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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.
| 151. |
Find the maximum value of `(sec^(-1) x) (cosec^(-1) x), x ge 1` |
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Answer» Correct Answer - `(pi^(2))/(16)` For `x ge 1, cosec^(-1) x gt 0` Using A.M. `ge` G.M., we have `(sec^(-1) x + cosec^(-1) x)/(2) ge sqrt(sec^(-1) x cosec^(-1) x)` `rArr ((pi//2))/(2) ge sqrt(sec^(-1) x cosec^(-1) x)` `rArr (pi)/(2) ge sqrt(sec^(-1) x cosec^(-1) x)` `:. sec^(-1) x cosec^(-1) x le (pi^(2))/(16)` Thus maximum value of `sec^(-1) x cosec^(-1) x " is " (pi^(2))/(16)` |
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| 152. |
Let `f(x)=sin^(-1)((2x)/(1+x^(2)))` and `g(x)=cos^(-1)((x^(2)-1)/(x^(2)+1))`. Then tha value of f(10)-g(100) is equal toA. `pi-2(tan^(-1)(10)+tan^(-1)(100))`B. 0C. `2(tan^(-1)(100)-tan^(-1)(10))`D. `2(tan^(-1)(10)-tan^(-1)(100))` |
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Answer» Correct Answer - C `f(x)=sin^(-1)((2x)/(1+x^(2)))=pi-2 tan^(-1)x`, for `x ge 1` and `g(x)=cos^(-1)((x^(2)-1)/(x^(2)+1))` `=pi-cos^(-1)((1-x^(2))/(1+x^(2)))` `=pi-cos^(-1)((1-x^())/(1+x^(2)))` `=pi -2 tan^(-1)x`, for `x ge 0` Now f(10)-g(100) `=(pi-2tan^(-1)(10))-(pi-2tan^(-1)(100))` `=2(tan^(-1)(100)-tan^(-1)(10))` |
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| 153. |
Find the minimum value of the function `f(x) = (pi^(2))/(16 cot^(-1) (-x)) - cot^(-1) x` |
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Answer» `f(x) = (pi^(2))/(16 cot^(-1) (-x)) - (pi - cot^(-1) (-x))` `= cot^(-1) (-x) + (pi^(2))/(16 cot^(-1) (-x)) - pi` `= (sqrt(cot^(-1) (-x)) - (pi)/(4sqrt(cot^(-1) (-x))))^(2) + (pi)/(2) - pi ge - (pi)/(2)` `f_("min") = -(pi)/(2)` |
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| 154. |
Find the value of `sin^(-1) (sin 5) + cos^(-1) (cos 10) + tan^(-1) {tan(-6)} + cot^(-1) {cot(-10)}` |
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Answer» `sin^(-1) (sin 5) + cos^(-1) (cos 10) + tan^(-1) {tan (-6)}` `= sin^(-1) (sin 5) + cos^(-1) (cos 10) - tan^(-1) {tan (6)}` `= (5 - 2pi) + (4 pi- 10) - (6 - 2pi) + pi (10 - 3 pi)` `= 8 pi - 21` |
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| 155. |
Find the value of `sin^(-1)(sqrt3/2)+cos^(-1)(-1)+tan^(-1)(-1)` |
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Answer» Since the principal values of `sin^(-1)x,cos^(-1)xandtan^(-1)x" lie in "[-pi/2,pi/2].[0,pi],(-pi/2,pi/2)` respectively and in these intervals `sin.(pi/3)=sqrt3/2, cospi=-1 tan.(-pi/4)=-1` `rArr sin^(-1)(sqrt3/2)=pi/3, cos^(-1)(-1)=pi,tan^(-1)(-1)=-pi/4` `rArr sin^(-1)(sqrt3/2)+cos^(-1)(-1)+tan^(-1)(-1)` `=pi/3+pi+(-pi/4)=(13pi)/12` |
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| 156. |
Prove that `tan^(-1) {(x)/(a + sqrt(a^(2) - x^(2)))} = (1)/(2) sin^(-1).(x)/(a), -a lt x lt a` |
