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| 151. |
If `f: Rvec[0,oo)i safu n c t ions u c ht h a tf(x-1)+f(x+1)=sqrt(3)f(x),`then prove that `f(x)`is periodic and find its period. |
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Answer» `f(x-1) +f(x+1)=sqrt(3) f(x) " (1)" ` Putting `x+2` for x in relation (1), we get `f(x+1) +f(x+3)=sqrt(3) f(x+2) " (1)" ` From (1) and (2), we get `f(x-1) +2f(x+1)+f(x+3) =sqrt(3)(f(x)+f(x+2))` `=sqrt(3)(sqrt(3)f(x+1))` `=3f(x+1)` `or f(x-1)+f(x+3)=f(x+1) " (3)" ` Putting `x+2` for x in (3), we get `f(x+1)+f(x+5)=f(x+3) " (4)" ` Adding (3) and (4), we get `f(x-1)= -f(x+5).` Now, put `x+1` for x. Then `f(x)= -f(x+6) " (5)" ` Put `x+6` in place of x in (5). Then `f(x+6)= -f(x+12).` Therefore, from (5) again, `f(x)= -[-f(x+12)]=f(x+12)`. Hence, the period of `f(x)` is 12. |
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| 152. |
If `f(x+y)=f(x)+f(y)-x y-1AAx , y in Ra n df(1)=1,`then the number of solution of `f(n)=n , n in N ,`is0 (b)1 (c) 2(d) more than 2 |
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Answer» Correct Answer - B Given `f(x+y)=f(x)+f(y)=xy=1 AA x, y in R` `f(1)=1,f(2)=f(1+1)=f(1)+f(1)-1-1=0` `f(3)=f(2+1)=f(2)+f(1)-2xx1 -1= -2` `f(n+1) = f(n)+f(1)-n-1=f(n)-n lt f(n)` Thus, `f(1) gt f(2) gt f(3) gt …, and f(1) =1.` Therefore, `f(1)=1 and f(n) gt 1, " for " n gt 1.` Hence, ` f(n)=n, n in N,` has only one solution `n=1.` |
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| 153. |
Let `f(x)=(x+1)^2-1, xgeq-1.`Then the set `{x :f(x)=f^(-1)(x)}`is`{0,1,(-3+isqrt(3))/2,(-3-isqrt(3))/2}`(b) `{0,1,-1``{0,1,1}`(d) `e m p t y`A. `{0,-1,(-3+isqrt(3))/(2),(-3-isqrt(3))/(2)}`B. `{0,1,-1}`C. `{0, -1}`D. empty |
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Answer» Correct Answer - C Since `f(x) = (x+1)^(2)-1` is continuous function, solution of `f(x)=f^(-1)(x)` lies on the line `y=x`. Therefore, `f(x)=f^(-1)(x)=x` ` or (x+1)^(2)-1=x` ` or x^(2)+x=0` i.e., ` x=0 or -1` Therefore, the required set is `{0,-1}.` |
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| 154. |
If `f(x+y+1)={sqrt(f(x))+sqrt(f(y))}^2`and `f(0)=1AAx ,y in R ,d e t e r m in ef(n),n in Ndot` |
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Answer» Correct Answer - `f(n)=(n+1)^(2)` Given `f(x+y+1)=(sqrt(f(x))+sqrt(f(y)))^(2)` Putting `x=y=0,` we get `f(1)=(sqrt(f(0))+sqrt(f(0)))^(2)=(1+1)^(2)=2^(2)` Again putting `x=0,y=1` we get `f(2)=(sqrt(f(0))+sqrt(f(1)))^(2)=(1+2)^(2)=3^(2)` and for `x=1,y=1,` we get `f(3)=(sqrt(f(1))+sqrt(f(1)))^(2)=(2+2)^(2)=4^(2)` Hence, `f(n)=(n+1)^(2)`. |
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| 155. |
Let `f:[-1,oo]vec[-1,)`is given by `f(x)=(x+1)^2-1,xgeq-1.`Show that `f`is invertible. Also, find the set `S={x :f(x)=f^(-1)(x)}dot`A. `{0,-1,(-3pm sqrt(3))/(2)}`B. {0,1,-1}C. {0,-1}D. {} |
| Answer» Correct Answer - C | |
| 156. |
Determine all functions `f: Rvecs u c ht h a tf(x-f(y))=f(f(y))+xf(y)+f(x)-1AAx , ygeq in Rdot` |
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Answer» Correct Answer - `f(x)=1-(x^(2))/(2)` Given `f(x-f(y))=f(f(y))+xf(y)+f(x)-1 " (1) " ` Putting `x=f(y)=0,` we get `f(0)=f(0)+0+f(0)-1,` i.e., `f(0)=1 " (2)" ` Again, putting `x=f(y)= lambda` in (1), we get `f(0)=f(lambda)+lambda^(2)+f(lambda)-1` or `1=2f(lambda)+lambda^(2)-1 " [From (2)]" ` `f(lambda)=(2-lambda^(2))/(2)=1-(lambda^(2))/(2)` Hence, `f(x)=1-(x^(2))/(2).` |
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| 157. |
Find the range of `f(x)=1/(1-3sqrt(1-sin^2x))` |
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Answer» Correct Answer - `(-oo,1//2] cup [1,oo)` `f(x)=(1)/(1-3sqrt(1-sin^(2)x))=(1)/(1-3sqrt(cos^(2)x))` `=(1)/(1-3|cosx|)` Now, `-3|cosx| in [-3,0]` or `1-3|cosx| in[-2,1]` or `(1)/(1-3|cosx|) in (-oo,-1//2] cup [1,oo)` |
