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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
One cubic meter piece of copper is melted and recast into a bar 36 meters long with a square cross-sectional area. An exact cube is cut off from this bar. The cost of this cube, if 1 m3 of copper costs Rs.216 will be : |
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Answer» `1m^3=l*b*h` `1m^3=l^2*h` `1/36=l^2=Area of square bar Side=1/6m. length of cube=1/6m 36m bar=216Rs. 1/6 m bar price=1/6*216/36=1 Rs. |
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| 2. |
If `(x-2,y+5)=(2,(1)/(3))` are two equal ordered paris, then `x=4, y=(-14)/(3)` |
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Answer» Correct Answer - False We have `(x-2,y+5)=(-2,(1)/(3))` `rArrx-2=-2,y+5=(1)/(3)rArrx=-2+2,y=(1)/(3)-5` `:. x=0,y=(-14)/(3)` |
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| 3. |
If `f(x)=(4x+3)/(6x-4), x !=2/3,`show that `fof(x)=x`for all `x!=2/3dot`What is the inverse of `f?` |
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Answer» `f(x) = ( 4x + 3)/( 6x - 4), x ne (2)/(3)` `therefore (fof)(x) = f{f(x)} = f((4x = 3)/(6x -4))` `" " = (4(( 4x = 3)/(6x - 4))+ 3)/(6((4x + 3)/(6x - 4))- 4)` `" " = ( 16x + 12 + 18x - 12 ) /( 24x + 18 - 24x +16) = ( 34x )/(34) = x` `therefore (fof) (x) = x` `rArr (fof) = 1 ` `rArr f = f^(-1)` Therefore, inverse of `f` is `f^(-1) = f` |
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| 4. |
The number of roots of `x^2-2=[sinx],w h e r e[dot]`stands for the greatest integer function is0 (b)1 (c) 2(d) 3. |
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Answer» Correct Answer - C Given equation is `x^(2) -2=[sinx].` There are three possibilities: `[sinx]= -1, [sinx] =0,[sin x] =1` Case I: If `[sinx] = -1,` `x^(2)=1` or `x= +-1` When `x=1, [sin x]=0.` But ` x=-1 implies [sinx]= -1` `x= -1` is a solution. Case II: If `[sinx] =0,` the equation is `x^(2) =2` or ` x= +-sqrt(2)` `[sin sqrt(2)]=0` But ` [sin (-sqrt(2))]=-1.` So, `x=sqrt(2)` is a solution. Case III: If `[sinx] =1` `x=(pi)/(2),(5pi)/(2),(9pi)/(2)` Clearly, these values do not satisfy the original equation. Thus, number of solutions `=2` |
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| 5. |
Domain of the function, `f(x)=[log_10 ((5x-x^2)/4)]^(1/2)` isA. `-oo lt x lt oo`B. `1 le x le 4 `C. `4 le x le 16`D. ` -1 le x le 1` |
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Answer» Correct Answer - B We have `f(x)=[log_(10)((5x-x^(2))/(4))]^(1//2) " (1)" ` From (1), clearly, `f(x)` is defined for those values of `x` for which `log_(10)[(5x-x^(2))/(4)] ge 0` or `((5x-x^(2))/(4)) ge 10^(0)` or `((5x-x^(2))/(4)) ge 1` or `x^(2)-5x+4 le 0` or `(x-1)(x-4) le 0` Hence, the domain of the function is [1, 4]. |
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| 6. |
The period of `f(x)=[x]+[2x]+[3x]+[4x]+[n x]-(n(n+1))/2x ,`where `n in N ,`is (where `[dot]`represents greatest integer function).`n`(b) 1(c) `1/n`(d) none of theseA. nB. 1C. `(1)/(n)`D. none of these |
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Answer» Correct Answer - B `f(x)=[x]+[2x]+[3x]+ …+[nx] -(x+2x+3x+ … + nx)` `= -({x}+{2x}+{3x}+ … +{nx})` Period of `f(x)=LCM (1,(1)/(2),(1)/(3), …, (1)/(n))=1.` |
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| 7. |
The derivative of `log_10 x` is ? |
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Answer» `log_b^a=log_c^a/log_c^b` `log_10^x=log_e^x/log_e^10` `=logx/log10` `d/(dx)(logx/log10)=1/log10*d/(dx)(logx)` `=1/(xlog10`. |
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| 8. |
Find the value of x `tan^-1 ((x-1)/(x+1)) + tan^-1 ((2x-1)/(2x+1)) = tan^-1 (23/36)` |
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Answer» We know, `tan^(-1)x+tan^(-1)y = tan^-1((x+y)/(1-xy))` ` tan^(-1)((x-1)/(x+1))+tan^(-1)((2x-1)/(2x+1))= tan^(-1)((x-1)/(x+1)+(2x-1)/(2x+1))/((1-((x-1)/(x+1))((2x-1)/(2x+1)))` `= tan^(-1)((((x-1)(2x+1)+(2x-1)(x+1))/((x+1)(2x+1)))/(((x+1)(2x+1)-(x-1)(2x-1))/((x+1)(2x+1))))` `=tan^(-1)(((2x^2-x-1)+(2x^2+x-1))/((2x^2+3x+1) - (2x^2-3x+1)))` `=tan^(-1)((4x^2-2)/(6x))` So, `tan^(-1)((4x^2-2)/(6x)) = tan^(-1)(23/36)` `:. (4x^2-2)/(6x) = 23/36` `=> 24x^2 -12 = 23x` `=>24x^2-23x-12 = 0` `=>24x^2-32x+9x-12 = 0` `=>8x(3x-4)+3(3x-4) = 0` `=>(3x-4)(8x+3) = 0` `:. x = 4/3 and x = -8/3` |
