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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. |
Find the equivalent definition of`f(x)=m a xdot{x^2,(-x)^2,2x(1-x)}w h r e0lt=xlt=1`A. `f(x)={{:(x^(2),0 le x le 1//3),(2x(1-x),1//3 le x le 2//3),((1-x)^(2), 2//3 le x le 1):}`B. `f(x)={{:((1-x)^(2), 0 le x le 1//3),(2x(1-x),1//3 le x le 2//3),(x^(2), 2//3 le x le 1):}`C. `f(x)={{:(x^(2),0 le x le 1//2),((1-x)^(2),1//2 le x le 1):}`D. none of these |
| Answer» Correct Answer - B | |
| 2. |
Domain of the function `f(x) = log(sqrt(x-4)+sqrt(6-x))`A. [4,6]B. `(-oo,6)`C. `(2,3)`D. none of these |
| Answer» Correct Answer - A | |
| 3. |
If `f(x)=sin(logx)` then `f(xy)+f(x/y)-2f(x)cos(logy)=` (A) `cos(logx)` (B) `sin(logy)` (C) `cos(log(xy))` (D) 0A. `-1`B. `0`C. 1D. none of these |
| Answer» Correct Answer - B | |
| 4. |
The domain of definition of the function `f(X)=x^((1)/(log_(10)x))`, isA. `(0,1) cup (1,oo)`B. `(0,oo)`C. `[0,oo)`D. `[0,1) cup (1,oo)` |
| Answer» Correct Answer - A | |
| 5. |
The domain of definition of the function `f(x)=log_(3){-log_(4)((6x-4)/(6x+5))}` , isA. `(2//3,oo)`B. `(-oo,-5//6) cup (2//3,oo)`C. `[2//3,oo)`D. `(-5//6,2//3)` |
| Answer» Correct Answer - A | |
| 6. |
The set of all real values of x for which the funciton `f(x) = sqrt(sin x + cos x )+sqrt(7x -x^(2) - 6)` takes real values isA. `[1, 3pi//4] uu [7pi//4, 6]`B. `[1, 3pi//4]uu [ 6- pi//4, 6]`C. `[1, 6]`D. none of these |
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Answer» Correct Answer - A We have, `f(x) = sqrt(sin x + cos x) + sqrt(7x- x^(2) - 6)` `rArr (x) = 2^(1//4) sqrt(sin (x+ pi//4)) + sqrt(7x-x^(2)-6)` For f(x) to be real, we must have `sin (x+(pi)/(4)) gt 0 and 7x - x^(2) - 6 gt 0` `rArr 2npi le x +(pi)/(4) le (2n+1) pi and 1 le x le 6`, where `n in Z` `rArr x in [1,(3pi)/(4)]uu[(7x)/(4) ,6]` |
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| 7. |
The equivalent definition of `f(x)=||x|-1|`, isA. `f(x)={{:( -x-1, x le -1),(x+1, -1 lt x le 0),( 1-x, 0 le x le 1), (x-1,x le 1):}`B. `f(x)={{:(x-1,x le -1),(x+1,-1 lt x le 0),(x-1, 0 le x le 1),(x+1, x ge 1):}`C. `f(x)={{:( x+1, x ge 0), (x+1, x le0):}`D. none of these |
| Answer» Correct Answer - A | |
| 8. |
`f(x)={(4, xlt-1) ,(-4x,-1lt=xlt=0):}` If `f(x)` is an even function in R then the definition of `f(x)` in `(0,oo)` is:(A) `f(x)={(4x, 0ltxle1),(4, xgt1):}` (B) `f(x)={(4x, 0ltxle1),(-4, xgt1):}` (C) `f(x)={(4, 0ltxle1),(4x, xgt1):}` (D) `f(x)={(4, xlt-1),(-4x, -1lexle0):}`A. `f(x)={:(4x, 0 lt x le 1),(4, x gt 1):}`B. `f(x)={{:(4x, 0 lt x le1),(-4, x gt 1):}`C. `f(x)={{:(4, 0 lt x le 1),(4x,x gt 1):}`D. none of these |
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Answer» Correct Answer - A The even extension of f(x) on `(0,oo)` is given by `f(x)=f(-x)` for all ` x in (0,oo)` `f(x)={{:(4, " for" -x lt -1),(4, " for " -1 le-x lt 0):}` `impliesf(x)={{:(4," for " x gt 1),(4x, " for " 0 lt x le 1):}` |
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| 9. |
If a funciton f(x) is defined for ` x in [0,1]`, then the function f(2x+3) is defined forA. `x in [0,1]`B. `x in [-3//2,-1]`C. `x in R`D. `x in [-3//2,1]` |