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Answer» Let `x = a sin theta, -a lt x lt a`. Then, `-a lt a sin theta lt a` or `-1 lt sin theta lt 1 rArr theta in (-(pi)/(2), (pi)/(2))` `rArr tan^(-1) {(x)/(a+ sqrt(a^(2) - x^(2)))} = tan^(-1) {(a sin theta)/(a + sqrt(a^(2) - a^(2) sin^(2) theta))}` `= tan^(-1) {(sin theta)/(1 + cos theta)}` `= tan^(-1) {(2 sin (theta//2) cos (theta//2))/(2 cos^(2) (theta//2))}` `= tan^(-1) {tan.(theta)/(2)}` `= (theta)/(2) = (1)/(2) sin^(-1). (x)/(a)` |
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| 157. |
The complete set of values of a for which the function `f(x)=tan^(-1)(x^(2)-18x +a)gt 0 AA x in R` isA. `(81,oo)`B. `[81,oo)`C. `(-oo,81)`D. `(-oo,81]` |
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Answer» Correct Answer - A `tan^(-1)(x-18x+a)gt 0 AA x in R` `rarr x^(2)-18x+a gt 0 AA x in R` `rArr D = (18)^(2) - 4a lt 0` `rArr a gt 81` `rArr a in (81, oo)` |
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| 158. |
Find the value of `sin^(-1)(sin.(5pi)/3)`. |
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Answer» Since `(5pi)/3 cancelin[-pi/2,pi/2]` `:.sin^(-1)(sin(5pi)/3)` `=sin^(-1)(sin(2x-pi/3))` `sin^(-1)(-sin.(pi)/3)` `sin^(-1)(sin.(-pi/3))` `=-pi/3` |
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| 159. |
`sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x`, then roots of theequation are-a.`0`b. `1`c.`-1 `d. `-2` |
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Answer» Correct Answer - D `"sin"^(-1)(3x)/(5)+"sin"^(-1)(4x)/(5)=sin^(-1)x` `rArr sin^(-1)((3x)/(5)sqrt(1-(16x^(2))/(25))+(4x)/(5)sqrt(1-(9x^(2))/(25)))=sin^(-1)x` `rarr (3x)/(5)(sqrt(25-16x^(2)))/(5)+(4x)/(5)(sqrt(25-9x^(2)))/(5)=x` `rArr x = 0` or `3sqrt(25-16x^(2))+4sqrt(25-9x^(2))=25` Now `sqrt(225-144x^(2))+sqrt(400-144x^(2))=25` .....(1) `rArr (175)/(sqrt(400-144x^(2))-sqrt(225-144x^(2)))=25` `rArr sqrt(400-144x^(2))-sqrt(225-144x^(2))=7` ....(2) From (1) and (2), `sqrt(400-144x^(2))=16` `rArr 400-144x^(2)=256` `rArr x^(2)=1` `rArr x = pm 1` |
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| 160. |
The number of roots of the equation `sin^(-1)x-(1)/(sin^(-1)x)=cos^(-1)x-(1)/(cos^(-1)x)` is |
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Answer» Correct Answer - C `sin^(-1) x-(1)/(sin^(-1)x)=cos^(-1)x-(1)/(cos^(-1)x)` `rArr (sin^(-1)x-cos^(-1)x)((sin^(-1)x.cos^(-1)x+1))/(sin^(-1)x.cos^(-1)x)=0` `rArr sin^(-1)x=cos^(-1)x` or `sin^(-1)x cos^(-1)x+1=0` `rArr x=(1)/(sqrt(2))` or `sin^(-1)x((pi)/(2)-sin^(-1)x)+1=0` `rArr x=(1)/(sqrt(2))` or `sin^(-1)x=((pi)/(2)pm sqrt(((pi^(2))/(4)+4)))/(2)` `rArr x = (1)/(sqrt(2))` or `sin^(-1)x=(pi)/(4)-sqrt((1+(pi^(2))/(16)))` |
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| 161. |
The value of x satisfying the equation `cos^(-1)3x+sin^(-1)2x=pi` isA. `x=(1)/(sqrt(3))`B. `x=(-1)/(sqrt(3))`C. `x=(-1)/(sqrt(3))`D. none of these |
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Answer» Correct Answer - D `sin^(-1)2x=pi-cos^(-1)3x` `rArr sin^(-1)2x=cos^(-1)(-3x)` `rArr sqrt(1-4x^(2))=-3x(xlt 0)` `rArr x^(2)=(1)/(13)` `rArr x=(-1)/(sqrt(13))` But for `x lt 0, pi//2 lt cos^(-1)3x lt pi` and `-pi//2 lt sin^(-1)2x lt 0` `therefore L.H.S. ne pi` Hence, equation has no solution. |
