Show that the sequence defined by an = 3n2 − 5 is not an A.P.
Question: Show that the sequence defined by $a_{n}=3 n^{2}-5$ is not an A.P. Solution: In the given problem, we need to show that the given sequence is not an A.P Here, $a_{n}=3 n^{2}-5$ Now, first we will find its few terms by substituting $n=1,2,3,4,5$ So, Substitutingn= 1,we get $a_{1}=3(1)^{2}-5$ $a_{1}=-2$ Substituting $n=2$, we get $a_{2}=3(2)^{2}-5$ $a_{2}=7$ Substitutingn= 3,we get $a_{3}=3(3)^{2}-5$ $a_{3}=22$ Substitutingn= 4,we get $a_{4}=3(4)^{2}-5$ $a_{4}=43$ Substitutingn= 5,we get...
Read More →All the letters of the word 'EAMCET' are arranged in different possible ways.
Question: All the letters of the word 'EAMCET' are arranged in different possible ways. The number of such arrangements in which two vowels are adjacent to each other, is (a) 360 (b) 144 (c) 72 (d) 54 Solution: Total number of words using letters of EAMCET is $\frac{6 !}{2}=\frac{6 \times 5 \times 4 \times 3 \times 1}{2}$ (Since $\mathrm{E}$ is respected twice) = 360 Total number of words with no two vowels together is :- Since vowels are $E, A, E=3$ and consonant is $M_{1} C_{1} T=3$. i.e numbe...
Read More →The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point
Question: The graph of the linear equation 2x+ 3y= 6 meets they-axis at the point(a) (2, 0)(b) (3, 0)(c) (0, 2)(d) (0, 3) Solution: If he graph of the linear equation 2x+ 3y= 6 meets they-axis, thenx= 0.Substituting the value ofx= 0 in equation2x+ 3y= 6, we get $2(0)+3 y=6$ $\Rightarrow 3 y=6$ $\Rightarrow y=\frac{6}{3}$ $\Rightarrow y=2$ So, the point of meeting is (0, 2).Hence, the correct answer is option (c)....
Read More →The graph of y + 2 = 0 is a line
Question: The graph ofy+ 2 = 0 is a line(a) making an intercept 2 on thex-axis(b) making an intercept 2 on they-axis(c) parallel to thex-axis at a distance of 2 units below thex-axis(d) parallel to they-axis at a distance of 2 units to the left ofy-axis Solution: As, the graph ofy+ 2 = 0 ory= 2is a line parallel tox-axis i.e.y= 0.⇒ The line represented by the equationy=2 is parallel tox-axis and intersectsy-axis aty= 2.So, the graph ofy+ 2 = 0 is a line parallel to thex-axis at a distance of 2 u...
Read More →There are four bus routes between A and B; and three bus routes between B and C.
Question: There are four bus routes betweenAandB; and three bus routes betweenBandC. A man can travel round trip in number of ways by bus fromAtoCviaB. If he does not want to use a bus route more than once, the number of ways he can make round trip, is (a) 72 (b) 144 (c) 14 (d) 19 Solution: There are 4 bus routes fromAtoBand 3 bus routes fromBtoC. Since, it is used trip so the man with travel back from $C$ to $A$ via $B$. it is restricted that man cannot use same bus routes fromCtoBandBtoAmore t...
Read More →If the function f : R→A given by
Question: If the function $f: R \rightarrow A$ given by $f(x)=\frac{x^{2}}{x^{2}+1}$ is a surjection, then $\mathrm{A}=$ (a) $R$ (b) $[0,1]$ (c) $[0,1)$ (d) $[0,1)$ Solution: As $f$ is surjective, r ange of $f=$ co-domain of $f$ $\Rightarrow A=$ range of $f$ $\because f(x)=\frac{x^{2}}{x^{2}+1}$ $y=\frac{x^{2}}{x^{2}+1}$ $\Rightarrow y\left(x^{2}+1\right)=x^{2}$ $\Rightarrow(y-1) x^{2}+y=0$ $\Rightarrow x^{2}=\frac{-y}{(y-1)}$ $\Rightarrow x=\sqrt{\frac{y}{(1-y)}}$ $\Rightarrow \frac{y}{(1-y)} \...
