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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.
| 201. |
STATEMENT -1 : If `[sin^(-1)x] gt [cos^(-1)x]`,where [] represents the greatest integer function, then `x in [sin1,1]` is and STATEMENT -2 : `cos^(-1)(cosx)=x,x in [-1,1]`A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -2B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -2C. Statement-1 is True, Statement -2 is FalseD. Statement -1 is False, Statement -2 is True |
| Answer» Correct Answer - C | |
| 202. |
The value of `sin^(-1){sin(-600^(@))}` isA. `(pi)/(3)`B. `-(pi)/(3)`C. `(2pi)/(3)`D. `-(2pi)/(3)` |
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Answer» `sin^(-1)(sin-6000^(@))=sin^(-1){(sin-600xx(pi)/(180))}` `=sin^(-1){(sin-10pi)/(3)=sin^(-1)(-sin(10pi)/(3))}` `=sin^(-1){-sin(3pi+(pi)/())}=sin^(-1)(-sin(10pi)/(33))` `=sin^(-1)(sin(pi)/(3))=(pi)/(3)` |
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| 203. |
Find the value of `4 tan^(-1).(1)/(5) - tan^(-1).(1)/(70) + tan^(-1).(1)/(99)` |
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Answer» `4 tan^(-1).(1)/(5) - tan^(-1). (1)/(70) + tan^(-1). (1)/(99)` `= 2 tan^(-1) [((2)/(5))/(1 - (1)/(25))] - tan^(-1).(1)/(70) + tan^(-1).(1)/(99)` `= 2 tan^(-1) ((5)/(12)) + tan^(-1) [((1)/(99) - (1)/(70))/(1 + (1)/(99) xx (1)/(70))]` `= tan^(-1) [((5)/(6))/(1 - (25)/(144))] + tan^(-1) ((-29)/(6931))` `= tan^(-1) ((120)/(119)) - tan^(-1) ((1)/(239))` `= tan^(-1) [((120)/(119) - (1)/(239))/(1 + (120)/(119) xx (1)/(239))] = tan^(-1) (1) = (pi)/(4)` |
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| 204. |
If `(x -1) (x^(2) + 1) gt 0`, then find the value of `sin((1)/(2) tan^(-1).(2x)/(1 - x^(2)) - tan^(-1) x)` |
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Answer» `(x -1) (x^(2) + 1) gt 0` `:. Sin[(1)/(2) tan^(-1) ((2x)/(1 - x^(2))) - tan^(-1) x]` `= sin [(1)/(2) (-pi + 2 tan^(-1) x) - tan^(-1) x] = sin (-(pi)/(2)) = -1` |
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| 205. |
Prove that `cot^(-1).(3)/(4) + sin^(-1).(5)/(13) = sin^(-1).(63)/(65)` |
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Answer» `cot^(-1).(3)/(4) + sin^(-1).(5)/(13) = sin^(-1).(4)/(5) + sin^(-1).(5)/(13)` `=sin^(-1) ((4)/(5) sqrt(1- ((5)/(13))^(2)) + (5)/(3) sqrt(1 - ((4)/(5))^(2)))` `= sin^(-1) ((4)/(5) (12)/(13) + (5)/(13) (3)/(5)) = sin^(-1).(63)/(65)` |
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| 206. |
The value of `cot^(-1)(-sqrt3)+cosec^(-1)(2)+tan^(-1)(sqrt3)` isA. `pi/6`B. `pi/3`C. `(5pi)/6`D. `(4pi)/3` |
| Answer» Correct Answer - D | |
| 207. |
Find the principal value of `tan^(-1)(-1/sqrt3)`.A. `-pi/6`B. `-pi/3`C. `(5pi)/6`D. Both (1) & (2) |
| Answer» Correct Answer - A | |
| 208. |
If `sec^(-1) x = cosec^(-1) y`, then find the value of `cos^(-1).(1)/(x) + cos^(-1).(1)/(y)` |
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Answer» `sec^(-1) x = cosec^(-1) y` or `cos^(-1).(1)/(x) = sin^(-1).(1)/(y)` or `cos^(-1).(1)/(x) = (pi)/(2) - cos^(-1).(1)/(y)` or `cos^(-1).(1)/(x) + cos^(-1).(1)/(y) = (pi)/(2)` |
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| 209. |
