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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.
| 551. |
The principal values of `sec^(-1)x` lie inA. `[0,pi]`B. `[0,pi]-{pi/2}`C. `(-pi/2,pi/2)`D. `{pi,2pi}-{(3pi)/2}` |
| Answer» Correct Answer - B | |
| 552. |
Find the principal value of `cos^(-1)(1/2)`. |
| Answer» Correct Answer - `pi/3` | |
| 553. |
Find the principal values of the following (i) `sin^(-1) (sin 1)` (ii) `sin^(-1) (sin 2)` (iii) `sin^(-1) (sin 10)` (iv) `sin^(-1) (sin 20)` (v) `sin^(-1) (sin 100)` (vi) `sin^(-1) (sin. (29 pi)/(5))` |
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Answer» (i) Since `1 in [-pi//2, pi//2], sin^(-1) (sin 1) = 1` (ii) Since `2 in [-pi//2, pi//2], sin^(-1) (sin 2) != 2` `:. Sin^(-1) (sin 2) = sin^(-1) (sin (pi) -2))` `= pi - 2 " as " (pi - 2) in [-pi//2, pi//2]` (iii) `sin^(-1) (sin 10) = sin^(-1) (sin (3 pi - 10))` `= 3 pi - 10 " as " (3 pi - 10) in [-pi//2, pi//2]` (iv) `sin^(-1) (sin (-20)) = sin^(-1) (sin (6 pi - 20))` (v) `sin^(-1) (sin 100) = sin^(-1) (sin (100 - 32 pi))` `= 100 - 32 pi " as " (100 - 32 pi) in [-pi//2, pi//2]` (vi) `sin^(-1) (sin.(29 pi)/(5)) = sin^(-1) (sin(6 pi - (pi)/(5)))` `= sin^(-1) (sin(-(pi)/(5)))` `= -(pi)/(5)` |
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| 554. |
The number of values of `x`for which `sin^(-1)(x^2-(x^4)/3+(x^6)/9)+cos^(-1)(x^4-(x^8)/3+(x^(12))/9ddot)=pi/2,`where `0lt=|x| |
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Answer» `sin^-1(x^2-x^4/3+x^6/9+...)+cos^-1(x^4-x^8/3+x^12/9+...) = pi/2` We know, `sin^-1y+cos^-1y = pi/2` It means, `x^2-x^4/3+x^6/9+... =x^4-x^8/3+x^12/9+...` these are two G.P.s with common ratio `-x^2/3 and -x^4/3.` `x^2/(1+x^2/3) = x^4/(1+x^4/3)->(1)` `=>3/(3+x^2) = (3x^2)/(3+x^4)` `=>9+3x^4 = 9x^2+3x^4` `=>x = +-1` `x = 0` also is a solution for `(1)`. So there are `3` solutions available for the given equation. |
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| 555. |
Solve `2 cos^(-1) x + sin^(-1) x = (2pi)/(3)` |
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Answer» Correct Answer - `(sqrt3)/(2)` `2 cos^(-1) x + sin^(-1) x = (2pi)/(3)` `rArr cos^(-1) x (cos^(-1) x + sin^(-1) x) = (2pi)/(3)` `rArr cos^(-1) x + (pi)/(2) = (2pi)/(3)` `rArr cos^(-1) x = (pi)/(6) rArr x = (sqrt3)/(2)` |
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| 556. |
Sove `sec^(-1) x gt cosec^(-1) x` |
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Answer» Correct Answer - `x in (-oo, -1] uu (sqrt2, oo)` `sec^(-1) x gt pi//2 - sec^(-1) x` `rArr sec^(-1) x gt pi//4` `rArr sec^(-1) x in ((pi)/(4), (pi)/(2)) uu ((pi)/(2) , pi]` `rArr x in (-oo, -1) uu (sqrt2, oo)` |
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| 557. |
Find the range of `f(x) = (sin^(-1) x)^(2) + 2pi cos^(-1) x + pi^(2)` |