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| 162. |
The exhaustive set of values of a for which `a - cot^(-1) 3x = 2 tan^(-1) 3x + cos^(-1) x sqrt3 + sin^(-1) x sqrt3` may have solution, isA. `[-(pi)/(4), (pi)/(4)]`B. `((pi)/(2), (3pi)/(2))`C. `[(2pi)/(3), (4pi)/(3)]`D. `[-(3pi)/(6), (7pi)/(6)]` |
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Answer» Correct Answer - C `a - cot^(-1) 3x = 2 tan^(-1) 3x + cos^(-1) x sqrt3 + sin^(-1) x sqrt3` Clearly, given equation is meaningful when `-1 le x sqrt3 le 1` Given equation becomes `a= tan^(-1) 3x + (tan^(-1) 3x+ cot^(-1) 3x) + (cos^(-1) x sqrt3 + sin^(-1) x sqrt3)` `= tan^(-1) 3x + pi//2 + pi//2` `pi + tan^(-1) 3x` Now `-1 le x sqrt3 le 1` `rArr -sqrt3 le 3x le sqrt3` `-(pi)/(3) le tan^(-1) 3x le (pi)/(3)` `rArr (2pi)/(3) le pi + tan^(-1) 3x le (4pi)/(3)` `rArr a in [(2pi)/(3), (4pi)/(3)]` |
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| 163. |
If `u=cot^-1 sqrt(tanalpha)-tan^-1 sqrt(tan alpha),` then `tan(pi/4-u/2)` is equal to (a) `sqrt(tan alpha)` (b) `sqrt(cos alpha)` (c) `tan alpha` (d) `cot alpha`A. `sqrt(tan alpha)`B. `sqrt(cot alpha)`C. `tan alpha`D. `cot alpha` |
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Answer» Correct Answer - A Let `sqrt(tan alpha) = tan x`. Then `u = cot^(-1) (tan x) - tan^(-1) (tan x)` `= (pi)/(2) - x -x = (pi)/(2) - 2x` or `2x = (pi)/(2) -u` or `x = (pi)/(4) - (u)/(2)` or `tan x = tan ((pi)/(4) -(u)/(2))` `rArr sqrt(tan alpha) = tan ((pi)/(4) - (u)/(2))` |
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| 164. |
The minimum integral value of `alpha` for which the quadratic equation `(cot^(-1)alpha)x^(2)-(tan^(-1)alpha)^(3//2)x+2(cot^(-1)alpha)^(2)=0` has both positive rootsA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `(cot^(-1)alpha)x^(2)-(tan^(-1)alpha)^(3//2)x + 2(cot^(-1)alpha)^(2)=0` Equation has real roots `therefore D ge 0 rArr (tan^(-1)alpha)^(3)-8(cot^(-1)alpha)^(2)=0` `therefore tan^(-1)alpha ge 2 cot^(-1)alpha = pi-2 tan^(-1)alpha` `rArr tan^(-1)alpha ge (pi)/(3)rArr alpha ge sqrt(3)` Sum of roots gt 0 `rArr ((tan^(-1)alpha)^(3//2))/(2cot^(-1)alpha)gt 0` Product of roots gt 0 `rArr 2cot^(-1)alpha gt 0 rArr alpha in R` `rArr alpha ge sqrt(3)` |
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| 165. |
Let `f(x) = tan^(-1) (x^2-18x + a) gt 0 x in R`. Then the value of a lies inA. `(81,oo)`B. `{81,oo)`C. `(-oo,81)`D. `(-oo,81]` |
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Answer» Correct Answer - A Given `f(x)=tan^(-1)(x^2-18x+a)gt0` `rArrtan^(-1)(x^2-18x+a) gt0` `rArrx^2-18x+agt0` `18^2-4a gt0` `rArr a gt 18^2/4=(18xx18)/4=81` `rArra gt 81` `rArr a in (81,oo)` |
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| 166. |
The number of triple satisfying `sin^(-1)x+cos^(-1)y+sin^(-1)z=2pi` is |
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Answer» We know that `sin^(-1)xle(pi)/(2) sin^(-1)zle(pi)/(2) and cos^(-1) y le pi` `therefore sin^(-1)x+cos^(-1)y+sin^(-1)zle2pi` `rarr sub^(-1)x=(pi)/(2) cos^(-1)y=pi and sin^(-1) =(pi)/(2)` `rarr x=1 , y=-1 =pi and sin^(-1) z=(pi)/(2)` `rarr x =1 ,y =-1 and z=1` |
