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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.
| 301. |
The solution of the inequality `"log"_(2) sin^(-1) x gt "log"_(1//2) cos^(-1) x` isA. `x in [0, (pi)/(sqrt2)]`B. `x in ((1)/(sqrt2), 1]`C. `x in (0, (1)/(sqrt2))`D. none of these |
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Answer» Correct Answer - C `"log"_(1//2) sin^(-1) x gt "log"_(1//2) cos^(-1) x` `rArr cos^(-1) x gt sin^(-1) x, 0 lt x lt 1` `rArr cos^(-1) x gt (pi)/(2) - cos^(-1) x, 0 lt x lt 1` `rArr cos^(-1) x gt (pi)//(4), 0 lt x lt 1` `rArr 0 lt x lt (1)/(sqrt2)` |
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| 302. |
Which of the following is/are the value of `cos [(1)/(2) cos^(-1) (cos (-(14pi)/(5)))]` ?A. `cos(-(7pi)/(5))`B. `sin((pi)/(10))`C. `cos((2pi)/(5))`D. `-cos((3pi)/(5))` |
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Answer» Correct Answer - B::C::D `cos (-(14pi)/(5)) = cos.(14pi)/(5) = cos.(4pi)/(5)` Hence, `cos.(1)/(2) cos^(-1) (cos.(4pi)/(5)) = cos.(2pi)/(5)` |
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| 303. |
Which of the following is not a rational number.A. `sin (tan^(-1) 3 + tan^(-1).(1)/(3))`B. `cos ((pi)/(2) - sin^(-1).(3)/(4))`C. `"log"_(2) (sin ((1)/(4) sin^(-1).(sqrt63)/(8)))`D. `tan ((1)/(2) cos^(-1).(sqrt5)/(3))` |
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Answer» Correct Answer - A::B::C (1) `sin (tan^(-1) 3 + tan^(-1).(1)/(3)) = sin.(pi)/(2) =1` (2) `cos((pi)/(2) - sin^(-1).(3)/(4)) = cos (cos^(-1).(3)/(4)) = (3)/(4)` (3) `sin((1)/(4) sin^(-1). (sqrt63)/(8))` Let `sin^(-1).(sqrt63)/(8) = theta`. Then, `sin theta = (sqrt63)/(8) rArr cos theta = (1)/(8)` We have `cos.(theta)/(2) = sqrt((1 + cos theta)/(2)) = (3)/(4)` `rArr sin.(theta)/(4) = sqrt((1 - cos.(theta)/(2))/(2)) = (1)/(2sqrt2)` Now, `"log"_(2) sin ((1)/(4) sin^(-1).(sqrt63)/(8)) = "log"_(2) (1)/(2sqrt2) = -(3)/(2)` (4) `cos^(-1).(sqrt5)/(3) = theta " or " cos theta = (sqrt5)/(3)` `:. tan.(theta)/(2) = sqrt(1 - cos theta)/(1 + cos theta)) = (3-sqrt5)/(2)`, which is irrational |
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| 304. |
If `cot^(-1)(n/(pi))>(pi)/6, n in N`, then the maximum value of n is :A. 1B. 5C. 9D. none of these |
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Answer» We have `cot^(-1)(n)/(pi)gt(pi)/(6)` `rarr cot(cot^(-1)(n)/(pi)}ltcot(pi)(6)` `rarr (n)/(pi)lt sqrt(3)rarr n lt sqrt(3)pi=5.5` So the maximum value of n is 5 |
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| 305. |
The value of `tan^(-1)tan(-6)`isA. `2pi-6`B. `2pi+6`C. `6-2pi`D. `3pi-6` |
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Answer» We know that `tan^(-1)(tan theta)=theta if -(pi)/(2)le theta le(pi)/(2)` Here`theta =-6` radians which does not lie between However `2pi-6` lies between `-(pi)/(2)and (pi)/(2)` such that `tan(2pi-6)=-tan 6 = tan(-6)` `therefore tan^(-1){tan^(-6)}=tan^(-1){tan(tan(2pi-6))}=2pi-6` |
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| 306. |
Evaluate: `cos(sin^(-1)1/4+sec^(-1)4/3)`A. `(3sqrt(15)-sqrt(7))/(16)`B. `(3sqrt(15)+sqrt(7))/(16)`C. `(sqrt(7)-3sqrt(15))/(16)`D. `(3sqrt(15)-sqrt(7))/(4)` |