| Answer» Correct Answer - B | |
| 10. |
If `b^2-4ac=0,a>0` then the domain of the function `f(x)=log(a x^3+(a+b)x^2+(b+c)x+c)` is :A. `R-{-(b)/(2a)}`B. `R-{{-(b)/(2a)} cup {x|x ge-1|}}`C. `R-{{-(b)/(2a)}cap(-oo,-1]}`D. none of these |
| Answer» Correct Answer - C | |
| 11. |
The domain of definition of `f(x)=log_(3)|log_(e)x|`, isA. `(1,oo)`B. `(0,oo)`C. `(e,oo)`D. none of these |
| Answer» Correct Answer - D | |
| 12. |
The function `f(x) = max {(1 - x), (1 + x), 2}, x in (- oo, oo)` is equivalent toA. `f(x)={{:(1-x,x le-1),(2,-1 lt x lt 1),(1+x, x ge1):}`B. `f(x)={{:(1+x","x le-1),(2","-1 lt x lt 1),(1-x"," x ge1):}`C. `f(x)={{:(1-x","x le-1),(1","-1 lt x lt 1),(1+x"," x ge1):}`D. none of these |
| Answer» Correct Answer - A | |
| 13. |
Let the function `f(x)=3x^(2)-4x+5log(1+|x|)` be defined on the interval [0,1]. The even extension of f(x) to the interval [0,1]. The even extension of f(x) to the interval [-1,1] isA. `3x^(2)+4x+8 log(1+|x|)`B. `3x^(2)-4x+8 log (1+|x|)`C. `3x^(2)+4x-8 log(1+|x|)`D. none of these |
| Answer» Correct Answer - A | |
| 14. |
The equivalent definition of the function `f(x)=lim_(n to oo)(x^(n)-x^(-n))/(x^(n)+x^(-n)), x gt 0` , isA. `f(x)={{:(-1,0 lt x le1),(1, x gt 1):}`B. `f(x)={{:(-1,0 lt x lt 1),(1, x ge 1):}`C. `f(x)={{:(-1,0 lt x lt 1),(0, x=1),(1,x gt 1):}`D. none of these |
| Answer» Correct Answer - C | |
| 15. |
If a funciton `f:[2,oo] to B` defined by `f(x)=x^(2)-4x+5` is a bijection, then B=A. RB. `[1,oo)`C. `[4,oo)`D. `[5,oo)` |
| Answer» Correct Answer - B | |
| 16. |
The domain of definition of the function `f(x)=log_(2)[-(log_(2)x)^(2)+5log_(2)x-6]` , isA. `(4,8)`B. [4,8]C. `(0,4) cup (8,oo)`D. `R-[4,8]` |
| Answer» Correct Answer - A | |
| 17. |
If f(x) is a real valued odd function , then which one of the following is incorrect ?A. `(f(x)-f-(-x))/(2)` is an odd function.B. `(f(x)+f(-x))/(2)` is an even function.C. `[|f(x)|+2]` is an even function, `[*]` denotes the greatest integer function.D. `(f(x)-f(-x))/(2)` is neither even nor odd. |
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Answer» Correct Answer - D We observe that `phi(x)=(f(x)+f(-x))/(2)` is an even function . `and Psi(x)=(f(x)-f(-x))/(2)` is an odd functions. Let `g(x)=[|f(x)|+2]`. Then, `g(-x)=[|f(-x)|+2]` `implies g(-x)=[|-f(x)|+2]` `=g(-x)=[|f(x)|+2]` `implies g(-x)=g(x)` for all ` x in R `. Therefore, g(x) is an even function. |
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| 18. |
If `f(n+2)=(1)/(2){f(n+1)+(9)/(f(n))}, n in N and f(n) gt0` for all ` n in N`, then ` lim_( n to oo)f(n)` is equal toA. 3B. `-3`C. `(1)/(2)`D. none of these |
| Answer» Correct Answer - A | |
| 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)`A. `[- pi//4, 2)`B. `[ 0, pi//2)`C. `[0, pi]`D. ` (-pi//2, pi//2)` |
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Answer» Correct Answer - B Clearly, `4^(-x^(2))` is defined for all `x in R` and `cos^(-1)((x)/(2)-1)` is defined for `-1 le (x)/(2) - 1 le 1` i.e., for `0 le (x)/(2) le 2` or, `x in [0,4]` Also ,log cos x is defined for all `x in (-pi//2,pi//2)` `therefore f(x)` is defined for all `x in (-(pi)/(2),(pi)/(2)) nn[0,4]nnR = [0,(pi)/(2))` |