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| 167. |
`tan^(-1)(C_(1)x-y)/(c_(1)+c_(3)c_(2))+..+tan^(-1)(1)/(c_(n))` is equal topA. `tan^(-1)(y)/(x)`B. `tan^(-1)(x)/(y)`C. `-tan^(-1)(x)/(y)`D. none of these |
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Answer» `tan^(-1)((c_(1)x-y)/(c_(1)y+x))+tan^(-1)(c_(2)-c_(1)/(12+c_(2)c_(1))+tan^(-1)(c_(3)-c_(2))/(1+c_(3)c_(2))+…+tan^(-1)(1)/(c_(n))` `=tan^(-1)(x)/(y)-tan^(-1)(1)/(c_(1))+(tan^(-1)(1)/(c_(1))-tan^(-1)(1)/(c_(2))+(tan^(-1))(1)/(c_(2))-tan^(-1)(1)/(c_(3))+..+(tan^(-12))(1)/(c_(n-1))-tan^(-1)(1)/(c_(n)))+tan^(-1)(1)/(c_(n))` `=tan^(-1)(x)/(y)` |
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| 168. |
If `cot(cos^(-1)x)=sec{tan^(-1)((a)/sqrt(b^(2)-a^(2)))}` then x equalsA. `(b)/sqr(2b^(2)-a^(2))`B. `(a)/sqr(2b^(2)-a^(2))`C. `sqrt(b^(2)-a^(2))/(a)`D. `sqrt(b^(2)-a^(2))/(b)` |
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Answer» We have `cot (cos^(-1)x) = see {tan^(-1)(a)/sqrt(b^(2)-a^(2))}` `rarr cot{cot^(-1)(x)/sqrt(1-x^(2))}=sec { sec^(-1)(b)/sqrt(b^(2)-a^(2))}` `rarr (1-x^(2))/(x^(2))=(b^(2)-a^(2))/(b^(2))` `rarr (1)/(x^(2))-1=1-(a^(2))/(b^(2) rarr (1)/(x^(2))= (2b^(2)-a^(2))/(b^(2)) rarr x= (b)/sqrt(2b^(2)-a^(2))` |
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| 169. |
If `sin^-1(sin 5)gtx^(2)-4x` thenA. `x in 2+sqrt(9-2pi , oo)`B. `x in 2-sqrt(9-2pi),2+sqrt(9-2pi)`C. `xin 2-sqrt(9-2pi)oo)`D. none of these |
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Answer» We have `sin^(-1)(sin5)gtx^(2)-4x` `rarr sin^(-1)sin(5-2pi)gtx^(2)-4x` `rarr 5-2 pi gt x^(2)-4x` `rarr x^(2)-4x-5+2pilt0 rarr -sqrt(9-2pi)ltxlt2+sqrt(9-2pi)` |
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| 170. |
The solution set of the equation `sin^(-1)x=2 tan^(-1)x` isA. `{1,2}`B. `{-1,2}`C. `{-1,1}`D. `{1,1//2,0}` |
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Answer» Clearly LHS of the given equation is meaningful for `x in [-1,1]` and RHS is defined for all x`x in R` So the given equation exists for `x in [-1,1]` Now `sin^(-1)x=2 tan^(-1)` `rarr sin^(-1) x= sin^(-1)((2x)/(1+x^(2)))` `rarr x^(3)=x rarr x=0,1,-1` |
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| 171. |
If `cos^(-1)(cos 4)gt3x^(2)-4x` thenA. `x in (-oo,(2-sqrt(6pi-8))/(3)`B. `x in ((2+sqrt(6pi-87))/(3),oo)`C. `x in ((2+sqrt(6pi-8))/(3),oo)`D. `x in ((2-sqrt(p6pi-8))/(3),(2+sqrt(6pi-8))/(3)` |
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Answer» Correct Answer - `oo` We have `cos^(-1)(cos4)gt3x^(2)-4x` `rarr cos^(-1)(cos(2pi-4))gt3x6(2)-4x` `rarr 2pi -4 gt3x^(2)-4x^(2)` `rarr 3x^(2)-4x-(2pi-4)lt0` `rarr (2-sqrt(6pi-8))/(3)ltxlt(2+sqrt(6pi-8))/(3)` |
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| 172. |
if `cos^(- 1)sqrt(p)+cos^(- 1)sqrt(1-p)+cos^(- 1)sqrt(1-q)=(3pi)/4` ,then the value of q is (A) 1 (B) `1/sqrt(2)` (C) `1/3` (D) `1/2`A. `(1)/sqrt(2)`B. 1C. `1/2`D. `1/3` |