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Answer» `cos(sin^(-1)1/4+sec^(-1)4/3)` `=cos(sin^(-1)1/4+cos^(-1)3/4)` `cos(sin^(-1)1/4)cos(cos^(-1)3/4)-sin^(-1)1/4)sin(cos^(-1)3/4)` `=cos(cos^(-1))sqrt(15)/(4)cos(cos^(-1)3/4)-sin(sin^(-1)1/4)sin(sin^(-1)sqrt(7)/(4))` `=sqrt(15)/(4)xx3/4-1/4xxsqrt(7)/(4)=3sqrt(15)-sqrt(7)/(16)` |
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| 307. |
Which of the following quantities is/are positive ?A. `cos (tan^(-1) (tan 4))`B. `sin(cot^(-1)(cot 4))`C. `tan (cos^(-1) (cos 5))`D. `cot(sin^(-1) (sin 4))` |
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Answer» Correct Answer - A::B::C (1) `cos (tan^(-1) (tan (4 - pi)))` `= cos(4 - pi) = cos (pi -4) = - cos 4 gt 0` (2) `sin(cot^(-1) (cot(4 - pi)))` `= sin(4 - pi) = - sin 4 gt 0 ("as " sin 4 lt 0)` (3) `tan (cos^(-1) (cos (2 pi - 5)))` `= tan (2pi - 5) = -tan 5 gt 0 ("as " tan 5 lt 0)` (4) `cot(sin^(-1) (sin (pi -4)) = cot (pi -4) = - cot 4 lt 0` |
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| 308. |
The value of`sin^(-1){cot(sin^(-1)sqrt((2-sqrt(3))/4)+cos^(-1)(sqrt(12))/4+sec^(-1)sqrt(2))}`is equal to`pi/4`(b) `pi/2`(c) 0(d) `-pi/2`A. `(pi)/(6)`B. `(pi)/(6)`C. 0D. `(pi)/(2)` |
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Answer» `sin^(-1){cot(sin^(-1)sqrt(2-sqrt(3))/(4)+cos^(-1)sqrt(12)/(4)+sec^(-1)sqrt(2))}` `=sin^(-1){cot(sin^(-1)sqrt(3)-1)/(2sqrt(2))+cos^(-1)sqrt(3)/(2)+sec^(-1)sqrt(2))}` `sin^(-1){cost(pi)/(12)+(pi)/(6)+(pi)/(4)}=sin^(-1)(cot(pi)/(2))=sin^(-1)0=0` |
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| 309. |
if `6sin_1(x^2-6x+12)=2pi`, then the value of `x`, isA. 1B. 2C. 3D. does not exist |
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Answer» We have `x^(2)-6x+12=(x-3)^(2)+3ge3e` for all x `therefore sin^(-1)(x^(2)-6x+12)` does not exist Thus there is no value of x satisfying the given equation |
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| 310. |
Which of the following is the solution set of the equation `sin^-1 x = cos ^-1 x+ sin ^-1(3x-2)`A. `[0,1/3]`B. `[1/3,2/3]`C. `[0,2/3]`D. none of these |
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Answer» For the existance of the given equation we must have `-1lexle1and-1le3x-1le1rarr0lexle2/3` Now `sin^(1)x=cps^(-1)x+sin^(-1)(3x-1)` `rarr sin^(-1)x-cos^(-1)x=sin^(-1)(3x-1)` `rarr 2sin^(-1)x-(pi)/(2)=sin^(-1)(3x-1)` `rarr sin(2sin^(-1)x-(pi)/(2))=sin(sin^(-1))(3x-1)` `rarr =cos(2sin^(-1)x)=3x-1` `rarr -{1-2sin^(2)(sin^(-1)x))}=3x-1` `rarr -(1-2x^(2)=3x-1)` `rarr 2x^(2)-3x=0rarr x=0 3/2rarrx=0` |
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| 311. |
If `(sin^(-1)x+sin^(-1)w)(sin^(-1)y+sin^(-1)z)=pi^2,`then`D=|x^(N_1)y^(N_3)z^(N_3)w^(N_4)|(N_1,N_2,N_3,N_4 in N)`has a maximum value of 2has a maximum value of 016 different D are possiblehas a minimum value of `-2` |
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Answer» `(sin^-1x+sin^-1w)(sin^-1y+sin^-1z) = pi^2` Maximum value of `sin^-1x` is `pi/2` and minimum value of `sin^-1x` is `-pi/2`. So, the solution for the given equation can be, `x = y = z = w = 1` or `x = y = z = w = -1` Now, `D = |[x^(N_1),y^(N_2)],[z^(N_3),w^(N_4)]|` Maximum value of `D` will be when `y^(N_2) = -1` `:. D = |[1,-1],[1,1]| = 2` Minimum value of `D` will be when `x^(N_1) = -1` `:. D = |[-1,1],[1,1]| = -2` Now, there are `2` possible values possible for each `x,y,z and w`. So, number of possible `D` are `2^4 = 16.` |
