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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.
| 451. |
Find the greatest and least values of the function `f(x)=(sin^(-1)x)^3+(cos^(-1)x)^3,-1 le x le 1` |
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Answer» Using `sin^(-1)x+cos^(-1)x=pi/2` we can write the function in terms of `sin^(-1)x(or cos^(-1)x)` alone. And then note that range of `sin^(-1)x " is "[-pi/2,pi/2] and cos^(-1)x " is "[0,pi]`. Put `cos^(-1)x=t, ` so that `sin^(-1)x=pi/2-t, and 0, let le pi` Now the trigonometry ends and algebra takes over. Finding the greatest and least values of the function f(x) amounts to finding the greatest and least values of function g(t). where `g(t)=(pi/2-t)^3+t^3,0 le t le pi` `=(pi/2-t+t^3)-3(pi/2-t)(t)(pi/2-t+t)` `=(pi/2)^3-3t(pi/2-t)pi/2` `pi^3/8-3pi^3/4t+(3pi)/2t^2` `(3pi)/2[t^2-pi/2t+pi^2/12]` `=(3pi)/2[(t0pi/4)^2+pi^2/12-pi^2/16]` `=(3pi)/2[(t-pi/2)^2+pi^2/48]` `g(pi/4)=(3pi)/2.pi^2/48=pi^3/32` `g(0)=pi^3/8` `g(pi)=(3pi)/2[((3pi)/4)^2+pi^2/48]` `=(3pi)/2[(9pi^2)/16+pi^2/48]` `=(3pi)/2.(28pi^2)/48` `=7/8pi^3` So the least value is `pi^3/32`, attained at `t=pi/4`, i.e., `x=1/sqrt2` and the greatest value is `(7pi^3)/8`, attained at `t=pi,i.e., x= -1` |
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| 452. |
Solve the equation `tan^(-1) 2x + tan^(-1) 3x = pi//4` |
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Answer» While solving such equations we use the following formulas: `tan^(1) x + tan^(-1) y = tan^(-1).(x + y)/(1 - xy)`...(i) We know that this formula is valid only when `xy lt 1`. So, we may get some extra solutions by solving equation with this formula. The extra solutions can be removed from the solution set by putting the values obtained in the original equation and checking whether it satisfies the equation or not `tan^(-1) 2x + tan^(-1) 3x = (pi)/(4)` or `tan^(-1) ((2x + 3x)/(1 - 6x^(2))) = (pi)/(4)` or `(5x)/(1 = 6x^(2)) = 1` or `6x^(2) + 5x -1 = 0` or `(6x -1) (x + 1) = 0` or `x = 1//6, -1` But for `x = -1, tan^(-1) 2x + tan^(-1) 3x lt 0` So, it does not satisfy Eq. (i) Hence, `x = (1)/(6)` |
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| 453. |
Find the values of `a`for which `sin^(-1)x=|x-a|`will have at least one solution. |
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Answer» `y=sinx` `y=|x-a|` 1)a>0 `y=a-x` `pi/2=a-1` `a=1+pi/2` 2)a<0 `y=x-a` `pi/2=1-a` `a=1-pi/2` `1-pi/2<=a<=1+pi/2`. |
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| 454. |
If `|cos^(-1) ((1 -x^(2))/(1 + x^(2)))| lt (pi)/(3)`, thenA. `x in [-(1)/(3), (1)/(sqrt3)]`B. `x in (-(1)/(sqrt3), (1)/(sqrt3))`C. `x in (0, (1)/(sqrt3))`D. none of these |