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Answer» `f(x) = (sin^(-1) x)^(2) + 2pi ((pi)/(2) - sin^(-1) x) + pi^(2)` `= (sin^(-1) x)^(2) + pi^(2) - 2pi sin^(-1) x + pi^(2)` `= (sin^(-1) x - pi)^(2) + pi^(2)` `f_("min") = ((pi)/(2) - pi)^(2) + pi^(2) = (pi^(2))/(4) + pi^(2) = (5 pi^(2))/(4)` `f_("max") = (-(pi)/(2) -pi)^(2) + pi^(2) = (9pi^(2))/(4) + pi^(2) = (13 pi^(2))/(4)` Therefore, range is `[(5 pi^(2))/(4), (13 pi^(2))/(4)]` |
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| 558. |
Find the principal value of the following (i) `cos^(-1) (cos 3)` (ii) `cos^(-1) (cos 4)` (iii) `cos^(-1) (cos 15)` (iv) `cos^(-1) (cos 30)` (v) `cos^(-1) (cos 50)` (vi) `cos^(-1) (cos.(48pi)/(7))` |
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Answer» (i) Since `3 in [0, pi], cos^(-1) (cos 3) = 3` (ii) Since `4 !in [0, pi], cos^(-1) (cos 4) != 4` `:. Cos^(-1) (cos 4) = 2pi - 4` (iii) `cos^(-1) (cos 15) = 15 - 4 pi` (iv) `cos^(-1) (cos 30) = 10 pi - 30` (v) `cos^(-1) (cos 50) = 16 pi - 50` (vi) `cos^(-1) (cos.(48pi)/(7)) = cos^(-1) (cos(6pi + (6pi)/(7)))` `= cos^(-1) (cos.(6pi)/(7)) = (6pi)/(7)` |
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| 559. |
The number of values of `x`for which `sin^(-1)(x^2-(x^4)/3+(x^6)/9)+cos^(-1)(x^4-((x^8)/3+(x^(12))/9ddot)=pi/2,`where `0lt=|x| |
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Answer» Correct Answer - 3 `sin^(-1) (x^(2) -- (x^(4))/(3) + (x^(6))/(9) -...) + cos^(-1) (x^(4) -(x^(8))/(3) + (x^(12))/(9) ...) = (pi)/(2)` `rArr (x^(2) - (x^(4))/(3) + (x^(6))/(9)..) = (x^(4) - (x^(8))/(3) + (x^(12))/(9)...)` or `(x^(2))/(1 + (x^(2))/(3)) = (x^(4))/(1 + (x^(4))/(3))` or `(3)/(3 + x^(2)) = (3x^(2))/(3 + x^(4)) " or "x = 0` or `9 + 3x^(4) = 9x^(2) + 3x^(4) " or " x = 0` or `x^(2) = 1 rArr x = 0, 1 " or " -1` |
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| 560. |
If `sin^(-1)x+sin^(-1)y=(2pi)/3` and `cos^(-1)x-cos^(-1)y=-pi/3` then the number of values of `(x,y)` is |
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Answer» Correct Answer - `x = (1)/(2), y = 1` Given equation are `sin^(-1) x + sin^(-1) y = (2pi)/(3)` `cos^(-1) x - cos^(-1) y = (pi)/(3)` `rArr ((pi)/(2) - sin^(-1) x) - ((pi)/(2) - sin^(-1) y) = (pi)/(3)` Let `sin^(-1) x = A` `sin^(-1) y = B` Then Eqs. (i) and (ii) become `A + B = (2pi)/(3)` `A - B = -(pi)/(3)` Solving Eqs. (iii) and (iv), we get `A = (pi)/(6), B = (pi)/(2)` `rArr sin^(-1) x = (pi)/(6), sin^(-1) y = (pi)/(2)` `rArr x = (1)/(2) and y = 1` |
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| 561. |
Find the principal values of the following (i) `tan^(-1) (tan. (2pi)/(3))` (ii) `tan^(-1) (tan (-6))` |
