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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.
| 451. |
`(1+tanA)/(1-tanA)=(cos2A)/(1-sin2A)` |
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Answer» LHS `=(1+tanA)/(1-tanA)` `=(1+(sinA/cosA))/(1-(sinA/cosA))` `=(cosA+sinA)/(cosA-sinA)*(cosA-sinA)/(cosA-sinA)` `=(cos^2A-sin^2A)/(cosA-sinA)^2` `=(cos2A)/(cos^2A+sin^2A-2sinAcosA)` `=(cos2A)/(1-sin2A)` RHS. |
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| 452. |
Consider the ellipse `E=x^2/16+y^2/b^2=1` . PQ is a variable chord of the ellipse `E=0` and PQ subtends an angle `90^@` at the origin. And if `1/(OP)^2+1/(OQ)^2=25/144` then `b^2`= |
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Answer» `1/(OP^2)+1/(OQ^2)=1/b^2+1/16` `1/b^2+1/16=25/144` `1/b^2=(25-9)/144=16/144` `b^2=9`. |
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| 453. |
Consider the ellipse `x^2/3 + y^2/1 = 1`. Let P,Q,R,S be four points on the ellipse such that the normal drawn from these points are con current at (2,2) then the centre of the conic on which these 4 points lie is (A) P,Q,R,S lie on the given ellipse (B) `(3,-1)` (C) `(3,1)` (D) `(0,0)` |
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Answer» `N=(a^2x)/x_1-(b^2y)/y_1=a^2-b^2` `=(2a^2)/x_1-(b^2 2/y_1)=a^2-b^2` `=3/x_1-1/y_1=1` `3y_1-x_1-x_1y_1-(1)` `3y_2-x_2=x_2y_2-(2)` `xy+x-3y=0` differentiate with respect to x y+1=0,y=-1 x+(-3)=0,x=3 (3,-1) option b is correct. |
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| 454. |
If `omega` is a cube root of unity then the value of the expression `2(1+w)(1+w^2)+3(2+w)(2+w^2)+.....+(n+1)(n+w)(n+w^2)` |
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Answer» We can write the given expression as , `S = sum_(m=1)^n(m+1)(m+omega)(m+omega^2)` So , general term in this expression can be given as, `T_m = (m+1)(m+omega)(m+omega^2)` `= (m+1)(m^2+(omega+omega^2)m+omega^3)` As, `1+omega+omega^2 = 0=>omega+omega^2 = -1` `:. T_m = (m+1)(m^2-m+1)` (As `omega^3 = 1`) `=> T_m = (m+1)(m(m-1)+1)` `=m(m^2-1)+m+1` `:. T_m=m^3+1` `:. S = sum_(m=1)^n (m^3+1) = sum_(m=1)^n m^3+ sum_(m=1)^n 1` `=> S= ((n(n+1))/2)^2+n = 1/4(n^2)(n+1)^2+n` |
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| 455. |
The maximum number of regions in which 10 circle can divide a plane is |
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Answer» Maximum number of regions in which `n` circles can divide a plane ` = 2^n` `:.` Maximum number of regions in which `10` circles can divide a plane ` = 2^10 = 1024` |
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| 456. |
If `|z + 2i| |
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Answer» `|z+2i|<=1` `z+2i<=e^i` `z<=-2i+e^(itheta)` `|iz+zi-4|=|iz+6-3i-4|` `=|iz+2-3i|` `=|2+e^(i(theta+pi/2) +2-3i|` `=|4-3i+e^(1(theta+pi/2)|` =6 option 2 is correct |
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| 457. |
If `z=re^i theta` then `|e^(iz)|` is equal to: |
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Answer» `z=re^(itheta),iz=ire^(itheta)` `=l8r{costheta+isintheta}` `=r(sintheta+icostheta)` `e^(iz)=e^r*(icostheta-sintheta)` `=e^(-rsintheta)*e^(ircostheta)` `|e^(iz) | =|e^(-rsintheta*e*^(ircostheta)` `=e^(-rsontheta)*|costheta|isintheta|^(rcostheta)` `=e^(-rsintheta)*(1)^rcostheta` `=e^(-rsintheta)`. |