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Answer» Correct Answer - B We have `|cos^(-1) ((1 -x^(2))/(1 + x^(2)))| lt (pi)/(3)` `rArr -(pi)/(3) lt cos^(-1) ((1 -x^(2))/(1 + x^(2))) lt (pi)/(3)` `rArr 0 le cos^(-1).(1 -x^(2))/(1 + x^(2)) lt (pi)/(3)` `rArr (1)/(2) lt (1 - x^(2))/(1 + x^(2)) le 1` `rArr 1 + x^(2) lt 2 (1 -x^(2)) le 2 (1 + x^(2))` `rArr 0 le x^(2) lt (1)/(3)` `rArr -(1)/(sqrt3) lt x lt (1)/(sqrt3)` |
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| 455. |
The value of `2cos^(-1)(-1/2)-2 sin^(-1)(-1/2)-cos^(-1)(-1)` isA. `2pi/3`B. `pi/2`C. `pi`D. `2pi` |
| Answer» Correct Answer - A | |
| 456. |
The value of `cos (2Cos^-1 0.8)` isA. 0.48B. 0.96C. 0.6D. none of these |
| Answer» Correct Answer - D | |
| 457. |
If `x^2+y^2+z^2=r^2,t h e ntan^(-1)((x y)/(z r))+tan^(-1)((y z)/(x r))+tan^(-1)((x z)/(y r))`is equal to`pi`(b) `pi/2`(c) 0(d) none of theseA. `pi`B. `(pi)/(2)`C. 0D. none of these |
| Answer» Correct Answer - B | |
| 458. |
A solution of the equation `tan^(-1)(1+x)+tan^(-1)(1-x)=(pi)/(2)` isA. x=1B. x=-1C. x=0D. `x=pi` |
| Answer» Correct Answer - C | |
| 459. |
The number of the solutions of the equation `2 sin^(-1) sqrt(x^(2) + x + 1) + cos^(-1) sqrt(x^(2) + x) = (3pi)/(2)` is |
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Answer» Correct Answer - B `0 le x^(2) + x + 1 le 1 and 0 le x^(2) + x le 1` `:. X = -1, 0` For `x = -1` `L.H.S. = 2 sin^(-1) 1 + cos^(-1) 0 = (3pi)/(2)` For `x = 0, L.H.S. = 2 sin^(-1) 1 + cos^(-1) 0 = (3pi)/(2)` Therefore, `x = 0` is a solution and sum of the solutions `= -1` |
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| 460. |
Number of integral solutions of the equation `2 sin^-1sqrt(x^2-x +1 )+ cos^-1sqrt(x^2-x) =3pi/ 2` is |
| Answer» Correct Answer - B | |
| 461. |
Sometimes we are just concerned with finding integral solutions to equations. Consider the equation `tan^(-1).(1)/m+tan^(-1).(1)/n=tan^(-1).(1)/lambda`, where `m,n, lambda in N` For `lambda=11`, an integral pair (m,n) satisfying the equation isA. `(12,72)`B. `(12, 133)`C. `(13,74)`D. `(13,136)` |
| Answer» Correct Answer - B | |
| 462. |
Sometimes we are just concerned with finding integral solutions to equations. Consider the equation `tan^(-1).(1)/m+tan^(-1).(1)/n=tan^(-1).(1)/lambda`, where `m,n, lambda in N` How many positive integral solutions (m,n) exist for the eqution if `lambda=3`?A. TwoB. FourC. SixD. Eight |
| Answer» Correct Answer - B | |
| 463. |
If the equation `sin^(-1)(x^2+x +1)+cos^(-1)(lambda x+1)=pi/2` has exactly two solutions, then the value of `lambda` isA. -1B. 0C. 1D. 2 |
| Answer» Correct Answer - A::B | |
| 464. |
If the equation `sin^(-1)(x^2+x+1)+cos^(-1)(lambda+1)=pi/2`has exactly two solutions for `lambda in [a , b]`, then the value of `a+b`is |