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| 158. |
Find the range of `f(x)=sec(pi/4cos^2x),`where `-oo |
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Answer» Correct Answer - `[1,sqrt(2)]` `f(x)="sec"((pi)/(4)"cos"^(2)x)` We know that `0 le cos^(x) le 1,` i.e., `0 le (pi)/(4)cos^(2)x le (pi)/(4)` For the above value of `theta =(pi)/(pi)/(4)cos^(2)x,sec x` is an increasing function. ` :. f(x) in [f(0),f((pi)/(4))] " or " f(x) in [1,sqrt(2)]` |
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| 159. |
If `f : [2,oo) to R` be the function defined by `f(x)=x^(2)-4x+5,` then the range of f isA. RB. `[1,oo)`C. `[4,oo)`D. `[5,oo)` |
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Answer» Correct Answer - B Given that, `f(x)= x^(2) -4x+5` Let ` y= x^(2) -4x+5` ` implies y= x^(2) -4x+4+1=(x-2)^(2)+1 ` ` implies (x-2)^(2)=y-1impliesx-2=sqrt(y-1)` `implies x=2+sqrt(y-1)` ` :. y-1 ge 0, y ge 1` Range `=[1, oo)` |
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| 160. |
Let `f(x)=sqrt(x)`and `g(x) = x`be two functions defined over the set of nonnegative real numbers. Find `(f + g) (x)`, `(f g) (x)`, `(fg) (x)`and `(f/g)(x)`. |
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Answer» We have `f(x )=sqrt(x) and g(x) =x` be two function defined in the domain `R^(+)uu{0}` (i) `(f+g)(x)=f(x)+g(x)=sqrt(x)+x " "` (ii) `(f-g)(x)=f(x)-g(x)=sqrt(x)-x` (iii) `(fg)(x)=f(x)*g(x)=sqrt(x)*x=x^(2/3) " "` (iv) `((f)/(g))(x)=(f(x))/(g(x))=(sqrt(x))/(x)=(1)/(sqrt(x))` |
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| 161. |
Let`f(x)={x^2-4x+3,x |
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Answer» Correct Answer - A::B::C `(f+g)(3.5)=f(3.5)+g(3.5)=(-0.5)+0.5=0` `f(g(3))=f(0)=3` `(fg)(2)=f(2)g(2)=(-1)xx(-1)=1` `(f-g)(4)=f(4)-g(4)=0-26 = -26` |
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| 162. |
The range of the following function is`f(x)=sqrt((1-cosx)sqrt((1-cosx)sqrt((1-cosx)sqrt(oo))))`(0,1) (b) `(0,1/2)`(c) `(0,2)`(d) `non eoft h e s e`A. [0, 1]B. [0, 1/2]C. [0, 2)D. None of these |
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Answer» Correct Answer - C Given `f(x)=sqrt((1-cosx)sqrt((1-cosx)sqrt((1-cosx)sqrt(...oo))))` ` =(1-cosx)^((1)/(2))(1-cosx)^((1)/(4))(1-cosx)^((1)/(8)) ...oo` `=(1-cosx)^((1)/(2)+(1)/(4)+(1)/(8)+ ...oo)` `=(1-cosx)^((1//2)/(1-(1//2)))` `=1-cosx` Thus, the range of `f(x)` is `[0,2).` |
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| 163. |
Let `f(x)=x^2`and `g(x) = 2x + 1`be two real functions. find `(f +g)(x)`, `(f-g)(x)`, `(fg)(x)`, `(f/g)(x)`. |
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Answer» Here, `f(x) = x^2` and `g(x) = 2x+1` `(f+g)(x) = f(x)+g(x) = x^2+2x+1` `(f-g)(x) = f(x)-g(x) = x^2-2x-1` `(fg)(x) = f(x)g(x) = x^2(2x+1) = 2x^3+x^2` `(f/g)(x) = f(x)/g(x) = x^2/(2x+1)` |
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| 164. |
If `fa n dg`are two distinct linear functions defined on `R`such that they map `{-1,1]`onto `[0,2]`and `h : R-{-1,0,1}vecR`defined by `h(x)=(f(x))/(g(x)),`then show that `|h(h(x))+h(h(1/x))|> 2.` |
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Answer» Let two linear functions be `f(x)=ax +b" and " g(x) =cx +d.` They map `[-1,1] to [0,2]` and mapping is onto. Therefore, `f(-1)=0 and f(1)=2` `and g(-1)=2 and g(1)=0,` i.e., `-a+b=0 and a+b=2 " (1) " ` `and -c+d=2 and c+d=0 " (2) " ` `or a=b=1 and c= -1, d=1` `or f(x)=x+1 and g(x)=-x+1` `or h(x)=(x+1)/(1-x) or h(h(x)) =((x+1)/(1-x)+1)/((x+1)/(1-x)-1)=(1)/(x)` ` or h(h(1//x))=x` `or |h(h(x))+h(h(1//x))|=|x+1//x| gt 2` |
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| 165. |
If `fa n dg`are one-one functions, then`f+g`is one one`fg`is one one`fog`is one one`non eoft h e s e`A. `f+g` is one-oneB. `fg` is one-oneC. `fog` is one-oneD. None of these |