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| 9. |
Let [ ] represents the greatest integer function and `[ x^3 + x^2 + 1 + x] = [ x^3 + x^2 + 1] + x`. The number of solution of the equation log|[x]| = 2 - |[x ]| |
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Answer» f(x)=log[[x]] g(x)=2-[x] [x]>o x`in`(0,1) [x]=0 at x=-1 g(x)=1 f(x)=0 at x=2 g(x)=0 f(x)=log2. |
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| 10. |
a bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag . What is the probability that she takes out(i) an orange flavored candy ? (ii) an lemon flavored candy ? |
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Answer» As the bag contains only lemon flavored candy. It does not have any orange flavored candy. (i)Let the total number of lemon candy is `n`. Then, the number of total candies will be `n` and the number of orange candies will be `0`. `:.` Probability of Malini taking out orange flavored candy ` = 0/n = 0` (ii)Probability of Malini taking out lemon flavored candy ` = n/n = 1` |
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| 11. |
The sum of last 3 digits of `3^100` is |
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Answer» As we know that `3^100 = 9^50 = (10-1)^50` by using binomial expansion we can write `(10-1)^50 = .^50C_0 10^50- .^50C_1 10^49 + .. .. .^50C_48(10)^2 - .^50C_49(10)1 + .^50C_50` `= (50 xx 49)/2 xx 100 - 50 xx 10 + 1` `= 122500 - 500 +1` `= 122001` So, sum of last 3 digit is `= 0 + 0 =1 = 1` option A is correct. |
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| 12. |
Let Q(n) denotes the number of seven digit numbers divisible by 9 which can be formed by using 7 distinct digits out of 1,2,3,4,5,6,7,8,9. Then which of the following is/are not the value of Q(n) |
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Answer» Sum of the given `9` digits ` = 1+2+3+4+5+6+7+8+9 = 45` Now, we have to select `7` digits. The number will be divisible by `9` if sum of those `7` digits is divisible by `9`. Minimum sum of two digits that we do not select `= 1+2 = 3` Maximum sum of two digits that we do not select `= 8+9 = 17` So, the sum of the `7` digits will lie between `45-17` and `45-3`, that is sum will be between `28` and `42`. Now, there is only number `36` that is divisible by `9`. So, the seven digits will have sum `36` in order to the number to be divisible by `9`. It means we have to exclude two numbers those have `9` as the sum. So, these two numbers can be any of the pair `(1,8),(2,7),(4,5),(6,3).` Total seven digits can be arranged in `7!` ways and there are `4` pairs of number that can be excluded. `:. Q(n) = 4(7!)` So, `Q(n)` can not take values `(A),(B) and (D).` |
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| 13. |
Water is pouring into a tank at a constant rate. when the tank is full, 10 pumps of equal capacity empty the tank in 12 hrs, while 15 pumps of the same capacity empty the tank in 6 hrs. The time which 25 pumps of the same capacity take to empty the tank, if the tank in initially full, will be : |
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Answer» Total work=12 hours. `10/x=1/y=1/12` `15/x=1/y=1/6`-(1) Subtracting equation 2 from 1 `5/x=1/6=1/12` `5/x=1/12` `-1/y=1/6-3/12` `=1/y=-1/12` `25/x=1/y=5/12-1/12=4/12=1/3` `1/x=1/3hours` `x=3 hours` |
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| 14. |
A three-digit number in which the digit in the units place is the smallest of the three digits is such that the sum of the squares of the digits is 149.If each of the digit is replaced by its predecessor,the new number formed has two digits common with the original number.The digit in the units place of the original number is : |
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Answer» 3 digit number:(100x+10y+z) `x^2+y^2+z^2=19` `x-1,y-1,z-1` (n,1),n,(n+1) `(n-1)^2+n^2+(n+1)^2=149` `n^2-2n+1+n^2+n^2+2n+1=149` `3n^2+2=149` `3n^2=147` `n^2=49` n=7 n-1=6 n+1=8 Min(6,7,8)=6. |
