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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.
| 201. |
` y = 2 sin^(-1) ((x-2)/sqrt6) - sqrt(2 + 4x - x^2)` then show that `(dy)/(dx) |_(x=2) = 2/sqrt6` |
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Answer» `y = 2sin^-1((x-2)/sqrt6)-sqrt(2+4x-x^2)` `dy/dx = 2(1/sqrt(1-((x-2)/sqrt6)^2))(1/sqrt6) - 1/2(1/(sqrt(2+4x-x^2)))(4-2x)` `=>dy/dx|_(x=2) = 2(1/(1-0))(1/sqrt6)-1/2(4-4) = 2/sqrt6 ` |
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| 202. |
Let f(x) be defined in the interval [-2,2] such that `f(x) = { -1; -2 |
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Answer» g(x)=f(|x|)+|f(x)| g(x)=-x-1+1 =-x-1+1-x =x-1+x-1 g(x)=-x =0 =2x-2 option b is correct. |
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| 203. |
A family has two children. What is the probability that both the children are boys given that atleast one of them is a boy ? |
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Answer» Let `B` represents a boy and `G` represents a girl. Then, Sample space `(S) = {(B,B),(G,G),(B,G),(G,B)}` So, Probability of both of children are boys`,P(A_1) = 1/4` Probability that at least one of them is a boy `,P(A_2) = 3/4` Here, we have to find `P(A_1/A_2)`. `=>P(A_1/A_2) = (P(A_1 nn A_2))/(P(A_2))` Here, `P(A_1 nn A_2) = 1/4` `=>P(A_1/A_2) = (1/4)/(3/4) = 1/3` So, required probability is `1/3`. |
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| 204. |
Consider functions f and g such that composite gof is defined and is one-one.Are f and g both necessarily one-one. |
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Answer» Let there are two functions `f` and `g` such that `f: A -> B and g: B -> C` Now `gof: A -> C` Given that `gof` is one-one. To prove that `f: A -> B` is one-one, we have to prove that if `f(x) = f(y)` then `x = y` for all `x, y in A`. Now let `x, y in A` such that `f(x) = f(y)` Then `gof(x) = g(f(x)) = g(f(y))` `=> gof(x) = gof(y)` `=> x = y` (since `gof(x)` is one-one) Since `gof` is one-one, hence it shows that `f` is one-one. Again `g` may or may not be one-one. So `gof` is one-one does not imply that both `f` and `g` has to be one-one. |
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| 205. |
Show that ` *: R xxR ->R`defined by `a*b = a +2b`is not commutative. |
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Answer» `a ** b = a+ 2b = 3 +10 = 13` `b ** a = b+ 2a = 5+6 = 11` `a=3, b=5` `a** b cancel(=) b ** a` hence proved |
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| 206. |
If `P = {1, 2}`form the set `P xxP xxP`. |
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Answer» `P ={1,2}` `P*P = {1,2}*{1,2}` `= {(1,1),(1,2),(2,4),(2,2)}` multiplying by P again `P*P*P = {(1,2)},{(1,1),(1,2),(2,1),(2,2)}` `= {(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)}` answer |
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| 207. |
Let R be the set of real numbers. Define the real function `f : R ->Rb y f(x) = x + 10`and sketch the graph of this function. |
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Answer» Here, `f(x) = x+10` `f(0) = 10` `f(1) = 11` `f(2) = 12` `f(3) = 13` `f(-10) = 0` Now, we can draw graph for all these points. Please refer video for the graph. |
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| 208. |
29. Which one of following best represents the graph of `y=x^(log_x pi)`A. B. C. D. |
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Answer» Correct Answer - C Given, `y=x^(log_(x)pi)=pi` Domain is ` x in (0,1) cup (1,oo)` Range is `{pi}` |