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Answer» (i) `tan^(-1) (tan.(2pi)/(3)) != (2pi)/(3), " as " (2pi)/(3)` does not lie between `- (pi)/(2) and (pi)/(2)` Now, `tan^(-1) (tan.(2pi)/(3)) = tan^(-1) (tan(pi - (pi)/(3)))` `= tan^(-1) (- tan. (pi)/(3))` `= tan^(-1) (tan(-(pi)/(3)))` `= -(pi)/(3)` (ii) `tan^(-1) (tan(-6)) = tan^(-1) (tan(2 pi -6))` `= 2pi - 6 " as " (2pi - 6) in (-pi//2, pi//2)` |
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| 562. |
For the equation `cos^(-1)x+cos^(-1)2x+pi=0`, the number of real solution isA. 1B. 2C. 0D. `oo` |
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Answer» Correct Answer - C We have `cos^(-1) x + cos^(-1) (2x) = -pi`, which is not possible as `cos^(-1) x and cos^(-1) 2x` never take negative values |
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| 563. |
The number of integral values of x satisfying the equation `tan^(-1) (3x) + tan^(-1) (5x) = tan^(-1) (7x) + tan^(-1) (2x)` is ____ |
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Answer» Correct Answer - 1 `tan^(-1) (3x) + tan^(-1) (5x) = tan^(-1) (7x) + tan^(-1) (2x)` or `tan^(-1) (3x) - tan^(-1) (2x) = tan^(-1) (7x) - tan^(-1) (5x)` or `tan^(-1) ((3x -2x)/(1 + 6x^(2))) = tan^(-1) ((7x -5x)/(1 + 35x^(2)))` or `(x)/(1 + 6x^(2)) = (2x)/(1 + 25 x^(2))` `rArr x = 0 " or " 1 + 35x^(2) = 2 + 12 x^(2)` `rArr x = 0 " or " x = (1)/(sqrt23) "or " -(1)/(sqrt23)` |
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| 564. |
The value of `a`for which `a x^2+sin^(-1)(x^2-2x+2)+cos^(-1)(x^2-2x+2)=1`has a real solution is`pi/2`(b) `-pi/2`(c) `2/pi`(d) `-2/pi`A. `(pi)/(2)`B. `-(pi)/(2)`C. `(2)/(pi)`D. `-(2)/(pi)` |
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Answer» Correct Answer - B The given equation is `ax^(2) + sin^(-1) ((x -1)^(2)+1) + cos^(-1) ((x -1)^(2) + 1) = 0` Now, `-1 le (x -1)^(2) + 1 le 1 rArr x = 1` So, we have `a + (pi)/(2) = 0 " or " a = -(pi)/(2)` |
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| 565. |
If `2^2pi//sin^((-1)x)-2(a+2)^pi//sin^((-1)x)+8a |
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Answer» Correct Answer - D `2^(2pi//sin^(-1)x) - 2 (a + 2) 2^(pi//sin^(-1) x) + 8a lt 0` `(2^(pi//sin^(-1)x) - 4) (2^(pi//sin^(-1)x) - 2a) lt 0` Now `2^(pi//sin^(-1) x) in (0, (1)/(4)] uu [4, oo)` For `2^(pi//sin^(-1)x) in (0, (1)/(4)]`. We have `(2^(pi//sin^(-1)x) -4) lt 0` `:. 2^(pi//sin^(-1)x) -2a gt 0` or `2a lt (1)/(4)` or `0 le a (1)/(8)` Similarly, for `2^(pi//sin^(-1) x) in [4, oo), a gt 2`, we get So, `a in [0, (1)/(8)) uu (2, oo)` |
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| 566. |
The sum `tan^(-1).(1)/(1+x+x^2)+tan^(-1).(1/(3+3x+x^2))+tan^(-1).(1/(7+5x+x^2))+tan^(-1)(1/(13+7x+x^2))` of first 100 terms of the series isA. `tan^(-1) (100/(1+x^2+100x))`B. `tan^(-1)((2x-100)/(1-x^2-100x))`C. `tan^(-1)(100/(1+x^2-100x))`D. `tan^(-1)((2x+100)/(1+x^2-100x))` |
| Answer» Correct Answer - A | |
| 567. |
The set of values of k for which `x^2 - kx + sin^-1 (sin 4) > 0` for all real x isA. `phi`B. `(-2,2)`C. RD. none of these |