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| 312. |
If `log_(2)xge0` then `log_(1//pi){sin^(-1)((2x)/(1+x^(2)))+2tan^(-1)x}` is equalA. `log_(1//pi)(4 tan^(-1)x)`B. 0C. `-1`D. none of these |
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Answer» We have `log_(2) xge 0rarr xge 2^(@)=1` for `xge1` we have `sin^(-1)(2x)/(1+x^(2))=pi-2 tan^(-1)x` `therefore log_(1//pi){sin^(-1)(2x)/(1+x^(2))+2 tan^(-1)x}` `=log_(1//pi)(pi-2tan^(-1)x+2 tan ^(-1)x)` `=log_(1//pi) pi=-1` |
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| 313. |
Let a, b and c be positive real numbers. Then prove that `tan^(-1) sqrt((a(a + b + c))/(bc)) + tan^(-1) sqrt((b (a + b + c))/(ca)) + tan^(-1) sqrt((c(a + b+ c))/(ab)) = pi` |
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Answer» Let `S = tan^(-1) sqrt((a(a + b + c))/(bc)) tan^(-1) sqrt((b(a + b + c))/(ca)) + tan^(-1) sqrt((c (a + b + c))/(ab))` Now, `sqrt((a(a + b + c))/(bc)) sqrt((b (a + b + c))/(ca)) = (a + b + c)/(c) = 1 + (b)/(c) + (a)/(c) gt 1` `S = pi + tan^(-1). (sqrt((a (a + b + c))/(bc)) + sqrt((b(a + b + c))/(ca)))/(1- sqrt((a(a + b + c))/(bc)) sqrt((b(a + b + c))/(ca))) + tan^(-1) sqrt((c(a + b + c))/(ab))` `= pi + tan^(-1). (sqrt((a + b + c)/(c)) (sqrt((a)/(b)) + sqrt((b)/(a))))/(1 - (a + b + c)/(c)) + tan^(-1) sqrt((c(a + b + c))/(ab))` `= pi + tan^(-1) (- sqrt((c (a + b + c))/(ab)) ) + tan^(-1) sqrt((c(a + b + c))/(ab))` `= pi` |
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| 314. |
prove that `2tan^-1 1/3+tan^-1 1/7=pi/4`A. `tan^(-1)(49/29)`B. `(pi)/(2)`C. 0D. `(pi)/(4)` |
| Answer» Correct Answer - D | |
| 315. |
If a,b,c are real positive numbers and `theta =tan^(-1)[(a(a+b+c))/(bc)]^(1/2)+tan^(-1)[(b(a+b+c))/(ca)]^(1/2)+tan^(-1)[(c(a+b+c))/(ab)]^(1/2)`, then `tantheta` equals |
| Answer» Correct Answer - A | |
| 316. |
The number of real solution of the equation `tan^(-1) sqrt(x^2-3x +7) + cos^(-1) sqrt(4x^2-x + 3) = pi` is |
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Answer» Since `sqrt(x^2-3x+7) ge 0 rArr 0 le tan^(-1)sqrt(x^2-3x+7) gtpi/2` `sqrt(4x^2-x+3) ge 0 rArr 0 lt cos^(-1)sqrt(4x^2-x+3) le pi/2` `rArr 0 lt LHS lt pi rArr `The given equation has no solution. |
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| 317. |
If `cot^(-1)((n^2-10 n+21. 6)/pi)>pi/6,`where `x y |
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Answer» `cot^-1((n^2-10n+21.6)/pi) gt pi/6` `=>(n^2-10n+21.6)/pi gt cot (pi/6)` `=>(n^2-10n+25 -25 + 21.6)/pi gt sqrt3` `=>(n-5)^2 - 3.4 gt sqrt3pi` `=>(n-5)^2 gt sqrt3pi+3.4` `=>-sqrt(sqrt3pi+3.4) lt n-5 lt sqrt(sqrt3pi+3.4)` `=>2.1 lt n lt 7.9` So, integral values of `n` can be `3,4,5,6,7`. So, options `a` and `c` are the correct options. |
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| 318. |
If `sin^(-1)x+sin^(-1)y=pi/2,t h e n(1+x^4+y^4)/(x^2-x^2y^2+y^2)`is equal to1 (b)2 (c) `1/2`(d) none of these |
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Answer» `sin^(-1)x=pi/2-sin^(-1)y` `sin^(-1)x=cos^(-1)y` `sin^(-1)x=sin^(-1)sqrt(1-y^2)` `x=sqrt(1-y^2` `x^2=1-y^2` `x^2+y^2=1-(1)` `(1+x^4+y^4)/((x^2-x^2y^2)+y^2)=(1+(x^2+y^2)^2-2x^2y^2)/(1-x^2y^2)` `(1+(x^2+y^2)^2-2x^2y^2)/(1-x^2y^2)` `(22-2x^2y^2)/(1-x^2y^2)` `2(1-x^2y^2)/(1-x^2y^2)` `2` Option B is correct. |
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| 319. |