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Answer» Clearly left hand side of the given equation is meaning if `0 le p le 1 and 0 le q le 1` Let `p cos^(2) theta` Then `cos^(-1)sqrt(p)+cos^(-1)sqrt(1-p) + cos^(-1) sqrt(1-q)=(3pi)/(4)` `rarr cos ^(-1)(cos theta) + cos^(-1)(sin theta)+cos^(-1)sqrt(1-q)=(3pi)/(4)` `rarr cos^(-1)sqrt(1-q)=(pi)/(4)` `rarr sqrt(1-q)=(1)/sqrt(2) rarr 1- q=1/2 rarr =1/2` |
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| 173. |
Find the range of `f(x) = cos^(-1) x + cot^(-1) x` |
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Answer» Correct Answer - `[(pi)/(4), (7pi)/(4)]` We have `f(x) = cos^(-1) x + cot^(-1) x` Clearly, domain of the function is `[-1 ,1]` Now, both `cos^(-1) x and cot^(-1) x` are decreasing function `:.` f(x) is also decreasing function `:.` Range of the function is `[f (1), f(-1)]` Now, `f(-1) = cos^(-1) + cot^(-1) (-1) = pi + 3 pi//4 = 7 pi//4` `f(1) = cos^(-1) (1) + cot^(-1) (1) = 0 + pi//4 = pi//4` Hence, range is `[pi//4, 7pi//4]` |
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| 174. |
Find the range of `y = (cot^(-1) x) (cot^(-1) (-x))` |
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Answer» `y = (cot^(-1) x) (cot^(-1) (-x))` `= cot^(-1) (x) (pi - cot^(-1) (x))` Now, `cot^(-1) (x) and (pi - cot^(-1) (x)) gt 0` Using A.M. `ge` G.M., we get `(cot^(-1) x + (pi - cot^(-1) x))/(2) ge sqrt((cot^(-1) x) (pi - cot^(-1) x))` `rArr 0 lt cot^(-1) (x) (pi - cot^(-1) (x))` `le ((cot^(-1) x + (pi - cot^(-1) x))/(2))^(2) = (pi^(2))/(4)` `rArr 0 lt y le (pi^(2))/(4)` |
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| 175. |
The number of integral values in the range of the function `f(x)=sin^(-1)x-cot^(-1)x+x^(2)+2x +6` isA. 10B. 11C. 12D. 8 |
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Answer» Correct Answer - D `f(x)=sin^(-1)x-((pi)/(2)-tan^(-1)x)+(x+1)^(2)+5` `=sin^(-1)x + tan^(-1)x-(pi)/(2)+(x +1)^(2)+5` Domain of f(x) is [-1,1] and `sin^(-1) x, tan^(-1)x` and `(x+1)^(2)` are all increasing in [-1,1] `therefore f(x)` is increasing `f(-1)=(-5pi)/(4)+5` and `f(1)=(pi)/(4)+9` `therefore` Range of f(x) is `[(-5pi)/(4)+5,(pi)/(4)+9]` |
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| 176. |
If `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi` then the value of `xsqrt(1-x^(2))+ysqrt(1-y^(2))+zsqrt(1-z^(2))`A. 2xyzB. `x^(2)+y^(2)+z^(2)`C. `xy+yz+zx`D. none of these |
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Answer» Let `sin^(-1) x =A sin^(-1) y = B and sin^(-1) z = C` Then `x= sin A, y sin B z = sin C` Also `sin^(-1)x+ sin^(-1) y + sin^(-1)z=pi` `rarr A+B+C=pi` `rarr sin 2A + sin 2B + sin 2C =4 A sin B sin C` `rarr sin A cos A + sin B cos B + sin C cos C =2 sin A sin B sin C` `rarr xsqrt(1-x^(2))+ysqrt(1-y^(2))+zsqrt(1-z^(2))=2xyz` |
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| 177. |
The domin of the function `cos^(-1) (2x-1)` isA. `[0,1]`B. `[-1,1]`C. `(-1,1)`D. `[0,pi]` |
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Answer» Correct Answer - A We have, `f(x) =cos^(-1)(2x-1)` `:. -1 le 2x le - 1 le 1` `rArr 0 le x le 1` `rArr 0 le x le 1` `x in[0,1]` |