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Answer» We have `sin^(-1)(sin4)=sin^(-1)sin(pi-4)=pi-4` `therefore x^(2)-lambdax+sin^(-1)(sin4)gt0` for all `x in R` `rarr x^(2)-lambdax+(pi-4)gt0` for all `x in R` `rarr lambda^(2)-4(pi-4)lt0rarr0lambda^(2)+16-4pi lt 0` But `lambda^(2)+16-4pigt0` But `lambda^(2)+16-4pigt0 for all lambda in R` So there is no value of `lambda` for which the given innequation holds true |
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| 568. |
The sum of the infinte series `sin^(-1)(1/sqrt(2))+sin^(-1)((sqrt(2)-1)/(sqrt(6)))+....sin^(-1)((sqrt(n)-sqrt(n-1))/(sqrt(n(n+1))))`A. `pi/3`B. `pi/2`C. `pi/4`D. `pi/6` |
| Answer» Correct Answer - B | |
| 569. |
The set of values of k for which `x^2 - kx + sin^-1 (sin 4) > 0` for all real x isA. No values of k is possibleB. `[-4(pi-4),4(pi-4)]`C. `[0,1]`D. All real values of k |
| Answer» Correct Answer - A | |
| 570. |
If `cot^(-1)((n^2-10 n+21. 6)/pi)>pi/6,`where `x y |
| Answer» Correct Answer - B | |
| 571. |
Find the value of x for which `sec^(-1) x + sin^(-1) x = (pi)/(2)` |
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Answer» We know that `sec^(-1) x` is defined for `x in (-oo, -1] uu [1, oo)`. But `sin^(-1) x` is defined for `x in [-1, 1]` Hence, `sec^(-1) x + sin^(-1) x = (pi)/(2) " for " x = +- 1` |
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| 572. |
Evaluate `tan(cosec^(-1)(5/3))`. |
| Answer» Correct Answer - `3/4` | |
| 573. |
Solution(s) of the equation `sin(-pi/3+tan^(-1)x+cot^(-1)x)=1/2` is/areA. `1/2`B. `-1/2`C. `sqrt3`D. `-sqrt3` |
| Answer» Correct Answer - A::B::C::D | |
| 574. |
If `sin^(-1) (x^(2) + 2x + 2) + tan^(-1) (x^(2) - 3x - k^(2)) gt (pi)/(2)`, then find the value of k |
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Answer» `sin^(-1) (x^(2) + 2x + 2) + tan^(-1) (x^(2) - 3x - k^(2)) gt (pi)/(2)` `rArr sin^(-1) ((x + 1)^(2) + 1) + tan^(-1) (x^(2) - 3x - k^(2) gt (pi)/(2)` Now, `sin^(-1) ((x + 1)^(2) + 1)` is defined if `(x + 1)^(2) = 0 " or " x = -1` Putting `x = -1`, we get `(pi)/(2) + tan^(-1) (4 - k^(2)) gt (pi)/(2)` `rArr tan^(-1) (4 - k^(2)) gt 0` `rArr 4 - k^(2) gt 0` `rArr -2 lt k lt 2` |
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| 575. |
If the maximum value of `(sec^-1 x)^2 + (cosec^-1 x)^2` approaches a, the minimum value of `(tan^-1 x)^3 +(cot^-1 x)^3` approaches b then `(a+b/pi)` is equal toA. `pi^2`B. `(41pi^2)/32`C. `pi^2/32`D. `(43pi^2)/32` |
| Answer» Correct Answer - B | |
| 576. |
Prove that `tan^(-1) x + tan^(-1).(1)/(x) = {(pi//2,"if" x gt 0),(-pi//2," if " x lt 0):}` |
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Answer» We know that `tan^(-1) ((1)/(x)) = {(cot^(-1) x,x gt 0),(-pi + cot^(-1) x,x lt 0):}` `rArr tan^(-1) x + "tan"^(-1)(1)/(x) = {(tan^(-1) x + cot^(-1) x,x gt 0),(-pi + cot^(-1) x + tan^(-1) x,x lt 0):}` `={((pi)/(2),x gt0),(-pi + (pi)/(2),x lt 0):}` `= {((pi)/(2),x gt 0),((-pi)/(2),x lt 0):}` |
