Find the arithmetic progression whose third term is
Question: Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12. Solution: Here, let us take the first term of the A.P asaand the common difference of the A.P asd Now, as we know, $a_{n}=a+(n-1) d$ So, for $3^{\text {rd }}$ term $(n=3)$, $a_{3}=a+(3-1) d$ $16=a+2 d$ $a=16-2 d$.......(1) Also, for 5thterm (n= 5), $a_{5}=a+(5-1) d$ $=a+4 d$ For 7thterm (n= 7), $a_{7}=a+(7-1) d$ $=a+6 d$ Now, we are given, $a_{7}=12+a_{5}$ $a+6 d=12+a+4 d$ $6 d-4 d=12$...
Read More →If f : R – {0} → R – {0} is defined as f(x)
Question: If $f: R-\{0\} \rightarrow R-\{0\}$ is defined as $f(x)=\frac{2}{3 x}$, then $r^{-1}(x)=$ Solution: Given: A function $f: R-\{0\} \rightarrow R-\{0\}$ is defined as $f(x)=\frac{2}{3 x}$ $f(x)=\frac{2}{3 x}$ $\Rightarrow y=\frac{2}{3 x}$ $\Rightarrow 3 x=\frac{2}{y}$ $\Rightarrow x=\frac{2}{3 y}$ Thus, $f^{-1}(x)=\frac{2}{3 x}$ Hence, if $f: R-\{0\} \rightarrow R-\{0\}$ is defined as $f(x)=\frac{2}{3 x}$, then $f^{-1}(x)=\frac{2}{3 x}$...
Read More →If n is even, then
Question: Ifnis even, thennCris maximum when ____________. Solution: Ifnis even, for anyn, The possible combinations are nC0,nC1,nC2.........nCr, .........nCn2,nCn1,nCn i.e totaln+ 1 combinations are possible the value of coefficient increase initially, then reaches its maximum and start decreasing. ${ }^{n} C_{r}$ is $r+1$ th term and middle term is $\frac{n}{2}+1$ $\Rightarrow r+1=\frac{n}{2}+1$ $\Rightarrow r=\frac{n}{2}$ i.erat whichnCris maximum....
Read More →The value of
Question: The value ofnPr+nCris ___________. Solution: ${ }^{n} P_{r} \div{ }^{n} C_{r}=\frac{{ }^{n} P_{r}}{{ }^{n} C_{r}}$ $=\frac{n !}{(n-r) !} \times \frac{(n-r) ! r !}{n !}$ Therefore, ${ }^{n} P_{r} \div{ }^{n} C_{r}=r$...
Read More →If an A.P. consists of n terms with first term a and nth term l
Question: If an A.P. consists ofnterms with first term a andnthtermlshow that the sum of the mthterm from the beginning and the mthterm from the end is (a + l). Solution: In the given problem, we have an A.P. which consists ofnterms. Here, The first term (a) =a The last term (an) =l Now, as we know, $a_{n}=a+(n-1) d$ So, for themthterm from the beginning, we take (n = m), $a_{m}=a+(m-1) d$ $=a+m d-d$........(1) Similarly, for themthterm from the end, we can takelas the first term. So, we get, $a...
Read More →The value of
Question: The value of15C8=15C915C615C7is ___________. Solution: 15C8+15C915C615C7 SincenCr=nCnr ⇒15C6=15C156=15C9 ...(1) and15C7=15C157=15C8 ...(2) 15C8+15C915C915C8= 0 (∵ from (1) and (2))...
Read More →If f : R → R is defined by
Question: If $f: R \rightarrow R$ is defined by $f(x)=8 x^{3}$ then, $f^{-1}(8)=$_________. Solution: Given: $f(x)=8 x^{3}$ $f(x)=8 x^{3}$ $\Rightarrow y=8 x^{3}$ $\Rightarrow x^{3}=\frac{y}{8}$ $\Rightarrow x=\left(\frac{y}{8}\right)^{\frac{1}{3}}$ Thus, $f^{-1}(x)=\left(\frac{x}{8}\right)^{\frac{1}{3}}$ $f^{-1}(8)=\left(\frac{8}{8}\right)^{\frac{1}{3}}$ $=1^{\frac{1}{3}}$ $(\because f: R \rightarrow R)$ Hence, if $f: R \rightarrow R$ is defined by $f(x)=8 x^{3}$ then $f^{-1}(8)=1$. $=1$...
