If the pth and qth terms of a G.P.
Question: If the pthand qthterms of a G.P. are q and p respectively, show that its (p + q)th term is Solution: The $n^{\text {th }}$ term of GP is given by $t_{n}=a r^{n-1}$ where $a$ is the first term and $r$ is the common difference $\mathrm{p}^{\text {th }}$ term is given as $\mathrm{q}$ $\Rightarrow t_{p}=a r^{p-1}$ The above equation can be written as $\Rightarrow q=a r^{p-1}$ $\Rightarrow \mathrm{q}=\frac{\mathrm{ar}^{\mathrm{P}}}{\mathrm{r}}$ On rearranging the above equation we get $\Rig...
Read More →Find the equation of the circle which passes through the points
Question: Find the equation of the circle which passes through the points (1, 3) and (2, - 1), and has its centre on the line 2x + y 4 = 0. Solution: The equation of a circle: $x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i) Putting $(1,3) \(2,-1)$ in (i) $2 g+6 f+c=-10 . .($ ii $)$ $4 g-2 f+c=-5 . .($ iii $)$ Since the centre lies on the given straight line, $(-g,-f)$ must satisfy the equation as $-2 g-f-4=0 \ldots$ (iv) Solving, $f=-1, g=-1.5, c=-1$ The equation is $x^{2}+y^{2}-3 x-2 y-1=0$...
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 →A man accepts a position with an initial salary of Rs 5200 per month.
Question: A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter. (a) Find his salary for the tenth month (b) What is his total earnings during the first year? Solution: Given the mans salary in first month is 5200rs and then it increases every month by 320rs Hence the sequence of his salary per month will be 5200, 5200 + 320, 5200 + 640 The sequence is in AP with...
Read More →Show that the points
Question: Show that the points $A(1,0), B(2,-7), c(8,1)$ and $D(9,-6)$ all lie on the same circle. Find the equation of this circle, its centre and radius. Solution: The general equation of a circle: $(x-h)^{2}+(y-k)^{2}=r^{2}$ ...(i), where $(\mathrm{h}, \mathrm{k})$ is the centre and $\mathrm{r}$ is the radius. Putting $(1,0)$ in (i) $(1-h)^{2}+(0-k)^{2}=r^{2}$ $\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}+1-2 \mathrm{~h}=\mathrm{r}^{2} . .$ (ii) Putting $(2,-7)$ in (i) $(2-h)^{2}+(-7-k)^{2}=r^{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 →A man saved Rs 66000 in 20 years.
Question: A man saved Rs 66000 in 20 years. In each succeeding year after the first year he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year? Solution: Given the total amount saved in 20 years is 66000 hence S20= 66000 Let the amount saved in first year be a Then as he increases 200 Rs every year then the amount in second year will be a + 200 then in third year a + 400 and so on The sequence will be a, a + 200, a + 400 The sequence is AP and the c...
Read More →The first term of an A.P.is a,
Question: The first term of an A.P.is a, and the sum of the first p terms is zero, show that the sum of its next q terms is $\frac{-a(p+q) q}{p-1}$ [Hint: Required sum $=\mathrm{S}_{\mathrm{p}+\mathrm{q}}-\mathrm{S}_{\mathrm{p}}$ ] Solution: Given first term is ' $\mathrm{a}$ ' and sum of first $\mathrm{p}$ terms is $\mathrm{S}_{p}=0$ Now we have to find the sum of next q terms Therefore, total terms are $p+q$ Hence, sum of all terms minus the sum of the first $p$ terms will give the sum of next...
Read More →Find the equation of the circle concentric with the circle
Question: Find the equation of the circle concentric with the circle $x^{2}+y^{2}+4 x+6 y+11$ = 0 and passing through the point P(5, 4). Solution: 2 or more circles are said to be concentric if they have the same centre and different radii. Given, $x^{2}+y^{2}+4 x+6 y+11=0$ The concentric circle will have the equation $x^{2}+y^{2}+4 x+6 y+c^{\prime}=0$ As it passes through P(5, 4), putting this in the equation $5^{2}+4^{2}+4 \times 5+6 \times 4+c^{\prime}=0$ $\Rightarrow 25+16+20+24+c^{\prime}=0...
Read More →Find the equation of the circle which is circumscribed about the triangle
Question: Find the equation of the circle which is circumscribed about the triangle whose vertices are A( - 2, 3), b(5, 2) and C(6, - 1). Find the centre and radius of this circle. Solution: The general equation of a circle: $(x-h)^{2}+(y-k)^{2}=r^{2}$ $\ldots(\mathrm{i})$, where $(\mathrm{h}, \mathrm{k})$ is the centre and $\mathrm{r}$ is the radius. Putting $A(-2,3), B(5,2)$ and $c(6,-1)$ in (i) we get $h^{2}+k^{2}+4 h-6 k+13=r^{2} \ldots$ (ii) $h^{2}+k^{2}-10 h-4 k+29=r^{2} \ldots$ (iii)and $...
