Find the equation of a circle with
Question: Find the equation of a circle with Centre $(-a,-b)$ and radius $\sqrt{a^{2}-b^{2}}$ 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. Substituting the centre and radius of the circle in he general form: $\left.\Rightarrow(x-(-a))^{2}+(y-(-b))^{2}=\sqrt{(} a^{2} 2-b^{2} 2\right)^{2}$ $\Rightarrow(x+a)^{2}+(y+b)^{2}=a^{2}-b^{2}$ $\Rightarrow x^{2}+2 x a+a^{2}+y^{2}+2 y a+b^{2}=a...
Read More →Find the number of integers greater than 7000
Question: Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated. [Hint: Besides 4-digit integers greater than 7000, five digit integers are always greater than 7000.] Solution: According to the question, Digits that can be used = 3,5,7,8,9 Since, no digits can be repeated, The number of integers is5P5= 5! = 120 For a four-digit integer to be greater than 7000, The four-digit integer should begin with 7,8 or 9. The number of...
Read More →If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.
Question: IfnCr 1= 36,nCr= 84 andnCr + 1= 126, then findrC2. [Hint: From equation usingnCr/nCr + 1andnCr/nCr 1to find the value of r.] Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ According to the question, nCr 1=36, nCr=84, nCr +1=126 $\frac{C_{r}^{n}}{C_{r+1}^{n}}=\frac{84}{126}$ $\Rightarrow \frac{\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}}{\frac{\mathrm{n} !}{(\mathrm{r}+1) !(\mathrm{n}-\mathrm{r}-1) !}}=\frac{84}{126}=\frac{2}{3}$ 2n-2r=3r+3 ⇒2n 3 = 5r (i) $\frac{C_...
Read More →A box contains two white,
Question: A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw.[Hint: Required number of ways =3C16C2+3C2x6C1+3C3.] Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ Drawing 1 black and 2 other ball =3C16C2 Drawing 2 black and 1 other ball =3C26C1 Drawing 3 black balls =3C3 Number of ways in which at least one black ball can be drawn = =(1 black and 2 other )or( 2 black and 1...
Read More →There are 10 lamps in a hall.
Question: There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.[Hint: Required number = 210 1]. Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ We also know that, $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{C}_{\mathrm{k}}^{\mathrm{n}}=2^{\mathrm{n}}-1$ According to the question, Number of lamps in a hall =10 Given that, One of the lamps can be switched on independently Hence, the number of ways in which...
Read More →There are 10 persons named P1,P2,P3, … P10.
Question: There are 10 persons named P1,P2,P3, P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4and P5do not occur. Find the number of such possible arrangements. [Hint: Required number of arrangement =7C4 5!] Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ According to the question, There are 10 person named P1, P2, P3, P10. Number of ways in which P1can be arranged =5! =120 Number of ways in which others can be arranged,...
Read More →If the solve the problem
Question: $f(x)=\sin 2 x, 0x\pi$ Solution: Given : $f(x)=\sin 2 x$ $\Rightarrow f^{\prime}(x)=2 \cos 2 x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 2 \cos 2 x=0$ $\Rightarrow \cos 2 x=0$ $\Rightarrow x=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$ Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{\pi}{4}, x=\frac{\pi}{4}$ is the point of maxima. The local maximum value of $f(x)$ at $x=\frac{\pi}{4}$ is given by $\sin \left(\...
Read More →Find the number of positive integers greater
Question: Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated. Solution: (i)Thousands Place can be fill with 6 alone. Hence, number of way =1 Unit place can be filled with either 0 or 5. Hence, number of way =2 Hundreds place can be filled with the remaining 8 digits. Hence, number of way =8 Tens place can be filled with 7 digits. Number of ways= 7. Thus, required number will be =1 x 8 x 7 x 2 =112...