Read More →Solve the following
Question: IfnPr= 840 andnCr= 35, thenr= ___________. Solution: GivennPr= 840 andnCr = 35 i. e $\frac{n !}{(n-r) !}=840$ and $\frac{n !}{r !(n-r) !}=35$ $\therefore\left(\frac{n !}{(n-r) !}\right) \frac{1}{r !}=35$ i. e $840\left(\frac{1}{r !}\right)=35 \quad\left(\because \frac{n !}{(n-r) !}=840\right.$ given $)$ i. e $r !=\frac{840}{35}=24$ $=4 \times 3 \times 2$ $=4 !$ i.e $r=4$...
Read More →Two lines AB and CD intersect at O.
Question: Two linesABandCDintersect atO. If AOC= 50, find AOD, BODand BOC. Solution: We know that if two lines intersect then the vertically-opposite angles are equal. Therefore, $\angle A O C=\angle B O D=50^{\circ}$ Let $\angle A O D=\angle B O C=x^{\circ}$ Also, we know that the sum of all angles around a point is360360.Therefore, $\angle A O C+\angle A O D+\angle B O D+\angle B O C=360^{\circ}$ $\Rightarrow 50+x+50+x=360^{\circ}$ $\Rightarrow 2 x=260^{\circ}$ $\Rightarrow x=130^{\circ}$ Henc...
Read More →In the adjoining figure, what value of x will make AOB a straight line?
Question: In the adjoining figure, what value ofxwill makeAOBa straight line? Solution: AOB will be a straight line if $3 x+20+4 x-36=180^{\circ}$ $\Rightarrow 7 x=196^{\circ}$ $\Rightarrow x=28^{\circ}$ Hence,x= 28 will make AOB a straight line....
Read More →If f : C → C is defined by
Question: If $f: C \rightarrow C$ is defined by $f(x)=8 x^{3}$, then $f^{-1}(8)=$. _________. Solution: Given: $f(x)=8 x^{3}$ $f(x)=8 x^{3}$ $\Rightarrow y=8 x^{3}$ $\Rightarrow x^{3}=\frac{y}{8}$ $\Rightarrow x=\left(\frac{y}{8}\right)^{\frac{1}{3}}$ Thus, $f^{-1}(x)=\left(\frac{x}{8}\right)^{\frac{1}{3}}$ $f^{-1}(8)=\left(\frac{8}{8}\right)^{\frac{1}{3}}$ $=1^{\frac{1}{3}}$ $=1, \omega, \omega^{2} \quad(\because f: C \rightarrow C)$ where, $\omega$ is the cube root of unity. Hence, if $f: C \r...
Read More →In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line,
Question: In the adjoining figure,x:y:z= 5:4:6. IfXOYis a straight line, find the values ofx,yandz. Solution: Let $x=5 a, y=4 a$ and $z=6 a$ XOY is a straight line. Therefore, $\angle X O P+\angle P O Q+\angle Y O Q=180^{\circ}$ $\Rightarrow 5 a+4 a+6 a=180^{\circ}$ $\Rightarrow 15 a=180^{\circ}$ $\Rightarrow a=12^{\circ}$ Therefore, $x \Rightarrow 5 \times 12^{\circ}=60^{\circ}$ $y \Rightarrow 4 \times 12^{\circ}=48^{\circ}$ and $z \Rightarrow 6 \times 12^{\circ}=72^{\circ}$...
Read More →The range of the function f : R –{–2) → R given by
Question: The range of the function $f: R-\{-2) \rightarrow R$ given by $f(x)=\frac{x+2}{|x+2|}$ is_________. Solution: Given: $f(x)=\frac{x+2}{|x+2|}$ $f(x)=\frac{x+2}{|x+2|}$ $= \begin{cases}\frac{x+2}{x+2} , x+2 \geq 0 \\ \frac{x+2}{-(x+2)} , x+20\end{cases}$ $= \begin{cases}1 , x+2 \geq 0 \\ -1 , x+20\end{cases}$ To find the range, we find the real values ofyobtained. Hence, the range of the function $f: R-\{-2) \rightarrow R$ given by $f(x)=\frac{x+2}{|x+2|}$ is $\{-1,1\}$...