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| 209. |
Explain transitive relation with suitable examples. |
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Answer» Let `R` is a relation such that`R = {(a,b): a ** b}`, where `**` is a binary operation. Let `(a,b) in R and (b,c) in R` Then, relation `R` will be transitive if `(a,c) in R`. Let us take an example. Let `R = {(a,b): a le b}` Let `(a,b) in R and (b,c) in R` It means, `a le b and b le c`. Therefore, ` a le c`. So, `(a,c) in R`. Thus, `R` is atransitive relation. |
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| 210. |
If `x`is real, then the value of the expression `(x^2+14 x+9)/(x^2+2x+3)`lies between(a) 5 and 4 (b) 5 and `-4`(c) `-5a n d4`(d) none of theseA. [4, 5]B. [-4, 5]C. [-5, 4]D. none of these |
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Answer» Correct Answer - C `(x^(2)+14x+9)/(x^(2)+2x+3)=y` or ` x^(2)+14x+9=x^(2)y+2xy+3y` or ` x^(2)(y-1)+2x(y-7)+(3y-9)=0` Since x is real, we have `4(y-7)^(2)-4(3y-9)(y-1) ge 0` or ` 4(y^(2)+49-14y)-4(3y^(2)+9-12y) ge 0` or ` (y+5)(y-4) le 0` ` :. y in [-5,4]` |
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| 211. |
For `x in R`, the expression `(x^2+2x+c)/(x^2+4x+3c)` can take all real values if `c in` |
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Answer» `f(x) = (x^2+2x+c)/(x^2+4x+3c)` When `c = 0`, `f(x) = (x^2+2x)/(x^2+4x) = (x(x+2))/(x(x+4))` But, it is given that `x in R`, so `x !=0`. When `c = 1`, `f(x) = (x^2+2x+1)/(x^2+4x+3) = ((x+1)^2)/((x+3)(x+1))` As, `x in R`, so `x!=-1 and x! = -3`. So, `c` can take any values between `0` and `1`. `:. c in (0,1)`. |
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| 212. |
`2tan^2 alpha tan^2 betatan^2gamma+tan^2 alpha tan^2beta+tan^2 betatan^2gamma+tan^2gamma tan^2 alpha=1` find the value of `sin^2 alpha+sin^2 beta +sin^2 gamma` |
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Answer» we know, `tan^2 alpha = sin^2alpha/cos^2alpha = sin^2alpha/(1-sin^2alpha)` Let `sin^2alpha = x` Then, `tan^2alpha = x/(1-x)` Similarly, if `sin^2beta = y, then, tan^2beta = y/(1-y)` Similarly, if `sin^2gamma= z, then, tan^2gamma = z/(1-z)` Now, putting these values in the given equation, `2(x/(1-x))(y/(1-y))(z/(1-z))+(x/(1-x))(y/(1-y))+(y/(1-y))(z/(1-z))+(z/(1-z))(x/(1-x)) = 1` `=>(xy)/((1-x)(1-y))+(yz)/((1-y)(1-z))+(zx)/((1-z)(1-x)) = 1- (2xyz)/((1-x)(1-y)(1-z))` `=>(xy-xyz+yz-xyz+zx-xyz)/((1-x)(1-y)(1-z)) = 1- (2xyz)/((1-x)(1-y)(1-z))` `=>(x+y+z) - (2xyz)/((1-x)(1-y)(1-z)) = 1 - (2xyz)/((1-x)(1-y)(1-z))` `=>x+y+z = 1` `:. sin^2alpha+sin^2beta+sin^2gamma = 1` |
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| 213. |
Let `f:[-1,oo] in [-1,oo]` be a function given `f(x)=(x+1)^(2)-1, x ge -1` Statement-1: The set `[x:f(x)=f^(-1)(x)]={0,1}` Statement-2: f is a bijection.A. Statement 1 is ture, statement 2 is true, statement 2 is a correct explanation for statement 1.B. Statement 1 is ture, statement 2 is true, statement 2 is not a correct explanation for statement 1.C. Statement 1 is ture, statement 2 is false.D. Statement 1 is false, statement 2 is true. |
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Answer» Correct Answer - C There is no information about co-domain therefore `f(x)` is not necessarily onto. Therefore statement 2 is false. However for `x ge -1, f(x) =(x+1)^(2)-1` is one-one. Assume that `f(x)` is onto. Now roots of `f(x)=f^(-1)(x)` lies on line `y=x`. So we have to find roots of `f(x)=x or (x+1)^(2)-1=x` ` :. x^(2) +x=0` ` :. x=0, -1` Hence, statement 1 is true. |
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| 214. |