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Answer» Correct Answer - C (a) `f(x)=sinx and g(x)=cosx, x in [0, pi//2]` Here, both f(x) and g(x) are one-one functions. But `h(x)=f(x)+g(x)=sinx+cosx` is many-one as ` h(0)=h(pi//2)=1.` (b) `h(x)=f(x) g(x)=sinx cosx=(sin 2x)/(2)` is many-one, as `h(0)=h(pi//2)=0.` (c) It is a fundamental property. |
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| 166. |
If `alpha` and `beta` are the zeros of the quadratic polynomial `p(t)=t^2-5t-1` , find the value of `alpha^2/beta^2 + beta^2/alpha^2 +2(alpha/beta+beta/alpha)-alphabeta`. |
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Answer» `alpha and beta` are the roots of polynomial `t^2-5t-1`. `:. alpha+beta = -(-5)/1 = 5` `alpha beta = -1/1 = -1` Now,`alpha^2/beta^2+beta^2/alpha^2+2(alpha/beta+beta/alpha)-alphabeta = (alpha^4+beta^4)/(alpha^2beta^2)+2(alpha^2+beta^2)/(alphabeta)-alphabeta` `=((alpha^2+beta^2)^2-2alpha^2beta^2)/(alphabeta)^2+(2((alpha+beta)^2-2alphabeta))/(alphabeta)-alphabeta` `=(((alpha+beta)^2-2alphabeta)^2-2(alphabeta)^2)/(alphabeta)^2+(2((alpha+beta)^2-2alphabeta))/(alphabeta)-alphabeta` `=((25+2)^2-2(1))/1 +(2(25+2))/(-1)+1` `=727-54+1 = 674` |
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| 167. |
The rate at which the population of a city varies as the population at that time in period of 20 year population increased from 10 lakhs to 14 lakhs. Find the population in another 20 years. |
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Answer» Let rate of increase is `R%`. Then, `1400000 = 1000000(1+R/100)^20` `=>(1+R/100)^20 = 14/10` So, after another `20` years population will be, `=1400000(1+R/100)^20` `= 1400000**14/10 = 1960000` So, population after another `20` years will be `19.6 Lakhs`. |
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| 168. |
Prove that `**: R xx R to R ` defined as `a ** b =a+2ab` is not commutative . |
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Answer» Let `a,b in R ` `:. A**b =a+2b and b**a =b+2a` `:. a**b ne ** a " "`(If a =2 and b=5 ) Therefore, `**` is not commutative . Hence Proved. |
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| 169. |
Show that subtraction and division are not binaryoperations on `N`. |
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Answer» Let a, b `in` N For subtraction , `a,b in N implies a-b notin N, ` for a=3 and b=5 `:.` Subtraction is not a binary operations on N. For division , `a,b in N implies(a)/(b) notin N`, for a=2 and b=3 `:.` Division is not a binary operation on N. Hence Proved. |
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| 170. |
Number of binary operations on the set {a, b} are(A) 10 (B) 16 (C) 20 (D) 8A. 10B. 16C. 20D. 8 |
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Answer» Number of binary operations in the set {a,b} `=2^4 =16` |
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| 171. |
The range of real number `alpha` for which the equation `z + alpha|z-1| + 2i = 0` has a solution is : |
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Answer» `Z+alpha|Z-1|+2i=0` `Z+2i=-alpha|Z-1|` `alpha=-(Z+2i)/|Z-1|` `alpha` is real `|z-1|` is real `Z=x+iy` `-iy*2i=0` `y=-2` `alpha=-x/|(x-1)-2i|` `|(x-1)-2i|^2=(-x/alpha)^2` `(x-1)^2+4=x^2/alpha^2` `x^2-2x+4=x^2/alpha^2` `x^2(1-1/alpha^2)-2x+5=0` `D>=0` `b^2-4ac>=0` `4-4(1-1/alpha^2)5>=0` `alpha^2<=5/4` `alpha^2-5/4<=0` `(alpha-sqrt5/2)(alpha+sqrt5/2)<=0` `alpha in [-sqrt5/2,sqrt5/2]` option a is correct. |
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| 172. |
If `m` and `x` are real numbers where `m in I` `e^(2micot^(-1)x)((x.i+1)^m)/((x.i-1)^m)` |
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Answer» `Cot^(-1)x=theta` `x=cottheta` `((x i+1)/(x i-1))^m=(i^n(x+1/i)^m)/(i^m(x-1/i)^m)` `=(x-1)^m/(x+i)^n` `=(cottheta-1)^m/(cottheta+1)^m` `=(costheta-sintheta)^m/(costheta+sintheta)^m` `=(e^(-itheta)/e^(1theta))=(e^(-i2theta))^m` `=e^(i2mtheta)` `=e^(i2mtheta-i2mtheta)=e^0=1` option c is correct. |