If the maximum possible value of `(sin^(- 1)x)^2+(tan^(- 1)y)^2+pisin^(- 1)x+pitan^(- 1)y+(pi^2)/2` is k, and [] represents the greatest integer function then the value of `[k/(2pi^2)]` is |
| Answer» Correct Answer - A | |
| 210. |
Simplify `cos(tan^(-1)(sin(cot^(-1)x))), x in (0,1)`. |
| Answer» Correct Answer - `sqrt((1+x^2)/(2+x^2))` | |
| 211. |
If `cos^(-1)(x)+cos^(-1)(y)+cos^(-1)(z)=pi(sec^2(u)+sec^4(v)+sec^6(w)),w h e r e u , v , w`are least non-negative angles such that `u |
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Answer» Correct Answer - 9 `sec^(2)u, sec^(4)v, sec^(6)w in [1, oo)` `:. Sec^(2) (u) + sec^(4)(v) + sec^(6) (w) in [3, oo)` `:. Pi (sec^(2) u + sec^(4)v + sec^(6)w) in [3pi, oo)` But `cos^(-1) x + cos^(-1) y + cos^(-1) z in [0, 3pi]` So equation is possible of L.H.S. = R.H.S. `= 3pi` `:. cos^(-1) x = cos^(-1) y = cos^(-1) z = pi` `:. x = y = z =-1` and `sec^(2) u = sec^(4) v = sec^(6) w = 1` `:. u = pi, v = 2pi, w = 3pi` `:. x^(2000) + y^(2002) + z^(2004) + (36pi)/(u + v + w) = 1 + 1 + 1 + (36pi)/(6pi) = 9` |
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| 212. |
Find the range of `f(x)=|3tan^(-1)x-cos^(-1)(0)|-cos^(-1)(-1)dot` |
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Answer» `f(x) = |3 tan^(-1) x - cos^(-1) (0)| - cos^(-1) (-1)` `= |3 tan^(-1) x - (pi//2)| - pi` Now, `-(pi)/(2) lt tan^(-1) x lt(pi)/(2)` or `-(3pi)/(2) lt 3 tan^(-1) x lt (3pi)/(2)` or `-2pi lt 3 tan^(-1) x-(pi)/(2) lt pi` or `0 le |3 tan^(-1) x -(pi)/(2)| lt 2pi` or `-pi |3 tan^(-1) x -(pi)/(2)| - pi lt pi` |
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| 213. |
Number of solutions of the equation `2(sin^(-1)x^2)-sin^(-1)x-6=0` isA. 2B. 1C. 0D. 3 |
| Answer» Correct Answer - B | |
| 214. |
If`p > q >0a n dp r |
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Answer» since `p, q gt 0`, we have `pq gt 0` `tan^(-1) (p -q)/(1 + pq) = tan^(-1) p - tan^(-1) q`....(i) Since `qr gt -1`, we have `tan^(-1) (q -r)/(1 + qr) = tan^(-1) q - tan^(-1) r`...(ii) Since `pr lt -1 and r lt 0`, we have `tan^(-1) (r - p)/(1 + rp) = pi + tan^(-1) t - tan^(-1) p`....(iii) On adding Eqs. (i), (ii), and (iii), we get `tan^(-1) (p -q)/(1 + pq) + tan^(-1) (q - r)/(1 + qr) + tan^(-1) (r - p)/(1 + rp) = pi` |
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| 215. |
The value of x satisfying `sin^(-1)(sqrt((3x-1)/(25)))+sin^(-1)(sqrt((3x+1)/(25)))=(pi)/(2)` lies in the intervalA. (1,2)B. (2,3)C. (3,4)D. (4,5) |
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Answer» Correct Answer - D `sin^(-1)(sqrt((3x-1)/(25)))+sin^(-1)(sqrt((3x + 1)/(25)))=(pi)/(2)` `rArr sin^(-1)(sqrt((3x-1)/(25)))=(pi)/(2)-sin^(-1)(sqrt((3x+1)/(25)))=cos^(-1)(sqrt((3x+1)/(25)))` `rArr sin^(-1)(sqrt((3x-1)/(25)))=sin^(-1)(sqrt(1-(3x+1)/(25)))` `rArr sin^(-1) (sqrt((3x-1)/(25)))=sin^(-1)(sqrt((24-3x)/(25)))` `rArr (3x-1)/(25)=(24-3x)/(25)` `rArr x = (25)/(6)` |
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| 216. |
The set of values of k for which the equation `sin^(-1)x+cos^(-1)x +pi(|x|-2)=k pi` possesses real solution is [a,b] then the value of a + b is |