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| 458. |
Let `A={1,2,..., n}`and `B={a , b`}. Then number of subjections from `A`into `B`isnP2 (b) `2^n-2`(c) `2^n-1`(d) nC2A. ` ""^(n)P_(2)`B. `2^(n)-2`C. `2^(n)-1`D. None of these |
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Answer» Correct Answer - D Given that, A = { 1, 2, 3, …, n} and B = {a, b}. We know that, if A and B are two non-empty finite sets containing m and n elements respectively, then the number of surjection from A into B is `""^(n)C_(m)xxm!, if n ge m` 0, if `n lt m` Here, m = 2 ` :. ` Number of surjection from A into B is `""^(n)C_(2)xx2! =(n!)/(2!(n-2)!)xx2!` ` " " =(n(n-1)(n-2)!)/(2xx1(n-2))xx2! =n^(2)-n` |
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| 459. |
If PSQ + PHR be two chords of an ellipse through its foci S and H, then prove that `(PS)/(SQ) + (PH)/(HQ) = 2(1 + e^2)/(1-e^2)`, Where e is the eccentricity of ellipse. |
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Answer» We can draw an ellipse with the given details. Please refer to video for the figure. Equation of ellipse with respect to focus `S` as a pole, `l/r = 1+ecos theta` Here, `r = SP`, `:. l/(SP) = 1+ e cos theta ->(1)` If vectorial angle of `P` is `theta`, then vectorial angle of `Q` will be `pi+theta`. If we write equation with respect to `SQ`, then, `:. l/(SQ) = 1+ ecos(pi+theta)` `=> l/(SQ) = 1-ecos theta->(2)` Adding (1) and (2), `l/(SP)+l/(SQ) = 2` `=> 1/(SP)+1/(SQ) = 2/l` `=>1+(SP)/(SQ) = (2(SP))/l->(3)` Similarly, for chord `PHR`, we can have, `1+(PH)/(HR) = (2(PH))/l->(4)` Adding (3) and (4), `2+((SP)/(SQ)+(PH)/(HR)) = 2/l(SP+PH)->(5)` As, `S` and `H` are foci of given ellipse.>br>`:. SP+PH =` length of major axis `= 2a` Also, we know, `2l = 2b^2/a => l =b^2/a` Putting these values in equation (5), `2+((SP)/(SQ)+(PH)/(HR)) = 4a^2/b^2` Als, we know, `b^2 = a^2(1-e^2)` So, our equation beecomes, `((SP)/(SQ)+(PH)/(HR)) = 4a^2/(a^2(1-e^2))-2` `=>((SP)/(SQ)+(PH)/(HR)) = (2+2e^2)/(1-e^2)` `=>((SP)/(SQ)+(PH)/(HR)) = (2(1+e^2))/(1-e^2)` |
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| 460. |
The tamer of wild animals has to bring one by one 5 lions & 4 tigers to the circus arena. The number of ways this can be done if no two tigers immediately follow each other |
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Answer» There are `6` spaces between `5` lions where `4` tiger can be arranged. So, there are `C(6,4)` ways of tigers arranging between lions so that two tigers do not follow each other immediately. Now, `4` tigers can be arranged in `4!` ways and `5` lions can be arranged in `5!` ways. `:.` Required number of ways ` = C(6,4)*4!*5! = 15**24**120 = 43200` |
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| 461. |
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation. |
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Answer» Set `A` is the set of all books in the library of a college. `R = {x, y): x` and `y` have the same number of pages}. Now, R is reflexive since `(x, x) in R` as `x` and `x` has the same number of pages. Let `(x, y) in R => x` and `y` have the same number of pages. `=> y` and `x` have the same number of pages. `=> (y, x) in R`. `:. R` is symmetric. Now, let `(x, y) in R` and `(y, z) in R`. `=> x` and `y` and have the same number of pages and `y` and `z` have the same number of pages. It means `x` and `z` have the same number of pages. => `(x, z) in R`. `:. R` is transitive. As `R` is reflexive, summetric and transitive, `R` is an equivalence relation. |