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| 577. |
Prove that `3 tan^(-1) x= {(tan^(-1) ((3x - x^(3))/(1 - 3x^(2))),"if " -(1)/(sqrt3) lt x lt (1)/(sqrt3)),(pi + tan^(-1) ((3x - x^(3))/(1 - 3x^(2))),"if " x gt (1)/(sqrt3)),(-pi + tan^(-1) ((3x - x^(3))/(1 - 3x^(2))),"if " x lt - (1)/(sqrt3)):}` |
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Answer» Let `x = tan theta`, where `theta in (-pi//2, pi//2)` `:. Tan^(-1). (3 x -x^(3))/(1 - 3x^(2)) = tan^(-1).(3 tan theta - tan^(3) theta)/(1 -3 tan^(2) theta)` `= tan^(-1)(tan 3 theta), " where " 3 theta in (-3pi//2, 3 pi//2)` `:. Tan^(-1).(3x - x^(3))/(1 - 3x^(2)) = {(3 theta,"if "-(pi)/(2) lt 3 theta lt (pi)/(2)),(3 theta - pi,"if " (pi)/(2) lt 3 theta lt (3pi)/(2)),(3 theta + pi,"if " -(3pi)/(2) lt 3 theta lt -(pi)/(2)):}` Now if `-(pi)/(2) lt 3 theta lt (pi)/(2)` `-(pi)/(2) lt 3 tan^(-1) x lt (pi)/(2)` `rArr -(pi)/(6) lt tan^(-1) x lt (pi)/(6)` `rArr -(1)/(sqrt3) lt x lt (1)/(sqrt3)` Similarly, from `(pi)/(2) lt 3 theta lt (3pi)/(2)`, we get `x gt (1)/(sqrt3)` And from `-(3pi)/(2) lt 3 theta lt -(pi)/(2)`, we get `x lt -(1)/(sqrt3)` Thus, tan^(-1).(3x -x^(3))/(1 - 3x^(2)) = {(3 tan^(-1) x,"if " -(1)/(sqrt3) lt x lt (1)/(sqrt3)),(3 tan^(-1) x - pi," if " (1)/(sqrt3) lt x lt oo),(3 tan^(-1) x + pi,"if " -oo lt x lt -(1)/(sqrt3)):}` |
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| 578. |
The value of `cot^(-1)(-1)+cosec^(-1)-sqrt(2)+sec^(-1)(2)` isA. `(5pi)/(6)`B. `(2pi)/(3)`C. `(7pi)/(6)`D. `(pi)/(6)` |
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Answer» `cot^(-1)(-1)+cosec^(-1)(-sqrt(2))+sec^(-1)(2)` `=(3pi)/(4)-(pi)/(4)+(pi)/(3)=(5pi)/(6)` |
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| 579. |
STATEMENT -1 : `tan^(-1)x=sin^(-1)yrArry in (-1,1)` and STATEMENT -2 : `-pi/2 lt tan^(-1)x lt pi/2`A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -1B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -1C. Statement-1 is True, Statement -2 is FalseD. Statement -1 is False, Statement -2 is True |
| Answer» Correct Answer - A | |
| 580. |
Evaluate the following:1. `sin^(-1)(s i n(2pi)/3)`2. `cos^(-1)(c o s(7pi)/6)`3. `tan^(-1)(t a n(2pi)/3)` |
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Answer» 1)`sin^(-1)sin(2/3pi)` `sin^(-1)sintheta=theta` `sin^(-1)sin(pi-pi/3)` `sin^(-1)sin(pi/3)` `pi/3` 2)`cos^(-1)cos(7/6pi)` `cos^(-1)cos(2pi-(2lambda)/6)` `cos^(-1)cos(5/6pi)` `5/6pi` 3)`tan^(-1)tan(2/3pi)` `tan^(-1)(tan(pi-2/3pi))` `tan^(-1)(-tanpi/3)` `-pi/3`. |
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| 581. |
The vlaue of `tan^(-1)sqrt(3)-sec^(-1)(-2)+cosec^(-1)(2)/sqrt(3)` isA. `(pi)/(3)`B. `-(pi)/(3)`C. 0D. `(4pi)/(3)` |