Read More →The two successive terms in the expansion of
Question: The two successive terms in the expansion of $(1+x)^{24}$ whose coefficients are in the ratio 1: 4 are (A) $3^{\text {rd }}$ and $4^{\text {th }}$ (B) $4^{\text {th }}$ and $5^{\text {th }}$ (C) $5^{\text {th }}$ and $6^{\text {th }}$ (D) $6^{\text {th }}$ and $7^{\text {th }}$ Solution: (C) $5^{\text {th }}$ and $6^{\text {th }}$ Explanation: Let the two successive terms in the expansion of $(1+x)^{24}$ be $(r+1)^{\text {th }}$ and $(r+$ 2) $^{\text {th }}$ term Now, $T_{r+1}={ }^{24}...
Read More →Given the integers r>1, n>2,
Question: Given the integers $r1, n2$, and coefficients of $(3 r)^{\text {th }}$ and $(r+2)^{\text {nd }}$ terms in the binomial expansion of $(1+x)^{2 n}$ are equal, then (A) $n=2 r$ (B) $n=3 r$ (C) $n=2 r+1$ (D) none of these Solution: (A) $n=2 r$ Explanation: Given $(1+x)^{2 n}$ $T_{3 r}=T_{(3 r-1)+1}={ }^{2 n} \mathrm{C}_{3 r-1} x^{3 r-1}$ $T_{r+2}=T_{(r+1)+1}={ }^{2 n} C_{r+1} x^{r+1}$ ${ }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}$ $3 r-1+r+1=2 n$ $n=2 r$ Hence option A is the correct answer....
Read More →Find the equation of the circle passing through the points
Question: Find the equation of the circle passing through the points (i) $(0,0),(5,0)$ and $(3,3)$ (ii) $(1,2),(3,-4)$ and $(5,-6)$ (iii) $(20,3),(19,8)$ and $(2,-9)$ Also, find the centre and radius in each case Solution: (i) The required circle equation Using Laplace Expansion $\Rightarrow 15\left(x^{2}+y^{2}\right)-75 x-15 y=0$ $\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-5 \mathrm{x}-\mathrm{y}=0$ is the equation with centre $=(2.5,0.5)$ Radius $=\sqrt{g^{2}+f^{2}-c}=\sqrt{\left(-2.5^{2}\right...
Read More →The total number of terms in the expansion
Question: The total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ after simplification is (A) 50 (B) 202 (C) 51 (D) none of these Solution: (C) 51 Explanation: Given $(x+a)^{100}+(x-a)^{100}$ $=\left({ }^{100} C_{0} x^{100}+{ }^{100} C_{1} x^{99} a+{ }^{100} C_{2} x^{98} a^{2}+\ldots\right)$ $+\left({ }^{100} C_{0} x^{100}-{ }^{100} C_{1} x^{99} a+{ }^{100} C_{2} x^{98} a^{2}+\ldots\right)$ $=2\left({ }^{100} C_{0} x^{100}+{ }^{100} C_{2} x^{98} a^{2}+\ldots++{ }^{100} C_{100} a^...
Read More →Find the term independent of
Question: Find the term independent of $x$ in the expansion of Solution: Given $\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Consider $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Using standard formula above expression can be written as $T_{r+1}={ }^{9} C_{r}\left(\frac{3}{2} x^{2}\right)^{9-r}\left(-\frac{1}{3 x}\right)^{r}={ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{18-3 r}$ Hence the general term in the expression of given exp...
Read More →Show that the equation
Question: Show that the equation $x^{2}+y^{2}-3 x+3 y+10=0$ does not represent a circle. Solution: Radius = $\sqrt{g^{2}+f^{2}-c}$ $=\sqrt{\left(-\frac{3}{2}\right)^{2}+\left(-\frac{3^{2}}{2}\right)-10}$ $=\sqrt{\frac{9}{2}-10}=\sqrt{-\frac{11}{2}}$ which implies that the radius is negative. (not possible) Therefore, $x^{2}+y^{2}-3 x+3 y+10=0$ does not represent a circle....
Read More →Prove the following
Question: If $\mathrm{x}^{p}$ occurs in the expansion of $x^{2}+\frac{1}{x}$ Prove that its coefficient is $\frac{\mid 2 n}{\frac{4 n-p}{3} \mid \frac{2 n+p}{3}}$ Solution: Given expression is $\left(x^{2}+\frac{1}{x}\right)^{2 n}$ Using the standard formula above expression can be written as $T_{r+1}={ }^{2 n} C_{r}\left(x^{2}\right)^{2 n-r}\left(\frac{1}{x}\right)^{r}={ }^{2 n} C_{r} x^{4 n-3 r}$ If $x^{\rho}$ occurs in the expansion, Let $4 n-3 r=p$ $3 r=4 n-p \Rightarrow r=\frac{4 n-p}{3}$ C...