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| 462. |
Let `f`be a function satisfying of `xdot`Then `f(x y)=(f(x))/y`for all positive real numbers `xa n dydot`If `f(30)=20 ,`then find the value of `f(40)dot` |
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Answer» Given `f(xy)=(f(x))/(y)` or `f(y)=(f(1))/(y) " "("Putting " x=1) ` or `f(30)=(f(1))/(30)` or `f(1)=30 xx f(30)=30xx20=600` ` :. f(40)=(f(1))/(40)` `=(600)/(40)=15` |
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| 463. |
If `[x]` stands for the greatest integer function, then `[1/2+1/1000]+[1/2+2/1000]+...+[1/2+999/1000]`= |
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Answer» `[0.5+1/1000]+[0.5+2/1000]+...+[0.5+999/1000]` There are total of 999 terms `[0.5+1/1000]+[0.5+2/1000]+...+[0.5+4999/1000]` There are total of 500 terms. Option 3 is correct. |
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| 464. |
A function `f`has domain `[-1,2]`and rang `[0,1]`. Find the domain and range of the function `g`defined by `g(x)=1-f(x+1)dot` |
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Answer» Correct Answer - Domain: [-2, 1]; Range: [0, 1] g(x) is defined if `f(x+1)` is defined . Hence, the domain of g is all x such that `(x+1) in [0, 2],` i.e., `-2 le x le 1`. Also, `f(x+1) in [0,1]` ` :. -f(x+1) in [-1,0]` ` :. 1-f(x+1) in [0,1]` Therefore, range of g(x) is [0, 1]. |
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| 465. |
The domain and range of the real function f defined by `f(x)=(4-x)/(x-4)` isA. Domain =R , Range ={-1,2}B. Domain =R -{1}, Range RC. Domain =R -{4}, Range ={-1}D. Domain =R -{-4}, Range ={-1,1} |
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Answer» Correct Answer - C We have, `f(x) =(4-x)/(x-4)` f(x) is defined, if `x-4 ne 0i.e.,xne4` `:.` Domain of f=R-{4} Let f(x)=y `:. Y=(4-x)/(x-4)rArrxy-4y=4-x` `rArr xy+x=4+4yrArrx(y+1)=4(1+y)` `:. x=(4(1+y))/(y+1)` x assumes real values, if, `y+1 ne0.i.e.,y=1` `:.` Range of f=R-{-1}` |
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| 466. |
Find the domain and the range of the real function f defined by `f(x)=sqrt((x-1))`.A. Domain`=(1,oo),` Range `= (0,oo)`B. Domain`=[1,oo),` Range `= (0,oo)`C. Domain`=(1,oo),` Range `= [0,oo)`D. Domain`=[1,oo),` Range `= [0,oo)` |
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Answer» Correct Answer - D We have, `f(x)=sqrt(x-1)` f(x) is defined if, `x-1ge0`. `rArr xge1` `:.` Domain of `f=[1,oo)` Let f(x)=y `:. y=sqrt(x-1)` `rArr y^(2)=x-1` `:. x=y^(2)+1` x assumes real values for `y in R.` but `yge0` `:.` Range of `f=[0,oo)` |
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| 467. |
Find the domain and the range of the real function/defined by `f(x)=|x-1|` |
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Answer» `f(x) = |x-1|` Domain = `R` Range = `{ f(x)}` {`y: y>=0` y is R} answer |
