The sides of a rectangle are given by the equations
Question: The sides of a rectangle are given by the equations x = - 2, x = 4, y = - 2 and y = 5. Find the equation of the circle drawn on the diagonal of this rectangle as its diameter. Solution: The intersection points in clockwise fashion are:( - 2, 5), (4, 5), (4, - 2), ( - 2, - 2). The equation of a circle passing through the coordinates of the end points of diameters is $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$ Substituting, values: $\left(x_{1}, ...
Read More →Find the coefficient of x 4 in the expansion
Question: Find the coefficient of $x 4$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{11}$. Solution: Given expression is $\left(1+x+x^{2}+x^{3}\right)^{11}$ $=\left[(1+x)+x^{2}(1+x)\right]^{11}=\left[(1+x)\left(1+x^{2}\right)\right]^{11}=(1+x)^{11} \cdot\left(1+x^{2}\right)^{11}$ $=\left({ }^{11} C_{0}+{ }^{11} C_{1} x+{ }^{11} C_{2} x^{2}+{ }^{11} C_{3} x^{3}+{ }^{11} C_{4} x^{4}+\ldots\right)\left({ }^{11} C_{0}+{ }^{11} C_{1} x^{2}+{ }^{11} C_{2} x^{4}+\ldots\right)$ Coefficient of $x^{...
Read More →If the coefficient of second,
Question: If the coefficient of second, third and fourth terms in the expansion of $(1+x)^{2 n}$ are in A.P. Show that $2 n^{2}-9 n+7=0$. Solution: Given $(1+x)^{2 n}$ Now, coefficient of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms are ${ }^{2 n} C_{1},{ }^{2 n} C_{2}$ and ${ }^{2 n} C_{3}$, respectively. Given that, ${ }^{2 n} C_{1},{ }^{2 n} C_{2}$ and ${ }^{2 n} C_{3}$ are in A.P. Then, $2 \cdot{ }^{2 n} C_{2}={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$ $2\left[\frac{2 n(2 n-1)(2 n-2)...
Read More →Find the equation of the circle, the coordinates of the end points of one
Question: Find the equation of the circle, the coordinates of the end points of one of whose diameters are $A(p, q)$ and $B(r, s)$ Solution: The equation of a circle passing through the coordinates of the end points of diameters is: $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$ Substituting, values: $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(\mathrm{p}, \mathrm{q}) \\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(\mathrm{r}, \mathrm{s})$ We get: $(x-p)(x-r...
Read More →Find the value of r,
Question: Find the value of $r$, if the coefficients of $(2 r+4)^{\text {th }}$ and $(r-2)^{\text {th }}$ terms in the expansion of $(1+x)^{18}$ are equal. Solution: Given $(1+x)^{18}$ Now, $(2 r+4)^{\text {th }}$ term, That is $T_{(2 r+3)+1}$ $T_{(2 r+3)+1}={ }^{18} C_{2 r+3}(x)^{2 r+3}$ And $(r-2)^{\text {th }}$ term, that is $T_{(r-3) \mid+1}$ $T_{(r-3)+1}={ }^{18} C_{r-3} x^{-3}$ Now according to the question, ${ }^{18} C_{2 r+3}={ }^{18} C_{r-3}$ $2 r+3+r-3=18$ $3 r=18 \quad \therefore \qua...
Read More →Find the equation of the circle, the coordinates of the end points of one
Question: Find the equation of the circle, the coordinates of the end points of one of whose diameters are $A(-2,-3)$ and $B(-3,5)$ Solution: The equation of a circle passing through the coordinates of the end points of diameters is: $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$ Substituting, values: $\left(x_{1}, y_{1}\right)=(-2,-3) \\left(x_{2}, y_{2}\right)=(-3,5)$ We get $(x+2)(x+3)+(y+3)(y-5)=0$ $\Rightarrow x^{2}+3 x+2 x+6+y^{2}-5 y+3 y-15=0$ $\Righ...
Read More →Find the sixth term of the expansion
Question: Find the sixth term of the expansion $\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^{n}$ if the binomial coefficient of the third term from the end is 45 . Solution: Given $\left(y^{1 / 2}+x^{1 / 3}\right)^{n}$ Also given that binomial coefficient of third term from the end $=45$ Therefore, ${ }^{n} C_{n-2}=45$ The above expression can be written as ${ }^{n} C_{2}=45$ $\Rightarrow \frac{n(n-1)(n-2) !}{2 !(n-2) !}=45$ $\Rightarrow \quad n(n-1)=90$ $\Rightarrow \quad n^{2}-n-90=0$ $\Right...
Read More →Find the equation of the circle, the coordinates of the end points of one
Question: Find the equation of the circle, the coordinates of the end points of one of whose diameters are $A(5,-3)$ and $B(2,-4)$ Solution: The equation of a circle passing through the coordinates of the end points of diameters is: $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$ Substituting, values: $\left(x_{1}, y_{1}\right)=(5,-3) \\left(x_{2}, y_{2}\right)=(2,-4)$ We get: $(x-5)(x-2)+(y+3)(y+4)=0$ $\Rightarrow x^{2}-2 x-5 x+10+y^{2}+3 y+4 y+12=0$ $\Righ...