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Answer» Correct Answer - 1 `sin^(-1) (x^(2) + x -1) + cos^(-1) (lamda x + 1) = (pi)/(2)` `rArr x^(2) + x + 1 = lamda x + 1` `rArr x^(2) + (1- lamda) x = 0` `rArr x = 0 " or " x = lamda -1` Also, `sin^(-1) (x^(2) + x + 1)` will exist if `-1 le x^(2) + x + 1 le 1` `rArr x^(2) + x le 0 rArr -1 le x lt 0` For exactly two solution, we have `-1 le lamda -1 lt 0` `rArr 0 le lamda lt 1` `:. a = 0 and b =1` |
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| 465. |
The value of`(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto``(alpha+beta)(alpha^2+beta^2)`(b) `(alpha+beta)(alpha^2-beta^2)``(alpha+beta)(alpha^2+beta^2)`(d) none of theseA. `(alpha-beta)(alpha^(2)-beta^(2))`B. `(alpha+beta)(alpha^(2)-beta^(2))`C. `(alpha+beta)(alpha^(2)+beta^(2))`D. none of these |
| Answer» Correct Answer - C | |
| 466. |
Equation `1+x^2+2x"sin"(cos^(-1)y)=0`is satisfied byexactly one value of `x`exactly two value of `x`exactly one value of `y`exactly two value of `y`A. exactly one value of xB. exactly two values of xC. exactly one value of yD. exactly two values of y |
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Answer» Correct Answer - A::C Given equation is `x^(2) + 2x sin(cos^(-1) y) + 1 =0` Since x is real, `D ge 0`. Therefore, `4(sin(cos^(-1)y))^(2) -4 ge 0` or `(sin(cos^(-1)y))^(2) ge 1` or `sin(cos^(-1)y) = +- 1` or `cos^(-1) y = (pi)/(2) rArr y = 0` Putting value of y in the original equation, we have `x^(2) + 2x + 1 = 0 rArr x -1` Hence, the equation has only one solution |
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| 467. |
For `x, y, z, t in R, sin^(-1) x + cos^(-1) y + sec^(-1) z ge t^(2) - sqrt(2pi t) + 3pi` The value of `x + y + z` is equal toA. 1B. 0C. 2D. `-1` |
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Answer» Correct Answer - D `sin^(-1) x in [-(pi)/(2), (pi)/(2)]` `cos^(-1) y in [0, pi]` `sec^(-1) z in [0, (pi)/(2)) uu ((pi)/(2), pi]` `rArr sin^(-1) x + cos^(-1) y + sec^(-1) z le (pi)/(2) + pi + pi = (5pi)/(2)` Also `t^(2) - sqrt(2pi) t + 3pi = t^(2) -2 sqrt((pi)/(2)) t + (pi)/(2) - (pi)/(2) + 3pi` `= (t -sqrt((pi)/(2)))^(2) + (5pi)/(2) ge (5pi)/(2)` The given inequation exists if equality holds, i.e., L.H.S. = R.H.S. `= (5pi)/(2)` `rArr x = 1, y = -1, z = -1 and t = sqrt((pi)/(2))` `rArr cos^(-1) (cos 5t^(2)) = cos^(-1) (cos 5 t^(2)) = cos^(-1) (cos ((5pi)/(2))) = (pi)/(2)` `cos^(-1) ("min") {x, y, z}) = cos^(-1) (-1) = pi` |
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| 468. |
`tan{(pi)/(4)+1/2cos^(-1)alpha}+tan{(pi)/(4)-1/2cos^(-1)alpha}` `alpha ne 0` is equal toA. `alpha`B. `2alpha`C. `(2)/(alpha)`D. none of these |
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Answer» We have `tan{(pi)/(4)+1/2cos^(-1)alpha}+tan{(pi)/(4)-1/2cos^(-1)alpha}` `tan(pi)/(4)+(theta)/(2)+tan((pi)/(4)-(theta)/(2))` where `theta = cos^(-1)alpha` `=(1+tan(theta)/(2))/(1-tan(theta)/(2))+(1-tan(theta)/(2))/(1+tan(theta)(theta)/(2))=2(1+tan^(2)(theta)/(2))/(1-tan^(2)(theta)/(2))=(2)/(cos theta)=(2)/(alpha)` |