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| 468. |
The domain of definition of f(x) =`(log_2(x+3))/(x^2+3x+2)` |
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Answer» `f(x) = (log_2(x+3))/(x^2+3x+2)` For `f(x)` to be defined: (i) `(x+3) gt 0 =>x gt -3` (ii) `x^2+3x+2 !=0 =>(x+1)(x+2) ! = 0 => x != -1 and x !=-2` `:. ` Domain of `f(x)` will be `(-3,oo)-{-1,-2}.` |
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| 469. |
Let `g(x)=int_0^x f(t).dt`,where f is such that `1/2 |
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Answer» `g(x)=int_0^2f(x)dt=int_0^1f(x)dt+int_1^2f(x)dt` `Ain(0,1)` `int_0^1 1/2dx<=int_0^1f(t)dt<=int_0^1dt` `1/2<=int_0^1f(t)dt<=1-(1)` `Ain(1,2)` `int_1^2dt<=int_1^2f(t)dt<=int_1^2 1/2dx` `0<=int_1^2f(t)dt<=1/2-(2)` adding equation 1 and 2 `1/2<=g(2)<=3/2` option 2 is correct. |
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| 470. |
Find the greatest integer x satisfying the inequality `(2x+1)/3-(3x-1)/2>1` |
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Answer» `(2x+1)/3 - (3x-1)/2 gt 1` `=>4x+2-9x+3 gt 6` `=>-5x+5-6 gt 0` `=>-5x-1 gt 0` `=>-5x gt 1` `=>x lt -1/5` So, `x lt -1/5`, will satisfy the given condition. |
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| 471. |
49. If `[x^2-2x + a] = 0 `has no solution thenA. `-oo lt a lt 1`B. `2 le a lt oo`C. `1 lt a lt 2`D. ` a in R` |
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Answer» Correct Answer - B Case (i) `x^(2)-2x+a lt 0 AA x in R` which not possible `x^(2)-2x +a ge 1` `x^(2)-2x +a-1 ge 0` Disc `le 0` `4-4(a-1) le 0` `2-a le 0` `a ge 2` ` a in [2, oo)` |
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| 472. |
If `[x]`and `{x}`represetnt the integral and fractional parts of `x ,`respectively, then the value of `sum_(r=1)^(2000)({x+r})/(2000)`is`x`(b) `[x]`(c) `{x}`(d) `x+2001`A. xB. [x]C. {x}D. x+2001 |
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Answer» Correct Answer - C `sum_(r=1)^(2000)({x+r})/(2000)=sum_(r=1)^(2000)({x})/(2000)=2000({x})/(2000)={x}.` |
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| 473. |
if the line 3x-4y-k=0 touches the circle `x^2+y^2-4x-8y-5=0` at (a,b),find (k+a+b)/5 is |
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Answer» Comparing the given equation of the circle with the general equation of the circle, `x^2+y^2+2gx+2fy+c = 0` We get, `g = -2 ` and `f = -4`. Any tangent that is touching the circle at `(x_1,y_1)` has the equation, `x x_1+yy_1+g(x+x_1)+f(y+y_1)+c = 0`Now, as the given line is touching the circle at `(a,b)` `:. ax+by-2(x+a)-4(y+b) -5 = 0` `=>(a-2)x+(b-4)y-(2a+4b+5) = 0` Comparing it with given equation, `3x-4y -k = 0`, `a-2 = 3, b-4 = -4 and k = 2a+4b+5` `a = 5, b = 0` `:. k = 10+0+5 = 15` So, `(k+a+b)/5 = (15+5+0)/5 = 20/5 = 4` |
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| 474. |
The number of points on the real line where the function f(x)-log- Ix-3) is not defined is `f(x) = log_(|x^2-1|)|x-3|` is not defined is |
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Answer» Correct Answer - 6 `log_(|x^(2)-1|)|x-3|` is defined when `x-3 ne 0` and `|x^(2) -1| gt 0 and |x^(2)-1| ne 1.` `implies x ne 3x^(2) -1 ne 0"i.e.," x ne 1 or -1` and `|x^(2)-1| ne 1` `implies x ne -sqrt(2) or sqrt(2) and x ne 0` ` :. ` the points at which the function is not defined are `x=0,1, -1sqrt(2), 3.` |
