 20 lines are drawn in a plane such that
Question: If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other? Solution: Let the number of intersection point of first line =0 Let the number of intersection point of 2ndline = 1 Let the number of intersection point of 3rdline = 2+1 Let the number of intersection point of 4thline = 3+2+1 . . . Let the number of intersection point of nthline =(n-1) +(n-2)..(3)(2)(1), where n=20 S = ( n 1) n/2 =1910 =...
Read More →If the solve the problem
Question: $f(x)=\sin x-\cos x, 0x2 \pi$ Solution: Given : $f(x)=\sin x-\cos x$ $\Rightarrow f^{\prime}(x)=\cos x+\sin x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow \cos x+\sin x=0$ $\Rightarrow \cos x=-\sin x$ $\Rightarrow \tan x=-1$ $\Rightarrow x=\frac{3 \pi}{4}$ or $\frac{7 \pi}{4}$ Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{3 \pi}{4}, x=\frac{3 \pi}{4}$ is the point of local maxima. The local maximum valu...
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