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| 20. |
If` f(x)` is an odd periodic function with period 2, then f(4) equals to-A. `-4`B. 4C. 2D. 0 |
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Answer» Correct Answer - D It is given that f(x) is an odd periodic function with period 2 . Therefore, `f-(x)=-f(x) and f(x+2)=f(x)` for all ` x in R ` `implies f(-x)+f(x)=0 and f(x+2)=f(x)` for all ` x in R ` `implies 2f(0)=0 and f(2)=f(0)` `implies f(0)=0 and f(2)=0` Since `f(x+2)=f(x)` for all ` x in R `. Therefore , `f(4)=f(2)implies f(4)=0` |
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| 21. |
If `f(x)` is a periodic function with peirodic function with period `lambda` then `f(lambda x + u)` where `mu` is any constant is periodic with period`(T)/(a)`.A. ` lamda `B. `1 `C. `(lamda )/( a)`D. none of these |
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Answer» Correct Answer - B We know that if f(x) is a priodic funntion with period T, then `f(ax +b), a gt 0` is perodic with period `(T)/(a)` . Therefore`,f(lambda x + mu)` is periodic with period `(lambda)/(lamda) = 1`. |
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| 22. |
The range of the function `y=(x+2)/(x^2-8x+4)`A. `(-oo,-(1)/(4)] cup [-(1)/(20),oo)`B. `(-oo,-(1)/(4)) cup (-(1)/(20),oo)`C. `(-oo,-(1)/(4)] cup (-(1)/(20),oo)`D. none of these |
| Answer» Correct Answer - B | |
| 23. |
The values of `ba n dc`for which the identity of `f(x+1)-f(x)=8x+3`is satisfied, where `f(x)=b x^2+c x+d ,a r e``b=2,c=1`(b) `b=4,c=-1``b=-1, c=4`(d) `b=-1,c=1`A. b=2,c=1B. b=4,c=-1C. b=-1,c=4D. b=-1,c=1 |
| Answer» Correct Answer - B | |
| 24. |
Find the period of`f(x)=sinx+tanx/2+sinx/(2^2)+t a n x/(2^3)++sinx/(2^(n-1))+tanx/(2^n)`A. `2pi `B. `2^(n-1)pi `C. `2^(n)pi`D. `n pi ` |
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Answer» Correct Answer - C We have, `f(x)=sum_(r=1)^(n)(sin""(x)/(2^(r-1))+tan""(x)/(2^(r)))` `implies f(x)=sum_(r=1)^(n)f_(r)(x), where" " f_(r)(x)= sin""(x)/(2^(r-1))+tan""(x)/(2^(r))` since each of ` sin""(x)/(2^(r-1)) and (x)/(2^(r ))` are periodic functions with period `2^(r)pi`. Therefore, `f_(r)(x)` is periodic with period ` 2^(r) pi`. But , `f(x)=sum_(r=1)^(n)f_(r)(x)` . Therefore, , f(x) is periodic with period T given by T=LCM of `(2pi, 2^(2)pi, ....,2^(n)pi)=2^(n)pi` |
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| 25. |
If `f(x)=x+(1)/(x)`, such that `[f(x)]^(3)=f(x^(3))+lambdaf((1)/(x))`, then `lambda=`A. 1B. 3C. -3D. -1 |
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Answer» Correct Answer - B we have, `f(x)=x+(1)/(x)implies f(x^(3))=x^(3)+(1)/(x^(3))` Now, `{f(x)}^(3)=(x+(1)/(x))^(3)` `implies{f(x)}^(3)=(x^(3)+(1)/(x^(3)))+3(x+(1)/(x))` `implies {f(x)}^(3)=f(x^(3))+3f(x)` `implies {f(x)}^(3)=f(x^(3))+3f((1)/(x))" " [ :. f(x)=f((1)/(x))]` `:. lambda=3` |
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| 26. |