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| 475. |
Let `f(x)=3x^2-7x+c ,`where `c`is a variable coefficient and `x >7/6`. Then the value of `[c]`such that `f(x)`touches `f^(-1)(x)`is (where [.] represents greatest integer function)_________ |
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Answer» Correct Answer - 5 `f(x) and f^(-1)(x)` can only intersect on the line `y=x` and therefore, `y=x` must be tangent at the common point of tangency. Hence, ` 3x^(2)-7x+c=x` ` or 3x^(2) -8x+c=0 " (1)" ` This equation must have equal roots. Therefore, `64-12c =0` `or c=(64)/(12)=(16)/(3)` |
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| 476. |
Let `f`be a real-valued invertible function such that `f((-32 x)/(x-2))=5x-2, x!=2.`Then value of `f^(-1)(13)`is________ |
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Answer» Correct Answer - 3 We have `f((2x-3)/(x-2))=5x-2` `or f^(-1)(5x-2)=(2x-3)/(x-2)` Let `5x-2=13." Then " x=3`. Here, `f^(-1)(13)=(2(3)-3)/(3-2)=3`. |
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| 477. |
If the relation `f(x)={(2x-3",",x le 2),(x^(3)-a",",x ge2):}` is a function, then find the value of a. |
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Answer» Correct Answer - 7 Given relation is function if `2(2)-3=2^(3)-a" or " a=7` |
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| 478. |
A real-valued functin `f(x)` satisfies the functional equation `f(x-y)=f(x)f(y)-f(a-x)f(a+y),` where a given constant and `f(0)=1.` Then prove that `f(x)` is symmetrical about point (a, 0). |
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Answer» We have `f(x-y)=f(x)f(y)-f(a-x)f(a+y)` Putting `x=a " and " y=x-a,` we get `f(a-(x-a))=f(a)f(x-a)-f(0)f(x) " (1)" ` Putting `x=0,y=0, ` we get `f(0)=f(0)(f(0))-f(a) f(a)` or ` f(0)=(f(0))^(2)-(f(a))^(2)` or `1=(1)^(2)-(f(a))^(2)` or `f(a)=0` ` :. f(2a-x)= -f(x) " (from (1))" ` Replacing x by `a-x,` we get `f(2a-(a-x))= -f(a-x)` or `f(a+x)= -f(a-x)` Therefore, `f(x)` is symmetrical about point (a, 0). |
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| 479. |
The number of integers in the domain of function, satisfying `f(x)+f(x^(-1))=(x^3+1)/x ,i s_____` |
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Answer» Correct Answer - 2 `f(x)+f((1)/(x))=x^(2)+(1)/(x)` Replacing x by `(1)/(x),` we get `f((1)/(x))+f(x)=(1)/(x^(2))+x` `or (1)/(x^(2))+x=x^(2)+(1)/(x)` `or x-(1)/(x)=x^(2)-(1)/(x^(2))` `or (x-(1)/(x))=(x-(1)/(x))(x+(1)/(x))` ` or (x-(1)/(x))(x+(1)/(x)-1)=0` `x=(1)/(x), x+(1)/(x)=1`(rejected) Hence, `x=1 or -1.` |
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| 480. |
If a polynomial function `f(x)`satisfies `f(f(f(x))=8x+21 ,`where `pa n dq`are real numbers, then `p+q`is equal to _______ |
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Answer» Correct Answer - 5 `f(x)` must be linear function. Therefore, `f(x)=px+q` ` :. f(f(x))=pf(x)+q=p(px+q)+q=p^(2)x+pq+q` ` :. f(f(f(x)))=p(p^(2)x+pq+q)+q` ` =p^(3)x+p^(2)q+pq+q` `=8x+21 ` (Given) ` :. P^(3)=8 or p=2` and `p^(2)q +pq+q=21 or q=3` ` :. p+q=5` |
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| 481. |