Read More →Show that the sequence defined by an = 5n −7 is an A.P,
Question: Show that the sequence defined by $a_{n}=5 n-7$ is an A.P, find its common difference. Solution: In the given problem, we need to show that the given sequence is an A.P and then find its common difference. Here, $a_{n}=5 n-7$ Now, to show that it is an A.P, we will find its few terms by substituting So, Substitutingn= 1,we get $a_{1}=5(1)-7$ $a_{1}=-2$ Substitutingn= 2, we get $a_{2}=5(2)-7$ $a_{2}=3$ Substitutingn= 3, we get $a_{3}=5(3)-7$ $a_{3}=8$ Substitutingn= 4, we get $a_{4}=5(4...
Read More →In a room there are 12 bulbs of the same wattage, each having a separate switch.
Question: In a room there are 12 bulbs of the same wattage, each having a separate switch. The number of ways to light the room with different amounts of illumination is (a) 122 1 (b) 212 (c) 212 1 (d) none of these. Solution: (c) $2^{12}-1$ Each of the bulb has its own switch, i.e each bulb will have two outcomes it will either glow or not glow. Thus, each of the 12 bulbs will have 2 outcomes. Total number of ways to illuminate the room = 21 Here, we have also considered the way in which all th...
Read More →The number of ways in which the letters of the word ARTICLE
Question: The number of ways in which the letters of the word ARTICLE can be arranged so that even places are always occupied by consonants is (a) 576 (b)4C3 4! (c) 2 4! (d) none of these. Solution: (a) 576 There are 3 even places in the 7 letter word ARTICLE. So, we have to arrange 4 consonants in these 3 places in4P3ways. And the remaining 4 letters can be arranged among themselves in 4! ways. $\therefore$ Total number of ways of arrangement $={ }^{4} P_{3} \times 4 !=4 ! \times 4 !=576$...
Read More →The graph of x + 3 = 0 is a line
Question: The graph ofx+ 3 = 0 is a line(a) making an intercept 3 on thex-axis(b) making an intercept 3 on they-axis(c) parallel to they-axis at a distance of 3 units to the left ofy-axis(d) parallel to thex-axis at a distance of 3 units below thex-axis Solution: As, the graph ofx+ 3 = 0 orx=-3is a line parallel toy-axis i.e.x= 0.⇒⇒The line represented by the equationx=-3 is parallel toy-axis and intersectsx-axis atx=-3.So, the graph ofx+ 3 = 0 is a line parallel to they-axis at a distance of 3 ...
Read More →The graph of x = 4 is a line
Question: The graph ofx= 4 is a line(a) making an intercept 4 on thex-axis(b) making an intercept 4 on they-axis(c) parallel to thex-axis at a distance of 4 units from the origin(d) parallel to they-axis at a distance of 4 units from the origin Solution: As, the graph ofx= 4 is a line parallel toy-axis i.e.x= 0.⇒⇒The line represented by the equationx= 4 is parallel toy-axis and intersectsx-axis atx= 4.So, the graph ofx= 4 is parallel toy-axis at a distance of 4 units from the origin making an in...
Read More →The number of different ways in which 8 persons can stand in a row so that between two particular persons A and B there are always two persons,
Question: The number of different ways in which 8 persons can stand in a row so that between two particular personsAandBthere are always two persons, is (a) 60 5! (b) 15 4! 5! (c) 4! 5! (d) none of these. Solution: (a) 60 5! The four people, i.e A, B and the two persons between them are always together. Thus, they can be considered as a single person. So, along with the remaining 4 persons, there are now total 5 people who need to be arranged. This can be done in 5! ways. But, the two persons th...
Read More →The graph of y = 5 is a line
Question: The graph ofy= 5 is a line(a) making an intercept 5 on thex-axis(b) making an intercept 5 on they-axis(c) parallel to thex-axis at a distance of 5 units from the origin(d) parallel to they-axis at a distance of 5 units from the origin Solution: As, the graph ofy= 5 is a line parallel tox-axis i.e.y= 0.⇒⇒The line represented by the equationy= 5 is parallel tox-axis and intersectsy-axis aty= 5.So, the graph ofy= 5 is a line parallel to thex-axis at a distance of 5 units from the origin m...