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| 178. |
All trigonometric functions have inverse over their respective domins. |
| Answer» We know that, all trigonometric function have inverse over their restricted domins. | |
| 179. |
The sum of the two angles `cot^(-1) 3 and cosec^(-1) sqrt(5)` isA. `(pi)/(2)`B. `(pi)/(3)`C. `(pi)/(4)`D. `(pi)/(6)` |
| Answer» Correct Answer - C | |
| 180. |
The value of `alpha`such that `sin^(-1)2/(sqrt(5)),sin^(-1)3/(sqrt(10)),sin^(-1)alpha`are the angles of a triangle is`(-1)/(sqrt(2))`(b) `1/2`(c) `1/(sqrt(3))`(d) `1/(sqrt(2))` |
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Answer» `sin^-1(2/sqrt5) = tan^-1 2` `sin^-1(3/sqrt10) = tan^-1 3` `sin^-1(alpha) = tan^-1 (alpha/(sqrt(1-alpha^2)))` So, given equation becomes, `tan^-1 2 + tan^-1 3 + tan^-1 (alpha/(sqrt(1-alpha^2))) = pi` As `2*3 >1`, `:. tan^-1 2 + tan^-1 3 = pi +tan^-1((2+3)/(1-(2)(3))) = pi+tan^-1(-1) ` So,our equation becomes, `pi+tan^-1(-1) + tan^-1 (alpha/(sqrt(1-alpha^2))) = pi` `=>tan^-1 (alpha/(sqrt(1-alpha^2))) = tan^-1(1)` `=> alpha/(sqrt(1-alpha^2))= 1` `=>alpha^2 = 1-alpha^2` `=>alpha = 1/sqrt2` So, option `d` is the correct option. |
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| 181. |
Prove that:`tan^(-1)(1/2tan2A)+tan^(-1)(cota)+tan^(-1)(cot^3A)={0,ifpi/4 |
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Answer» For `0 lt A lt (pi//4), cot A gt 1` `rArr (cot A) (cot^(3) A) gt 1` Then, `tan^(-1) ((1)/(2) tan 2A) + tan^(-1) (cotA) + tan^(-1) (cot^(3) A)` `=tan^(-1) ((tanA)/(1 - tan^(2)A)) + pi + tan^(-1) ((cot A + cot^(3) A)/(1 - cot^(4)A))` `= tan^(-1) ((tan A)/(1 - tan^(2)A)) + pi + tan^(-1) ((cot A)/(1- cot^(2)A))` `= tan^(-1) ((tanA)/(1- tan^(2)A)) + pi + tan^(-1) ((tanA)/(tan^(2) A -1)) = pi` |
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| 182. |
`2"tan"(tan^(-1)(x)+tan^(-1)(x^3)),w h e r ex in R-{-1,1},`is equal to`(2x)/(1-x^2)``t(2tan^(-1)x)``tan(cot^(-1)(-x)-cot^(-1)(x))``"tan"(2cot^(-1)x)`A. `(2x)/(1 -x^(2))`B. `tan(2 tan^(-1) x)`C. `tan (cot^(-1) (-x) - cot^(-1) x))`D. `tan (2 cot^(-1) x)` |
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Answer» Correct Answer - A::B::C Let `tan^(-1) x = alpha and tan^(-1) x^(3) = beta` `tan alpha = x and tan beta = x^(3)` `:. 2 tan (alpha + beta) = (2(tan alpha + tan beta))/(1-tan alpha tan beta) = 2[(x + x^(3))/(1 -x^(4))] = (2x)/(1 -x^(2))` Also, `(2x)/(1 -x^(2)) = (2 tan alpha)/(1 -tan^(2) alpha) = tan 2 alpha = tan (2 tan^(-1) x)` `= tan (2 ((pi)/(2) - cot^(-1) x))` `= tan (pi -cot^(-1) x - cot^(-1) x)` `= tan (cot^(-1) (-x) - cot^(-1) (x))` |
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| 183. |
Statement-1: `cosec^(-1)(3)/(2)+cos^(-1)(2/3)-2cot^(-1)(1/7)-cot^(-1)7=cot^(-1)7` Statement-2: `cos^(-1)x=sin^(-1)((1)/(x)) and for xgt0cot^(-1)x=tan^(-1)((1)/(x))`A. Statement-1 is is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for 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» statement -2 is true (see theory) using statement -2 we have `cosec^(-1)3/2+cos^(-1)2/3-2cos^(-1)1/7-cos^(-1)7` `=(sin^(-1)2/3+cos^(-1)2/3)-cot^(-1)1/7-(tan^(-1)7+cot^(-1)7)` `=(pi)/(2)-cot^(-1)1/7-(pi)/(2)=-cot^(-1)1/7=-tan^(-1)7` so statement -1 is not correct |