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| 469. |
The value of `tan[1/2cos^(-1).sqrt5/3]` isA. `(3-sqrt5)/2`B. `(3+sqrt5)/2`C. `(sqrt5-3)/2`D. `(-sqrt5-3)/2` |
| Answer» Correct Answer - A | |
| 470. |
The value of `tan[1/2cos^(-1).sqrt5/3]` isA. `(3pmsqrt5)/2`B. `(3+sqrt5)/2`C. `(3-sqrt5)/2`D. `1/2` |
| Answer» Correct Answer - C | |
| 471. |
`1/2cos^(-1)(3/5)` equalsA. `pi/2-cos^(-1).(4)/5`B. `tan^(-1).(1)/2`C. `pi/4-1/2cos^(-1).(4)/5`D. `pi/2-tan^(-1).(1)/2` |
| Answer» Correct Answer - B::C | |
| 472. |
To the equation `2^2pi//cos^((-1)x)-(a+1/2)2^pi//cos^((-1)x)-a^2=0`has only one real root, then`1lt=alt=3`(b) `ageq1``alt=-3`(d) `ageq3`A. `1 le a le 3`B. `a ge 1`C. `a le -3`D. `a ge 3` |
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Answer» Correct Answer - B::C `1 le (pi)/(cos^(-1) x) lt oo` `rArr 2 le 2^((pi)/(cos^(-1)x)) lt oo` Hence, 2 should lie between or on the roots of `t^(2) - (a + (1)/(2)) t - a^(2) = 0, " where " t = 2^(pi//cos^(-1)x)` `rArr f(2) le 0 rArr a^(2) + 2a -3 ge 0` `rArr a in (-oo, -3] uu [1, oo)` |
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| 473. |
The expression `tan.(pi/4+1/2cos^(-1)x)+tan(pi/4-1/2cos^(-1)x)` equalsA. `1/x`B. xC. `2/x`D. 2x |
| Answer» Correct Answer - C | |
| 474. |
`tan(pi/4+1/2cos^-1x)+tan(pi/4-1/2cos^-1x)`, `x!=0` is equal toA. `x`B. `2x`C. `(2)/(x)`D. none of these |
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Answer» Correct Answer - C `tan ((pi)/(4) + (1)/(2) cos^(-1) x) + tan ((pi)/(4) - (1)/(2) cos^(-1) x)` `= (1 + tan ((1)/(2) cos^(-1) x))/(1 - tan((1)/(2) cos^(-1) x)) + (1- tan ((1)/(2) cos^(-1) x))/(1 + tan ((1)/(2) cos^(-1) x))` `= ((1 + tan ((1)/(2) cos^(-1) x))^(2)+ (1- tan ((1)/(2) cos^(-1) x))^(2))/(1 - tan^(2) ((1)/(2) cos^(-1) x))` `= 2(1 + tan^(2) ((1)/(2) cos^(-1) x))/(1 - tan^(2) ((1)/(2) cos^(-1) x))` `= (2)/(cos(cos^(-1) x)) = (2)/(x)` |
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| 475. |
Write `cot^(-1)(1/(sqrt(x^2-1))),|x|>1`in the simplest form. |
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Answer» let` x=sectheta` then `theta=sec^(-1)x` =`cot^(-1)(1/sqrt(sec^2theta-1))` =`cot^(-1)(1/tantheta)` =`cot^(-1)(cottheta)` =`theta` from (1) =`sec^(-1)x` |
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| 476. |
Find the value of `tan^(-1)(tan(3pi)/4)` |
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Answer» `tan^-1(tan((3pi)/4))` can be written as `=tan^-1(tan(pi-pi/4))` `=tan^-1(tan(-pi/4)` `=-pi/4`, which is the required value. |
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| 477. |
Prove that`tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))=pi/4-1/2cos^(-1)x,-1/(sqrt(2))lt=xlt=1` |