Read More →In which of the following situations, the sequence of numbers formed will form an A.P.?
Question: In which of the following situations, the sequence of numbers formed will form an A.P.?(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.(ii) The amount of air present in the cylinder when a vacuum pump removes each time1414of their remaining in the cylinder.Q Solution: (i) In the given problem, Cost of digging a well for the first meter = Rs 150 Cost of digging a well for subsequent meter is increased by Rs 20 So,Cost of digging ...
Read More →Let A={x : −1≤x≤1} and f : A→A such that f(x)=x|x|, then f is
Question: Let $A=\{x:-1 \leq x \leq 1\}$ and $f: A \rightarrow A$ such that $f(x)=x|x|$, then $f$ is (a) a bijection(b) injective but not surjective(c) surjective but not injective(d) neither injective nor surjective Solution: Injectivity:Letxandybe any two elements in the domainA. Case-1: Letxandybe two positive numbers, such that $f(x)=f(y)$ $\Rightarrow x|x|=y|y|$ $\Rightarrow x(x)=y(y)$ $\Rightarrow x^{2}=y^{2}$ $\Rightarrow x=y$ Case-2: Letxandybe two negative numbers, such that $f(x)=f(y)$...
Read More →The number of words that can be made by re-arranging the letters of the word APURBA
Question: The number of words that can be made by re-arranging the letters of the word APURBA so that vowels and consonants are alternate is (a) 18 (b) 35 (c) 36 (d) none of these. Solution: (c) 36 The word APURBA is a 6 letter word consisting of 3 vowels that can be arranged in 3 alternate places, in $\frac{3 !}{2 !}$ ways. The remaining 3 consonants can be arranged in the remaining 3 places in 3! ways. $\therefore$ Total number of words that can be formed $=\frac{3 !}{2 !} \times 3 !=18$ But t...
Read More →The equation 2x + 5y = 7 has a unique solution, if x and y are
Question: The equation 2x+ 5y= 7 has a unique solution, ifxandyare(a) natural numbers(b) rational numbers(c) positive real numbers(d) real numbers Solution: Since, every point on the line represented by the equation 2x+ 5y= 7 is its solution.Therefore, there are infinite solutions of the equationthe equation 2x+ 5y= 7 in which the values ofxandyare rational numbers, positive real numbers or real numbers.But, as 2 + 5 = 7, i.e.x= 1 andy= 1 are the only pair of natural numbers that are the solutio...
Read More →The linear equation 3x − 5y = has
Question: The linear equation 3x 5y= has(a) a unique solution(b) two solutions(c) infinitely many solutions(d) no solution Solution: (c) infinitely many solutions Given linear equation: $3 x-5 y=15$ Or, $x=\frac{5 y+15}{3}$ When $y=0, x=\frac{15}{3}=5$. When $y=3, x=\frac{30}{3}=10$. When $y=-3, x=\frac{0}{3}=0$ Thus, we have the following table: Plot the points $A(5,0), B(10,3)$ and $C(0,-3)$. Join the points and extend them in both the directions. We get infinite points that satisfy the given ...
Read More →The number of arrangements of the letters of the word BHARAT taking 3 at a time is
Question: The number of arrangements of the letters of the word BHARAT taking 3 at a time is (a) 72 (b) 120 (c) 14 (d) none of these. Solution: (a) 72 When we make words after selecting letters of the word BHARAT, it could consist of a single A, two As or no A. Case-I: A is not selected for the three letter word. Number of arrangements of three letters out of $\mathrm{B}, \mathrm{H}, \mathrm{R}$ and $\mathrm{T}=4 \times 3 \times 2=24$ Case-II: One A is selected and the other two letters are sele...