| Answer» `tan^(-1)sqrt(3)-sec^(-1)(-2)+cosec^(-1)(2)/sqrt(3)=(pi)/(3)-(2pi)/(3)+(pi)/(3)=0` | |
| 582. |
If `tan^(-)(x/(pi))lt (pi)/(3) ,x in N` then the maximum vlaue of x isA. 2B. 5C. 7D. none of these |
| Answer» Correct Answer - B | |
| 583. |
If x takes negative permissible vlaue then `sin^(-1)x=`A. `cos^(-1)sqrt(1-x^(2))`B. `-cos^(-1)sqrt(1-x^(2))`C. `cos^(-1)sqrt(x^(2)-1)`D. `pi-cos^(-1)sqrt(1-x^(2))` |
| Answer» Correct Answer - B | |
| 584. |
The principal values of `cos^(-1)(-sin(7pi)/(6))` isA. `(5pi)/(3)`B. `(7pi)/(6)`C. `(pi)/(3)`D. none of these |
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Answer» We have `cos^(-1)-sin(7pi)/(6)=cos^(-1)-sin(pi+(pi)/(6))=cos^(-1)sin(pi)/(6)` `=(pi)/(2)-sin^(-1)sin(pi)/(6)=(pi)/(2)-(pi)/(6)=(pi)/(3)` |
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| 585. |
The vlaue of `tan^(-1)1/2+tan^(-1)1/3` is |
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Answer» We have `tan^(-1)1/2+tan^(-1)1/3=tn^(-1)(1//2+1//3)/(1-1//2xx1//3)=tan^(-1)1=(pi)/(4)` |
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| 586. |
If `tan^(-1)x+2cot^(-1)x=(2pi)/(3)` then x =A. 3B. `sqrt(3)`C. `sqrt(2)`D. `(sqrt(3)-1)/(sqrt(3)+1)` |
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Answer» We have `tan^(-1)x+2cot(-1)x=(2pi)/(3)` `rarr tant^(-1)x+2(pi)/(2)-tan^(-1)x=(2pi)/(3)rarr tan^(-1)x=(n)/(3)rarrx=sqrt(3)` |
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| 587. |
prove that `tan^-1(cosx/(1-sinx))-cot^-1(sqrt(1+cosx)/sqrt(1-cosx))=pi/4 , xepsilon(0,pi/2)` |
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Answer» `tan^(-1)A -tan^(-1)B=tan^(-1)((A-B)/(1+AB))`.......(1) Also `cot^(-1)x=tan^(-1)(1/x)` `tan^(-1)(cosx/(1-sinx))-tan^(-1)sqrt(((1-cosx)(1-cosx))/((1+cosx)(1-cosx))` `tan^(-1)(cosx/(1-sinx))-tan^(-1)((1-sinx)/cosx)` using (1) we get`tan^(-1)1=pi/4` |
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| 588. |
Prove that `cot(pi/4-2cot^(- 1)3)=7` |
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Answer» Let `cot^(-1)3=theta` the equation reduces to=>`Cot(pi/4-2theta)=1/(tan(pi/4-2theta))=(1+tan2theta)/(1-tan2theta)` since `cottheta=3 ` then `tantheta=1/3` `tan2theta=(2tantheta)/(1-tan^2(theta))`=> substituting value we get, `tan2theta=3/4` thus`(1+tan2theta)/(1-tan2theta)=7` hence proved |
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| 589. |
The value of `sec^(2)(tan^(-1)2)+cosec^(2)(cot^(-1)3)` isA. 5B. 10C. 15D. 20 |
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Answer» We have `sec^(2)(tan^(-1)2)+cosec^(2)(cot^(-1)3)` `{sec(tan^(-1)2)}^(2)+{cosec(cot^(-1)3)` `=[sec(tan^(-1)2)^(2)+{cosec(cot^(-1)3)^(2)` `={sec(sec^(-1))sqrt(5)^(2)+cosec (cot^(-1))3/1}^(2)` `{sec(sec^(-1)2/1)}^(2)+(cosec(cosec^(-1))sqrt(10))^(2)` `=sqrt(5)^(2)+sqrt(10)^(2)=15` |
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| 590. |