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| 15. |
The number of solutions of `x+2+tanx=pi/2` in `[0, 2pi]` is |
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Answer» `x+2+tanx = pi/2` `=> tanx = pi/2-x-2` Now, if we draw graphs of `tanx` and `pi/2-x-2`, we find that these two graphs intersects at two points from `[0,pi]`. Please refer to video to see the graphs. Thus, there are two solutions for the given equation. |
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| 16. |
If `A={-1,2,3}and B={1,3}`, then determine (i) `AxxB` (ii) `BxxA` (iii) `BxxB` (iv)`AxxA` |
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Answer» `A={-1,2,3,} and B ={1,3}` (i) `AxxB={(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3)}` (ii) `BxxA={(1,-1),(1,2),(1,3),(3-1),(3,2),(3,3)}` (iii) `BxxB={(1,-1) ,(1,3),(3,1),(3,3)}` (iv) `AxxA={(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1), (3,2),(3,3)}` |
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| 17. |
Let the function `f(x)=x^2+x+s in x-cosx+log(1+|x|)`be defined on the interval `[0,1]`.Define functions `g(x)a n dh(x)in[-1,0]`satisfying `g(-x)=-f(x)a n dh(-x)=f(x)AAx in [0,1]dot` |
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Answer» Correct Answer - `g(x)= -x^(2)+x+sinx +cosx -log(1+|x|)` `h(x)=x^(2)-x-sinx-cosx+log(1+|x|)` Clearly `g(x)` is the odd extension of the function `f(x)` and `h(x)` is the even extension. Since `x^(2),cosx, log(1+|x|)` are even functions and `x, sin x` and odd functions. `g(x)= -x^(2)+x+sinx +cosx-log(1+|x|)` and `h(x)=x^(2)-x-sinx-cosx +log(1+|x|)` Clearly this function satisfies the restriction of the problem. |
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| 18. |
Let `f: R^+vecR`be a function which satisfies `f(x)dotf(y)=f(x y)+2(1/x+1/y+1)`for `x , y > 0.`Then find `f(x)dot` |
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Answer» We have `f(x)xxf(y)=f(xy)+2((1)/(x)+(1)/(y)+1) " …(1)" ` To get `f(x)` we put `y=1`, ` :. f(x)xxf(1)=f(x)+2((1)/(x)+2)` `=f(x)+2((2x+1)/(x))` `impliesf(x)(f(1)-1)=(2(2x+1))/(x)` `implies f(x)=(2(2x+1))/(x(f(1)-1)) " ...(2)" ` Now, we need the value of f(1). Put `x=1` and `y=1` in (1), we get `(f(1))^(2)-f(1)-6=0` `implies f(1)=3 " or "f(1)= -2` For `f(1)=3,f(x)=(2x+1)/(x)` and for `f(1)= -2, f(x)=(2(2x+1))/(-3x)` |
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| 19. |
The largest interval lying in `(-pi/2,pi/2)`for which thefunction `[f(x)=4^-x^2+cos^(-1)(x/2-1)+log(cosx)]`is defined,is(1) `[0,pi]`(2) `(-pi/2,pi/2)`(3) `[-pi/4,pi/2)`(4) `[0,pi/2)` |
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Answer» `F(x)=4^(-x^2)+cos^-1(x/2-1)+log(cosx)` `-1<=x/2-1<=1` `0<=x/2<=2` `0<=x<=4........(1)` `log (cosx) when cosx>0` `cos>0` `x=(0,pi/2).....(2)` final`x=(0,pi/2)` option`4` |
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| 20. |
The function `t` which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by `t(C)=(9C)/5+32`.Find (i) `t (0)` (ii) `t(28)` (iii) `t(-10)` (iv) The value of C, when `t(C) = 212`. |
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Answer» `t(c) = (9C)/5 + 32` (i) `t(0) = 9(0)/5 + 32= 32 F` (ii) `t(28) = (9*98)/5 + 32 = 82.4F` (iii) `t(-10) = (9*(-10))/5 + 32 = 14 F` (iv) `C=? , t(C) = 212` `(9C)/5 + 32 = 212` `(9C)/5 = 180` `C= 100^@` answer |
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| 21. |
If 30 men working 7 hours a day can make 12 tables in 18 days, how many days will 45 women working 9 hours a day take to make 32 chairs? Given, 4 men can make 3 tables in the same time as 3 women can make 4 chairs. a. 12 days b. 14 days C. 28 days d. 21 days |
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Answer» `4` men can make `3` tables in the same time as `3` women can make `4` chairs. It means `1` man makes `3/4` tables and `1` woman makes `4/3` tables. Now, we can use the formula, `(W_1)/(M_1H_1D_1) = (W_2)/(M_2H_2D_2) ` Putting the given values, `=>12/(30xx(3/4)xx7xx18) = 32/(45xx(4/3)xx9xxD_2)` `=>1/(15xx7xx9) = 2/(45xx3xxD_2)` `=>D_2 = 2xx7 = 14` So, option `b` is the correct option. |
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| 22. |