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Answer» `L.H.S. = tan^-1[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]` `=tan^-1[(sqrt(1+x)(1-sqrt(1-x)/sqrt(1+x)))/(sqrt(1+x)(1+sqrt(1-x)/sqrt(1+x)))]` `=tan^-1[(1-sqrt(1-x)/sqrt(1+x))/(1+sqrt(1-x)/sqrt(1+x))]` As, `tan^-1x-tan^-1y = (x-y)/(1+xy)` So, our expression becomes, `=tan^-1(1)+tan^-1(sqrt(1-x)/sqrt(1+x))` `=pi/4+1/2(2tan^-1(sqrt(1-x)/sqrt(1+x)))` Also, `2tan^-1y = cos^-1((1-y^2)/(1+y^2))` So, our expression becomes, `= pi/4+1/2cos^-1((1-(sqrt(1-x)/sqrt(1+x))^2)/(1+(sqrt(1-x)/sqrt(1+x))^2))` `=pi/4+1/2cos^-1((1+x-1+x)/(1+x+1-x))` `=pi/4+1/2cos^-1((2x)/2)` `=pi/4+1/2cos^-1(x)=R.H.S.` |
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| 478. |
If two angles of a triangle are `tan^(-1) (2) and tan^(-1) (3)`, then find the third angles |
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Answer» Given two angles are `tan^(-1) (2) and tan^(-1) (3)`. Now `(2) (3) gt 1` `rArr tan^(-1) (2) and tan^(-1) (3) = pi + tan^(-1) ((2+3)/(1 - 2 xx 3))` `= pi + tan^(-1) = pi - (pi)/(4) = (3pi)/(4)` Hence, the third angle is `pi - (3pi)/(4) = (pi)/(4)` |
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| 479. |
Find the range of `cot^(-1)(2x-x^2)` |
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Answer» Let `theeta = cot^(-1) (2 x - x^(2))`, where `theta in (0, pi)` `rArr cot theta = 2x - x^(2), " where" theta in (0, pi)` `= 1- (1 - 2x + x^(2)), " where " theta in (0, pi)` `= 1 - (1 - x)^(2), " where " theta in (0, pi)` `rArr cot theta le 1, " where " theta in (0, pi)` `rArr (pi)/(4) le theta le pi` `rArr` Range of `f(x) "is " [(pi)/(4), pi)` |
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| 480. |
If `cos^(-1) alpha + cos^(-1) beta + cos^(-1) gamma = 3pi`, then `alpha (beta + gamma) + beta(gamma + alpha) + gamma(alpha + beta)` equal to |
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Answer» Correct Answer - C We have, `cos^(-1) alpha + cos^(-1) beta + cos^(-1) gamma = 3 pi` We know that `0 le cos^(-1) x le pi` `rArr cos^(-1) alpha + cos^(-1) beta + cos^(-1) gamma = 3 pi` If and only if, `cos^(-1) alpha + cos^(-1) beta + cos^(-1) gamma = pi` `rArr cos pi = alpha = beta = gamma` `rArr -1 = alpha = beta = gamma` `rArr alpha = beta = gamma = - 1` `:. alpha(beta + gamma) + beta(gamma + alpha) + gamma (alpha + beta)` `= - 1 (-1-1)-1(-1-1)-1(-1-1)` `= 2+2+2 = 6` |
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| 481. |
Simplify`[(3sin2alpha)/(5+3cos2alpha)]+tan^(-1)[(tanalpha)/4],w h e r e-pi/2 |
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Answer» `tan^(-1)[(3sin2alpha)/(5+3cos2alpha)]+tan^(-1)[tanalpha/4]` `tan^(-1)[(6tanalpha)/(8+2tan^2alpha)]+tan^(-1)(tanalpha/4)` `(6tan^2alpha)/(4(8+2tan^2alpha))<1` `tan^(-1)[((3tanalpha)/(4ttan^2alpha)+tanalpha/4)/(1-(3tanalpha)/(4+tan^2alpha)*tanalpha/4)]` `tan^(-1)((12tanalpha+4tanalpha+tan^3alpha)/(16+tan^2alpha))` `tan^(-1)((16tanalpha+tan^3alpha)/(16+tan^2alpha))` `tan^(-1)((tanalpha(16+tan^2alpha))/(16tan^2alpha))` `tan^(-1)tanalpha=alpha`. |