If `f(x) =ax^(2) + bx + c` satisfies the identity `f(x+1) -f(x)= 8x+ 3` for all `x in R` Then (a,b)=A. `(2,1)`B. `(4,-1)`C. `(-1,4)`D. `(-1,1)` |
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Answer» Correct Answer - B We have, `f(x)= ax^(2) + bx + c` `therefore f(x+1) -f(x) = 8x +3` for all `x in R` `rArr {a(x+1)^(2) +b(x+1) +c} - {ax^(2) + bx + c} = 8x + 3` for all `x in R` `rArr x (2a-8) + (a+b-c) =0` for all `x in R` `rArr 2a - 8 =0 and a+b -3=0` `rarr a = 4and b=-1` |
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| 27. |
Suppose `f:[-2,2] to R ` is defined by `f(x)={{:(-1 " for " -2 le x le 0),(x-1 " for " 0 le x le 2):}` , then ` {x in [-2,2]: x le 0 and f(|x|)=x}=`A. `{-1}`B. `{0}`C. `{-1//2}`D. `phi` |
| Answer» Correct Answer - C | |
| 28. |
The domain of the function `f(x) = log_10 log_10 (1 + x ^3)` isA. `(-1,oo)`B. `(0,oo)`C. `[0,oo)`D. `(-1,oo)` |
| Answer» Correct Answer - B | |
| 29. |
The period of f(x)=5 sin 3x-7 sin 8x , isA. `pi`B. `2pi`C. `3pi`D. `8 pi` |
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Answer» Correct Answer - B We have, Period of ` 5 sin 3x ` is ` (2pi)/(3)` , Period of ` 7 sin 8x` is `(2pi)/(8)=(pi)/(4)` `:.` Period of `f(x)=("LCM of " 2pi and pi)/("HCF of" 3 and 4)=(2pi)/(1)=2pi` |
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| 30. |
If `f:[-pi//2,pi//2] cupR` given by `f(x)=cosx+sin[(x+1)/(lambda)]` is an even function. Then the set of values of ` lambda(lambda gt0)` is Here, `[*]` denotes the greatest integer function.A. `((-pi)/(2),(pi)/(2))`-{0}B. `((pi+2)/(2),oo)`C. `(0,(pi+2)/(2)) cup ((pi+2)/(2),oo)`D. `((-pi)/(2),(pi+2)/(2))-{0}` |
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Answer» Correct Answer - B For `f(x)` to be even , we must have ` sin[(x+1)/(lambda)]=0` `implies [(x+1)/(lambda)]=0` `implies o le (x+1)/(lambda) lt 1 implies 0 le x +1 lt lambdaimplies x+1 gt0 and lambda gt x+1`. Now , `x in [-(pi)/(2),(pi)/(2)]implies -(pi)/(2) +1 le x+1 le(pi)/(2)+1` Therefore , for `x+1 ge0` , we must have `0 le x +1 le (pi)/(2) +1 implies ` maximum value of `x+1` is `(pi)/(2)+1` . Now, `lambda gt x+1 implies lambda gt ` maximum value of `x+1 implies lambda gt pi//2+1` Hence , `lambda in ((pi)/(2)+1,oo)` |
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| 31. |
If `f:R->R` and `g:R->R` is given by f(x) =|x| and g(x)=[x] for each `x in R` then `{x in R:g(f(x))le f(g(x))}`A. `Z cup (-oo,0)`B. `(-oo,0)`C. ZD. R |
| Answer» Correct Answer - A | |
| 32. |
Find the domain of `log_10 (1-log_10(x^2-5x+16))`A. (1,3)B. (2,3)C. [2,3]D. none of these |
| Answer» Correct Answer - B | |
| 33. |
Domain of the function, `f(x)=[log_10 ((5x-x^2)/4)]^(1/2)` isA. [1,4]B. (1,4)C. (0,5)D. [0,5] |
| Answer» Correct Answer - A | |
| 34. |
The period of `f(x)=tan 3x+ cos""(5x)/(2)` ,isA. `2pi`B. `6pi`C. `4pi`D. `10 pi ` |
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Answer» Correct Answer - C We have , `f(x) = tan 3x+ cos""(5x)/(2)` Clearly , period of ` tan 3x is (pi)/(3)` and that of ` cos 5x` is ` (2pi)/(5//2)=(4pi)/(5)` `:.` Period of `f(x)=("LCM of " pi and 4pi)/("HCF of " 3 and 5 ) =4pi` |