Let A and B be sets. Show that `f" ":" "AxxB`,`BxxA`such that `f" "(a ," "b)" "=" "(b ," "a)`is bijective function. |
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Answer» In `f: AxxB to BxxA` `f(a, b) = (b, a) AA (a, b) in A xx B` Let `" " (a, b), (c, d) in AxxB` and `" " f(a, b) = f(c,d)` `rArr (b, a) = (d, c)` `rArr b =d and a = c ` `rArr (a, b) = (c, d)` `therefore f ` is one-one. Again, for each `(a, b) in B xx A, (b, a) in A xx B` is such that `therefore " " f(b, a) = (a, b)` `therefore f` is onto. Therefore, `f` is one-one onto function. |
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| 23. |
Let A be a factor of 120. The number of positive integral solution of `x_1 x_2 x_3 = A `is : |
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Answer» `120=2*2*2*5*3` `A=n_1n_2n_3` A=1 times 2, 2=1 A=2 times 2, 2*2=2 A=3 times 2, 2*2*2=1 there are 4 possible ways `2*2*2,2*2*3,2*2*5,2*3*5` |
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| 24. |
The number of even proper divisors of `1008` |
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Answer» `1008 = 2^4 xx 3^2 xx 7` so `2^1 xx(3^0 + 3^1 + 3^2) xx (7^0 + 7^1)` `= 2^2 xx (3^0+3^1+3^2) xx (7^0+ 7^1)` `= 2^3 xx (3^0 + 3^1 + 3^2) xx (7^0 + 7^1)` `= 2^4 xx (3^0 + 3^1 + 3^2) xx (7^0 + 7^1)` so, `1xx 3 xx 2 + 1xx3xx2 + 1xx3xx2 + 1xx3xx2` `= 4 (1xx 2xx3)` `=24` option D is correct |
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| 25. |
Let ` x_1 , x_2 , x_3,....., x_k` be the divisors of positive integer n (including 1 and n). If `x_1 + x_2 + x_3 + ...... + x_k = 75` Then ` sum_(i=1)^k (1/x_i)` is equal to (A) `75/k` (B)`75/n` (C) `1/n` (D)`1/75` |
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Answer» `sum_(i=1)^k(1/x_i)` `1/x_1+1/x_2+1/x_3+1/x_4+...+1/x_k` `x_k/n+x_(k-1)/n+...+x_2/n+x_1/n` `x_k+x_(k-1)+...+x_2+x_1` `=75/n` Option B is correct. |
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| 26. |
If `f(x)` is a real-valued function defined as `f(x)=In (1-sinx),` then the graph of `f(x)` isA. symmetric about the line `x =pi`B. symmetric about the y-axisC. symmetric and the line ` x=(pi)/(2)`D. symmetric about the origin |
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Answer» Correct Answer - C `f((pi)/(2)-x)=In (1-cos x)` ` and f((pi)/(2)+x)=In(1-cosx)` Thus, `f((pi)/(2)+x)=f((pi)/(2)-x)` Thus f(x) is symmetrical about line `x=(pi)/(2).` |
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| 27. |
The Cartesian product `A xxA ` has 9 elements among which are found `(-1, 0) " and " (0, 1)`. Find the set `A` and the remaining elements of `A xxA`. |
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Answer» `(-1,0) & (0,1) in A*A` `(-1,0) in A` `(0,1) in A` so, `(-1,0,1) in A` `A*A = {-1,0,1}*{-1,0,1}` `= {(0,-1),(-1,0),(-1,1),(0,-1),(0,0),(0,1),(1,-1) ,(1,0),(1,1)}` `` Answer |
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| 28. |
A box contains 4 defective and 6 non defective bulls. Find the probability that atleast 3 bulbs are defective when 4 bulbs are selected at random. |
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Answer» We have `4` defective bulbs and `6` non-defective bulbs. Now, we have to select `4` bulbs out of which at least `3` are defective. So, we have to select either `3` defective bulbs or `4` defective bulbs. When we have to select `3` defective bulbs, then probability `= (C(4,3)**C(6,1))/(C(10,4)` When we have to select `3` defective bulbs, then probability `= (C(4,4)**C(6,0))/(C(10,4))` So, required probability `P(E) = (C(4,3)**C(6,1)+C(4,4)**C(6,0))/(C(10,4))` `=>P(E) = (4**6+1**1)/((10**9**8**7)/(4**3**2**1)) = 25/210 = 5/42` So, required probability is `5/42`. |
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| 29. |
A company sells two different products A and B. the two products are produced in a common production proce is which has a total capacity of 500 hours ofusing man power. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The demand in the market shows that the maximum number of units of a that can be sold is 70 and that of B is 125. Profit on each of unit a is 20 Rs and on B is 125. How many units of A and B should be produced to to maximise the profit. |