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| 173. |
On set `A= {1,2,3}`, relation `R and S` are given by `R={(1,1),(2,2),(3,3),(1,2),(2,1)} and S={(1,1),(2,2),(3,3),(1,3),(3,1)}` Then |
| Answer» Correct Answer - (i) Many-one into `" " ` (ii) One-one onto | |
| 174. |
`f: R to R ` is a function where f(x)= 2x-3 . Check whether f is noe -one ? |
| Answer» Correct Answer - Yes | |
| 175. |
Suppose `f(x)=(x+1)^2forxgeq-1.`If `g(x)`is the function whose graph is the reflection of the graph of `f(x)`with respect to the line `y=x ,`then `g(x)`equal.`a-sqrt(x)-1,xgeq0`(b) `1/((x+1)^2),x >-1``sqrt(x+1,)xgeq-1`(d) `sqrt(x)-1,xgeq0`A. `1-sqrt(x)-1, x ge 0`B. `(1)/((x+1)^(2)),x gt -1 `C. `sqrt(x+1), x ge -1`D. `sqrt(x)-1, x ge 0` |
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Answer» Correct Answer - D Given that `f(x)=(x+1)^(2), x ge -1.` Now, if g(x) is the reflection of f(x) in the line y = x, then g(x) is an inverse function of `y = f(x).` Given `y=(x+1)^(2) (x ge -1 and y ge 0)` ` or x = +- sqrt(y)-1` ` or g(x) =f^(-1)(x)=sqrt(x)-1, x ge 0` |
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| 176. |
Let `f={(2,4),(5,6),(8,-1),(10,3)` and `g={(2,5),(7,1),(8,4),(10,13),(11,5)}` be two real functions. Then, match the following . The domain of `f-g,f+g,f*g,(f)/(g)` is domain of f `nn` domain of g. Then, find their images. |
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Answer» We have `f={(2,4),(5,6),(8,1),(10,-3)` and `g={(2,5),(7,1),(8,4),(10,13),(11,5)}` So, `f-g,f+g,f*g,(f)/(g)` are defined in the domain (domain of f`nn` domain of g) `i.e., {2,5,8,10}nn{2,7,8,10,11} rArr {2,8,10}` (i) `(f-g)(2)=f(2)-g(2)=4-5=-1` `(f-g)(8)=f(8)-g(8)=-1-4=-5` `(f-g)(10)=f(10)-g(10)=-3-13=-16` `f-g={(2,-1),(8,-5),(10,-16)}` (ii) `(f+g) (2)=f(2)=f(2)+g(2)=4+5=9` `(f+g)(8)=f(8)+g(8)=-1+4=3` `(f+g)(10)=f(10)+g(10)=-3+13=10` `:. f-g={(2,9).(8,3),(10,10)}` (iii) `(f-g )(2)=f(2)*g(2)=4xx5=20` `(f*g)(8)=f(8)*(8)=-1xx4=-4` `(f*g)(10)=f(10)*g(10)=-3xx13=-39` `:. fg={(2,20),(8,-4),(10,-39)}` (iv) `((f)/(g))(2)=(f(2))/(g(2))=(4)/(5)` `((f)/(g))(8)=(f(8))/(g(8))=(-1)/(4)` `((f)/(g))(10)=(f(10))/(g(10))=(-3)/(13)` `:. (f)/(g)={(2,(4)/(d)),(8,-(1)/(4)),(10,(-3)/(13))}` Hence , the correct matches are `(i) rarr(c),(ii) rarr(d),(iii)rarr(b), (iv) rarr(a).` |
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| 177. |
The domain and range of the function f given by` f(x) =2 - |x-5|` isA. Domain =`R^(+)`, Range =`(-oo,1]`B. Domain =R, Range =`(-oo,2]`C. Domain =R, Range =`(-oo,2)`D. Domain =`R^(+)`, Range =`(-oo,2]` |
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Answer» Correct Answer - B We have, f(x)=2-|x-5| f(x) is defined for all `x in R` `:.` Domain of f=R We know, that, `|x-5|ge0rArr-|x-5|le0` `rArr 2-|x-5|le2` `:. f(x) le2` `:.` Range of `f=[-oo,2]` |
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| 178. |
Let f and g be two real functions given by `f={(0,1),(2,0),(3,-4),(4,2),(5,-1)}` and `g={(1,0),(2,2),(3,-1),(4,4),(5,3)}` then the domain of `f*g` is given by .......... |
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Answer» We have `f={(0,1),(2,0),(3,-4),(4,2),(5,-1)}` and `g={(1,0),(2,2),(3,-1),(4,4),(5,3)}` `:.` Domain of `f={0,2,3,4,5}` and Domain of g ={1,2,3,4,5} `:.` Domain of `(f*g)` =Domain of `fnn` Domain of g={2,3,4,5} |
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| 179. |
Find the values of `x` for which the functions `f(x)=3x^2-1` and `g(x)=3+x` are equalA. `[-1,(4)/(3)]`B. `[1,(4)/(3)]`C. `[-1,-(4)/(3)]`D. `[-2,-(4)/(3)]` |