If `u=x^2tan^(-1)(y/x)-y^2tan^(-1)(x/y)`,prove that `(del^2u)/(delxdely)=(x^2-y^2)/(x^2+y^2)` |
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Answer» `u = x^2 tan^-1(y/x)-y^2tan^-1(x/y)` `=> (del u)/ (del x) = 2x tan^-1(y/x) + x^2(x^2/(x^2+y^2))(-y/x^2) - y^2(y^2/(x^2+y^2))(1/y) ` `=> (del u)/ (del x) = 2x tan^-1(y/x) - (x^2y)/(x^2+y^2) - (y^3)/(x^2+y^2)` Now, `(del^2u)/(delxdely) = 2x(x^2/(x^2+y^2))(1/x) - [(x^2(x^2+y^2)-x^2y(2y))/(x^2+y^2)^2] - [(3y^2(x^2+y^2)- y^3(2y))/(x^2+y^2)^2]` `(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) - [(x^4+x^2y^2-2x^2y^2+3x^2y^2+3y^4-2y^4)/(x^2+y^2)^2]` `(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) - [(x^4+2x^2y^2+y^4)/(x^2+y^2)^2]` `(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) - [(x^2+y^2)^2/(x^2+y^2)^2]` `(del^2u)/(delxdely) = (2x^2)/(x^2+y^2) -1` `(del^2u)/(delxdely) = (x^2 - y^2)/(x^2+y^2)` |
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| 215. |
Let `f(x)` be a polynomial of degree 5 such that `f(|x|)=0` has 8 real distinct , Then number of real roots of `f(x)=0` is ________. |
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Answer» Correct Answer - 5 Since `f(|x|)=0` has 8 roots `implies f(x)=0` has 4 real positive roots but since `f(x)` is polynomial of degree 5 `implies` fifth root is also real. ` :. ` Number of real roots of `f(x)=0` is 5 |
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| 216. |
Solve` (5x+1)/((x+1)^2)` |
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Answer» Correct Answer - `x lt 0 " or " x gt 3, x ne -1` `(5x+1)/((x+1)^(2))-1 lt 0` or `(5x+1-(x+1)^(2))/((x+1)^(2)) lt0` or `(-x^(2)+3x)/((x+1)^(2)) lt 0` or `(x(x-3))/((x+1)^(2)) gt 0` `implies x lt 0" or " x gt 3, x ne -1` |
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| 217. |
In an examination out of 100 students,75 passed in English, 60 passed in Mathematics and 45 passed in both English and Mathematics. What isthe number of students passed in exactly one of the two subjects?(a) 45 (b) 60 (C) 75 (d) 90 |
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Answer» `n(E uu M) = n(E) + n(M) - n(E nn M)` `= 75 + 60 - 2 xx 45` `= 135 - 90` `=45` Answer |
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| 218. |
Show that the Modulus Function `f : R->R ,`given by `f (x) = | x |`, is neither one-one nor onto, where `| x |`is x, if x is positive or 0 and `| x |`is ` x`, if x is negative. |
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Answer» `f: R to R and f(x) = |x|` Let `x, y in R` and` " " f(x) = f(y)` `rArr " " |x| = |y|` `rArr " " x = pm y` `therefore f` is not one-one. Again `-1 in R` and there does not exist `x in R` for which `f(x)=-1` `therefore f ` is not onto. Therefore, `f` is neither one-one nor onto. |
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| 219. |
81. A mapping f : R -> R which is defined as f(x) = cos x; `x in R` is:(a) oniy one-one(b) only onto(C) one-one onto(d) Neither one-one nor onto |
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Answer» option `D` is correct when `f: R ->R` as `cos x` Answer |
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| 220. |
Let A denotes the set of values of x for which `(x+2)/(x-4) le0` and B denotes the set of values of x for which `x^2-ax-4 le 0`. If B is the subset of A then a cannot take integral value (a) 0, (b) 1 (c) 2 (d) 3 |
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Answer» `(x+2)/(x-4)<=0` `x in [-2,4)` `A in [-2,4)` `x^2-4<=0` `f(-2)*f(4)>=0` `(4+2a-4)(16-4a-4)>=0` `2a(12-4a)>=0` `(4a-12)(2a)<=0` `a in[0,3)`. |