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Answer» 5x+3y=500 x`<=`70 y`<=`125 Profit=Z=20x+15y at B=Z=20*25+15*125=2375. at C=Z=20*70+15*50=2150 profit at B is greater than profit at C A=25,B=125. |
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| 30. |
If `A={x:x inW,xle2},B={x:x inN,1ltxlt5 and C={3,5}`, then find `(i) Axx(BnnC)` (ii) `Axx(BuuC)` |
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Answer» We, have `A={x:x in W,xlt2}={0,1}` and `B={x:x in N,1ltxlt5}` `:. ={2,3,4}and C={3,5}` `(i) :. BnnC={3}` `:. Axx(BnnC)={0,1}xx{3}={(0,3),(1,3)}` (ii) `:. (BuuC)={2,3,4,5}` `:. Axx(BuuC)={0,1}xx{2,3,4,5}` `={(0,2,),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}` |
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| 31. |
Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.(i) On `Z^+`, define ∗ by `a ∗ b = a – b`(ii) On `Z^+`, define ∗ by `a ∗ b = ab`(iii) On `R`, define ∗ by `a ∗ b = ab^2` (iv) On `Z^+`, define ∗ by `a ∗ b = |a – b|`(v) On `Z^+`, define ∗ by `a ∗ b = a ` |
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Answer» (i) On `Z^+, a**b = a-b` In this case, `a` and `b`, both are positive and `a` is less than `b`. Then, `a-b` will be negative and will not fall in the set of positive integers. `:.` This is not a binary operation. (ii) On `Z^+, a**b = ab` In this case, `a` and `b`, both are positive and multiplication of two positive numbers is always positive number and fall into the set `Z^+`. `:.` This is a binary operation. (iii)On `Z^+, a**b = ab^2` In this case, `a` and `b`, both are positive. So, `b^2` will also be positive and multiplication of two positive numbers is always positive number and fall into the set `Z^+`. `:.` This is a binary operation. (iv) On `Z^+, a**b = |a-b|` In this case, `a` and `b`, both are positive and modulus of any number is always positive and fall into the set `Z^+`. `:.` This is a binary operation. (v) On `Z^+, a**b = a` In this case, `a` and `b`, both are positive. So, result of this operation will also positive and fall into the set `Z^+`. `:.` This is a binary operation. |
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| 32. |
A relation `R` on the set of complex numbers is defined by `z_1 R z_2` if and only if `(z_1-z_2)/(z_1+z_2)` is real. Show that R is an equivalence relation. |
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Answer» (i) For Z in C `_(z)R_(z)hArr (Z-Z)/(Z+Z)` is a real number. `rArr` o is a real number which is true. `:.` R is symmetric. (ii) Let `Z_(1), Z_(2)in C and _(Z_(1))R_(Z_(2))` Now, ` _(Z_(1))R_(Z_(2))rArr (Z_(1)-Z_(2))/(Z_(1)+Z_(2))`is a real number. `rArr-((Z_(2)-Z_(1))/(Z_(2)+Z_(1)))` is a real number. `rArr(Z_(2)-Z_(1))/(Z_(2)+Z_(1))` is a real number. ` rArr_(Z_(2))R_(Z_(1))` `:.` R is symmetric. (iii) `Let Z_(1), Z_(2), Z_(3) in C and _(Z_(1))R_(Z_(2)) and _(Z_(2))R_(Z_(3))` . Now, `_(z_(1))R_(z_(2))rArr (Z_(1)-Z_(2))/(Z_(1)+Z_(2))` is a real number. `rArr ((x_(1)+iy_(1))-(x_(2)+iy_(2)))/((x_(1)+iy_(1))+(x_(2)+iy_(2)))` is a real number. Where `Z_(1) = x_(1) + iy_(1) and Z_(2)=x_(2)+iy_(2)` `rArr ((x_(1)-x_(2))+i(y_(1)-y_(2)))/((x_(1)+x_(2))+i(y_(1)+y_(2)))` is a real number. `rArr([(x_(1)-x_(2))+i(y_(1)-y_(2))][(x_(1)+x_(2))-i(y_(1)+y_(2))])/([(x_(1)+x_(2))+i(y_(1)+y_(2))][(x_(1)+x_(2))-i(y_(1)+y_(2))])` is a real number `((x_(1)^(2)-x_(2)^(2))+(y_(1)^(2)-y_(2)^(2))+i{(x_(1)+x_(2))(y_(1)-y_(2))-(x_(1)-x_(2))-i(y_(1)+y_(2))})/((x_(1)+x_(2))^(2)+(y_(1)+y_(2))^(2))` is a real number. `rArr` Coefficient of i = 0 `rArr (x_(1)+x_(2))(y_(1)-y_(2))-(x_(1)-x_(2))(y_(1)+y_(2))= 0` `rArr x_(2)y_(1)-x_(1)y_(2)=0` `rArr (x_(1))/(y_(1))=(x_(2))/(y_(2))` `:. ""_(z_(1))R_(z_(2)) rArr (x_(1))/(y_(1))=(x_(2))/(y_(2))` Similarly, ` ""_(z_(1))R_(z_(2)) rArr (x_(2))/(y_(2))=(x_(3))/(y_(3))` `:.""_(z_(1))R_(z_(2)) rArr (x_(2))/(y_(2))=(x_(3))/(y_(3))` and `""_(z_(2))R_(z_(2)) ` `rArr (x_(1))/(y_(1))-(x_(2))/(y_(2)) and (x_(2))/(y_(2))=(x_(3))/(y_(3))` `rArr (x_(1))/(y_(1))= (x_(3))/(y_(3))rArr "" _(z_(1))R_(z_(3))` `:.` R is transitive. Thus, the given relation R is reflexive, symmetric. and transitive. i.e., R is an equivalence relation. Hence Proved. |