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Answer» Correct Answer - B `sin^(-1)x + cos(-1)x + pi(|x|-2)= k pi, x in [-1,1]` `rArr (pi)/(2)+pi(|x|-2)= k pi` `rArr (1)/(2)+|x|-2=k` `rArr |x|=k + (3)/(2) in [0,1]` `rArr 0 le k + (3)/(2) le 1` `rArr -(3)/(2)le k le -(1)/(2)` |
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| 217. |
If `sin^(-1)(x-1)+cos^(-1)(x-3)+tan^(-1)(x/(2-x^2))=cos^(-1)k+pi,`then the value of `k`isA. 1B. `-(1)/(sqrt2)`C. `(1)/(sqrt2)`D. none of these |
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Answer» Correct Answer - C `sin^(-1) (x -1)` `rArr -1 le x -1 le 1` `rArr 0 le x le 2` `cos^(-1) (x -3)` `rArr -1 le x -3 le 1` `rArr 2 le x le 4` `:. X = 2` So, `sin^(-1) (2 -1) + cos^(-1) (2 -3) + tan^(-1).(2)/(2 -4) = cos^(-1) k + pi` or `sin^(-1) 1 + cos^(-1) (-1) + tan^(-1) (-1) = cos^(-1) + pi` `rArr (pi)/(2) + pi - (pi)/(4) = cos^(-1) k + pi` `rArr cos^(-1) k = (pi)/(4) " or " k = (1)/(sqrt2)` |
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| 218. |
If maximum and minimum values of `|sin^(-1)x|+|cos^(-1)x|` are Mand m, then M+m isA. `pi//2`B. `pi`C. `2pi`D. `3pi` |
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Answer» Correct Answer - C `f(x)={{:(sin^(-1)x+cos^(-1)x,,0le x le 1),(cos^(-1)x-sin^(-1)x,,-1le x le x lt 0):}` `f(x)={{:(pi//2",",0le x le 1),((pi)/(2)-2sin^(1)x",",-1le x lt 0):}` For `-1 le x le 0` `-pi//2 le sin^(-1)x le 0` `therefore -pi le 2 sin^(-1)x le 0` `therefore 0 le -2 sin^(-1)x le pi` `therefore pi//2 le pi//2 -2 sin^(-1) x le 3pi//2` `therefore` Max. value `= 3pi//2` and Min. value `= pi//2` |
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| 219. |
If the function `f(x)=sin^(-1)x+cos^(-1)x` and g(x) are identical, then g(x) can be equal toA. `sin^(-1)|x|+|cos^(-1)x|`B. `tan^(-1)x+cot^(-1)x`C. `|sin^(-1)x|+cos^(-1)|x|`D. `(sqrt(sin^(-1)x))^(2)+(sqrt(cos^(-1)x))^(2)` |
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Answer» Correct Answer - C `f(x)=sin^(-1)x + cos^(-1)x=(pi)/(2), AA x in [-1,1]` If `g(x)=|sin^(-1)x|+cos^(-1)|x|={{:(-sin^(-1)x +pi-cos^(-1)x, -1le x lt 0),(=pi-(pi)/(2)=(pi)/(2)),(sin^(-1)x + cos^(-1)x, 0le x le 1),(=(pi)/(2)):}` |
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| 220. |
If `alpha, beta (alpha lt beta)` are the roots of the equation `6x^(2) + 11x + 3 = 0`, then which of the following are real ?A. `cos^(-1) alpha`B. `sin^(-1) beta`C. `cosec^(-1) alpha`D. Both `cot^(-1) alpha and cot^(-1) beta` |
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Answer» Correct Answer - B::C::D `6x^(2) + 11x + 3 = 0` or `(2x + 3) (3x + 1) = 0` or `x = -3//2, -1//3` For `x = -3//2, cos^(-1) x` is not defined as domain of `cos^(-1) x` is `9-1, 1] and " for " x = -1//3, cosec^(-1) x` is not defined as domain of `cosec^(-1)x " is " R -(-1, 1)`. However, `cot^(-1) x` is defined for both of these values as domain of `cot^(-1) x " is " R` |
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| 221. |
The solution set of the equation`sin^(-1)sqrt(1-x^2)+cos^(-1)x=cot^(-1)(sqrt(1-x^2))/x-sin^(-1)x`is`[-1,1]-{0}`(b) `(0,1)uu{-1}``(-1,0)uu{1}`(d) `[-1,1]`A. `[-1, 1] -{0}`B. `(0, 1] uu {-1}`C. `[-1, 0) uu {1}`D. `[-1, 1]` |