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| 482. |
If `u=cot^(-1)sqrt(cos theta) -tan^(-1)sqrt(cos theta)` then sin u=A. `tan theta //2`B. `tan^(2) theta//2`C. `cot theta//2`D. `cot^(2) theta//2` |
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Answer» We have `u= cot^(-1)sqrt(cos theta)-tan^(-1)sqrt(cos theta)` `rarr u = (pi)/(2) -2 alpha where alpha =tan^(-1)(sqrt(cos theta))` `rarr sin u =cos 2 alpha` `rarr sin u =(1-tan^(2)alpha)/(1+tan^(2) alpha)=(1-cos theta)/(1+cos theta)=tan^(2)(theta)/(2)` |
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| 483. |
If `u=cot^(-1)sqrt(cos theta) -tan^(-1)sqrt(cos theta)` then sin u=A. `sin^(2) theta`B. `cos^(2) theta`C. `tan^(2) theta `D. `tan^(2) 2theta` |
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Answer» We have `u = cot^(-1) sqrt(cos2 theta)-tan^(-1) sqrt(cos2 theta)` `rarr u = (pi)/(2) - tan^(-1)sqrt(cos 2 theta)-tan^(-1)sqrt(cos 2 theta)` `rarr u = (pi)/(2) -tan^(-1)sqrt(cos 2 theta)/(sin^(2)theta)` ltrbgt `rarr tan(pi)/(2)-u=sqrt(cos 2 theta)/(sin^(2)theta)` `rarr cot u =sqrt(cos^(2) theta)/(sin^(2)theta)` `rarr sin u = (sin^(2) theta)/sqrt(sin^(4)theta + 1 -2 sin^(2) theta)` `rarr sin u =(sin^(2)theta)/(|1-sin^(2)theta|)=tan^(2)theta` |
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| 484. |
For `x in(0,1)`, let `alpha=sin^(-1)x,beta=x,gamma=tan^(-1)x, delta=cot^(-1)x-(pi)/(2)`. Which of the following is true ?A. `alpha gt beta gt gamma`B. `beta gt alpha gt gamma gt delta`C. `alpha gt beta gt gamma gt delta`D. `beta gt alpha gt delta gt gamma` |
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Answer» Correct Answer - C For `x in (0,1)` `sin^(-1)x gt x gt tan^(-1)x gt cot^(-1)x-(pi)/(2)` as `cot^(-1)x-(pi)/(2)=-(tan^(-1)x)` |
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| 485. |
Find the value of `tan^(-1)(-x)-cot^(-1)(-1/x),x gt0` |
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Answer» Since, `tan^(-1)(-x)=-tan^(-1)x` `and cot^(-1)(-1/x)=pi-cot^(-1)1/x` Also, `cot^(-1)(-1/x)=tan^(-1)x,x gt 0` `:.tan^(-1)(-x)-cot^(-1)(-1/x)` `-tan^(-1)x-(pi-cot^(-1)1/x)` `=-tan^(-1)x-pi+tan^(-1)x` `=-pi` |
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| 486. |
If `x in [1, 0)`, then find the value of `cos^(-1) (2x^(2) - 1) - 2 sin^(-1) x` |
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Answer» Let `x = cso theta, " for " x in [-1, 0), theta in (pi//2, pi]` `rArr cos^(-1) (2x^(2) -1) - 2 sin^(-1) x` `= cos^(-1) (2 cos^(2) theta -1) -2 sin^(-1) (cos theta)` `= cos^(-1) (cos 2 theta) - 2 sin^(-1) (sin(pi//2 - theta))` `= cos^(-1) (cos 2 theta) - 2 ((pi)/(2) - theta)` `= 2 pi - 2 theta + 2 theta - pi` `= pi` |