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| 33. |
A relation R is defined on the set of integers as follows : `""_(a)R_(b) hArr(a - b)` , is divisible by 6 where a, b, `in` I. prove that R is an equivalence relation. |
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Answer» (i) For each a `in` I `a- a = 0 = 0 xx 6` `rArr a - a` , is divisible by 6 `rArr _(a)R_(a), AA a in I` `:.` R is reflexive. (ii) Let a, b `in` I such that `""_(a)R_(b)` `:. ""_(a)R_(b) rArr a-b`, is divisible by 6 `rArr - (b - a)`, is divisible by 6 `rArr ""_(b)R_(a)` `:.` R is symmetric. (iii) Let a, b, c `in` I such that `""_(a)R_(b) and"" _(b)R_(c)` Now `""_(a)R_(b) and ""_(b)R_(c)` `rArr a - b `, and b - c, both are divisible by 6 `rArr [(a - b) + (b - c)]`, is divisibe by 6 `rArr(a - c)`, is divisible by 6 `rArr ""_(a)R_(c)` `:.` R is transitive. Hence, R is reflexive, symmetric and transitive `rArr` R is an equivalence relation. Hence Proved. |
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| 34. |
If `sin^6theta+cos^6theta+kcos^2(2theta)=1`, then `k` is |
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Answer» `sin^6theta=cos^6theta=(sin^2theta+cos^2theta)(sin^4theta+costheta-sin^2thetacostheta)=1` `(sin^2theta+cos^2theta)^2-3sin^2thetacos^2theta)=1` `=1-3sin^2thetacos^2theta` `=1-3/4*sin^2(2theta)` `=1/4+3/4cos^2 2theta` `3cos^2 2theta+4kcos^2 2 theta=3` `cos^2 2 theta(3+4k)=3` `k=3/4+tan^2theta` `cos^2 2theta*(3+3tan^2 2theta)=3` option d is correct. |
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| 35. |
In an examination 30% of the students failed in Physics, 25% in Mathematics and 12% in both Physics and Mathematics. A student is selected at random. Find the probability that the student has failed in Physics, if it is known that he has failed in Mathematics. (ii) the student has failed at least one of the two subjects (ii) the student has passed at least one of the two subjects (iv) the student has passed in Mathematics if he failed in Physics. |
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Answer» let the event of student failed in physics be P let the event of student failed in maths be M `P(P) = 3/10` `P(M)= 1/4` `P(P nn M) =3/25` (I)`P(P/M)= (P(P nn M))/(P(M)) = (3/25)/(1/4) = 12/25` (ii)`P(P uu M) = P(P) + P(M) -P(P nn M)` `= 3/10 + 1/4 - 3/25 = 43/100 = 0.43` (iii) `P=1 - P(P nn M) = 1 - 12/100 = 88/100 =22/25` (iv)`p-> 30 , m-> 25` `P nn M -> 12` `P= 18/30 = 3/5` |
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| 36. |
Let `a_i^2+b_i^2+c_i^2=1` (for i=1,2,3) and `a_ia_j+b_ib_j+c_ic_j=0 (i != j;i,j=1,2,3).` Then the absolute value of determinant `|(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)|` (A) `1/2` (B) 0 (C) 1 (D) 2 |
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Answer» Let `vecp = a_1hati+b_1hatj+c_1hatk`, `vecq = a_2hati+b_2hatj+c_2hatk`, `vecr = a_3hati+b_3hatj+c_3hatk` Here, we are given, for `(i=1,2,3)`, `a_i^2+b_i^2+c_i^2 = 1` It means, `|vecp|=|vecp|=|vecp| = 1` Also, we are given,`a_ia_j+b_ib_j+c_ic_j = 0` `:. vecp*vecq = a_1a_2+b_1b_2+c_1c_2 = 0` So, `vecp _|_vecq`. Similarly, `vecq _|_vecr` and `vecr _|_vecp`. Now, we have to find the value of `A=|[a_1,a_2,a_3],[b_1,b_2,b_3],[c_1,c_2,c_3]|` Transpose of this determinant will be same. So, we can write, `A = |[a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_2,c_3]|` `A = [vecp*vecq*vecr] ` As `|vecp|=|vecp|=|vecp| = 1` `:. A =1` |
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| 37. |
P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of the cases they are likely to agree in the same fact ? Do you think, when the agree, means both are speaking truth. |
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Answer» Here, probability of `P` speaking truth, `P(T) = 70% = 7/10` `:.` probability of `P` speaking lie, `P(L) = 1- 7/10 = 3/10` Probability of `Q` speaking truth, `Q(T) = 80% = 4/5` `:.` probability of `Q` speaking lie, `Q(L) = 1- 4/5 = 1/5` Probability of both `P` and `Q` agree in same fact `P(S) = P(T)nnQ(T) +P(L)nnQ(L)` As, all these events are indepependent event, `:. P(S) = P(T)Q(T)+P(L)Q(L)` `P(S) = 7/10**4/5+3/10**1/5 = 28/50+3/50 = 31/50` So, percentage of both of them agree on same fact `= 31/50**100 = 62%` Also, we can clearly see that when both of them agree, they also speak lie and probability of that is `3/50`. |