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| 221. |
Let A = { x | x |
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Answer» `A/3 => 0,1,2 ` rem `(a+b+c)` is a multiple of 3 `a/3 , b/3, c/3` now re`-> 0 , 0, 0 = .^3C_3= 1` `1,1,1 = .^3C_3= 1` `1,2,0 = .^3C_1.^3C_1.^3C_1= 27` `2,2,2= .^3C_3= 1` `A = { 1,2,3,4,5,6,7,8,9}` total subset=`27+1+1+1= 30` option d is correct answer |
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| 222. |
A trust invested some money in two types of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received Rs. 2,800 as interest. However, if trust had interchanged money in bonds, they would have got Rs. 100 less as interest. Using matrix method, find the amount invested by the trust. |
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Answer» Let the money invested`In first =Rs.x` 2nd`=Rs.y` 1st Interest`=10%=0.1` 2nd Interest`=12%=0.12` `0.1x+0.12y=2800` `x+1.2y=28000-----1` `0.1y+0.12x=2700` `y+1.2x=27000----2` `1.2xx1.2x+1.2y=27000xx1.2` `- x +1.2y=28000` `(1.44-1)x=1000(27xx1.2-28)` `0.44x=1000(32.4-28)` `0.44x=1000xx4.4/10` `x=Rs10000` |
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| 223. |
Find the range fo the function `f(x)=(x^(2)-x-6)/(x-3)` |
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Answer» Correct Answer - R - {5} `f(x)=(x^(2)-x-6)/(x-3)=((x-3)(x+2))/(x-3)=x+2,x ne 3` Hence range is `R-{5}` |
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| 224. |
If the relation `f(x)={(1",",x in Q),(2",",x notin Q):}` where Q is set of rational numbers, then find the value `f(pi)+f((22)/(7))`. |
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Answer» Correct Answer - 3 ` pi` is irrational number, so `f(pi)=2.` `(22)/(7)` is rational number. So, `f((22)/(7))=1.` ` :. f(pi)+f((22)/(7))=2+1=3` |
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| 225. |
Consider the following relations:R = {(x, y) | x, y are real numbers and x = wy forsome rational number w};`S={(m/n , p/q)"m , n , pandqa r ei n t e g e r ss u c ht h a tn ,q"!="0andq m = p n"}`. Then(1)neither R nor Sis an equivalence relation(2)S is anequivalence relation but R is not an equivalence relation(3)R and S both areequivalence relations(4)R is anequivalence relation but S is not an equivalence relationA. R and S both are equivalence relations.B. R is an equivalence relation but S is not an equivalence relation.C. Neither R nor S is an equivalence relation.D. S is an equivalence relation but R is not an equivalence relation. |
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Answer» Correct Answer - D xRy need not imply yRx. `(m)/(n) S(p)/(q)hArrqm=pn` `(m)/(n)S(m)/(n)impliesS` is reflexive `(m)/(n)S(p)/(q)` `implies (p)/(q) S(m)/(n)implies S` is symmetric `(m)/(n)S(p)/(q),(p)/(q) S(r)/(s)` `implies qm=pn,ps=rqimpliesms=rn` `implies (m)/(n) S(r)/(s)implies R ` is transitive. Hence, S is an equivalence relation. |
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| 226. |
Three relations `R_1, R_2` and `R_3` are defined on a set `A ={a, b, c}`: as for `R_1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c c)}` (a Find whether or not each of the relations R1,is Symmetric , reflexive or transitive. |
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Answer» `R_1 = {(a,a),(a,b),(a,c),(b,b),(b,c),(c,a),(c,b),(c,c)}` As `R_1` contains `(a,a),(b,b) and (c,c)`, so, it is reflexive. As `R_1` contains `(a,b)` but do not contain `(b,a)`, so it is not symmetric. As `R_1` contains `(a,b),(b,c)` and `(a,c)`, so it is transitive. `R_2 = {(a,a)}` As `R_2` contains `(a,a)`,but, do not contain `(b,b) and (c,c)`, so, it is not reflexive. As `R_2` do not contain `(a,b)` and `(b,a)`, so it is not symmetric. Also, `R_2` is not transitive. `R_3 = {(b,c)}` Based on the above explanation, `R_3` is not symmetric, reflexive and transitive. |