Read More →Find the number of different words
Question: Find the number of different words that can be formed from the letters of the word TRIANGLE so that no vowels are together. Solution: We know that, nPr $=\frac{n !}{(n-r) !}$ According to the question, Total number of vowels letter =3, Total number of consonants letter =5 The vowels can be placed in 6P3= 6!/3! = 120 The number of way consonants can be arranged placed =5! =120 Therefore, total number of ways it can be arranged =5!6P3=120120 =14400...
Read More →solve this
Question: $f(x)=x^{3}-6 x^{2}+9 x+15$ Solution: Given : $f(x)=x^{3}-6 x^{2}+9 x+15$ $\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+9$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}-12 x+9=0$ $\Rightarrow x^{2}-4 x+3=0$ $\Rightarrow(x-1)(x-3)=0$ $\Rightarrow x=1$ or 3 Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $3, x=3$ is the point of local minima. The local minimum value of $f(x)$ at $x=3$ is given by $(3)^{3}-6(3)^{2}...
Read More →Find the equation of a circle with
Question: Find the equation of a circle with Centre $(a \cos \propto, a \sin \propto)$ and radius a 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: $(x-(a \cos \propto))^{2}+(y-(a \sin \alpha))^{2}=a^{2}$ $\Rightarrow(x-a \cos \alpha)^{2}+(y-a \sin \alpha)^{2}=a^{2}$ $\Rightarrow x^{2}-2 x a \cos \alpha+a^{2} \cos ^{...
Read More →Find the number of permutations
Question: Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together. Solution: Permutations of n distinct things taken r together =nCr And when 3 particular things must occur together, we get, =n 3Cr 3 =n 3Cr 3 (r 2)! 3!...
Read More →A bag contains 5 black and 6 red balls.
Question: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot. Solution: We know that, nCr $=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$ According to the question, Number of black balls = 5 Number of red balls = 6 Number of ways in which 2 black balls can be selected =5C2 $=\frac{5 !}{2 ! 3 !}=10$ Number of ways in which 3 red balls can be selected =5C3 $=\frac{6 !}{3 ! 3 !}=20$ Number of ways in which...
Read More →How many automobile license plates
Question: How many automobile license plates can be made if each plate contains two different letters followed by three different digits? Solution: According to the question, Number of letters in automobile license plates = 2 We know that, There are 26 alphabets So, Letter can be arranged without repetition in the following number of ways, = 2625 =650 Number of digits in automobile license plates =3 We know that, there 10 digits Hence the number of digits without repetitions =1098=720 Therefore,...
Read More →Find the equation of a circle with
Question: Find the equation of a circle with Centre $(a, a)$ and radius $\sqrt{2}$ 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. Substituting the centre and radius of the circle in he general form: $\Rightarrow(x-a)^{2}+(y-a)^{2}=(\sqrt{2})^{2}$ $\Rightarrow(x-a)^{2}+(y-a)^{2}=2$ Ans; equation of a circle with Centre $(a, a)$ and radius $\sqrt{2}$ is: $(x-a)^{2...
Read More →If the solve the problem
Question: $f(x)=\frac{1}{x^{2}+2}$ Solution: Given : $f(x)=\frac{1}{x^{2}+2}$ $\Rightarrow f^{\prime}(x)=\frac{-2 x}{\left(x^{2}+2\right)^{2}}$ For the local maxima or minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0$ $\Rightarrow x=0$ Now, for values close to $x=0$ and to the left of $0, f^{\prime}(x)0$. Also, for values close to $x=0$ and to the right of $0, f^{\prime}(x)0$. Therefore, by first derivative test, $x=0$ is a point of local maxima and the...
Read More →Find the equation of a circle with
Question: Find the equation of a circle with Centre $(-3,-2)$ and radius 6 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: $\Rightarrow(x-(-3))^{2}+(y-(-2))^{2}=6^{2}$ $\Rightarrow(x+3)^{2}+(y+2)^{2}=36$ Ans; equation of a circle with Centre $(-3,-2)$ and radius 6 is: $\Rightarrow(x+3)^{2}+(y+2)^{2}=36$...