Read More →In the adjoining figure, AOB is a straight line. Find the value of x.
Question: In the adjoining figure,AOBis a straight line. Find the value ofx. Also, find AOC, CODand BOD. Solution: AOB is a straight line. Therefore, $\angle A O C+\angle C O D+\angle B O D=180^{\circ}$ $\Rightarrow(3 x+7)^{\circ}+(2 x-19)^{\circ}+x^{\circ}=180^{\circ}$ $\Rightarrow 6 x=192^{\circ}$ $\Rightarrow x=32^{\circ}$ Therefore, $\angle A O C=3 \times 32^{\circ}+7=103^{\circ}$ $\angle C O D=2 \times 32^{\circ}-19=45^{\circ}$ and $\angle B O D=32^{\circ}$...
Read More →Find n if the given value of x is the nth term of the given A.P.
Question: Find n if the given value of x is the nth term of the given A.P. (i) $25,50,75,100, \ldots, x=1000$ (ii) $-1,-3,-5,-7, \ldots ; x=-151$ (iii) $5 \frac{1}{2}, 11,16 \frac{1}{2}, 22, \ldots ; x=550$ (iv) $1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, \ldots, x=\frac{171}{11}$ Solution: In the given problem, we need to find the number of terms in an A.P (i) 25, 50, 75, 100 We are given, $a_{n}=1000$ Let us take the total number of terms asn. So, First term (a) = 25 Last term (an) = 1000...
Read More →The range of the function f : R → R given by
Question: The range of the function $f: R \rightarrow R$ given by $f(x)=x+\sqrt{x^{2}}$ is_________. Solution: Given: $f(x)=x+\sqrt{x^{2}}$ $f(x)=x+\sqrt{x^{2}}$ $=x+|x|$ $= \begin{cases}x+x , x \geq 0 \\ x-x , x0\end{cases}$ $= \begin{cases}2 x , x \geq 0 \\ 0 , x0\end{cases}$ To find the range, we find the real values ofy obtained. $y=2 x$ when $x \geq 0$ $\Rightarrow x=\frac{y}{2} \geq 0$ $\Rightarrow y \geq 0$ $\Rightarrow y \in[0, \infty)$ $y=0$ when $x0 \quad \ldots(1)$ Thus, from (1) and ...
Read More →The domain of the real function
Question: The domain of the real function $f(x)=\frac{x}{\sqrt{9-x^{2}}}$ is___________. Solution: Given: $f(x)=\frac{x}{\sqrt{9-x^{2}}}$ To find the domain, we find the real values ofx for which the function is defined. $x \in R$ and $9-x^{2}0$ $\Rightarrow x \in R$ and $9x^{2}$ $\Rightarrow x \in R$ and $x^{2}9$ $\Rightarrow x \in R$ and $-3x3$ $\Rightarrow-3x3$ $\Rightarrow x \in(-3,3)$ Hence, the domain of the real function $f(x)=\frac{x}{\sqrt{9-x^{2}}}$ is $(-3,3)$....
Read More →The domain of the real function
Question: The domain of the real function $f(x)=\sqrt{16-x^{2}}$ is_________. Solution: Given: $f(x)=\sqrt{16-x^{2}}$ To find the domain, we find the real values ofxfor which the function is defined. $16-x^{2} \geq 0$ $\Rightarrow 16 \geq x^{2}$ $\Rightarrow x^{2} \leq 16$ $\Rightarrow x \leq 4$ and $x \geq-4$ $\Rightarrow-4 \leq x \leq 4$ $\Rightarrow x \in[-4,4]$ Hence, the domain of the real function $f(x)=\sqrt{16-x^{2}}$ is $[-4.4]$....
Read More →The domain of the real function
Question: The domain of the real function $f(x)=\sqrt{16-x^{2}}$ is_________. Solution: Given: $f(x)=\sqrt{16-x^{2}}$ To find the domain, we find the real values ofxfor which the function is defined. $16-x^{2} \geq 0$ $\Rightarrow 16 \geq x^{2}$ $\Rightarrow x^{2} \leq 16$ $\Rightarrow x \leq 4$ and $x \geq-4$ $\Rightarrow-4 \leq x \leq 4$ $\Rightarrow x \in[-4,4]$ Hence, the domain of the real function $f(x)=\sqrt{16-x^{2}}$ is $[-4.4]$....