Prove that:`tan^(-1)(63/16)=sin^(-1)(5/13)+cos^(-1)(3/5)` |
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Answer» `R.H.S -> sin^-1(5/13)+cos^-1(3/5)` We can create two triangles with angle `theta` and `phi` as expalined in video. In that case, `sin^-1(5/13)` can be written as `tan^-1(5/12)` `cos^-1(3/5)` can be wriiten as `tan^-1(4/3)` So, we can rewrite R.H.S. as `tan^-1(5/12)+tan^-1(4/3)`As, `tan^-1x+tan^-1y = tan^-1((x+y)/(1-xy))`, above can be wtitten as `tan^-1(5/12+4/3)/(1-5/12**4/3)` `=tan^-1((7/4)/(4/9)) = tan^-1(63/16) = L.H.S.` |
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| 591. |
The value of cot`[sin^(-1){cos(tan^(-1)1)}]` isA. `2/3`B. `sqrt(12)/(3)`C. `(1)/sqrt(2)`D. `sqrt(3)/(2)` |
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Answer» `sin[cot^(-1){cos(tan^(-1)1)}]` `sin{cot^(-1)(cos(pi)/(4))}` `=sin(cot^(-1)(1)sqrt(2))` `=sin(sin^(-1)sqrt(2)/sqrt(3))=sqrt(2/3)"before"^(-1)(1)/(sqrt(2)=sin^(-1)sqrt(2)/sqrt(3)]` |
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| 592. |
Prove that: `sin^(-1)(4/5)+sin^(-1)(5/(13))+sin^(-1)((16)/(65))=pi/2`A. `(3pi)/2`B. `pi/2`C. `pi-sin^(-1).(3713)/4225`D. `pi-tan^(-1).(3713)/2016` |
| Answer» Correct Answer - B | |
| 593. |
Prove the followingresults:`tan(sin^(-1)(15)/(13)+cos^(-1)3/5)=(63)/(16)`(ii) `sin(cos^(-1)3/5+sin^(-1)5/(13))=(63)/(65)`A. `48/65`B. `15/65`C. `33/65`D. `63/65` |
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Answer» `sin(cos^(-1)3/5+cosec^(-1)13/5)` `=sin(cos^(-1)3/5)cos(cosec^(-1)13/5)+cos(cos^(-1)3/5)` `sin(cosec^(-1)13/5)` `=sin(sin^(-1)4/5)cos(cos^(-1)12/13)+cos(cos^(-1)3/5)sin(sin^(-1)5/13)` `4/5xx12/13xx3/5xx5/13=63/65` |
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| 594. |
If `x_1,x_2,x_3,x_4,x_5,x_6` all are independent then the maximum and minimum values of `[sin^(-1)x_1]+[cos^(-1)x_2]+[tan^(-1)x_3]+[cot^(-1)x_4]+[sec^(-1)x_5]+[cosec^(-1)x_6]`, where [] represents greatest integer function, respectively areA. 9,3B. 11,5C. 12,-6D. 12,-3 |
| Answer» Correct Answer - D | |
| 595. |
If `a_1, a_2,a_3, ,a_n`is an A.P. with common difference `d ,`then prove that`"tan"[tan^(-1)(d/(1+a_1a_2))+tan^(-1)(d/(1+a_2a_3))+tan^(-1)(d/(1+a_(n-1)a_n))]=((n-1)d)/(1+a_1a_n)` |
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Answer» As `a_1,a_2,a_3...a_n` are in A.P., So, the left hand side can be written as, `L.H.S. = tan[tan^-1((a_2-a_1)/(1+a_1a_2))+tan^-1((a_3-a_2)/(1+a_2a_3))+...+tan^-1((a_n-a_(n-1))/(1+a_(n-1)a_n))]` `=tan[tan^-1a_2-tan^-1a_1+tan^-1a_3-tan^-1a_2+tan^-1a_4-tan^-1a_3...+tan^-1a_n-tan^-1a_(n-1)]` `=tan(tan^-1(a_n) - tan^-1(a_1))` `=tan(tan^-1((a_n-a_1)/(1+a_1a_n)))` As, `a_n - a_1 = (n-1)*d` So, it becomes, `=tan(tan^-1(((n-1)d)/(1+a_1a_n)))` `=((n-1)d)/(1+a_1a_n) = R.H.S.` |
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