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| 38. |
Let R be the relation in the set N given by `R = {(a , b) : a = b 2, b > 6}`. Choose the correct answer.(A) `(2, 4) in R` (B) `(3, 8) in R` (C) `(6, 8) in R`(D) `(8, 7) R`A. `(2, 4) in R`B. `(3, 8) in R`C. `(6, 8) inR`D. `(8, 7) in R` |
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Answer» Correct Answer - C `because " " b gt 6` `therefore " " (2, 4) notin R` `" " (3, 8) notin R because 3 ne 8-2` `" " (8, 7) notin R because 8 ne 7-2` Therefore, `(6, 8) in R ` |
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| 39. |
Check the injectivity and surjectivity of the following functions:(i) `f : N ->N`given by `f(x)=x^2`(ii) `f : Z-> Z`given by `f(x)=x^2`(iii) `f : R ->R`given by `f(x)=x^2`(iv) `f : N-> N`given by `f(x)=x^3`(v) `f : Z -> |
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Answer» (i) `f:N to N and f(x) = x^(2)` Let `x, y in N and f(x) -f(y)` `rArr " " x(2) =y^(2)` `rArr " " x=y ( because x,y in N)` `therefore f` is one-one. Let `f(x) = y` where `y in N` `rArr x ^(2) =y ` `rArr x = pm sqrty notin N ` if `y =2` `therefore f` in not onto. Therefore, `f` is one-one but not onto. (ii) `f: Z to Z and f(x) = x^(2)` Let `x, y in Z and f(x) = f(y)` `rArr x ^(2) = y^(2) rArr x = pm y ` `therefore f `is not one-one. Again, let `f(x) = y` where `y in Z` `rArr x ^(2) =y` `rArr x = pm sqrty notin Z ` If ` y = 2 ` ` therefore f` is not onto. Therefore, `f` is neither one-one nor onto. (iii) `f : R to R and f(x) = x^(2)` Let `x, y in R and f(x) = f(y)` `rArr x^(2) = y^(2) rArr x = pm y ` `therefore f ` is not one-one Again, let `f(x) = y ` where `y in R` `rArr x^(2) = y` `rArr x = pm sqrty notin R` if ` y = -2` ` therefore f` is not onto. Therefore, `f` is neither one-one nor onto. (iv) `f: N to N and f(x) = x^(3)` Let `x , y in N and f(x) = f(y)` `rArr " "x ^(3) = y^(3) rArr x =y ` `therefore f` is one-one. Let `" " f(x) =y " " ` where `y in N ` `rArr x ^(3) = y ` ` rArr " "x = y ^(1//3) notin N ` if ` y = -2 ` `therefore f ` is not onto. Therefore, `f` is one-one but not onto. (v) `f: Z toZ and f(x) = x^(3)` Let ` x , y in Z and f(x) = f(y)` `rArr x ^(3) = y^(3) rArr x =y` `therefore f` is one-one. Again let `f(x) -y ` where `y in Z` `rArr x ^(3) = y` `rArr x = y^(1//3) notin Z ` if `y= 2` `therefore f ` is not onto. Therefore, `f` is one-one but not onto. |
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| 40. |
Show that the function `f: R_0->R_0`, defined as `f(x)=1/x`, is one-one onto,where `R_0`is the set of allnon-zero real numbers. Is the result true, if the domain `R_0`is replaced by `N`with co-domain beingsame as `R_0`? |
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Answer» `f: R_(**) to R_(**)` and `" " f(x) = (1)/(x) AA x in R_(**)` Let `x, y in R_(**) and f(x) = f(y) ` `rArr " " (1)/(x) = (1)/(y) rArr x =y` `therefore f` is one-one. For each `y in R_(**)`, ` " "x = (1)/(y) in R_(**)` such that `" " f(x) = f((1)/(y)) = (1)/((1)/(y)) = y ` `therefore f `is onto. Therefore, function `f` is one-one onto function. Again, let `" " g: N to R_(**)` then, `" " g(x) = (1)/(x) AA x in N` Let `x, y in N and g(x) = g(y)` `rArr (1)/(x) = (1)/(y) rArr x =y` `therefore g ` is one-one. `because ` For 12`inR_(**)` there does not exist `x in N ` such that `g(x) = (1)/(12)` `therefore g` is not onto. Therefore, g is one-one but not onto. |
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| 41. |
Find the number of all onto functions from the set `{1, 2, 3, , n)`to itself. |
| Answer» The number of all onto functions from the set { 1, 2, 3, ... , n} to itself will be equal to number of permutations of 1, 2, 3 , .. , n i.e.,`""^(n)P_n = lfloor n.` | |
| 42. |
Total number of onto function from a set of n elements to a set of another r elements. |