Read More →Find the coefficient of
Question: Find the coefficient of $1 / x^{17}$ in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ Solution: Given $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ From the standard formula of $T_{r+1}$ we can write given expression as $T_{r+1}={ }^{15} C_{r}\left(x^{4}\right)^{15-r}\left(-\frac{1}{x^{3}}\right)^{r}={ }^{15} C_{r} x^{60-4 r}(-1)^{r} x^{-3 r}={ }^{15} C_{r} x^{60-7 r}(-1)^{r}$ For the coefficient $\mathrm{x}^{-17}$, we have $60-7 r=-17$ Therefore, $r=11$ Then above expression ...
Read More →Find the equation of the circle, the coordinates of the end points of one
Question: Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(3, 2) and B(2, 5) Solution: The equation of a circle passing through the coordinates of the end points of diameters is: $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$ Substituting, values: $\left(x_{1}, y_{1}\right)=(3,2) \\left(x_{2}, y_{2}\right)=(2,5)$ We get: $(x-3)(x-2)+(y-2)(y-5)=0$ $\Rightarrow x^{2}-2 x-3 x+6+y^{2}-5 y-2 y+10=0$ $\Rightarrow ...
Read More →Find the coefficient of
Question: Find the coefficient of $x^{15}$ in the expansion of $\left(x-x^{2}\right)^{10}$. Solution: Given $\left(x-x^{2}\right)^{10}$ $T_{r+1}={ }^{10} \mathrm{C}_{r} x^{10-r}\left(-x^{2}\right)^{r}=(-1)^{r 10} \mathrm{C}_{r} x^{10-r} x^{2 r}=(-1)^{r 10} \mathrm{C}_{r} x^{10+r}$ For the coefficient of $x^{15}$, we have $10+r=15 \Rightarrow r=5$ $T_{5+1}=(-1)^{510} C_{5} x^{15}$ Coefficient of $x^{15}=-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times ...
Read More →Find the middle term (terms) in the expansion of
Question: Find the middle term (terms) in the expansion of (i) $\frac{x}{a}-\frac{a}{x}$ (ii) $3 x-\frac{x^{3}}{6}$ Solution: a. Given $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$ Here index $n=10$ which is even number. So, there is one middle term which is $(10 / 2+1)^{\text {th }}$ term that is $6^{\text {th }}$ term $\therefore \quad T_{6}=T_{5+1}={ }^{10} C_{5}\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^{5}$ $=-{ }^{10} C_{5}\left(\frac{x}{a}\right)^{5}\left(\frac{a}{x}\right)^{5...
Read More →If two diameters of a circle lie along the lines
Question: If two diameters of a circle lie along the lines x y = 9 and x 2y = 7, and the area of the circle is 38.5 sq cm, find the equation of the circle. Solution: The point of intersection of two diameters is the centre of the circle. $\therefore$ point of intersection of two diameters $x-y=9$ and $x-2 y=7$ is $(11,2)$. $\therefore$ centre $=(11,2)$ Area of a circle $=\pi r^{2}$ $38.5=\pi r^{2}$ $\Rightarrow r^{2}=\frac{38.5}{\pi}$ $\Rightarrow \mathrm{r}^{2}=12.25$ sq.cm the equation of the ...
Read More →Find the term independent of
Question: Find the term independent of $\mathrm{x}$ in the expansion of $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ Solution: Given $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ From the standard formula of $T_{r+1}$ we can write given expression as $\mathrm{T}_{r+1}={ }^{15} C_{r}(3 x)^{15-r}\left(\frac{-2}{x^{2}}\right)^{r}={ }^{15} C_{r} 3^{15-r} x^{15-3 r}(-2)^{r}$ For the term independent of $x$, we have $15-3 r=0$ Which implies $r=5$ The term independent of $x$ is $T_{5+1}={ }^{15} C_{5} 3^{15-5}(-2...
Read More →If the solve the problem
Question: (i) $f(x)=x^{4}-62 x^{2}+120 x+9$ (ii) $f(x)=x^{3}-6 x^{2}+9 x+15$ (iii) $f(x)=(x-1)(x+2)^{2}$ (iv) $f(x)=2 / x-2 / x^{2}, x0$ (v) $f(x)=x e^{x}$ (vi) $f(x)=x / 2+2 / x, x0$ (vii) $f(x)=(x+1)(x+2)^{1 / 3}, x \geq-2$ (viii) $f(x)=x \sqrt{32-x^{2}}, \quad-5 \leq x \leq 5$ (ix) $f(x)=x^{3}-2 a x^{2}+a^{2} x, a0, x \in R$ (x) $\mathrm{f}(\mathrm{x})=x+\frac{a 2}{x}, a0, \mathrm{x} \neq 0$ (xi) $f(x)=x \sqrt{2-x^{2}}-\sqrt{2} \leq x \leq \sqrt{2}$ (xii) $f(x)=x+\sqrt{1-x}, x \leq 1$ Solutio...