Which of the following functions is/are identical to `|x-2|` ? (a) `f(x)=sqrt(x^(2)-4x+4) " (b) " g(x)=|x|-|2|` (c ) `h(x)=(|x-2|^(2))/(|x-2|) " (d) "t(x)=|(x^(2)-x-2)/(x+1)|` |
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Answer» Correct Answer - a `f(x)=sqrt(x^(2)-4x+4)=sqrt((x-2)^(2))=|x-2|` `g(x)=|x|-|2|=|x|-2=={(-x-2",",x lt 2),(x-2",",x ge2):}`. Thus `g(x)` is not same as `|x-2|` `h(x)=(|h-2|^(2))/(|x-2|)=|x-2|,x ne 2.` This is not same as `|x-2|` as `h(x)` has domain `R-{2}` `t(x)=|(x^(2)-x-2)/(x+1)|=|((x-2)(x+1))/(x+1)|=|x-2|, x ne -1`. Thus `t(x)` is not same as `|x-2|` as `t(x)` has domain `R-{-1}.` |
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| 482. |
Which of the following function from Z to itself are bijections?`f(x)=x^3`(b) `f(x)=x+2``f(x)=2x+1`(d) `f(x)=x^2+x` |
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Answer» Correct Answer - b (b) Clearly, `f(x)` must be `x+2` as for this function, each image has its preimage and each image has one and only one preimage. |
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| 483. |
For `f(x)=(kcosx)/(pi-2x)`, if `x!=pi/2`, `3`, if `x=pi/2`, then find the value of `k` so that `f` is continous at `x=pi/2` |
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Answer» `f(x)=(kcosx)/(pi-2x)` if `x!=pi/2` `=k/2(cosx)/(pi/2-x)=k/2(sin(pi/2-x)/(pi/2-n))` `=lim_(x->pi/2)f(x)=lim_(x->pi/2)k/2sin(pi/2-n)/(pi/2-x)` `=k/2 lim_(x->pi/2)(sin(pi/2-x))/(pi/2-x)` `=k/2*1=k/2` if f is continous at x=`pi/2` `lim_(x->pi/2)f(x)=f(pi/2)` `k/2-3,k=6`. |
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| 484. |
`f:R->R` defined by `f(x) = x^2+5`A. `(x+5)^((1)/(3))`B. `(x-5)^((1)/(3))`C. `(5-x)^((1)/(3))`D. `5-x` |
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Answer» Correct Answer - B Given that, `f(x)= x^(3) +5` Let ` " " y=x^(3)+5 implies x^(3)=y-5` `x=(y-5)^((1)/(3))implies f(x)^(-1)=(x-5)^((1)/(3))` |
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| 485. |
`A={1,2,3,4,5},S={(x,y):x in A,y, in A}`, then find the ordered which satisfy the conditions given below. (i) `x+y=5` (ii) `x+ylt5` (iii) `x+y=gt8` |
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Answer» We have `A={1,2,3,4,5},S={(x,y): x in A,y, in A}`, (i) The set of ordered pairs stistying x+y=5 is `{(1,4),(2,3),(3,2),(4,1)}` (ii) The set of ordered paris satisfying `x+ylt "is" (1,1),(1,2),(1,3),(2,1), (2,2),(3,1)` (iii) The set of ordered pairs satisfying `x+y gt 8 "is" {(4,5),(5,4),(5,5)}` |
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| 486. |
If `f:A->B, g:B->C` are bijective functions show that `gof:A->C` is also a bijective function.A. `f^(-1)og^(-1)`B. `fog`C. `g^(-1)of^(-1)`D. `gof` |
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Answer» Correct Answer - A Given that, `f:A to B` and `g :B to C` be the bijective functions. `(gof)^(-1)=f^(-1)og^(-1)` |
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| 487. |
If `R={(x,y): x,y in W, x ^(2)+y^(2)=25}`, then find the domain and range or R. |
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Answer» We have, `R={(x,y): x,y in W, x ^(2)+y^(2)=25}` `={(0,5),(3,4),(4,3),(5,0)}` Domain of R= Set of first element of ordered pairs in R `={0,3,4,5}` Range of `" "` R= Set of second element of ordered pairs in R `={5,4,3,0}` |