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Answer» Correct Answer - C `sin^(-1). sqrt(1 -x^(2)) + cos^(-1) x = cot^(-1). (sqrt(1-x^(2)))/(x) - sin^(-1)x` or `(pi)/(2) + sin^(-1). sqrt(1 -x^(2)) = cot^(-1). (sqrt(1-x^(2)))/(x)` `tan^(-1). (sqrt(1 -x^(2)))/(x) = - sin^(-1) sqrt(1 -x^(2))` `rArr x in [-1, 0) uu {1}` |
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| 222. |
`tan""(2pi)/(5)-tan""(pi)/(15)-sqrt3tan""(2pi)/(5)tan""(pi)/(15)` is equal toA. `-sqrt(3)`B. `(1)/sqrt(3)`C. 1D. `sqrt(3)` |
| Answer» Correct Answer - D | |
| 223. |
`f(x)=tan^(-1)x+tan^(-1)(1/x);g(x)=sin^(-1)x+cos^(-1)x`are identical functions if`x in R`(b) `x >0`(c) `x in [-1,1]`(d) `x in [0,1]`A. `x in R`B. `x gt 0`C. `x in [-1, 1]`D. `x in (0,1]` |
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Answer» Correct Answer - D `f(x) = tan^(-1) x + tan^(-1).(1)/(x) = {((pi)/(2),x gt 0),((-pi)/(2),x lt 0):}` `g(x) = sin^(-1) x + cos^(-1) x = (pi)/(2) AA x in [-1, 1]` Therefore, common domain is `x in (0, 1]` |
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| 224. |
The number of real solution of the equation `tan^(-1) sqrt(x^(2) - 3x + 2) + cos^(-1) sqrt(4x - x^(2) -3) = pi` isA. oneB. twoC. zeroD. infinite |
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Answer» Correct Answer - C Since `sqrt(x^(2) - 3x + 2) ge 0` `rArr 0 le tan^(-1) sqrt(x^(2) - 3x + 2) lt (pi)/(2)` and `sqrt(4x -x^(2) -3) le 0 rArr 0 lt cos^(-1) sqrt(4x - x^(2) -3) le (pi)/(2)` Adding, we have `0 lt L.H.S. lt pi` Therefore, the given equation has no solution |
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| 225. |
If `[cot^(-1)x]+[cos^(-1)x]=0`, where `[]`denotes the greatest integer functions, then the complete set of valuesof `x`is`(cos1,1)`(b) `cos1,cos1)``(cot1,1)`(d) none of theseA. `(cos 1,1]`B. `(cos 1, cot1)`C. `(cot 1, 1]`D. none of these |
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Answer» We have `cot^(-1) x in [0,pi] and cos^(-1) x in [0,pi]` `therefore [cost(-1)x]gt0 and [cos^(-1)x]ge0` Now `[cos^(-1)x]=0 =[cos^(-1)xlt1` `rarr cot 1 ltx lt infty and cos 1 lt x le 1` `rarr fx in (cot1,infty) and x in (cos 1,1]rarr x in (cot1,1]` |
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| 226. |
The set of values of `x` satisfying `|sin^(- 1)x|lt|cos^(- 1)x|,`isA. `[-1,1/sqrt2]`B. `[-1,-1/sqrt2] uu [1/sqrt2,1]`C. `(-1,1/sqrt2)`D. `[1/sqrt2,1]` only |
| Answer» Correct Answer - A | |
| 227. |
If `cot^(-1)(n/(pi))>(pi)/6, n in N`, then the maximum value of n is :A. 6B. 7C. 5D. none of these |
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Answer» Correct Answer - C `cot^(-1).(n)/(pi) gt (pi)/(6)` or `(n)/(pi) lt cot.(pi)/(6)` [as `cot^(-1) x` is a decreasing function] or `(n)/(pi) lt sqrt3` or `n lt sqrt3pi` or `n lt 5.46` or maximum value of `n` is 5 |
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| 228. |
The set of values of x, satisfying the equation `tan^2(sin^-1x) > 1` is -A. `[-1,1]`B. `[-(1)/sqrt(2),(1)/sqrt(2)]`C. `(-1,1)-[-(1)/sqrt(2),(1)/sqrt(2)]`D. `[-1,1]-(1)/sqrt(2),(1)/sqrt(2)` |
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Answer» We have `tan^(2)(som^(-1)x)gt1` `rarr tan^(2)(sin^(-1)x)-1)gt0` `rarr tan (sin^(-1)x)lt-1 or tan (sin^(-1)x)gt1` `rarr -infty lt tan(sin^(-1)x)lt-1 or 1 lt tan (sin^(-1)x)lt infty` `rarr -(pi)/(2)ltsin^(-1)xlt(pi)/(2)` `rarr x in(-1,-(1)/sqrt(2) or x in (1)sqrt(2),1)` `rarr x in (-1,-1)-[-(1)/sqrt(2)),(1)/sqrt(2)]` |