Read More →Write the expression an- ak for the A.P. a, a + d, a + 2d, ...
Question: Write the expression $a_{n}-a_{k}$ for the A.P. $a, a+d, a+2 d, \ldots$ Hence, find the common difference of the A.P. for which (i) $11^{\text {th }}$ term is 5 and $13^{\text {th }}$ term is 79 . (ii) $a_{10}-a_{5}=200$ (iii) $20^{\text {th }}$ term is 10 more than the $18^{\text {th }}$ term. Solution: A.P:a, a+d, a+2d Here, we first need to write the expression for $a_{n}-a_{k}$ Now, as we know, $a_{v}=a+(n-1) d$ So, for $n^{\text {th }}$ term, $a_{n}=a+(n-1) d$ Similarly, for $k^{\...
Read More →The total number of onto functions from the set A = (1, 2, 3, 4, 5) to the set B = {x, y} is
Question: The total number of onto functions from the setA= (1, 2, 3, 4, 5) to the setB= {x,y} is _________. Solution: Given: $f: A \rightarrow B$ where $A=\{1,2,3,4,5\}$ and $B=\{x, y\}$ Number of elements inA= 5Number of elements inB= 2Each Element of Ahave 2 options to form an image.Thus, Total number of functions that can be formed = 2 2 2 2 2 = 32Number of functions having only one image i.e., {x} = 1Number of functions having only one image i.e., {y} = 1Thus, Number of onto functions that ...
Read More →The total number of one-one functions from the set A = {a, b, c} to the set B = {x, y, z, t} is
Question: The total number of one-one functions from the setA= {a,b,c} to the setB= {x,y,z,t} is _________. Solution: Given: $f: A \rightarrow B$ where $A=\{a, b, c\}$ and $B=\{x, y, z, t\}$ Number of elements inA= 3Number of elements inB= 4To form aone-one function,ElementaAhave 4 options to form an image.ElementbAhave 3 options to form an image.ElementcAhave 2 options to form an image.Thus, Number of one-one functions that can be formed = 4 3 2 = 24Hence, thetotal number of one-one functions ...
Read More →In the adjoining figure, AOB is a straight line.
Question: In the adjoining figure,AOBis a straight line. Find the value ofx. Hence, find AOCand BOD. Solution: As $A O B$ is a straight line, the sum of angles on the same side of $A O B$, at a point $O$ on it, is $180^{\circ}$. Therefore, $\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180^{\circ}$ $\Rightarrow(3 x-7)^{\circ}+55^{\circ}+(x+20)^{\circ}=180$ $\Rightarrow 4 x=112^{\circ}$ $\Rightarrow x=28^{\circ}$ Hence, $\angle \mathrm{AOC}=3 x-7$ $=3 \times 28-7$ $=77^{\circ}$ and ...
Read More →The total number of functions from the set A = (1, 2, 3, 4 to the set B = a, b, c) is
Question: The total number of functions from the setA= (1, 2, 3, 4 to the setB=a,b,c) is _________. Solution: Given: $f: A \rightarrow B$ where $A=\{1,2,3,4\}$ and $B=\{a, b, c\}$ Number of elements inA= 4Number of elements inB= 3Each element ofAhave 3 options to form an image.Thus, Number of functions that can be formed = 3 3 3 3 = 81Hence, thetotal number of functions from the setA= {1, 2, 3, 4} to the setB= {a,b,c} is81....
Read More →Let f : N → R be the function defined by
Question: Let $f: N \rightarrow R$ be the function defined by $f(x)=\frac{2 x-1}{2}$ and $g: Q \rightarrow R$ be another function defined by $g(x)=x+2$. Then, $(g \circ f)(3 / 2)$ is (a) 1 (b) 2 (c) $\frac{7}{2}$ (d) none of these Solution: Given: $f(x)=\frac{2 x-1}{2}$ and $g(x)=x+2$ $(g o f)(x)=g(f(x))$ $=g\left(\frac{2 x-1}{2}\right)$ $=\frac{2 x-1}{2}+2$ $=\frac{2 x-1+4}{2}$ $=\frac{2 x+3}{2}$ $(g o f)\left(\frac{3}{2}\right)=\frac{2\left(\frac{3}{2}\right)+3}{2}$ $=\frac{3+3}{2}$ $=\frac{6}...
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