Read More →The number of arrangements of the letters of the word BHARAT taking 3 at a time is
Question: The number of arrangements of the letters of the word BHARAT taking 3 at a time is (a) 72 (b) 120 (c) 14 (d) none of these. Solution: (a) 72 When we make words after selecting letters of the word BHARAT, it could consist of a single A, two As or no A. Case-I: A is not selected for the three letter word. Number of arrangements of three letters out of $\mathrm{B}, \mathrm{H}, \mathrm{R}$ and $\mathrm{T}=4 \times 3 \times 2=24$ Case-II: One A is selected and the other two letters are sele...
Read More →Solve the following
Question: If ${ }^{k+5} P_{k+1}=\frac{11(k-1)}{2} \cdot{ }^{k+3} P_{k}$, then the values of $k$ are (a) 7 and 11 (b) 6 and 7 (c) 2 and 11 (d) 2 and 6 Solution: (b) 6 and 7 $k+5 P_{k+1}=\frac{11(k-1)}{2} \cdot{ }^{k+3} P_{k}$ $\Rightarrow \frac{(k+5) !}{(k+5-k-1) !}=\frac{11(k-1)}{2} \times \frac{(k+3) !}{(k+3-k) !}$ $\Rightarrow \frac{(k+5) !}{4 !}=\frac{11(k-1)}{2} \times \frac{(k+3) !}{3 !}$ $\Rightarrow \frac{(k+5) !}{(k+3) !}=\frac{11(k-1)}{2} \times \frac{4 !}{3 !}$ $\Rightarrow(k+5)(k+4)=2...
Read More →Solve the following
Question: If ${ }^{k+5} P_{k+1}=\frac{11(k-1)}{2} \cdot{ }^{k+3} P_{k}$, then the values of $k$ are (a) 7 and 11 (b) 6 and 7 (c) 2 and 11 (d) 2 and 6 Solution: (b) 6 and 7 $k+5 P_{k+1}=\frac{11(k-1)}{2} \cdot{ }^{k+3} P_{k}$ $\Rightarrow \frac{(k+5) !}{(k+5-k-1) !}=\frac{11(k-1)}{2} \times \frac{(k+3) !}{(k+3-k) !}$ $\Rightarrow \frac{(k+5) !}{4 !}=\frac{11(k-1)}{2} \times \frac{(k+3) !}{3 !}$ $\Rightarrow \frac{(k+5) !}{(k+3) !}=\frac{11(k-1)}{2} \times \frac{4 !}{3 !}$ $\Rightarrow(k+5)(k+4)=2...
Read More →The point of the form (a, −a), where a ≠ 0, lies on
Question: The point of the form (a,a), wherea 0, lies on(a) thex-axis(b) they-axis(c) the liney=x(d) the linex+y= 0 Solution: (d) the line $x+y=0$ Given, a point of the form $(a,-a)$, where $a \neq 0$ When $a=1$, the point is $(1,-1)$. When $a=2$, the point is $(2,-2)$. When $a=3$, the point is $(3,-3) \ldots \ldots$ and so on. Plot these points on a graph paper. Join these points and extend them in both the directions. You will get the equation of the line $x+y=0$....
Read More →The product of r consecutive positive integers is divisible by
Question: The product ofrconsecutive positive integers is divisible by (a)r! (b) (r 1) ! (c) (r+ 1) ! (d) none of these. Solution: (a)r! The product ofrconsecutive integers is equal tor!, so it will be divisible byr!....
Read More →Which of the following functions from
Question: Which of the following functions from $A=\{x:-1 \leq x \leq 1\}$ to itself are bijections? (a) $f(x)=\frac{x}{2}$ (b) $g(x)=\sin \left(\frac{\pi x}{2}\right)$ (c) $h(x)=|x|$ (d) $k(x)=x^{2}$ Solution: (a) Range of $f=\left[\frac{-1}{2}, \frac{1}{2}\right] \neq A$ So, $f$ is not a bijection. (b) Range $=\left[\sin \left(\frac{-\pi}{2}\right), \sin \left(\frac{\pi}{2}\right)\right]=[-1,1]=A$ So, $g$ is a bijection. (c) $h(-1)=|-1|=1$ and $h(1)=|1|=1$ $\Rightarrow-1$ and 1 have the same i...
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