Read More →Find the coefficient of x
Question: Find the coefficient of $x$ in the expansion of $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$ Solution: Given $\left(3 x-\frac{2}{x^{2}}\right)^{15}$ $=\left(1-3 x+7 x^{2}\right)\left({ }^{16} C_{0}-{ }^{16} C_{1} x^{1}+{ }^{16} C_{2} x^{2}+\ldots+{ }^{16} C_{16} x^{16}\right)$ $=\left(1-3 x+7 x^{2}\right)\left(1-16 x+120 x^{2}+\ldots\right)$ Coefficient of $x=-19$...
Read More →If the term free from
Question: If the term free from $\mathrm{x}$ in the expansion of $\sqrt{x}-\frac{k}{x^{2}}$ is $\mathbf{4 0 5}$, find the value of $k$. Solution: Given $\sqrt{x}-\frac{k}{x^{2}}$ From the standard formula of Tro1 we can write given expression as $T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}={ }^{10} C_{r}(x)^{\frac{1}{2}(10-r)}(-k)^{r} x^{-2 r}$ $={ }^{10} C_{r}(x)^{5-\frac{r}{2}-2 r}(-k)^{r}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}$ For the term free from $x$ we h...
Read More →Find the term independent of
Question: $\mathrm{x}, \mathrm{x} \neq 0$, in the expansion of $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$. Solution: Given $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$ From the standard formula of $T_{r+1}$ we can write given expression as $T_{r+1}={ }^{15} C_{r}\left(\frac{3 x^{2}}{2}\right)^{15-r}\left(-\frac{1}{3 x}\right)^{r}$ $T_{r+1}={ }^{15} C_{r}(-1)^{r} 3^{15-2 r} 2^{r-15} x^{30-3 r}$ For the term independent of $x$, we have $30-3 r=0$ Which implies $r=10$ By substitutin...
Read More →Find the equation of the circle passing through the point
Question: Find the equation of the circle passing through the point $(-1,-3)$ and having its centre at the point of intersection of the lines $x-2 y=4$ and $2 x+5 y+1$ $=0$. Solution: The intersection of the lines: $x-2 y=4$ and $2 x+5 y+1=0$. is $(2,-1)$ $\therefore$ This problem is same as solving a circle equation with centre and point on the circle given. 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...
Read More →Find the equation of the circle whose centre
Question: Find the equation of the circle whose centre is (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4. Solution: The intersection of the lines: 3x + 2y = 11 and 2x + 3y = 4 Is $(5,-2)$ $\therefore$ This problem is same as solving a circle equation with centre and point on the circle given. 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 q...
Read More →Everybody in a room shakes hands with everybody else.
Question: Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is A. 11 B. 12C. 13D. 14 Solution: B. 12 Explanation: We know that, nCr $=\frac{n !}{r !(n-r) !}$ Let total time of handshakes=nC2= 66 $\frac{n !}{2 !(n-2) !}=66$ $\Rightarrow \frac{n(n-1)}{2}=66$ ⇒ n2-n=132 ⇒ (n-12) (n+11) = 0 n=12 or n = 11 Therefore, n = 12 Hence, Option (B) 12 is the correct answer....
Read More →Find the equation of the circle of radius 5 cm
Question: Find the equation of the circle of radius 5 cm, whose centre lies on the y - axis 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. Since, centre lies on $Y$-axis, $\therefore$ it's $X$ - coordinate $=0$, i.e.h $=0$ Hence, $(0, k)$ is the centre of the circle. Substituting the given values in general form of the equation of a circle w...
Read More →A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4
Question: A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done isA. 216B. 600C. 240D. 3125[Hint:5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.] Solution: A. 216 Explanation: 5-digit numbers that can be formed using digits 0, 1, 2, 4, 5 44321=96 5-digit numbers can be formed using digits 1, 2, 3, 4, 5 = ...
Read More →Total number of words formed by 2 vowels
Question: Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal toA. 60B. 120C. 7200D. 720 Solution: C. 7200 Explanation: We know that, nCr $=\frac{n !}{r !(n-r) !}$ Total number of given vowels=4 Total number of given consonant=5 Total number of words that can be formed by 2 vowels and 3 consonants =4C2.5C3 =4!/2!2! 2 vowel and 3 consonant =5! Total number of word = 5! 4!/2!2! = 7200 Hence, Option (C) 7200 is the correct answer....
Read More →If the solve the problem
Question: $f(x)=\frac{x}{2}+\frac{2}{x}, x0$ Solution: Given: $f(x)=\frac{x}{2}+\frac{2}{x}$ $\Rightarrow f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$ For the local maxima or minima, we must have $f^{\prime}(x)=0$ $\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0$ $\Rightarrow \frac{1}{2}=\frac{2}{x^{2}}$ $\Rightarrow x^{2}=\pm 2$ Since $x0, f^{\prime}(x)$ changes from negative to positive when $x$ increases through $2 .$ So, $x=2$ is a point of local minima. The local minimum value of $f(x)$ at $x=2$ is...
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