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| 229. |
The value of x that satisfies `tan^(-1)(tan -3)=tan^(2)x` isA. `(pi)/(3)`B. `-(pi)/(3)`C. `sqrt(tan^(-1)3)`D. `none of these |
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Answer» We have `tan^(-1)(tan-3)=tan^(-1)tan(3-pi))=3-pilt0` and `tan^(2)ge0` `therefore tan^(-1)(tan3)=tan^(-1) x` has no solution |
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| 230. |
Solve : `tan^(-1)(x-1)/(x-2)+tan^(-1)(x+1)/(x+2)=pi/4` |
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Answer» Correct Answer - `x = sqrt((5)/(2))` `tan^(-1).(x -1)/(x + 2) + tan^(-1).(x + 1)/(x + 2) = (pi)/(4)` `rArr tan^(-1) [((x -1)/(x + 2) + (x + 1 )/(x + 2))/(1 - ((x -1)/(x + 2)) ((x +1)/(x + 2)))] = (pi)/(4)` `rArr [(2x (x + 2))/(x^(2) + 4 + 4x -x^(2) + 1)] = tan.(pi)/(4)` `rArr (2x (x + 2))/(4x + 5) = 1` `rarr 2x^(2) + 4x = 4x + 5` `:. x = +- sqrt((5)/(2))` But for `x = -sqrt5//2`, L.H.S. is negative Hence, `x = sqrt(5//2)` |
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| 231. |
If `cosec^(-1) (cosec x) and cosec (cosec^(-1) x)` are equal function then the maximum range of value of x isA. `[-(pi)/(2), -1] uu [1, (pi)/(2)]`B. `[-(pi)/(2), 0)uu[0, (pi)/(2)]`C. `(-oo, -1] uu [1, oo)`D. `[-1, 0) uu [0, 1)` |
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Answer» Correct Answer - A `cosec (cosec^(-1) x) = x AA x in R -(-1, 1)` Also range of `cosec^(-1) (cosec x) in [-(pi)/(2), 0) uu (0, (pi)/(2]` So combing these two, we get `x in [-(pi)/(2), -1] uu [1, (pi)/(2)]` |
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| 232. |
Let `cos^(-1)(x)+cos^(-1)(2x)+cos^(-1)(3x)b epidot`If `x`satisfies the equation `a x^3+b x^2+c x-c_1=0,`then the value of `(b-a-c)`is_________ |
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Answer» Correct Answer - 3 `cos^(-1) (x) + cos^(-1) (2x) + cos^(-1) cos^(-1) (3x) = pi` or `cos^(-1) (2x) + cos^(-1) (3x)= pi - cos^(-1) (x) = cos^(-1) (-x)` or `cos^(-1) [(2x) (3x) - sqrt(1 -4x^(2)) sqrt(1 -9x^(2))] = cos^(-1) (-x)` or `6x^(2) - sqrt(1-4x^(2)) sqrt(1 -9x^(2)) = -x` or `(6x^(2) + x)^(2) = (1-4x^(2)) (1-9x^(2))` or `x^(2) + 12x^(3) = 1 - 13x^(2)` or `12x^(3) + 14x^(2) -1 = 0` `rArr a = 12, b = 14, c =0` `rArr a + b + c = 12 + 14 -1 = 25` |
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| 233. |
The number of real values of x satisfying `tan^-1(x/(1-x^2))+tan^-1 (1/x^3)` is |
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Answer» Correct Answer - no solution Given equation is `tan^(-1) ((x)/(1 -x^(2))) + tan^(-1) ((1)/(x^(3))) = (3pi)/(4)` Clearly `x != +- 1` `tan^(-1) ((x)/(1 - x^(2))) + tan^(-1) ((1)/(x^(3))) = tan.((x)/(1 - x^(2)) + (1)/(x^(3)))/(1 - (x)/(x^(3) (1 - x^(2))))` `= tan^(-1).(x^(4) + 1 -x^(2))/((x^(2) -x^(4) -1) x)` `= tan^(-1).(-(1)/(x))` `:. tan^(-1) (-(1)/(x)) = (3pi)/(4)` `rArr x = 1, " but " x != 1` So the given equation has no solution |
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| 234. |