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| 227. |
If `f:{1,2,3,4} ->{1,2,3,4}`, y = f(x) be a function such that `|f(alpha) - alpha| |
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Answer» Here, `f(alpha) = |f(alpha) - alpha| le 1` for `alpha in {1,2,3,4}` `:. f(1) = |f(1) - 1| le 1 = {1,2}` `:. f(2) = |f(2) - 2| le 1 = {1,2,3}` `:. f(3) = |f(3) - 3| le 1 = {2,3,4}` `:. f(4) = |f(4) - 4| le 1 = {3,4}` Here, `f(1)` and `f(4)` contains `2` functions each and `f(2)` and `f(3)` contains `3` functions each. `:.` Total number of functions ` = 2**3**3**2 = 36`. |
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| 228. |
If the graph of the function `f(x)=(a^x-1)/(x^n(a^x+1))`is symmetrical about the `y-a xi s ,t h e nn`equals2 (b)`2/3`(c) `1/4`(d) `1/3`A. 2B. `(2)/(3)`C. `(4)/(3)`D. `-(1)/(3)` |
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Answer» Correct Answer - D `f(x)=(a^(x)-1)/(x^(n)(a^(x)+1))` `f(x)` is symmetrical about y-axis. Thus, `f(x)=f(-x)` `or (a^(x)-1)/(x^(n)(a^(x)+1))=(a^(-x)-1)/((-x)^(n)(a^(-x)+1))` `or (a^(x)-1)/(x^(n)(a^(x)+1))=(1-a^(x))/((-x)^(n)(1+a^(x)))` `or x^(n)= -(-x)^(n)` Hence, the value of n which satisfies this relation is ` -(1)/(3)`. |
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| 229. |
If `f(2x+y/8,2x-y/8)=x y`, then `f(m , n)=0`only when `m=n`only when `m!=n``on l yw h e nm=-n`(d) `fora l lma n dn`A. only when `m = n`B. only when `m ne n`C. only when `m = -n`D. for all m and n |
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Answer» Correct Answer - D Let `2x+(y)/(8)=alpha" and " 2x-(y)/(8)=beta.` Then ` x=(alpha+beta)/(4) and y=4(alpha-beta).` Given `f(2x+(y)/(8), 2x-(y)/(8))=xy` ` :. f(alpha, beta)=alpha^(2)-beta^(2)` `or f(m,n)+f(n,m)=m^(2)-n^(2)+n^(2)-m^(2)=0` for all m,n |
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| 230. |
If `f(x+1/2)+f(x-1/2)=f(x)fora l lx in R ,`then the period of `f(x)`is1 (b)2 (c) 3(d) 4A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `f(x+(1)/(2))+f(x-(1)/(2))=f(x) " (1)" ` Replacing x by `(x+(1)/(2)),` we get `f(x+1)+f(x)=f(x+(1)/(2)) " (2)" ` ` or f(x+1)+f(x-(1)/(2))=0 " (from (1) and (2))" ` ` or f(x+(3)/(2))= - f(x) " "("Replcing x by "(x+(1)/(2)))` ` or f(x+3) = -f(x+(3)/(2))=f(x) " " ("Replacing x by "(x+(3)/(2)))` Therefore, f(x) is periodic with period 3. |
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| 231. |
If `a f(x+1)+b f(1/(x+1))=x,x !=-1,a != b,`then `f(2)` is equal toA. `(2a+b)/(2(a^(2)-b^(2)))`B. `(a)/(a^(2)-b^(2))`C. `(a+2b)/(a^(2)-b^(2))`D. none of these |
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Answer» Correct Answer - A `af(x+1)+bf((1)/(x+1))=(x+1)-1 " (1)" ` Replacing `x+1" by " (1)/(x+1),` we get `af((1)/(x+1))+bf(x+1)=(1)/(x+1)-1 " (2)" ` `Eq. (1)xx a-Eq.(2) xx b` `implies (a^(2)-b^(2))f(x+1)=a(x+1)-a-(b)/(x+1)+b` Putting `x=1, (a^(2)-b^(2))f(2)=2a-a-(b)/(2)+b=a+(b)/(2)=(2a+b)/(2)` |
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| 232. |
If `f(3x+2)+f(3x+29)=AAx in R ,`then the period of `f(x)`is7 (b) 8(c) 10 (d) none of theseA. 7B. 8C. 10D. none of these |
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Answer» Correct Answer - D `f(3x+2)+f(3x+29)=0 " (1)" ` Replacing x by `x+9`, we get `f(3(x+9)+2)+f(3(x+9)+29)=0` ` or f(3x+29)+f(3x+56)=0 " (2)" ` From (1) and (2), we get `f(3x+2)=f(3x+56)` ` or f(3x+2)=f(3(x+18)+2)` Therefore, f(x) is periodic with period 54. |