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Answer» `r^n--rC_1(r-1)^n-[rC_2(r-2)^n-[rC_3(r-3)^n-[` `r^n-rC_1(r-1)^n+rC_2(r-2)^n-rC_3(r-3)^n+rC_4(r-4)^2...rC_(r-1)C_(r-1)(1)^n`. |
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| 43. |
Domain of fuction f(x) = `sqrt(cos x)` |
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Answer» `f(x)=sqrtcosx` `cosx>=0` `x in[2npi-pi/2,2npi+pi/2]`. |
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| 44. |
Let `A={1,2,3ddot, 14}`. Define a relation on a set A by `R={(x , y):3x-y=0. w h e r e x , y in A}`.Depict this relationship using an arrow diagram. Write down its domain,co-domain and range. |
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Answer» Correct Answer - `R={(3,1),(6,2),(9,3),(12,4)}` Domain: {3, 6, 9, 12}, Range: {1, 2, 3, 4}, Codomain: A Given relation R from set A to itself is `R={(x,y):x=3y, " where " x,y in A} ` ` :. R={(3,1),(6,2),(9,3),(12,4)}` Domain of R is `{3,6,9,12}` and range of R is `{1,2,3,4}` Codomain of R is set A. |
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| 45. |
If ` f: x ->y` defined by f(x) = ` sqrt3 sinx + cos x + 4` is one-one and onto, then x and y are given by |
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Answer» `f(x)=sqrt3sinx+cosx+4` `=(sqrt(3+1)(sqrt3sinx)/(sqrt(3+1))+cosx/sqrt(3+1))+4` `=2(sinx*sqrt3/2+cosx*1/2)+4` `y=f(x)=2sin(x+30^@)+4` Option 2 is correct. |
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| 46. |
If `f : R -> R` be a function such that `f(x) = { x|x| -4; x in Q, x|x| - sqrt3; x !in Q` then f(x) is |
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Answer» `f(2)=f(3^(1//4))=0.implies` many-to-one function Also, `f(x) ne sqrt(3) AA x in R implies ` into function |
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| 47. |
If w is a non real cube root of unity, then minimum value of `| a+ bw + c w^2|`is ( If a,b,c are not equal): (A) 0 (B) `(sqrt3)/2` (C) 1 (D) 2 |
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Answer» Let `z = a+b omega + c omega^2` We know, `1+ omega +omega^2 = 0` `:. omega^2 = -1-omega` `:. z = a+b omega+c (-1-omega)` ` z = a+ b omega-c - c omega` `z = (a-c)+(b-c) omega` Putting value of `omega`, `z = (a-c) +(b-c)(-1/2+sqrt3/2i)` `z = (a-b/2-c/2)+sqrt3/2(b-c)i` We know, `|z| = sqrt(x^2+y^2` So, here,`|z| = sqrt((a-b/2-c/2)^2+3/4(b-c)^2)` `|z| = sqrt(1/2[(a-b)^2+(b-c)^2+(c-a)^2])` To find minimum value of `|z|`, we will put, `a = b = k`and `c = k+1` Then,`|z| = sqrt(1/2(0+1^2+1^2)) = 1` So, minimum value of `|z|` will be `1`. |
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| 48. |
Which one of the following relations on R is an equivalence relation?A. `a R_(1) b hArr |a|=|b|`B. `a R_(2) b hArr a ge b`C. `a R_(3) b hArr a" divides " b`D. `a R_(4) b hArr a lt b` |
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Answer» Correct Answer - A `aR_(1)b hArr |a|=|b|` satisfies. |
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| 49. |
Let `A = {1, 2, 3}`. Then number of equivalence relations containing (1, 2) is (A) 1 (B) 2 (C) 3 (D) 4A. 1B. 2C. 3D. 4 |
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Answer» A= {1,2,3} equivalence relation which includes (1,2) ` R_1 ={(1,1),(2,2),(3,3),(2,1),(1,2)}` and `R_2 ={(1,1),(2,2),(3,3),(2,1),(1,2)(3,1),(1,2)}` |
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| 50. |
Let `A = {1, 2, 3}`. Then number of equivalence relations containing (1, 2) is (A) 1 (B) 2 (C) 3 (D) 4 |
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Answer» Here, `A = {1,2,3}` `:.` Total possible pairs `= {(1,1)(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}` `:.` Smallest equivalence relation containing `(1,2), (R_1) = {(1,1),(2,2),(3,3),(1,2),(2,1)}` Now, if we add `(2,3)`, then we have to add `(3,2)` to make it symmetric. As `(1,2),(1,3)` are there, we have to add `(1,3)` also to make it transitive. As we are adding `(1,3)`, we need to add `(3,1)` also to make it symmetric.`:. R_2 = {(1,1)(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}` These are the `two` equivalence relations are possible. So, `B` is the correct option. |
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