If `0 le x le 1 then cos^(-1)(2x^(2)-1)` equalsA. `2 cos^(-1)x`B. `pi-2 cos^(-1)x`C. `2pi -2 cos^(-1)x`D. none of these |
| Answer» Correct Answer - B | |
| 320. |
`sin[1/2cos^(- 1)4/5]`A. `-(1)/sqrt(10)`B. `(1)/sqrt(10)`C. `-(1)/(10)`D. `1/10` |
| Answer» Correct Answer - B | |
| 321. |
solve `sin^(-1) (sin 5) gt x^(2) - 4x` |
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Answer» Correct Answer - `2 0 sqrt(9 - 2pi) lt x lt 2 + sqrt(9 - 2pi)` We have `sin^(-1) (sin 5) gt x^(2) - 4x` Since, `(3pi)/(2) lt 5 lt (5pi)/(2)` `:. Sin^(-1) (sin 5) = 5 - 2pi` `rArr 5 - 2pi gt x^(2) - 4x` `rArr x^(2) - 4x + 4 lt 9 - 2pi` `rArr (x - 2)^(2) lt 9 - 2pi` rArr -sqrt(9 - 2pi) lt x x - 2 lt sqrt(9 - 2pi)` `rArr 2- sqrt(9 - 2 pi) lt x lt 2 + sqrt(9 - 2 pi)` |
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| 322. |
If `alpha le sin^(-1)x+cos^(1)x +tan^(-1)xlebeta` thenA. `alpha=(pi)/(4),beta =(3pi)/(4)`B. `alpha=-pi, beta=2pi`C. `alpha=0, beta=pi`D. none of these |
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Answer» We have `sin^(-1)x+cos^(-1)x=(pi)/(2)` for all `x in [-1,1]` Also `-(pi)/(4)le tan x le(pi)/(4)` for all `x in [-1,1]` `therefore -(pi)/(4)+(pi)/(2)lesin^(-1)x+cos^(-1)x+tan^(-1)xle(pi)/(41)+(pi)/(2)` `rarr (pi)/(4)lesin^(-1)x+cos^(-1)x+tan^(-1)xle(3pi)/(4)` |
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| 323. |
The equation `sin^-1x-cos^-1x=cos^-1(sqrt3/2)` hasA. No solutionsB. Unique solutionC. Infinite number of solutionD. Two solutions |
| Answer» Correct Answer - B | |
| 324. |
`cos^(-1)(15/17)+2 tan^(-1)(1/5)=`A. `(pi)/(2)`B. `cos^(-1)(171)/(221)`C. `(pi)/(4)`D. none of these |
| Answer» Correct Answer - D | |
| 325. |
Find the possible values of `sin^(-1) (1 - x) + cos^(-1) sqrt(x -2)` |
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Answer» We have `f(x) = sin^(-1) (1 - x) + cos^(-1) sqrt(x - 2)` We must have `- 1 le (1 - x) le 1` `rArr -2 le - x le 0` `rArr 0 le x le 2 and x -2 ge 0` `rArr 0 le x le 2 and x le 2` `:. X = 2` Thus, expression is defined only for `x = 2` `:. f(2) = sin^(-1) (-1) + cos^(-1) (0)` `= (-pi//2) + (pi//2)` `=0` |
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| 326. |
If `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi,t h e nx^4+y^2+z^4+4x^2y^2z^2=K(x^2y^2+y^2z^2+z^2x^2),`where `K`is equal to1 (b)2 (c) 4(d) none of theseA. 1B. 2C. 4D. none of these |
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Answer» We have `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi` `rarr sin^(-1)x +sin^(-1)y=pi -sin^(-1)z` `rarr cos(sin^(-1)x+sin^(-1)y)=cos(pi-sin^(-1)z)` `rarr cos(sin^(-1)x)cos(sin^(-1)y)-sin^(-1)x)sin(sin^(-1)y)=-cos(sin^(-1)z)` `rarr sqrt(1-x^(2))sqrt(1-y^(2))-xy=-sqrt(1-z^(2))` `rarr (1-x^(2))(1-y^(2))=x^(2)y^(2)-2xy sqrt(1-z^(2))` ltrbgt `rarr 1-x^(2)-y^(2)+x^(2)y^(2)=x^(2)y^(2)-2xysqrt(1-z^(2))` `rarr x^(2)+y^(2)-z^(2)=x^(2)=4x^(2)yu^(2)(1-a^(2))` `rarr x^(4)+y^(4)+z^(4)+4x^(2)y^(2)z^(2)=2 x^(2)y^(2)z^(2)+y^(2)z^(2)+z^(2)x^(2)` |
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| 327. |
lf `x >=0` and `theta = sin^(-1)x + cos^(-1)x-tan^(-1) x`, thenA. `(pi)/(2) lt theta le (3pi)/(4)`B. `0 le theta le (pi)/(4)`C. `-(pi)/(4) le theta le 0`D. `(pi)/(4) le theta le (pi)/(2)` |
| Answer» Correct Answer - D | |
| 328. |
Find the minimum value of `2sin^2theta+3cos^2theta` |