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| 233. |
Identify the type of the function `f:R to R,` `f(x)=e^(x^(2))+cosx.` |
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Answer» Correct Answer - many-one,into `f(x)=e^(x^(2))+cosx` Clearly, `f(1)=f(-1),f(2)=f(-2)` etc. So, `f(x)` is many-one. Range of `e^(x^(2))` is `[1,oo)` and range of ` cos x " is " [-1,1].` Thus `f(x)` does not take negative values. Hence `f(x)` is into. |
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| 234. |
Find the range of `f(x)=(x^2+34 x-71)/(x^3+2x-7)` |
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Answer» Correct Answer - `(-oo, 5] cup [9,oo)` Let `(x^(2)+34x-71)/(x^(2)+2x-7)=y` or ` x^(2)(1-y)+2(17-y)x+(7y-71)=0` For the real value of x, `b^(2)-4ac ge0` or `4(17-y)^(2)-4(1-y)(7y-71) ge 0` or ` y^(2)-14y+45 ge 0` i.e., `y le 5 " or " y ge 9` Hence, range is `(-oo,5] cup [9,oo).` |
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| 235. |
Let `f(x) =ax^2+bx+c` where a,b,c are real numbers. Suppose `f(x)!=x` for any real number x. Then the number of solutions for f(f(x))=x in real numbers x is |
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Answer» `f(x)=ax^2+bx+c` `g(x)=x` `f(f(x))=x` x=d is solution `f(f(d))=d` `f(d)=e` `y=f(x)` When`!=`e (d,p)(,p,d) When d=p MI with respect to t=x which is not possible `f(d)=d->f(x)!=x` option c is correct. |
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| 236. |
The entire graph of the equation `y=x^2+k x-x+9`in strictly above the `x-a xi s`if and only if`k |
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Answer» Correct Answer - `-5 lt k lt 7` `y=x^(2)+(k-1)x+9=(x+(k-1)/(2))^(2)+9-((k-1)/(2))^(2)` For entire graph to be above x-axis, we should have `9-((k-1)/(2))^(2) gt 0` `implies k^(2)-2k-35 lt 0` `implies (k-7)(k+5) lt 0` `implies -5 lt k lt 7` |
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| 237. |
Let g(x+y),g(x).g(y) and g(x-y) are in AP. For all x,y and `g(0)!=0` then a) g(2)=0 (b) the graph of g is symmetry about y-axis (c) g ia an odd unction (d) g(0)=-1 |
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Answer» As `g(x+y),g(x).g(y), g(x-y)` are in `AP`. `:. g(x+y)+g(x-y) = 2g(x)*g(y)` When `y = 0`, `=>g(x)+g(x) = 2g(x)*g(0)` `=>g(0) = 1` It means, option `(d)` is correct. When, `x = 0`, `=>g(-y)+g(y) = 2g(0)*g(y)` `=>g(-y)+g(y) = g(y)->(1)` If we replace, `y` with `-y`, then, `=>g(-y)+g(y) = g(-y)->(2)` From (1) and (2), we can see that `g(y) and g(-y)` are equal. It means `g(x)` is an even function. So, graph of `g` will be symmetrical about `y`-axis. `:.` Option `(b)` and `(d)` are the correct options. |
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| 238. |
If I is the incenter of a triangle ABC, then the ratio IA IB IC is equal to |
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Answer» `sinB/2=r/(IB)` `IB=rcosecB/2` `IA:IB:IC=cosecA/2:cosecB/2:cosecC/2`. |
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| 239. |
If `A`,`B`, `C` have position vectors `(0,1,1)`,`(3,1,5)`,`(0,3,3)`, respectively, then show that `Delta ABC` is right angled at `C`. |
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Answer» Here, Position vector of `A = hatj+hatk` Position vector of `B = 3hati+hatj+5hatk` Position vector of `C = 3hatj+3hatk` `vec(AC) = (3hatj+3hatk)-(hatj+hatk) = 2hatj+2hatk` `vec(BC) = (3hatj+3hatk) - (3hati+hatj+5hatk) = -3hati+2hatj-2hatk` `vec(AC).vec(BC) = 0+2(2)+2(-2) = 4-4 = 0` As, dot product of AC and BC is `0`, it mean they are perpendicular. `:. Delta ABC` is a right angled triangle with `/_C =90^@` |