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| 35. |
The period of the fuction ` f(x)=sin""((pix)/(n!))+ cos((pix)/((n+1)!))` , isA. `2xx(n+1)!`B. `2(n!)`C. `n+1`D. none of these |
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Answer» Correct Answer - A Since ` sin ((pix)/(n!)) and cos((pix)/((n+1)!))` are periodic functions with period `2xx(n!) and 2(n+1)!` respectively. Therefore, f(x) is periodic with period equal to LCM of `2(n!)` and `2xx(n+1) ! i.e., 2xx(n+1)!` |
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| 36. |
Statement-1: Every function can be uniquely expressed as the sum of an even function and an odd function. Statement-2: The set of values of parameter a for which the functions f(x) defined as ` f(x)=tan(sinx)+[(x^(2))/(a)]` on the set [-3,3] is an odd function is , `[9,oo)`A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False , Statement-2 is True. |
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Answer» Correct Answer - B Clearly , statement-1 is true .(see theory ) It is given that `f(x)=tan(sinx)+[(x^(2))/(a)]` is an odd function. `:.f(-x)=-f(x)` `implies -tan (sinx)+[(x^(2))/(a)]=-tan(sinx)-[(x^(2))/(a)]` `implies 2[(x^(2))/(a)=0` `implies 0le (x^(2))/(a) lt 1` `implies a gt 0 and 0 lex^(2) lt a " for all " x in[-3,3]` `implies a gt 9, i.e., a in [9,oo)` so, statement-2 is also true. |
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| 37. |
Let `f(n)=1+1/2+1/3++1/ndot`Then`f(1)+f(2)+f(3)++f(n)`is equal to`nf(n)-1`(b) `(n+1)f(n)-nn``(n+1)f(n)+n`(d) `nf(n)+n`A. `n f (n )-1 `B. `(n + 1 ) f(n) - n`C. `(n+ 1 ) f(n) +n `D. ` n f(n) +n ` |
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Answer» Correct Answer - B We have, `f(n) = 1+(1)/(2)+(1)/(3) +……+(1)/(n)` `therefore f(1) + f(2) + …….+f(n)` `= 1+(1+(1)/(2)) +(1+(1)/(2)+(1)/(3)) +.....+(1+(1)/(2)+(1)/(3)+.....+(1)/(n))` `= n +((n-1))/(2)+((n-2))/(3) +....+(n-(n-1))/(n)` `=n(1+(1)/(2)+(1)/(3)+....+(1)/(n))-((1)/(2) +(2)/(3)+.....+(n-1)/(n))` `=n(1+(1)/(2)+(1)/(3)+.....+(1)/(n))-{(1-(1)/(2))+(1-(1)/(3))+....+(1-(1)/(n))}` `=n (1+(1)/(2)+(1)/(3)+.....+(1)/(n))-{(n-1)-((1)/(2)+(1)/(3)+....+(1)/(n))}` `= n f (n) - {(n-1)-(f(n) -1)}` `= (n+1) f(n) -n` |
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| 38. |
If `N` denotes the set of all positive integers and if `f : N -> N` is defined by `f(n)=` the sum of positive divisors of `n` then `f(2^k*3)`, where k is a positive integer isA. `2^(k+1) -1`B. `2(2^(k+1)-1)`C. `3(2^(k+1)-1)`D. `4(2^(k+1)-1)` |
| Answer» Correct Answer - C | |
| 39. |
Let `f(x) = x(2-x), 0 le x le 2`. If the definition of `f(x)` is extended over the set `R-[0,2]` by `f (x+1)= f(x)`, then f is aA. periodic function with period 1B. non-periodic functionC. periodic function with period 2D. periodic function with period `1//2` |
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Answer» Correct Answer - C For any `x in R -[0,2]` we have `f(x+2) = f((x+ 1)+1)` `rArr f(x+2) = f(x+1)" "[because f(x+1) = f(x)]` `rArr f(x+2) = f(x+1) " "[because f(x+1) = f(x)]` Therefore `f(x)` is periodic with period 2. |