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Answer» `f(theta) = 2sin^2theta+3cos^2theta` `= 2(cos^2theta+sin^2theta)+cos^2theta` `=2+cos^2theta` `=2+1/2(1+cos2theta)` `=2+1/2+1/2cos2theta` `f(theta)=5/2+1/2cos2theta` `f(theta)` will be minimum when `cos2theta = -1` `:f(theta)_min = 5/2-1/2 = 2 ` |
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| 329. |
find the maximum value of `f(x) = (sin^(-1) (sin x))^(2) - sin^(-1) (sin x)` |
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Answer» Correct Answer - `(pi)/(4) (pi + 2)` `f(x) = (sin^(-1) (sin x))^(2) - sin^(-1) (sin x)` `= (sin^(-1) (sin x) - (1)/(3))^(2) - (1)/(4)` For maximum value of `f(x), sin^(-1) (sinx) = (-pi)/(2)` `:. F_("max") = (-(pi)/(2) -(1)/(2))^(2) - (1)/(4) = (pi)/(4) (pi + 2)` |
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| 330. |
`sin^(-1){sin(2x^(2)+4)/(x^(2)+1)=ltpi-3` ifA. `x in [-1,0]`B. `x in [0,1]`C. `x in (-1,1)`D. `x in (1,oo)` |
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Answer» We have `(2x^(2)+4)/(x^(2)+1)=(2(x^(2)+1)+2)/(x^(2)+1)=2+(2)/(x^(2)+1)` `therefore 2lt(2x^(2)+4)/(x^(2)+1)le 4 forall x in R` `rarr pi-4ltpi-(2x^(2)+45)/(x^(2)+1)ltpi-2 forall x in R` `rarr -(pi)/(2)ltpi-(2x^(2)+4)/(x^(2)+1 lt(pi)/(2) forall x in R` `=pi-(2x^(2)+4)/(x^(2)+1)` Hence `sin^(-1){sin(2x^(2)+4)/(x^(2)+1)}ltpi-3` `rarr pi-(2x^(2)+4)/(x^(2)+1)ltpi-3` `rarr (2x^(2)+4)/(x^(2)+1)gt3` `rarr 2+(2)/(x^(2)+1)rarr2gt3gtx^(2)+1rarrx^(2)1lt0rarrx in (-1,1)` |
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| 331. |
If `(sin^(-1)x)^2+(sin^(-1)y)^2+(sin^(-1)z)^2=3/4pi^2`, find the value of `x^2+y^2+z^2`. |
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Answer» Correct Answer - `-3` `(sin^(-1) x)^(2) + (sin^(-1) y)^(2) + (sin^(-1) z)^(2) = (3pi^(2))/(4)` `rArr sin^(-1) x = sin^(-1) y = sin^(-1) z = +- (pi)/(2)` ltrbgt for minimum value of `x + y + z, x = y =z = -1` Hence, `(x + y + z)_("min") = -3` |
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| 332. |
If `0< x< 1`,then `tan^(-1)(sqrt(1-x^2)/(1+x))` is equal toA. `1/2 cos^(-1)x`B. `cso^(-1)sqrt(1+x)/(2)`C. `sin^(-1)sqrt(1-x)/(2)`D. all the above |
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Answer» Let `x = cos theta` Then `0ltxlt1 rarr 0ltcos theta lt1 rarr0lttheta lt(pi)/(2)` Now `tan^(-1)sqrt(1-x^(2))/(1+x)` `tan^(-1)sqrt(1-x^(2)/(1+x)` Thus option (a)is true `cos^(-1)sqrt(1+x)//(2)=cos^(-1)(cos(theta)/(2))=1/2theta=1/2 cos^(-1)x` so option (b) is true `sin^(-1)sqrt(1-x)/(2)=sin^(-1)(sin(theta)/(2))=1/2theta=1/2cos^(-1)x` `sin^(-1)sqrt(1-x)/(2)=sin^(-1)(sin(theta)/(2))=(theat)/(2)=(1)/(2)cosT^(-1)x` so option (c ) is true |
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| 333. |
`cosec^(-1)(cosx)` is defined ifA. `x in [-1,1]`B. `x in R`C. `x=(2n+1)(pi)/(2), n in z`D. `x=n pi , n in z` |
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Answer» We know that `cosec^(-1)x` is defined for `x in(-infty,-1]cup[1,infty)` Therefore `cosec^(-1) (cosx)` is defined for `cosx=px1rarr x n pi n in z` |
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| 334. |
If f(x) `=cos^(-1)x+cos^(-1){(x/2+1/2sqrt(3-3x^(2)))}` then `f(2/3)` equalsA. `(pi)/(3)`B. `2 cos^(-1) 2/3 -(pi)/(3)`C. `(2pi)/(3)`D. `2cso^(-1)2/3` |