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| 240. |
`sin20^@(4+sec20^@)=` |
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Answer» `sin20^@(4+sec20^@)` `=sin20^@(4+1/cos20^@)` `=(4sin20^@cos20^@+sin20^@)/(cos20^@)` `=(2sin40^@+sin20^@)/(cos20^@)` `=(2sin(60-20)^@+sin20^@)/(cos20^@)` `=(2sin60^@cos20^@-2cos60^@sin20^@+sin20^@)/(cos20^@)` `=(2(sqrt3/2)cos20^@-2(1/2)sin20^@+sin20^@)/(cos20^@)` `=(sqrt3cos20^@-sin20^@+sin20^@)/(cos20^@)` `=sqrt3` So, `sin20^@(4+sec20^@) = sqrt3` |
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| 241. |
Let R ={(a, b):a,b in Z and (a-b) is divisible by 5 }. Show that R is an equivalence relation on Z. |
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Answer» Here, `R = {(a,b):a,b in Z and (a-b)` is divisible by `5}` For all `a in R`, `=> (a-a) =0` and `0` is divisible by `5`. `:. R` is reflexive. Since in `R` for every `(a,b) in R` `=> (a-b)` is divisible by `5`. `=> (-(b-a))` is divisible by `5`. `=> (b-a)` is also divisble by `5`. `:. (b,a) in R`. `:. R` is symmetric. Since `(a,b) in R and (b,c) in R` `=> (a-b)` is divisible by `5` & `(b-c)` is divisible by `5`. `=> (a-b+(b-c))` is divisible by `5`. `=> (a-c)` is divisible by `5`. `:. (a,c) in R`. `:. R` is transitive. As `R` is reflexive, symmetric and transitive, `R` is an equivalence relation. |
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| 242. |
Prove m:n theoremin a `Delta ABC`, a point D is taken on side BC such that BD:DC is m:n.Then prove that(1) `(m+n)cottheta= mcotalpha-ncotbeta`(2) `(m+n)cottheta= ncotB-mcotC` |
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Answer» 1)`(BD)/(DC)=m/n` `/_ADB=180-theta` `/_ABD=180-(alpha+180-theta)` `=theta-alpha`. `In/_ABD` `(BD)/sinalpha=(AD)/(sin(theta-alpha))` `In/_ADC` `(DC)/sinbeta=(AD)/(sin(theta-beta))` `((BD)/sinalpha)/((DC)/sinbeta)=((AD)/(sin(theta+alpha)))/((AD)/(sin(theta+beta)` `(BDsinbeta)/(DCsinalpha)=(sin(theta+beta))/(sin(theta-alpha))` `(msinbeta)/(nsinalpha)=(sinthetacosbeta+costhetasinbeta)/(sinthetacosalpha-costhetasinalpha)` Divide by`sinthetasinbetasinalpha` `m(cotalpha-cottheta)=n(cotbeta+cottheta)` `mcotalpha-mcottheta=ncotbeta+ncottheta` `mcotalpha-ncotalpha=cottheta(m+n)` Hence proved. 2)`/_ABD=/_ABC=B` `/_ACD=c` `/_BAD=(theta-beta)` `msin(theta+c)sinbeta=nsincsin(theta-beta)` `msinbeta(sinthetacosc+costhetasinc)=nsinc(sinthetacosbeta-costhetasinbeta)` Divide by`sinthetasinbetasinc` `m(cotc+cottheta)=n(cotbeta-cottheta)` `mcotc+mcottheta=ncotbeta-ncottheta` `cottheta(m+n)=ncotbeta-mcotc` Hence Proved. |
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| 243. |
`sqrt(2)cosA=cosB+cos^3B`, `sqrt(2)sinA=sinB-sin^3B` , find `sin(A-B)` |
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Answer» `sqrt2cos A = cosB + cos^3B` `=>sqrt2cos A = cosB(1 + cos^2B)->(1)` `sqrt2sinA = sinB - sin^3B` `=>sqrt2sinA = sinB(1 - sin^2B)` `=>sqrt2sinA = sinBcos^2B` `=>cos^2B = sqrt2sinA /sinB` Putting value of `cos^2B` in (1), `=>sqrt2cos A = cosB(1 + (sqrt2sinA) /sinB) ` `=>sqrt2cosAsinB = sinBcosB+sqrt2sinAcosB` `=>sqrt2(sinAcosB-cosAsinB) = -sinBcosB` `=>sqrt2(sin(A-B)) = (-sin2B)/2` `=>sin(A-B) = (-sin2B)/(2sqrt2).` |
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| 244. |
`tan(pi/4+1/2cos^-1x)+tan(pi/4-1/2cos^-1x)`, `x!=0` is equal to |