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| 184. |
`2"tan"(tan^(-1)(x)+tan^(-1)(x^3)),w h e r ex in R-{-1,1},`is equal to`(2x)/(1-x^2)``t(2tan^(-1)x)``tan(cot^(-1)(-x)-cot^(-1)(x))``"tan"(2cot^(-1)x)` |
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Answer» Let `tan^-1x = alpha => tanalpha = x` Let `tan^-1x^3 = beta => tan beta = x^3` Then, `2tan(tan^-1x+tan^-1x^3) = 2tan(alpha+beta)` `=2[(tanalpha+tanbeta)/(1-tanalphatanbeta)]` `=2[(x+x^3)/(1-x(x^3))]` `=2[(x(1+x^2))/(1-x^4)]` `=2[(x(1+x^2))/((1-x^2)(1+x^2))]` `=(2x)/(1-x^2)->(1)` `=(2tanalpha)/(1-tan^2alpha)` `=tan2alpha` `=tan(2tan^-1x)->(2)` `=tan(2(pi/2-cot^-1x)` `=tan(pi-2cot^-1x)` `=tan(cot^-1(-x)-cot^-1(x))->(3)` From (1),(2), (3), options, `a`,`b` and `c` are correct options. |
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| 185. |
If `tan^(-1)x+2cot^(-1)x=(5pi)/6`, then find x. |
| Answer» Correct Answer - `1/sqrt3` | |
| 186. |
`tan(cot^(-1)x)` is equal toA. `(pi)/(2)-x`B. `cot(tan^(-1)x)`C. tan xD. none of these |
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Answer» We have `tan(cot^(-1)x)=tan (p[i)/(2)-tan^(-1)x)=cot(tan^(-1)x)` |
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| 187. |
The principal values of `cost^(-1)(-sin(7pi)/(6))` isA. `(5pi)/(3)`B. `(7pi)/(6)`C. `(pi)/(3)`D. none of these |
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Answer» Correct Answer - C `cos^(-1)(-sin.(7pi)/(6))=cos^(-1)(-sin(pi+(pi)/(6)))` `=cos^(-1)(sin.(pi)/(6))` `=cos^(-1)((1)/(2))=(pi)/(3)` |
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| 188. |
The least and the greatest values of `(sin^(-1)x)^3+(cos^(-1)x)^3`are`(-pi)/2,pi/2`(b) `(-pi^3)/8,(pi^3)/8``(pi^3)/(32),(7pi^3)/8`(d) none of theseA. `-(pi)/(2),(pi)/(2)`B. `-(pi^(3))/(8),(pi^(3))/(8)`C. `(pi^(3))/(32),(7 pi^(3))/(8)`D. none of these |
| Answer» Correct Answer - C | |
| 189. |
`sin^(-1)(sin((7pi)/6))` isA. `-pi/6`B. `pi/6`C. `(7pi)/6`D. `-(5pi)/6` |
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Answer» Correct Answer - A Sin^-1(sin∆)=∆,if -π/2<∆<π/2 sin^-1(sinπ+π/6) sin^-1(sin-π/6) -π/6 |
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| 190. |
If `a le 1/32` then the number of solution of `(sin^(-1) x)^(3) +(cos^(-1) x)^(3) = a pi^(3)` is |
| Answer» Correct Answer - A | |
| 191. |
If `-1 le x le -(1)/sqrt(2)` then `sin^(-1)2xsqrt(1-x^(2))` equalsA. `2 sin^(-1)x`B. `pi-2 sin^(*-1)x`C. `-pi-2 sin^(-1)x`D. `none of these |
| Answer» Correct Answer - C | |
| 192. |
Range of the function `f (x) = cos^-1 (-{x})`, where `{.}` is fractional part function, is:A. `(pi)/(2),1)`B. `(pi),(pi)/(2)`C. `[(pi)/(2),pi]`D. `(0,(pi)/(2))]` |
| Answer» Correct Answer - C | |
| 193. |
Let `cos(2 tan^(-1) x)=1/2` then the value of x isA. `sqrt(3)`B. `(1)/sqrt(3)`C. `1-sqrt(3)`D. `1-(1)/sqrt(3)` |