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Answer» We have `cos^(-1){xy+sqrt(1-x^(2))sqrt(1-y^(2))}` `={:(cos^(-1)x-cosf^(-1)y, if -1 le x,y le 1 and x le y),(cos^(-1)y-cos^(-1)x if -1 le y le 0, 0 le x le 1 and y le x):}` ` rarr f(2/3)= cos^(-1)1/2=(pi)/(3)` |
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| 335. |
If `0 |
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Answer» Correct Answer - 9 `1 + sin (cos^(-1) x) + sin^(2) (cos^(-1) x) + ...oo = 2` `rArr (1)/(1-sin (cos^(-1)x)) = 2` or `(1)/(2) = 1 - sin (cos^(-1) x)` or `sin (cos^(-1)x) = (1)/(2)` or `sin (cos^(-1) x) = (1)/(2)` or `cos^(-1) x = (pi)/(6)` or `x = (sqrt3)/(2) " or " 12x^(2) = 9` |
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| 336. |
If `tan^(-1)(sqrt(1+x^2-1))/x=4^0`then`x=tan2^0`(b) `x=tan4^0``x=tan1/4^0`(d) `x=tan8^0`A. `x = tan 2^(@)`B. `x = tan 4^(@)`C. `x = tan (1//4)^(@)`D. `x = tan 8^(@)` |
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Answer» Correct Answer - D `tan^(-1). (sqrt(1 + x^(2)) -1)/(x) = 4^(@) " " (x != 0)` `rArr (sqrt(1 + x^(2)) -1)/(x) = tan4^(@)` `rArr sqrt(1 + x^(2)) = 1 + x tan 4^(@)` `rArr 1 + x^(2) = 2x tan 4^(@) + 1 + x^(2) tan^(2) 4^(@)` `rArr x = 0 " or " x = (2 tan 4^(@))/(1 - tan^(2) 4^(@)) = tan 8^(@)` Since `x != 0`, we have `x = tan 8^(@)` |
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| 337. |
The value of `tan^(-1)((xcostheta)/(1-xsintheta))-cot^(-1)((costheta)/(x-sintheta))i s``2theta`(b) `theta`(c) `theta/2`(d) independent of `theta`A. `2 theta`B. `theta`C. `theta//2`D. independent of `theta` |
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Answer» Correct Answer - B `tan^(-1) ((x cos theta)/(1 - x sin theta)) - cot^(-1) ((cos theta)/(x - sin theta))` `= tan^(-1) ((x cos theta)/(1 - x sin theta)) - tan^(-1) ((x - sin theta)/(cos theta))` `= tan^(-1) (((x cos theta)/(1-x sin theta) -(x - sin theta)/(cos theta))/(1 + ((x cos theta)/(1 -x sin theta)) ((x-sin theta)/(cos theta))))` `= tan^(-1) ((x cos^(2) theta - x + sin theta + x^(2) sin theta - x sin^(2) theta)/(cos theta - x cos theta sin theta + x^(2) cos theta - x cos theta sin theta))` `= tan^(-1) ((-x sin^(2) theta + sin theta + x^(2) sin theta - x sin^(2) theta)/(cos theta - 2x cos theta sin theta + x^(2) cos theta))` `= tan^(-1) ((-2 x sin^(2) theta + sin theta + x^(2) sin theta)/(cos theta -2 x cos theta sin theta+ x^(2) cos theta))` `= tan^(-1) ((sin theta(-2 x sin theta + 1 + x^(2)))/(cos theta (1-2x sin theta + x^(2))))` `= tan^(-1) (tan theta)` `= theta` |
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| 338. |
If `tan^(-1)(sin^2theta-2sintheta+3)+cot^(-1)(5^sec^(2y)+1)=pi/2,`then value of `cos^2theta-sintheta`is equal to0 (b) `-1`(c) `1`(d) none of these |
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Answer» Correct Answer - C From the given equation `sin^(2) theta - 2 sin theta + 3 = 5^("sec"^(2)y) + 1`, we get `(sin theta -1)^(2) + 2 = 5^("sec^(2)y) + 1` `L.H.S. le 6, R.H.S. ge 6` Possible solution is `sin theta = -1` when L.H.S. = R.HgtS. `rArr cos^(2) theta = 0` `rArr cos^(2) theta - sin theta = 1` |
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| 339. |
The value (s) of `theta` satisfying the equation `theta=tan^(- 1)(2tan^2theta)-1/2(sin^(- 1)((3sin2theta)/(5+4cos2theta)))` isA. `npi+pi/4`B. `npi+tan^(-1)(-2)`C. `npi+pi/3`D. `npi+pi/6` |
| Answer» Correct Answer - A::B | |