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Answer» Correct Answer - A We have `f(x)=3x^(2)-1 and g(x) =3+x` f(x) =g(x) `rArr 3x^(2)-1=3+x` `rArr 3x^(2)-x-4=0` `rArr 3x^(2)-4x+3x-4=0` `rArr x(3x-4)+1(3x-4)=0` `rArr (3x -4)(x+1)=0` `:. x=-1,(4)/(3)` So domain for which f(x) and g(x) are equal to `[-1,(4)/(3)]` |
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| 180. |
Each of the following defines a relation on `N :``x > y , x , y in N`(ii) `x+y=10 , x , y in N`(iii) `x y`is square of an integer, `x , y in N`(iv) `x+4y=10 , x , y in N`Determine which of the above relations arereflexive, symmetric and transitive. |
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Answer» (i) x is greater than `y,x,y in N` `(x,x) in R` For `xRx " " x gt x ` is not true for any `x in N`. Therefore, R is not reflexive. Let ` (x,y) in R implies xRy` `x gt Y` but `y gt x ` is not true for any `x, y in N` Thus, R is not symmetric. Let `xRY and yRz` `x gt y and y gt z implies x gt z` `implies xRz` So, R is transitive. (ii) `x+y =10, x, y in N` ` R = {(x,y), x+y=10, x,y in N }` `R ={(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)} (1,1) notin R ` So, R is not reflexive, `(x,y) in R implies (y,x) in R` Therefore, R is symmetric, `(1,9) in R, (9,1) in R implies (1,1) notin R` Hence, R is not transitive. (iii) Given xy, is square of an integer `x, y in N. ` `implies R={(x,y):xy` is square of an integer `x,y in N } ` `(x,x) in R, AA x in N` As `x^(2)` is square of an integer for any `x in N` Hence, R is reflexive. If `(x,y) in R implies (y,x) in R` Therefore, R is xymmetric. If `(x,y) in R , (y,z) in R` So, xy is square of an integer and yz is square of an integer. Let `xy=m^(2)` and `yz=n^(2)` for some `m, n in Z` `x=(m^(2))/(y)` and `z=(x^(2))/(y)` `xz =(m^(2)n^(2))/(y^(2)),` which is square of an integer. So, R is transitive. (iv) `x+4y=10, x,y in N` `R={(x,y):x+4y=10, x,y in N} ` `R={(2, 2),(6,1)}` `(1,1),(3,3), ... notin R` Thus, R is not reflexive. `(6,1) in R` but `(1,6) notin R` Hence, R is not symmetric. `(x,y) in R implies x+4y =10` but `(y,z) in R` `y+4z =10 implies (x,z) in R` So, R is transitive. |
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| 181. |
Is `g={(1,1),(2,3),(3,5,),(4,7)}`a function? If this is described by the formula, `g(x)=alphax+beta,`then what values should be assigned to `alpha` and `beta?` |
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Answer» Given that, `g={(1,1),(2,3),(3,5),(4,7)}.` Here, each element of domain has unique image. So, g is a function. Now given that, `g(x) =alpha x + beta ` `g(1)=alpha + beta` `alpha + beta =1 " " ` ...(i) `g(2)=2 alpha + beta` `2 alpha + beta=3 " " `...(ii) From Eqs. (i) and (ii), `2(1 - beta) +beta =3` `implies 2-2beta + beta=3` `implies 2 -beta=3` ` " " beta= -1` If `beta = -1,` then `alpha = 2` `alpha=2, beta = -1` |
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| 182. |
Is `g={(1,1),(2,3),(3,5,),(4,7)}`a function? If this is described by the formula, `g(x)=alphax+beta,`then what values should be assigned to `alphaa n dbeta?` |
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Answer» We have `={(1,1),(2,3),(3,5),(4,7),}` Since, every element has unique image under g. So g is a function. Now,`g(x)=alpha x+beta` When x =1 then, `g(1)=alpha(1) +beta " " ....(i)` `rArr 1=alpha+beta` When x=2, then, `g(2)=alpha(2)+beta` `rArr 3=2alpha+beta" " ....(ii)` On solving Eqs. (i) and (ii), we get `alpha=2,beta=-1` |
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| 183. |
Consider `f : {1, 2, 3}->{a , b , c}`given by `f(1) = a`, `f(2) = b`and `f(3) = c`. Find `f^(-1)`and show that `(f^(-1))^(-1)= f`. |