Express `sin^(-1).(sqrtx)/(sqrt(x + a))` as a function of `tan^(-1)` |
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Answer» Correct Answer - `tan^(-1) (sqrt((x)/(a)))` Putting `x = a tan^(2) theta` `:. Sin.^(-1) (sqrtx)/(sqrt(x + a) = sin^(-1) (sqrta sqrt(tan^(2)theta))/(sqrt(a tan^(2) theta + a))` `= sin.^(-1) (sqrta tan theta)/(sqrta sec theta)` `= sin^(-1) sin theta = theta = tan^(-1) (sqrt((x)/(a)))` |
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| 235. |
`sec^(2) (tan^(-1) 2) + cosec^(2) (cot^(-1) 3)` is equal toA. 5B. 13C. 15D. 6 |
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Answer» Correct Answer - C Let `tan^(-1) 2 = alpha " or " tan alpha = 2` and `cot^(-1) 3 = beta " or " cot beta = 3` `sec^(2) (tan^(-1) 2) + cosec^(2) (cot^(-1) 3)` `= sec^(2) alpha + cosec^(2) beta` `= 1 + tan^(2) alpha + 1 + cot^(2) beta` `= 2 + (2)^(2) + (3)^(2) = 15` |
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| 236. |
The maximum value of `f(x)=tan^(-1)(((sqrt(12)-2)x^2)/(x^2+2x^2+3))`isA. `18^(@)`B. `36^(@)`C. `22.5^(@)`D. `15^(@)` |
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Answer» Correct Answer - D `f(x) = tan^(-1). (((sqrt12 - 2)x^(2))/(x^(4) + 2x^(2) + 3))` `= tan^(-1) ((2(sqrt3 -1))/(x^(2) + (3)/(x^(2)) + 2))` As `x^(2) + (3)/(x^(2)) ge 2 sqrt3` [using A.M. `ge` G.M.] `rArr x^(2) + (3)/(x^(2)) + 2 ge 2 + 2 sqrt3` `:. (f(x))_("max") = tan^(-1) ((2(sqrt3 -1))/(2(sqrt3 + 1)))` `= (pi)/(12)` |
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| 237. |
If the roots of the equation `x^(3) -10 x + 11 = 0` are u, v, and w, then the value of `3 cosec^(2) (tan^(-1) u + tan^(-1) v + tan^(-1 w)` is ____ |
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Answer» Correct Answer - 6 Let `tan^(-1) u = alpha " or " tan alpha = u` `tan^(-1) v = beta " or " tan beta = v` `tan^(-1) w = gamma " or " tan gamma = w` `tan (alpha + beta + gamma) = (s_(1) -s_(3))/(1 - s_(2)) = (0 -(-11))/(1-(-10)) = (11)/(11) = 1` `:. alpha + beta + gamma = tan^(-1) (1) = (pi)/(4)` `rArr 3 cosec^(2) (tan^(-1) u + tan^(-1) v + tan^(-1) w) = 6` |
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| 238. |
Find the value of`sin^(-1)(2^x)`(ii) `cos^(-1)sqrt(x^2-x+1)``tan^(-1)(x^2)/(1+x^2)`(iv) `sec^(-1)(x+1/x)` |
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Answer» (i) We have that `2^(@) gt 0` But for `sin^(-1) (2^(x))` to be defined, we must have `2^(x) le 1` `:. 0 lt 2^(x) le 1` `rArr 0 sin^(-1) (2^(x)) le pi//2` (ii) `cot^(-1) sqrt(x^(2)-x+1)= cos^(-1) sqrt((x-1/2)^(2)+3/4)` We have `sqrt((x-1/2)^(2)+3/4) ge sqrt(3)/2` `:. sqrt(3)/2 le sqrt((x-1/2)^(2)+3/4) le 1` `:. 0 le cos^(-1) sqrt((x-1/2)^(2)+3/4) le pi/6` (iii) `"tan"^(-1) x^(2)/(1+x^(2))=tan^(-1) (1-1/(1+x^(2)))` Now, `1 le 1+ x^(2) lt oo` `:. 0 lt 1/(1+x^(2)) lt 1` `:. -1 lt -1/(1+x^(2)) lt 0` `:. 0 lt 1-1/(1+x^(2)) lt 1` `:. 0 lt tan^(-1) (1-1/(1+x^(2))) lt pi/4` (iv) Let `x+1/x=y` `:. x^(2)-yx+1=0` Since x is real, `D ge 0` `:. y^(2) -4 ge 0` `:. y le -2` or `y ge 2` `:. sec^(-1) (x+1/x) in [pi/3, pi/2) uu (pi/2, (2pi)/3]` |
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| 239. |
Find the value of x for which `f(x) = 2 sin^(-1) sqrt(1 - x) + sin^(-1) (2 sqrt(x - x^(2)))` is constant |