| 340. |
If `x_(1) = 2 tan^(-1) ((1 + x)/(1 -x)), x_(2) = sin^(-1) ((1 - x^(2))/(1 + x^(2))), " where " x in (0, 1)`, then `x_(1) + x_(2)` is equal to |
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Answer» Correct Answer - C `x_(1) = 2 tan^(-1) ((1 + x)/(1 - x))` and `x_(2) = sin^(-1) ((1 - x^(2))/(1 + x^(2)))` `= tan^(-1) ((1 - x^(2))/(2x))` Now `(1 + x)/(1 -x) gt 1` `rArr x_(1) = pi + tan^(-1) ((2((1 + x)/(1 -x)))/(1 -((1+ x)/(1 -x))^(2)))` `= pi + tan^(-1) ((1 -x^(2))/(-2x))` `= pi - tan^(-1) ((1 -x^(2))/(2x))` `rArr x_(1) + x_(2) = pi` |
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| 341. |
If `Sigma_(r=1)^(2n) sin^(-1) x^(r )=n pi, then Sigma__(r=1)^(2n) x_(r )` is equal toA. nB. 2nC. `(n(n+1))/(2)`D. none of these |
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Answer» We have `(pi)/(2)lt sin^(-1)x_(e)le(pi)/(2)for i=1,2…,2n` `therefore underset(r=1)overset(2n)Sigma sin^(-1)x_(r )=npi` `rarr sin^(-1) pi_(r ),r=1,2..,2n` `rarr underset(r=1)overset(2n)Sigma x_(r )=underset(r=1)overset(2n)Sigma 1=2n` |
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| 342. |
The value of `Sigma_(n=1)^(3) tan^(-1)(1)/(n)` is |
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Answer» We have `underset(n=1)overset(3)Sigma tan^(-1)(1)/(n)=tan^(-1)+tan^(-1)1/2+tan^(-1)1/3` `rarr underset(n=1)overset(3)Sigma tan^(-1)(1)/(n)=(pi)/(4)O+tan^(-1)((1/2+1/3)/(1-1/2xx1/3))` `rarrunderset(n=1)overset(3)Sigma tan^(-1)(1)/(n)=(pi)/(4)+tan^(-1)1=(pi)/(4)+(pi)/(4)=(pi)/(2)` |
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| 343. |
Find the value of `sum_(r = 1)^(10) sum_(s = 1)^(10) tan^(-1) ((r)/(s))` |
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Answer» `S = underset(r = 1) overset(10)sum underset(s = 1) overset(10)sum tan^(-1) ((r)/(s))` `:. S = underset(r = 1)overset(10)sum underset(s = 1)overset(10)sum tan^(-1) ((s)/(r))` (As r and s are independent) On adding, we get `2S = underset(r =1)overset(10)sum underset(s =1) overset(10)sum (tan^(-1) ((r)/(s)) + tan^(-1) ((s)/(r)))` `rArr 2S = underset(r = 1)overset(10)sum underset(s =1) overset(10)sum (pi)/(2) = (pi)/(2) underset(r =1)overset(10)sum 10 = (100pi)/(2)` `:. S = 25 pi` |
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| 344. |
`Sigma_(r=1)^(infty) tan^(-1)(1)/(2r^(2))=t` then tan t is equal toA. `2/3`B. 1C. `sqrt(5)/(3)`D. none of these |
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Answer» we have `t=underset(oo)overset(r=1)Sigma tan^(-1)((1)/(2r)^(2))` `rarr t=underset(r=1)overset(oo)Sigma tan^(-1)(2)/(1+4r^(2)-1)` `underset(r=1)overset(oo)Sigma{tan^(-1)(2r+1)-tan^(-1)(2r-1)}` `=underset(n rarr oo) lim {(tan^(-1) 3-tan^(-1)(2n+1)-tan^(-1)(2r-1)}` `=underset(n rarr oo) lim {(tan^(-1) 3-tan^(-1)+(tan^(-1)5-tan^(-1)3)+(tan^(-1)(2n+1)-tan^(-1)(2n-1)}` `rarr t= underset(n rarr oo) lim (tan^(-1)(2n+1)-tan^(-1)1)=(pi)/(2)-(pi)/(4)=(pi)/(4)` `therefore tan t=1` |
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| 345. |
If `Sigma_(r=1)^(n) cos^(-1)x_(r)=0, then Sigma_(r=1)^(n) x_(r)` equals |
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Answer» We know that `0lecos^(-1)x_(r )lepi,r=1,2,.n` `therefore underset(r=1)overset(n)Sigma cos^(-1) x_(r ) =0` `rarr cos^(-1)x_(r ) =0 for r=1,2..n` `rarr cos^(-1)x_(r)=0 for r=1,2..n` `rarr x_(r ) =1 for r=1,2,…n ` `therefore underset(r=1)overset(n)Sigma x_(r )=underset(r=1)overset(n)Sigma1=n` |