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| 488. |
Let f(x) and g(x) be bijective functions where `f:{a,b,c,d} to {1,2,3,4} " and " g :{3,4,5,6} to {w,x,y,z},` respectively. Then, find the number of elements in the range set of `g(f(x)).` |
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Answer» The range of `f(x)` for which `g(f(x))` is defined is `{3,4}`. Hence, the domain of `g{f(x)}` has two elements. Therefore, the range of `g(f(x))` also has two elements. |
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| 489. |
In a `Delta ABC` the value of `/_A` is given by 5 CosA - 3 = 0 then the equation whose roots are sinA and tanA is : |
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Answer» `5cosA-3=0` `cosA=3/5, sinA=sqrt(1-cos^2A)=4/5` `tanA=4/3` Sum of roots=`4/3+4/5=32/15` Product of roots=`4/3*4/5=16/15` `x^2-(32/15)x+16/15=0` `15x^2-32x+16=0` option d is correct. |
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| 490. |
If `R_(1)={(x,y)|y=2x+7,` where `x in R and -5lexle5}` is a relation. Then, find the domain and range of `R_(1)` |
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Answer» We have, `R_(1)={(x,y)|y=2x+7,` where `x in R and -5lexle5}` Domain of `R_(1)={-5le x le, x in R}` `=[-5,5]` `:. y=2x +7` When `x=-5, then y=2(-5)+7=-3` When x =5, then `y=2(5) + 7 =17` `:.` Range of `R_(1)={-3 leyle 17, y in R}` =[-3,17] |
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| 491. |
If `R_(3)={(x,|x|),|x` is a real number } is a releation, then find domain and range of `R_(3)` |
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Answer» We have, `R_(3)={(x,|x|)x` is real number} Clearly, domain of `R_(3)=R` Since, image of any real number `R_(3)` is positive real number of zero, `:.` Range of `R_(3)=R^(+)uu{0}or (0,oo)` |
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| 492. |
Let `f:{2,3,4,5} to {3,4,5,9}and g:{3,4,5,9} to {7,11,15}` be functions defined as `f(2)=3,f(3)=4, f(4)=f(5)=5, g(3)=g(4)=7, " and " g(5)=g(9)=11. " Find " gof.` |
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Answer» We have `gof(2)=g(f(2))=g(3)=7,` `gof(3)=g(f(3))=g(4)=7,` `gof(4)=g(f(4))=g(5)=11,` and `gof(5)=g(5)=11.` |
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| 493. |
Find the fundamental period of `f(x)=cosxcos2xcos3xdot` |
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Answer» Correct Answer - `pi` `f(x)=cosx cos2x cos 3x` `cosx,cos2x,` and `cos 3x` are periodic with period `2pi,pi,` and `(2pi)/(3)` L.C.M. of `(pi,2pi, (2pi)/(3))=2pi` But `f(x+pi)=cos(x+pi) cos(2x+2pi)cos(3x+3pi)` `=(-cosx)(cos 2x)(-cos 3x)` `=cosx cos 2x cos 3x` `=f(x)` Hence, period is `pi`. |
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| 494. |
The difference between the maximum and minimum value of the function `f(x)=3sin^4x-cos^6x` is : |
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Answer» `f(x) = 3sin^4x-cos^6x` `= 3sin^4x-(1-sin^2x)^3` `= 3sin^4x-(1-sin^6x-3sin^2x(1-sin^2x))` `= 3sin^4x-(1-sin^6x-3sin^2x+3sin^4x)` `=sin^ 6x + 3sin^2x -1` `:. f(x) = sin^2x(sin^4x+3) - 1` Now, `f(x)` will be maximum, when `sinx = 1` and will be minimum when `sin x = 0`. `:. f(x)_max = 1(1+3)-1 = 3` `f(x)_min = 0(0+3)-1 = -1` `:. f(x)_max-f_min = 3-(-1) = 4`, which is the required difference. |