Read More →How many committee of five persons
Question: How many committee of five persons with a chairperson can be selected from 12 persons? [Hint: Chairman can be selected in 12 ways and remaining in11C4.] Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ Number of ways a chairperson can be selected =12 Selection of 4 other people =11C4 $=\frac{11 !}{4 ! 7 !}=330$ Selection of 5 people = 330 12 =3960...
Read More →If the solve the problem
Question: $f(x)=(x-1)(x+2)^{2}$ Solution: Given: $f(x)=(x-1)(x+2)^{2}$ $\Rightarrow f^{\prime}(x)=(x+2)^{2}+2(x+2)(x-1)$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow(x+2)^{2}+2(x+2)(x-1)=0$ $\Rightarrow(x+2)(x+2+2 x-2)=0$ $\Rightarrow(x+2)(3 x)=0$ $\Rightarrow x=0,-2$ Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $0, x=0$ is the point of local minima. The local minimum value of $f(x)$ at $x=0$ is given by $(0-1)(0+2)^{2}...
Read More →We wish to select 6 persons from 8,
Question: We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can selections be made? Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ According to the question, Case 1: If both A and B are selected =1x1x6C4 $=\frac{6 !}{4 !(6-4) !}=\frac{6 !}{4 ! 2 !}=15$ Case 2: If neither A nor B are selected =6C6= 1 If B is selected but A is not selected = 1x6C5 $=\frac{6 !}{5 !(6-4) !}=6$ Adding the results of both A and B being selected, neither A...
Read More →Find the equation of a circle with
Question: Find the equation of a circle with Centre $(2,4)$ and radius 5 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: $\Rightarrow(x-2)^{2}+(y-4)^{2}=5^{2}$ $\Rightarrow(x-2)^{2}+(y-4)^{2}=25$ Ans; equation of a circle with Centre (2, 4) and radius 5 is: $\Rightarrow(x-2)^{2}+(y-4)^{2}=25$...
Read More →Out of 18 points in a plane,
Question: Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.[Hint: Number of straight lines =18C25C2 + 1] Solution: We know that, nCr $=\frac{\mathrm{n} !}{\Gamma !(\mathrm{n}-\mathrm{r}) !}$ According to the question, Number of points = 18 Number of Collinear points = 5 Number of lines form by 18 points =18C2 For 5 points to be collinear =5C2 The number of lines that can be formed join...
Read More →If the solve the problem
Question: $f(x)=x^{3}(x-1)^{2}$ Solution: Given : $f(x)=x^{3}(x-1)^{2}$ $\Rightarrow f^{\prime}(x)=3 x^{2}(x-1)^{2}+2 x^{3}(x-1)$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 3 x^{2}(x-1)^{2}+2 x^{3}(x-1)=0$ $\Rightarrow x^{2}(x-1)\{3 x-3+2 x\}=0$ $\Rightarrow x^{2}(x-1)(5 x-3)=0$ $\Rightarrow x=0,1, \frac{3}{5}$ Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $1, x=1$ is the point of local minima. The local minimum value ...
Read More →A candidate is required to answer 7 questions
Question: A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions. Solution: We know that, nCr $=\frac{n !}{r !(n-r) !}$ No of questions in group A=6 No of questions in group B=6 According to the question, The different ways in which the questions can be attempted are, Hence, the number of different...
Read More →If the letters of the word RACHIT are arranged in
Question: If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rank of the word RACHIT?[Hint: In each case number of words beginning with A, C, H, I is 5!] Solution: According to the question, Arranging in alphabetical order, we get, A C H I R T Number of words that can start with A=5! Number of words that can start with C=5! Number of words that can start with H=5! Number of words that can start with I=5! Total = 5! +5!+5!+5! = 120+120+12...
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