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Answer» `f: {1, 2, 3} to {a, b, c}` and `f(1) = a, f(2) = b and f(3) = c ` Let, in `g: {a, b, c} to { 1, 2, 3}, g(a)=1, g(b) = 2 and g(c ) = 3` `therefore " " (fog) (a) = f{g(a)} = f(1) =a` `" " (fog) (b) = f{g(b)} = f(2) = b` `" " (fog) (c ) = f{g(c)} = f(3) = c` and `" " (gof) (1) = g{f(1)} = g(a) =1 ` `" " (gof)(2) = g{f(2)}= g(b) = 2` `" "(gof) (3) = g{f(3)} = g(c) = 3` `therefore " " gof = I_X and fog = I_Y` where `X = { 1, 2, 3} and Y = {a, b, c}` `therefore " " f^(-1) = g ` and `f^(-1) : {a, b, c} to {1, 2, 3 }` and `f^(-1)(a) = 1, f^(-1) (b) = 2, f^(-1) (c) = 3` Again let, in `h: {1, 2, 3} to {a, b, c}` `" " h(1) = a, h(2) = b, h(3) = c` `" " (goh) (1) = g{h(1)} = g(a)= 1` `" " (goh)(2) = g{h(2)} = g(b) = 2 ` `" " (goh)(3) = g{h(3)}= g(c) = 3` and `" " (hog)(a) = h{g(a)} = h(1) =a` `" " (hog )(b) = h{g(b)} = h(2) = b` `" " (hog) (c) = h{g(c)} = h(3) =c ` `therefore " " goh = I_X and hog = I_Y` where `" " X= {1, 2, 3} and y = {a, b, c}` `thererfore " "g^(-1) = h` `rArr " " (f^(-1)) = f` |
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| 184. |
Let `f: X -> Y`be an invertible function. Show that the inverse of `f^(-1)`is f, i.e., `(f^(-1))^(-1)= f`. |
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Answer» `because f : X to Y ` is an invertible function. ` therefore g : Y to X` will be a function such that `" " gof = I_X and fog = I_Y` `rArr " " f^(-1) = g ` Now `" " gof = I_X and fog = I_Y` `rArr " " f^(-1) of = I_X and fof ^(-1) = I_Y` `rArr f^(-1) : Y to X ` will be invertible and `(f^(-1))^(-1) = f ` |
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| 185. |
Let `f(x)`be periodic and `k`be a positive real number such that `f(x+k)+f(x)=0fora l lx in Rdot`Prove that `f(x)`is periodic with period `2kdot` |
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Answer» We have `f(x+k)+f(x)=0 AA x in R` or ` f(x+k)= -f(x) AA x in R` Put `x=x+k.` Then, or `f(x+2k)= -f(x+k) AA x in R` or ` f(x+2k) = f(x), AA x in R " "[ "As "f(x+k)= -f(x)]` which clearly shows that f(x) is periodic with period 2k. |
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| 186. |
Let `f(x)=x^2-2x ,x in R ,a n dg(x)=f(f(x)-1)+f(5-(x))dot`Show that `g(w)geqoAAx in Rdot` |
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Answer» Given `f(x)=x^(2)-2x =(x-1)^(2)-1` or `g(x)=f(f(x)-1)+f(5-f(x))` `=f((x-1)^(2)-2)+f(6-(x-1)^(2))` `=[(x-1)^(2)-2-1]^(2)-1+[6-(x-1)^(2)-1]^(2)-1` `=(x-1)^(4)-6(x-1)^(2)+9-1+(x-1)^(4)-10(x-1)^(2)+25-1` `=2(x-1)^(4)-16(x-1)^(2)+32` `=2[(x-1)^(4)-8(x-1)^(2)+16]` `=2[(x-1)^(2)-4]^(2) ge 0 AA x in R` |
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| 187. |
Consider a differentiable `f:R to R` for which `f(1)=2 and f(x+y)=2^(x)f(y)+4^(y)f(x) AA x , y in R.` The minimum value of `f(x)` isA. 1B. ` -(1)/(2)`C. `-(1)/(4)`D. None of these |
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Answer» Correct Answer - C `f(x+y)=2^(x) f(y)+4^(y)f(x) " …(i)" ` Interchanging x and y, we get `f(x+y)=2^(y)f(x)+4^(x)f(y) " (ii)" ` `implies 2^(x) f(y)+4^(y)f(x)=2^(y)f(x)+4^(x)f(y)` `implies (f(x))/(4^(x)-2^(x))=(f(y))/(4^(y)-2^(y))=k` `implies f(x)=k(4^(x)-2^(x))` `f(1)=2` `implies k=1.` Hence, `f(x)=4^(x)-2^(x).` `f(4)=4^(4)-2^(4)=240` `f(x)=(2^(x))^(2)-2^(x)=(2^(x)-1//2)^(2)-1//4` Thus `f(x)` has least value `-1//4` Also `4^(x)-2^(x)=2` `implies (2^(x))^(2)-2^(x)-2=0` `implies (2^(x)-2)(2^(x)+1)=0` `implies2^(x)-2=0` `implies x=1` |
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| 188. |
What is the probability that the sum of any two different single digit natural numbers is a prime number? |
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Answer» We know single digit natural number are from `1` to `9`. Now, the maximum sum that we can get is `9+8=17` as we have to use different numbers. Now, possible two digits having a sum equal to prime number are, `(1+2),(1+4),(2+3),(1+6),(2+5),(3+4),(2+9),(3+8),(4+7),(5+6),` `(4+9),(5+8),(6+7),(8+9)`. So, there are `14` possible values. Now, number of ways selecting these `2` digits from `9` digits is `C(9,2)`. `:.` the required probability ` = 14/(C(9,2)) = 14/36 = 7/18` |