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| 495. |
Find the values of x for which the following function is defined: `f(x)=sqrt((1)/(|x-2|-(x-2)))` |
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Answer» Correct Answer - `(-oo,2)` `|x-2|={(x-2",",x ge2),(2-x",",x lt 2):}` `implies|x-2|-(x-2)={(0",",x ge2),(4-2x",",x lt 2):}` Therefore, given expression is defined for `(-oo,2)` |
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| 496. |
If `R_2={(x,y)| x and y` are integers and `x^2 +y^2=64`} is a relation. Then find `R_2`. |
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Answer» We have `R_(2)={(x,y)x and y` are integers and `x^(2)+y^(2)=64}` Since, 64 is the sum of squares of `0and +-8` When x=0 then `y^(2)=64rArry=+-8` x =8, then `y^(2)=64-8^(2)rArr64-64=0` x=-8, then `y^(2)=64-(-8)^(2)=64-64=0` `:. R_(2)={(0,8),(0,-8),(8,0),(-8,0)}` |
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| 497. |
If `f(x)=lambda|sinx|+lambda^2|cosx|+g(lambda)` has a period = `pi/2` then find the value of `lambda` |
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Answer» Correct Answer - 1 Since the period of `|sin x|+|cosx|` is `pi//2`, it is possible when `lambda=1.` |
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| 498. |
Find the period of the following function (i) `f(x) =|sinx|+|cosx|` (ii) `f(x)=cos(cosx)+cos(sinx)` (iii) `f(x)= (|sinx+cosx|)/(|sinx|+|cosx|)` |
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Answer» (i) `f(x) =|sinx|+|cosx|` Period of both `|sinx|` and `|cosx|` is `pi`. So, we can consider period of f(x) to be `pi`. But `f((pi)/(2)+x)=|sin((pi)/(2)+x)|+|cos((pi)/(2)+x)|` `=|cosx|+|-sinx|` `=|cosx|+|sinx|` `=f(x)` So, period of f(x) is `Pi//2`. (ii) `f(x)=cos(cosx)+cos(sinx)` Period of `cos(cosx) " is " pi.` Period of `cos(sinx) " is " pi.` So, we can consider period of f(x) to be `pi`. But `f((pi)/(2)+x)=cos(cos((pi)/(2)+x))+cos(sin((pi)/(2)+x))` `=cos(-sinx)+cos(cosx)` `=cos(sinx)+cos(cosx)` So, period of f(x) is `pi//2` (iii) `f(x)= (|sinx+cosx|)/(|sinx|+|cosx|)` `|sinx +cosx|=sqrt(2)|sin(x+(pi)/(4))|`, which is has period `pi`. Period of `|sinx|+|cosx| " is " pi//2`. Therefore, period of f(x) is L.C.M. of `(pi,pi//2)=pi` |
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| 499. |
Find the period `f(x)=sinx+{x},`where {x} is the fractional part of `xdot` |
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Answer» Here, `sinx` is periodic with period `2pi` and [x} is periodic with 1. The LCM of `2pi` (irrational) and 1 (rational ) does not exist. Thus, f(x) is not periodic. |
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| 500. |
The number of value(s) of x satisfying the equation `(2011)^x+ (2012)^x + (2013)^x -(2014)^x=0` is/are : |
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Answer» `(2011)^x+(2012)^x+(2013)^x - (2014)^x = 0` `=> (2011)^x+(2012)^x+(2013)^x = (2014)^x` If we draw the graph of `2011^x,2012^x and 2013^x`, we will find that their sum will match exactly one time with the `2014^x`. Please refer to video to see the graph. Thus, there will be exactly one solution fo the given equation. |
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