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| 40. |
If `f: RvecR`is a function satisfying the property `f(2x+3)+f(2x+7)=2AAx in R ,`then find the fundamental period of `f(x)dot`A. 2B. 4C. 8D. 12 |
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Answer» Correct Answer - C We have , `f(2x+3)+f(2x+7)=2 ` for all ` x in R `. `implies f(y)+f(y+4)=2` for all `y in R " "[ :. Y=2x+3]` `implies f(y-4)+f(y)=2` for all ` y in R ` [ Replacing y by(y-4)] Thus , we have `f(y)+f(y+4)=2 and f(y-4) +f(y)=` for all ` y in R `. On subtracting , we get `f(y+4)-f(y-4)=0` for all `y in R `. ` implies f(y+4)-f(y-4)=0` for all ` y in R `. `implies f(y)=f(y+8) ` for all ` y in R ` [ Replacing y by ( y+4)] `implies f(x)` is periodic with period 8 units. |
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| 41. |
Statement-1: The domain of definition of the function `f(x)=e^(2x)+cos^(-1)((x)/(2)-1)` , is `(0,1) cup (1,2) cup (2,3) cup (3,4)` Statement:- The domain of ` cos^(-1)""((x)/(2)-1) is [0,4]`.A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False , Statement-2 is True. |
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Answer» Correct Answer - C Clearly , ` cos^(-1)""((x)/(2)-1)` is defined for all x satisfying `-1 le ((x)/(2)-1) le1 ` i.e., ` o le (x)/(2) or x in [0,4]` So, Statement-2 is not true. The domain of definition of ` e^(2x), cos^(-1)((x)/(2)-1)` and ` log sqrt({x})` are `R cap [0,4] cap R-Z=(0,1) cup (,2) cup (2,3) cup (3,4)` So, statement-1 is true. |
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| 42. |
Let `f(x)=x+1 and phi(x)=x-2.` Then the value of x satisfying `|f(x)+phi(x)|=|f(x)|+|phi(x)|` are :A. `(-oo,1]`B. `[2,oo)`C. `(-oo,-2]`D. `[1,oo)` |
| Answer» Correct Answer - B | |
| 43. |
The function `f(x)=2 cos 5x+3 sin sqrt(5x)` , isA. a periodic function with period ` 2pi`B. a periodic function with period `(2pi)/(5)`C. a periodic function with period `(2pi)/(sqrt(5))`D. not a periodic function |
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Answer» Correct Answer - D Clearly , ` 2 cos 5x and 3 sin sqrt(5x)` are periodic functions with periods `(2pi)/(5) and (2pi)/(sqrt(5))` respectively . But , `(2pi)/(5) and (2pi)/(sqrt(5))` do not have a common multiple. Hence , f(x) is not a periodic function. |
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| 44. |
Which one of the following is not periodic ?A. `|sin 3x|+sin^(2)x`B. `cos sqrt(x)+cos^(2)x`C. `cos 4x+tan^(2)x`D. `cos 2x + sinx ` |
| Answer» Correct Answer - B | |
| 45. |
The period of `f(x)=(1)/(2){(| sinx|)/(cos x)+(|cos|)/(sinx)}`, isA. `pi`B. `(pi)/(2)`C. `2pi`D. none of these |
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Answer» Correct Answer - C `f(x)=(1)/(2){(sinx)/(|cosx|)+(|sinx|)/(cosx)}` Since sin x and | cosx| are periodic with period ` 2pi and pi` respectively . Therefore ` ,(sinx)/(|cosx|)` is periodic with period ` 2pi`. Similarly , `(|sinx|)/(cosx)` is periodic with period `2pi`. Hence , f(x) is periodic with period ` 2pi`. |
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| 46. |