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| 189. |
Find the range of the following functions given by `f(x)=(3)/(2-x^(2)) " "` (ii) `f(x)=1-|x-2|` (iii) `f(x)=|x-3| " "` (iv) `f(x)=1+3 cos 2x` |
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Answer» We have, `f(x)=(3)/(2-x^(2))` Let y=f(x) Then, `y=(3)/(2-x^(2))rArr 2-x^(2)=(3)/(4)` `rArr x^(2)=2-(3)/(y)rArrx=sqrt((2y-3)/(y))` x assums real values, if `2y-3ge0 and ygt0 rArr y ge (3)/(2)` `:.` Range of `f=[(3)/(2),oo)` (ii) We known that, `|x-2|ge or rArr-|x-2|le0` `rArr 1-|x-2|le1 rArrf(x)le1` `:.` Range of `f=(-oo,1]` (iii) We know, that, `|x-3|leorrArrf(x)ge0` `:.` Rang of `f=0,oo]` (iv) We hnow that, `-1le cos2xle1rArr-3le3cos2xle3` `rArr 1-3le1+3cos2xle1+3rArr-2le1+3cos2xle1+3` `rArr -2lef(x)le4` `:.` Range of `f=[-2,4]` |
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| 190. |
If a and c are positive real number and ellipse `x^2/(4c^2)+y^2/c^2=1` has four distinct points in comman with the circle `x^2+y^2=9a^2`, then (A) `6ac+9a^2-2c^2 gt 0` (B) `6ac+9a^2-2c^2 lt 0`(C) `6ac-9a^2-2c^2 gt 0` (D) `6ac-9a^2-2c^2 lt 0` |
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Answer» 2C>3a>C `4C^2>9a^2>c^2` `6ac+9a^2-2c^2=y` `c(6a-c)lty<2c(3a+c)` `6c^2>9a^2+2c^2>3c^2` `9ac-9a^2-2c^2=Z` `9ac-(9a^2+2c^2)=z` `9ac-6c^2ltz<90c-3c^2` `3c(3a-3c)ltz<3c(3a-3c)` option 1 is correcct. |
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| 191. |
`|z-4| < |z-2|` represents the region given by: (a) `Re(z) > 0` (b) `Re(z) < 0` (c) `Re(z) > 3` (d) None of these |
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Answer» z=x+iy `|(x-4)+iy|<|(x-2)+iy|` `(x-4)<(x-2)` option c is correct. |
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| 192. |
Re define the function `f(x)=|x-2|+|2+x|,-3lexle3` |
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Answer» Since, `|x-2|=-(x-2),xlt2` `x-2,xge2` and `|2+x|=-(2+x ),xlt2` `(2+x),xle2` `:. "f"(x)=|x-2|+|2+x|,x-3lexle3` `{{:(-(x2)-(2+x),-3lexle3),((x-2)+2+x,-2lexle2),(x-2+2+x,2lexle3):}` `{{:(-2x,-3lexlt-2),(4,-2lexlt2),(2,2lexle3):}` |
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| 193. |
Let f(x) = maximum { x + |x| , x - [x]}, where [x] is the greatest integer less than or equal to x, The ` int_(-2)^2 f(x) dx` = |
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Answer» `x+|x|={(x+x,2x,x>0),(x-x,=0,x<0):}` `int_-2^2f(x)=2*1/2*1+1+6/2=5` option a is correct. |
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| 194. |
Consider a curve `ax^2 + 2hxy + by^2-1=0` and a point P not on the curve.A line is drawn from the point P intersects the curve at the point Q and R.If the product PQ.PR is independent of the sIope of the line, then the curve is: |
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Answer» The given curve is a circle as it is a property of circle if `QR` is a chord from an outside point `P`, then, `PQ*PR = PT^2,` where `PT` is the tangent to the circle. So, we can see that `PQ*PR` is independent of the slope of the line. So, option `(B)` is the correct option. |
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| 195. |
n-similar balls each of weight w when weighed in pairs the sum of the weights of all the possible pairs is 120 when they are weighed in triplets the sum of the weights comes out to be 480 for all possible triplets,then n is: |
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Answer» Weight of a pair of balls ` = 2w` Weight of triplets of balls `= 3w` Number of possible pairs ` = C(n,2)` Number of possible triplets ` = C(n,3)` Now, we are given,`C(n,2)**2w = 120->(1)` `C(n,3)**3w = 480->(2)` Dividing `(1)` by `(2)`, `(2C(n,2))/(3C(n,3)) = 1/4` `=>((n!)/((2!)(n-2)!))/((n!)/((3!)(n-3)!)) = 3/8` `=>3/(n-2) = 3/8` `=>n-2 = 8` `=> n = 10` |