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Answer» `f(x) = 2 sin^(-1) sqrt(1 - (sqrtx)^(2)) + sin^(-1) (2 sqrt((x)^(2) {1 - (sqrtx)^(2)})` Put `sqrtx = cos theta, " where " theta = cos^(-1) sqrtx in [0, (pi)/(2)]` `:. F(x) = 2 sin^(-1) sqrt(1 - cos^(2) theta) + sin^(-1) 2sqrt((cos^(2) theta) (1 - cos^(2) theta))` `= 2 sin^(-1) (sin theta) + sin^(-1) (2 sin theta cos theta)` `= 2 sin^(-1) (sin theta) + sin^(-1) (sin 2 theta)` `= 2 theta + sin^(-1) (2 theta)` `= 2 theta + (pi - 2 theta) " if " 2 theta in [(pi)/(2), pi]` `= pi, theta in [(pi)/(4), (pi)/(2)]` So, `sqrtx = cos theta in [0, (1)/(sqrt2)]` or `x in [0, (1)/(2)]` |
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| 240. |
Find the minimum value of `(sec^(-1) x)^(2) + (cosec^(-1) x)^(2)` |
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Answer» Let `I = (sec^(-1) x)^(2) + (cosec^(-1) x)^(2)` `= (sec^(-1) x + cosec^(-1) x)^(2) - 2 sec^(-1) x cosec^(-1) x` `= (pi^(2))/(4) - 2 sec^(-1) x ((pi)/(2) - sec^(-1) x)` `= (pi^(2))/(4) + 2 (sec^(-1) x)^(2) - pi sec^(-1) x` `= (pi^(2))/(4) + 2 [(sec^(-1) x)^(2) - 2(pi)/(4) sec^(-1) x + ((pi)/(4))^(2)] - (pi^(2))/(8)` `= 2 (sec^(-1) x - (pi)/(4))^(2) + (pi^(2))/(8) rArr I ge (pi^(2))/(8)` |
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| 241. |
Solve `sin^(-1). (14)/(|x|) + sin^(-1).(2 sqrt15)/(|x|) = (pi)/(2)` |
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Answer» `sin^(-1).(14)/(|x|) + sin^(-1). (2sqrt15)/(|x|) = (pi)/(2)` `rArr sin^(-1).(14)/(|x|) = (pi)/(2) - sin^(-1).(2 sqrt15)/(|x|)` `= cos^(-1).(2 sqrt15)/(|x|) = sin^(-1)sqrt(1 - ((2 sqrt15)/(|x|))^(2))` `rArr ((14)/(|x|))^(2) = 1 - ((2 sqrt15)/(|x|))^(2)` `rArr |x| = 16 " or " x = +- 16`, which satisfies `|x| ge 2 sqrt15` |
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| 242. |
Solve `tan^(-1) x gt cot^(-1) x` |
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Answer» Correct Answer - `x gt 1` `tan^(-1) x gt cot^(-1) x` `rArr tan^(-1) x gt pi//2 - tan^(-1) x` `rArr tan^(-1) x gt pi//4` `rArr x gt 1` |
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| 243. |
Sove `2 cos^(-1) x = sin^(-1) (2 x sqrt(1 - x^(2)))` |
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Answer» Let x = cos y, where `0 le y le pi, |x| le 1` `2 cos^(-1) x = sin^(-1) (2 x sqrt(1 -x^(2)))`...(i) `rArr 2 cos^(-1) (cos y) = sin^(-1) (2 cos y sqrt(1 - cos^(2) y))` `= sin^(-1) (2 cos y sin y)` `= sin^(-1) (sin 2 y)` `rArr sin^(-1) (sin 2 y) = 2y " for " -pi//2 le y le pi//4` and `2 cos^(-1) (cos y) = 2y " for " 0 le y le pi` Thus, Eq. (i) holds only when `y in [0, pi//4]` `rArr x in [1//sqrt2, 1]` |
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| 244. |
Sove `[cot^(-1) x] + [cos^(-1) x] =0`, where `[.]` denotes the greatest integer function |
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Answer» We have `[cot^(-1) x] + [cos^(-1) x] =0` Since `cos^(-1)x, cot^(-1) x ge 0`, we must have `[cot^(-1) x] = [cos^(-1) x] =0` Now, `[cot^(-1) x] = 0` `rArr 0 lt cot^(-1) x lt 1` `rArr x in (cot 1, oo)`...(1) Also, `[cos^(-1) x] = 0` ltrbgt `rArr 0 le cos^(-1) x lt1` `rArr x in (cos 1, 1]`...(2) From (1) and (2) `x in (cot 1,10` |