| Answer» Correct Answer - B | |
| 194. |
If `cosec^(-1)x+cosec^(-1)y+cosec^(-1)z=-(3pi)/(2),then x/y+y/z+z/x=`A. 1B. `-3`C. 3D. `3/2` |
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Answer» We know that the minimum value of `cosec^(-1)x is -(pi)/(2)` which is attained at x=-1 `therefore cosec^(-1)+cosec^(-1)y+cosec^(-1)z=-(3pi)/(2)` `rarr cos^(-1)x+cosec^(-1)+cosec^(-1)z=(-pi)/(2)+(-pi)/(2)+(pi)/(2)` `rarr x=-1,y=-1z=-1` `therefore x/y+y/z+z/x=(-1)/(-1)+(-1)/(-1)+(-1)/(-1)=3` |
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| 195. |
`tan^(- 1)(a/x)+tan^(- 1)(b/x)=pi/2` then `x=`A. `sqrt(ab)`B. `sqrt(2ab)`C. 2abD. ab |
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Answer» We have `tan^(-1_(a)/(x)+tan^(-1)(b)/(x))=(pi)/(2)` `rarrtan^(-1)((a)/(x)+(b)/(x)-(ab)/(x^(2)))=(pi)/(2)rarr1-(ab)/(x^(2))=0rarrx=sqrt(ab)` |
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| 196. |
If `x gt y gt 0`, then find the value of `tan^(-1).(x)/(y) + tan^(-1) [(x + y)/(x -y)]`A. `(pi)/(2)`B. `(pi)/(3)`C. `(pi)/(4)`D. `(pi)/(4) or -(3pi)/(4)` |
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Answer» We have `tan^(-1)(x/y)-tan^(-1)(x-y)/(x+y)` `=tan^(-1)x/y-tan^(-1)(1-y//x)/(1+y//x)` `=tan^(-1)x/y-(ta^(-1)1-tan^(-1)y/x)` `tan^(-1)x/y+tan^(-1)y/x=(pi)/(4)=ta^(-1)x/y_+cot^(-1)x/y-(pi)/(2)-(pi)/(4)=(pi)/(4)` |
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| 197. |
If `x gt (1)/sqrt(3) then tan^(-1) (3x-x^(3))/(1-3x^(2))` equalsA. `3 tan^(-1)x`B. `-pi+3tan^(-1)`C. `pi+3tan^(-1)x`D. none of these |
| Answer» Correct Answer - B | |
| 198. |
If `alpha and beta (alpha gt beta)` are the roots of `x^(2) + kx - 1 =0`, then find the value of `tan^(-1) alpha - tan^(-1) beta` |
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Answer» Correct Answer - `pi` We have `alpha beta = -1 and alpha gt beta` `rArr tan^(-1) alpha gt tan^(-1) beta` `:. Tan^(-1) alpha - tan^(-1) beta = tan^(-1).(alpha - beta)/(1 + alpha beta)` `= tan^(-1) oo = (pi)/(2)` |
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| 199. |
If `x + y + z = xyz and x, y, z gt 0`, then find the value of `tan^(-1) x + tan^(-1) y + tan^(-1) z` |
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Answer» Correct Answer - `pi` `xy = 1 _ (x)/(z) + (y)/(z) gt 1` `rArr tan^(-1) x + tan^(-1) y + tan^(-1) z` `= pi + tan^(-1) [(x + y)/(1 - xy)] + tan^-(1) z` `= pi + tan^(-1) [(xyz -z)/(1 - xy)] + tan^(-1) z` `= pi + tan^(-1) [(z (xy -1))/(1 -xy)] + tan^(-1) z` `= pi + tan^(-1) (-z) + tan^(-1) z = pi` |
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| 200. |
If `cos^(-1)lambda+cos^(-1)mu+cos^(-1)gamma=3pi,`then find the value of `lambdamu+mugamma+gammalambda` |
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Answer» `cos^(-1)lambda+cos^(-1)mu+cos^(-1)gamma=3pi` `0<=cos^(-1)x<=x` `cos^(-1)lambda=cos^(-1)mu=cos^(-1)gamma=pi` `cos^(-1)lambda=pi` `cospi=lambda=-1` `lambda=mu=gamma=-1` `lambdamu+mulambda+gammalambda=3` `1+1+1=3`. |
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