The value of integer n for which the function `f(x)=(sinx)/(sin(x / n)` has `4pi` its period isA. 2B. 3C. 5D. 4 |
| Answer» Correct Answer - A | |
| 47. |
The period of `f(x)=(1)/(2){(| sinx|)/(cos x)+(|cos|)/(sinx)}`, isA. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False , Statement-2 is True. |
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Answer» Correct Answer - B We have. `f(x)=(1)/(2)((|sin|)/(cosx)+(sinx)/(|cosx|))` `impliesf(x)={{:(tanx,x in [0,pi//2)),(0,x in(pi//2,pi]),(-tanx,x in [pi,3pi//2)),(0,x in[3pi//2,2pi]):}` Clearly , f(x) is preiodic with period ` 2pi`. So, Statement-1 is ture. Also , Statement-2 is true. |
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| 48. |
Statement-1: The period of sinx , cos x is `2pi` and period of f(x)+g(x) is the LCM of the periods of f(x) and g(x)A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False , Statement-2 is True. |
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Answer» Correct Answer - A Clearly, Statement-2 is true . Now, ` sin""(pi)/(4)[x]=sin(2pi+(pi)/(4)[x])=sin""(pi)/(4)(8+[x])=sin""(pi)/(4)[x+8]` `implies sin""(pi)/(4)[x]` is periodic with period 3. `cot""(pi)/(3)[x]=cot(pi+(pi)/(3)[x])=cot""(pi)/(3)[3+[x])=cot""(pi)/(3)[x+3]` `implies cot""(pi)/(3)[x]` is periodic with period 3. and , ` cos""(pix)/(2)` is periodic with period `(2pi)/(pi//2)=4` Using statement-2, period of f(x) is LCM of (8,3,4)=24 |
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| 49. |
Consider the function f defined on the set of all non-negative interger such that `f(0) = 1, f(1) =0` and `f(n) + f(n-1) = nf(n-1)+(n-1) f(n-2)` for `n ge 2`, then f(5) is equal toA. 40B. 44C. 45D. 60 |
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Answer» Correct Answer - B We have, `f(n) + f(n-1) = nf(n-1) + n(n-1) f(n-2)` `rArr f(n) - nf(n-1) = - { f(n-1) - (n-1) f(m-2)}` `=(-1)^(2) {f(n-2) = - {f(n-1) -(n-1) f(n-2)}` ` = (-1)^(n-1) {(f(1) - f(0)}` `rArr f(n) - n f(n-1) = - (1)^(n)` `rArr (f(n))/(n!) = (f(n-1))/((n-1)!) +((-1)^(n-1))/((n-1)!)` `rArr (f(n-2))/((n-2)!) = (f(n-3))/((n-3)!)+((-1)^(n-2))/((n-2)!)` `(f(2))/(2!) = (f(1))/(1!) +((-1)^(2))/(2!)` `(f(1))/(1!) = (f(0))/(0!) +((-1)^(1))/(1!)` Adding all these equalities, we obtain `(f(n))/(n!) = ((-1)^(n))/(n!)) +((-1)^(n-1))/((n-1)!) +((-1)^(n-2))/((n-2)!) +......+((-1)^(2))/(2!)+((-1)^(1))/(1!)` ` f(n) = n! {1-(1)/(1!)+(1)/(2!) -(1)/(3!)+....+((-1)^(n))/(n!)}` `therefore f(5) = 5! (1-(1)/(1!)+(1)/(2!) -(1)/(3!) + (1)/(4!)-(1)/(5!))= 44` |
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| 50. |
The peroid of the function `f(x) =(|sinx|-|cosx|)/(|sin x + cosx|)` isA. `(pi)/(2)`B. `2pi`C. `pi `D. none of these |
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Answer» Correct Answer - C We observe that `f(x+pi)=(|sin (pi+x)|-|cos(pi+x)|)/(|sin(pi x - cos x)|+cos(x+pi)|)` `rArr f(x +pi)=(|sin x|-|cosx|)/(|-sin x - cosx|)=(|sin x|-|cosx|)/(|sin x + cos x)|` `rArr f(x + pi)= f(x) "for all " x in R` Therefore , `f(x)` is periodic with period `pi`. |
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