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| 196. |
Let A be a `2 xx 2` matrix with non-zero entries and let A^2=I, where i is a `2 xx 2` identity matrix, Tr(A) i= sum of diagonal elements of A and `|A|` = determinant of matrix A.Statement 1:Tr(A)=0Statement 2:`|A|`=1 |
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Answer» Let `A = [[a,b],[c,d]]` It is given that, `A^2 = I` `:. [[a,b],[c,d]][[a,b],[c,d]] = [[1,0],[0,1]]` `=>[[a^2+bc,ab+bd],[ac+cd,bc+d^2]]= [[1,0],[0,1]]` `=>a^2+bc = 1->(1)` `=>ab+bd = 0 =>b(a+d) = 0 => a = -d->(2) ...[As b!=0]` So, we can write, `A = [[a,b],[c,-a]]` `:. Tr(A) = a+(-a) = 0` `|A| = -a^2-bc =-(a^2+bc) = -1` So, first statement is true but second statement is false. |
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| 197. |
If `f(x)=(sin([x]pi))/(x^2+x+1)`, where `[dot]`denotes the greatest integer function, then`fi son eon e``fi snoton e-on ea n dnon-con s t a n t``fi sacon s t a n tfu n c t ion``non eoft h e s e`A. `f` is one-oneB. `f` is not one-one and non-constantC. `f` is a constant functionD. None of these |
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Answer» Correct Answer - C `f(x)=(sin[x]pi)/(x^(2)+x+1)` Let `[x]=n ~= "integer" ` ` :. sin[x] pi=0` or `f(x)=0` Hence, `f(x)` is constant function. |
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| 198. |
Match the following lists: |
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Answer» Correct Answer - `a to p; b to q; c to q,s; d to r,s` a. `f(x) ={(sgn x)^(sgn x)}^(n)` `={([(1)^(1)]^(n)",",x gt 0),([(-1)^(-1)]^(n) ",",x lt 0):}={(1",",x gt 0),(-1",",x lt 0):}.` Hence, `f(x)` is an odd function. b. `f(x)=(x)/(e^(x)-1)+(x)/(2)+1` `or f(-x)=(-x)/(e^(-x)-1)-(x)/(2)+1` `=(xe^(x))/(e^(x)-1)-(x)/(2)+1` `=(xe^(x)-x+x)/(e^(x)-1)-(x)/(2)+1` `=x+(x)/(e^(x)-1)-(x)/(2)+1` `=(x)/(e^(x)-1)+(x)/(2)+1` `=f(x)` c. `f(x)={(0",","if x is rational"),(1",","if x is irrational"):}` `f(-x)={(0",","if -x is rational"),(1",","if -x is irrational"):}` `={(0",","if x is rational"),(1",","if x is irrational"):}` `=f(x)` d. `f(x)=max [tanx, cotx}` From the graph of function, it can be verified that `f(x)` is neither odd nor even. Also, `f(x+pi)=max {tan(x+pi),cot(x+pi)` `=max {tanx, cot x}` Hence, `f(x)` is periodic with period `pi`. |
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| 199. |
Consider a differentiable `f:R to R` for which `f(1)=2 and f(x+y)=2^(x)f(y)+4^(y)f(x) AA x , y in R.` The number of solutions of `f(x)=2` is |
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Answer» Correct Answer - B `f(x+y)=2^(x) f(y)+4^(y)f(x) " …(i)" ` Interchanging x and y, we get `f(x+y)=2^(y)f(x)+4^(x)f(y) " (ii)" ` `implies 2^(x) f(y)+4^(y)f(x)=2^(y)f(x)+4^(x)f(y)` `implies (f(x))/(4^(x)-2^(x))=(f(y))/(4^(y)-2^(y))=k` `implies f(x)=k(4^(x)-2^(x))` `f(1)=2` `implies k=1.` Hence, `f(x)=4^(x)-2^(x).` `f(4)=4^(4)-2^(4)=240` `f(x)=(2^(x))^(2)-2^(x)=(2^(x)-1//2)^(2)-1//4` Thus `f(x)` has least value `-1//4` Also `4^(x)-2^(x)=2` `implies (2^(x))^(2)-2^(x)-2=0` `implies (2^(x)-2)(2^(x)+1)=0` `implies2^(x)-2=0` `implies x=1` |
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| 200. |
The function `f(x)` is defined on the interval [0, 1]. Now, match the following lists: |
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Answer» Correct Answer - `a to s; b to r; c to p; d to q` `f(tanx) " is defined if " 0 le tan x le 1` `or x in [n pi, npi +(pi)/(4)], n in Z` `f(sinx) " if defined if " 0 le sinx le 1` `or x in [2n pi, (2n+1) pi], n in Z` `f(cosx)" is defined if " 0 le cosx le 1` `or x in [2n pi-(pi)/(2),2n pi+(pi)/(2)], n in Z` `f(2sinx) " is defined if " 0 le 2sinx le 1 or 0 le sinx le 1//2` `or [2n pi, 2n pi + (pi)/(6)] cup [2npi+(5pi)/(6),(2n+1)pi], n in Z` |
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