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| 346. |
If `xgt0x,ygt0 and x gty` then `tan^(-1){(x)/(y)}+tan^(-1){(x+y)/(x-y)}` is equal toA. `(pi)/(4)`B. `(pi)/(4)`C. `(3pi)/(4)`D. none of these |
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Answer» `tan^(-1)(x)/(y)+tan^(-1)(x+y)/(x-y)` `tan^(-1)(x)/(y)+tan^(-1)(1+(x)/(y))/((x)/(y)-1)` `tan^(-1)(x)/(y)-tan^(-1)(1+(x)/(y))/(1-(x)/(y))` `tan^(-1)(x)/(y)-(-pi+tan^(-1)1+tan^(-1)(x)/(y))` `=pi-(pi)/(4)=(3pi)/(4)` |
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| 347. |
The number of positive solution satisfying the equation `tan^(-1)((1)/(2x+1))+tan^(-1)((1)/(4x+1))=tan^(-1)(2/(x^2))` isA. 1B. 2C. 8D. 9 |
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Answer» We have `tan^(-1)(1)/(2x+1)+tan^(-1)(1)/(4x+1)=tan^(-1)(2)/(x^(2))` `rarr tan^(-1)(3x+1)/(4x^(2)+3x)=tan^(-1)(2)/(x^(2))` `rarr (3x+1)/(4x^(2)+3x)=(2)/(x^(2))rarr3x^(2)-7x-6-0rarrx-2/3,3` |
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| 348. |
The value of `cot[{Sigma_(n=1)^(23) {cot^(-1))1+Sigma_(k=1)^(n) 2k}]` isA. `23/25`B. `25/23`C. `23/24`D. `24/23` |
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Answer» Clearly `cot^(-1)(1+underset(k=1)overset(n)Sigma 2k)=cot^(-1)1+2underset(k=1)overset(n_Sigmak)=cot^(-1)1+n(n+1)` `=tan^(-1)(1)/(1+n(n+1)}=(tan^(-1)(n+1)-n)/(1+n(n+1))` `tan^(-1)(n1)-tan^(-1)n` `therefore underset(n=1)overset(23)Sigma(tan^(-1)(n+1)-tan^(-1)n)` `tan^(-1)24-tan^(-1)1` `tan^(-1)(24-1)/(1+24xx1)=tan^(-1)(23)/(25)=cot^(-1)25/23` Hence `cot[underset(n=1)overset(23)Sigma{cot^(-1)(1+underset(k=1)overset(n)Sigma 2k)}]=cot^(-1)25/23=25/23` |
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| 349. |
The value of sum `sum_(n=1)^(oo)cot^(-1)(((n^(2)+2n)(n^(2)+2n+1)+1)/(2n+2))` is equal toA. `cos^(-1)((1)/(sqrt(5)))`B. `sec^(-1)((sqrt(5))/(2))`C. `sin^(-1)((1)/(sqrt(5)))`D. `cot^(-1)(1)` |
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Answer» Correct Answer - C `T_(n)tan^(-1)((2n+2)/(1+(n^(2)+2n)(n^(2)+2n+1)))` `=tan^(-1)((2n+2)/(1+n(n+2)(n+1)(n+1)))` `=(tan^(-1)(n+1)(n+2)-tan^(-1)n(n+1))` `therefore S_(n)sum_(n=1)^(n)T_(n)=(tan^(-1)(n+1)(n+2)-tan^(-1)2)` So, `lim_(n rarr oo)S_(n)=((pi)/(2)-tan^(-1)2)=cot^(-1)2=sin^(-1)((1)/(sqrt(5)))` |
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| 350. |
`sum_(m=1)^(n) tan^(-1) ((2m)/(m^(4) + m^(2) + 2))` is equal toA. `tan^(-1) ((n^(2) + n)/(n^(2) + n + 2))`B. `tan^(-1) ((n^(2) -n)/(n^(2) - n + 2))`C. `tan^(-1) ((n^(2) + n + 2)/(n^(2) + n))`D. none of these |
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Answer» Correct Answer - A We have `underset(m=1)overset(n)sum tan^(-1) ((2m)/(m^(4) + m^(2) + 2))` `= underset(m =1)overset(n)sum tan^(-1) ((2m)/(1+(m^(2) + m +1)(m^(2) -m +1)))` `= underset(m=1)overset(n)sum tan^(-1) (((m^(2) + m + 1) -(m^(2) - m + 1))/(1 + (m^(2) + m + 1) (m^(2) - m + 1)))` `= underset(m=1)overset(n)sum [tan^(-1) (m^(2) + m+1) - tan^(-1) (m^(2) - m + 1)]` `= (tan^(-1) 3 - tan^(-1)1) + (tan^(-1) 7 - tan^(-1) 3) + (tan^(-1) 13 - tan^(-1) 7) +....+ [tan^(-1) (n^(2) + n + 1) -tan^(-1) (n^(2) -n + 1)]` `= tan^(-1) (n^(2) + n + 1) - tan^(-1) 1` `= tan^(-1) ((n^(2) + n)/(2 + n^(2) + n))` |
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