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| 245. |
solve the equation `cot^-1 x + tan^-1 3 = pi/2` |
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Answer» Correct Answer - `x =3` `cot^(-1) x tan^(-1) 3 = (pi)/(2)` `rArr cot^(-1) x = (pi)/(2) - tan^(-1) 3` `rArr cot^(-1) x = cot^(-1) 3` `:. x = 3` |
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| 246. |
The solution set of inequality `(cot^(-1)x)(tan^(-1)x)+(2-pi/2)cot^(-1)x-3tan^(-1)x-3(2-pi/2)>0`is `(a , b),`then the value of `cot^(-1)a+cot^(-1)b`is____ |
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Answer» `(cot^(-1)x)(tan^(-1)x)-:(2-pi/2)cot^(-1)x-3tan^(-1)x-3(2-pi/2)>0` `cot^(-1)x(tan^(-1)x-pi/2)+2cot^(-1)x-6-3(tan^(-1)x-pi/2)>0` `(cot^(-1)x)^2+5cot^(-1)x-6>0` `(cot^(-1)x-3)(1-cot^(-1)x)>0` `2ltcot^(-1)xlt3` `cot3ltxltcot2` `cot^(-1)(cot3)+cot^(-1)(cot2)=5`. |
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| 247. |
Which of the following pairs of function/functions has same graph?`y=tan(cos^(-1)x); y=(sqrt(1-x^2))/x``y=t a n(cot^(-1)x);y=1/x``y="sin"(tan^(-1)x); y=x/(sqrt(1-x^2))``y="cos"(tan^(-1)x); y=s in(cot^(-1)x)` |
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Answer» (i) `y = tan(cos^-1x) and y = sqrt(1-x^2)/x` `=>y = tan(tan^-1(sqrt(1-x^2)/x)) and y = sqrt(1-x^2)/x` `=>y = sqrt(1-x^2)/x and y = sqrt(1-x^2)/x` `:.` Domain and range of both functions are same, so there graphs will be same. (ii) `y = tan(cot^-1x) and y = 1/x` `=>y = tan(tan^-1(1/x)) and y = 1/x` `=>y = 1/x and y = 1/x` `:.` Domain and range of both functions are same, so there graphs will be same. (iii) `y = sin (tan^-1x) and y = x/sqrt(1-x^2)` `=>y = sin(sin^-1( x/sqrt(1-x^2))) and y = x/sqrt(1-x^2)` `=>y = x/sqrt(1-x^2) and y = x/sqrt(1-x^2)` `:.` Domain and range of both functions are same, so there graphs will be same. (iv) `y = cos (tan^-1x) and y = sin(cot^-1x)` `=>y = cos(cos^-1(1/(sqrt(1+x^2)))) and y = sin(sin^-1(1/(sqrt(1+x^2))))` `=> y = 1/(sqrt(1+x^2)) and y = 1/(sqrt(1+x^2))` `:.` Domain and range of both functions are same, so there graphs will be same. |
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| 248. |
If `cot^(-1)((n^2-10 n+21. 6)/pi)>pi/6,`where `x y |
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Answer» Correct Answer - A::C We have `cot^(-1) ((n^(2) -10n + 21.6)/(pi)) gt (pi)/(6)` `rArr (n^(2) - 10n + 21.6)/(pi) lt cot.(pi)/(6)` (as cot x decreasing for `0 lt x lt pi`) or `n^(2) - 10n + 21.6 lt pisqrt3` or `n^(2) - 10n + 25 + 21.6 - 25 lt pi sqrt3` or `(n-5)^(2) lt pi sqrt3 + 3.4` or `-sqrt(sqrt3 po + 3.4) lt n - 5 lt sqrt(sqrt3 pi + 3.4)` or `5 -x sqrt(sqrt5 pi + 3.4) lt n - 5 lt sqrt(sqrt3 pi + 3.4)` ...(i) Since `sqrt3 pi = 5.5`, nearly , `sqrt(sqrt3 pi + 3.4) ~ sqrt(8.9) ~ 2.9` `rArr 2.1 lt n lt 7.9` `:. n = 3, 4, 5, 6, 7 " " {"as " n in N}` |
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| 249. |
The principal value of `sin^(-1)(1/2)` isA. `pi/6`B. `(5pi)/6`C. `-pi/6`D. Both (1) & (2) |
| Answer» Correct Answer - A | |
| 250. |
Find the principal value of `cot^(-1)(-sqrt3)`A. `-pi/3`B. `-pi/6`C. `(2pi)/3`D. `(5pi)/6` |
| Answer» Correct Answer - D | |