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Answer» Let `1/2 cos^-1x = theta=> cos2theta = x` Then, our expression becomes, `tan(pi/4+theta)+tan(pi/4-theta)` `=(1+tantheta)/(1-tantheta) + (1-tantheta)/(1+tantheta)` `(1+tan^2theta+2tantheta+1+tan^2theta-2tantheta)/(1-tan^2theta)` `=2((1+tan^2theta)/(1-tan^2theta))` As `cos2theta = (1-tan^2theta)/(1+tan^2theta)` So, our expression becomes, `=2/(cos 2theta) = 2/x`, which is the desired value for our expression. |
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| 245. |
The equation of line are 5x-3 = 15y + 7 = 3 - 10z. Write the direction cosines of the line. |
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Answer» `5x-3=15y+7=3-10z` `5(x-3/5)=15(y+7/5)=-10(z-3/10)` `(x-3/5)/(1/5)=(y+7/15)/(1/15)=(z-3/10)/(-1/10)` `l:m:n=1/5:1/15:-1/10=6:2:-3` `sqrt(l^2+m^2+n^2)=sqrt(6^2+2^2+3^2)=7` `cosalpha=6/7,alpha=cos^(-1)(6/7)` `cosbeta-2/7=beta=cos^(-1)(2/7)` `cosgamma=-3/7=gamma=cos^(-1)(-3/7)` `=(6/7,2/7,-3/7)`. |
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| 246. |
Sum of first 20 terms of ` 3/1^2 + 5/(1^2 + 2^2) + 7/(1^2 + 2^2 + 3^2) `+... upto 20 terms is : |
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Answer» `T_n=(2n+1)/(n(n+1)(2n+1)` `=6/(n(n+1)` `=6(1/n-1/(n+1))` `T_1=6(1-1/2)` `T_2=6(1/2-1/3)` `T_3=6(1/3-1/4)` `t_20=6(1/20-1/21)` `T_1+T_2+T_3...T_20=6(1-1/21)` `=6*20/21=40/7`. |
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| 247. |
The foot of the perpendicular drawn from the origin O to a plane is N(12,-4,-3). Find the equation of the plane in cartesian form and vector form. |
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Answer» `(r - bar A).OA = 0` `(vec r - (12 hat i - 4 hat j - 3 hat k))*(12 hat i - 4 hat j - 3 hat k) = 0` `vec r * (12 hat i - 4 hat j - 3 hat k) = |12 hat i - hat j - 3 hat k|^2` `vec r (12 hat i - 4 hat j - 3 hat k ) = 169 ` it is in the form of vector so , in cartesian form `(x hat i + y hat j + z hat k)*(12 hat i - 4 hat j - 3 hat k) = 169` answer |
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| 248. |
If the graph of non-constant function is symmetric about the point (3,4), then the value of `sum_(r = 0)^6 f(r) + f(3)` is equal to |
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Answer» Here, we are given, `f(3) = 4` As, given function is symmetric. `:. f(3+h) = 3+k` `f(3-h) = 3-k` Here, `h` and `k` are constants. Now, `sum_(r=0)^6f(r) +f(3) = f(0)+f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(3)` `=2f(3)+f(1)+f(5)+f(2)+f(4)+f(0)+f(6)` `=2f(3)+f(3-2)+f(3+a)+f(3-1)+f(3+1)+f(3-3)+f(3+3)` From the property of symmetric functions, `=2(4)+4-a+4+a+4-b+4+b+4-c+4+c` (`a,b,c` are constants.) ` = 8+8+8+8 =32` `:.` Value of given expression will be `32`. |
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| 249. |
The number of three digit numbers of the form xyz where x>y >z is |
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Answer» There can be two cases. Case 1: When `0` is any of the numbers. In this case, `z` will be `0`. So, number of ways to select `x` and `y` will be `C(9,2)`. Case 2: When `0` is not included in any of the numbers. Number of ways to select `x`,`y` and `z` will be `C(9,3)`. `:.` Total number of three digits number `= C(9,2)+C(9,3) = 36+84 = 120` |
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| 250. |
What is the coefficient of `x^(100)` in the expansion of `(1+x+x^2+x^3+.....+x^100)^3` |
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Answer» `(1+x+x^2+x^3+...+x^100)^3` Coefficient of `x^n` in an expression `(x_1+x_2+...+x_m)` can be given as `C(n+m-1,n)`. Here, `n =100, m = 3`. `:.` Coefficient of 100th term in given expression ` = C(100+3-1,100) = C(102,100) = C(102,2)`. so, option `C` will be the correct option. |
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