Read More →In the expansion of
Question: In the expansion of $(x+a)^{n}$ if the sum of odd terms is denoted by 0 and the sum of even term by E. Then prove that (i) $\mathrm{O}^{2}-\mathrm{E}^{2}=\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)^{n}$ (ii) $40 E=(x+a)^{2 n}-(x-a)^{2 n}$ Solution: (i) We know that $(x+a)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{2} x^{n-2} a^{2}+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots$ Sum of odd terms, $O={ }^{n} C_{0} x^{n}+{ }^{n} C_{2} x^{n-2} a^{2}+\ldots$ And also sum of even term...
Read More →Show that the equation
Question: Show that the equation $x^{2}+y^{2}+2 x+10 y+26=0$ represents a point circle. Also, find its centre. Solution: The general equation of a circle: $x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i) where $c, g, f$ are constants. Given, $x^{2}+y^{2}+2 x+10 y+26=0$ Comparing with (i) we see that the equation represents a circle with $2 \mathrm{~g}=2 \Rightarrow \mathrm{g}=1,2 \mathrm{f}=$ $10 \Rightarrow f=5$ and $c=26$. Centre $(-g,-f)=(-1,-5)$. Radius $=\sqrt{g^{2}+f^{2}-c}$ $=\sqrt{1^{2}+5^{2}-26...
Read More →Show that the equation
Question: Show that the equation $3 x^{2}+3 y^{2}+6 x-4 y-1=0$ represents a circle. Find its centre and radius. Solution: The general equation of a conic is as follows $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ where $a, b, c, f, g, h$ are constants For a circle, $a=b$ and $h=0$. The equation becomes: $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{~g} \mathrm{x}+2 \mathrm{fy}+\mathrm{c}=0 \ldots(\mathrm{i})$ Given, $3 x^{2}+3 y^{2}+6 x-4 y-1=0 \Rightarrow x^{2}+y^{2}+2 x-\frac{4}{3} y-\frac{1}{3}=0$ Co...
Read More →Find n in the binomial
Question: Find $n$ in the binomial $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$, if the ratio of $7^{\text {th }}$ term from the beginning to the $7^{\text {th }}$ term from the end is $\frac{1}{6}$. Solution: Given expression is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$ Now, $7^{\text {th }}$ term from beginning, $T_{7}=T_{6+1}={ }^{n} C_{6}(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^{6}$ And $7^{\text {th }}$ term from end is same as $7^{\text {th }}$ term from the...
Read More →Show that the middle term in the expansion
Question: Show that the middle term in the expansion of $\left(x-\frac{1}{x}\right)^{2 n}$ is $\frac{1 \times 3 \times 5 \times \ldots \times(2 n-1)}{n !} \times(-2)^{n}$ Solution: Given, expression is $\left(x-\frac{1}{x}\right)^{2 n}$ Since the index is $2 n$, which is even. So, there is only one middle term, i.e., $\left(\frac{2 n}{2}+1\right)$ th term $=(n+1)$ th term $T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(-\frac{1}{x}\right)^{n}={ }^{2 n} C_{n}(-1)^{n}=(-1)^{n} \frac{(2 n !)}{n ! \cdot n ...
Read More →Show that the equation
Question: Show that the equation $x^{2}+y^{2}+x-y=0$ represents a circle. Find its centre and radius. Solution: The general equation of a conic is as follows $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ where $a, b, c, f, g, h$ are constants For a circle, a = b and h = 0. The equation becomes: $x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i) Given, $x^{2}+y^{2}+x-y=0$ Comparing with (i) we see that the equation represents a circle with $2 \mathrm{~g}=1$ $\Rightarrow g=\frac{1}{2}, 2 f=-1 \Rightarrow f=-\fr...
Read More →Show that the equation
Question: Show that the equation $x^{2}+y^{2}-4 x+6 y-5=0$ represents a circle. Find its centre and radius. Solution: The general equation of a conic is as follows $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ where $a, b, c, f, g, h$ are constants For a circle, a = b and h = 0. The equation becomes: $x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i) Given, $x^{2}+y^{2}-4 x+6 y-5=0$ Comparing with (i) we see that the equation represents a circle with $2 \mathrm{~g}=-4 \Rightarrow \mathrm{g}=-2$, $2 f=6 \Right...
Read More →If p is a real number and if the middle
Question: If $p$ is a real number and if the middle term in the expansion of $\frac{p}{2}+2$ Is 1120 , find $p$. Solution: Given expansion is $\left(\frac{p}{2}+2\right)^{8}$ Since index is $n=8$, there is only one middle term, i.e., $\left(\frac{8}{2}+1\right)$ th $=5^{\text {th }}$ term $T_{5}=T_{4+1}={ }^{8} C_{4}\left(\frac{p}{2}\right)^{8-4} \cdot 2^{4}$ $\Rightarrow \quad 1120={ }^{8} C_{4} p^{4}$ By substituting the values, we get $\Rightarrow \quad 1120=\frac{8 \times 7 \times 6 \times 5...
Read More →