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| 487. |
If `alpha in (-(pi)/(2), 0)`, then find the value of `tan^(-1) (cot alpha) - cot^(-1) (tan alpha)` |
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Answer» Correct Answer - `-pi` `tan^(-1) (cot alpha) - cot^(-1) (tan alpha)` `= tan^(-1) ((1)/(tan alpha)) - ((pi)/(2) - tan^(-1) (tan alpha))` `= -(pi)/(2) + (tan^(-1) ((1)/(tan alpha)) + tan^(-1) (tan alpha))` `= -(pi)/(2) - (pi)/(2) " " ("as " (-pi)/(2) lt alpha lt 0)` |
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| 488. |
If `cos^(-1)((1-x^(2))/(1+x^(2)))+sin^(-1)((2x)/(1+x^(2)))=p` for all `x in[-1,0]`, then p is equal toA. `(-pi)/(2)`B. 0C. `(pi)/(2)`D. `(2pi)/(3)` |
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Answer» Correct Answer - B For `x in [-1,0],cos^(-1)((1-x^(2))/(1+x^(2)))=-2 tan^(-1)x` and `sin^(-1)((2x)/(1+x^(2)))=2 tan^(-1)x` `therefore p=0 AA x in [-1,0]` |
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| 489. |
Let (x,y) be such that `sin^(-1)(ax)+cos^(-1)y+cos^(-1)(bxy)=pi/2` |
| Answer» Correct Answer - A(p), B( q) , C(p) , D(s) | |
| 490. |
STATEMENT -1 : The value of `tan^(-1)x+tan^(-1)(1/x)=pi/2, AA x in R -{0}`. and STATEMENT -2 : The value of `tan^(-1).(1/x)={:{(cot^(-1)x,x gt0),(-pi+cot^(-1)x,x lt0):}`A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -4B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -4C. Statement-1 is True, Statement -2 is FalseD. Statement -1 is False, Statement -2 is True |
| Answer» Correct Answer - D | |
| 491. |
If `(1)/(sqrt2) lt x lt 1`, then prove that `cos^(-1) x + cos^(-1) ((x + sqrt(1 - x^(2)))/(sqrt2)) = (pi)/(4)` |
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Answer» `cos^(-1) x + cos^(-1) ((x + sqrt(1 - x^(2)))/(sqrt2))` Let `cos^(-1) x = theta " or " x = cos theta` For `(1)/(sqrt2) lt x lt 1, 0 lt theta lt (pi)/(4)` `rArr cos^(-1) x + cos^(-1) (x + sqrt(1 - x^(2))/(sqrt2))` `= theta + cos^(-1) ((cos theta + sqrt(1 - cos^(2) theta))/(sqrt2))` `= theta + cos^(-1) ((cos theta + sin theta)/(sqrt2))` `= theta + cos^(-1) (cos (theta - (pi)/(4)))` Now, `(theta - (pi)/(4)) in (-(pi)/(4), 0)`, which is not the principal values of `cos^(-1)` function But `((pi)/(4) - theta) in (0, (pi)/(4))` `rArr theta + cos^(-1) (cos (theta - (pi)/(4))) = theta + cos^(-1) (cos ((pi)/(4) - theta))` `= theta + ((pi)/(4) - theta)` `= (pi)/(4)` |
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| 492. |
The solution of `sin^(-1)x-sin^(-1)2x=pm(pi)/(3)` isA. `pm(1)/(3)`B. `pm(1)/(4)`C. `pm(sqrt(3))/(2)`D. `pm(1)/(2)` |
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Answer» Correct Answer - D `sin^(-1)x-sin^(-1)2x=pm (pi)/(3)` `rArr sin^(-1)x-sin^(-1)(pm(sqrt(3))/(2))=sin^(-1)2x` `rArr sin^(-1)[x sqrt(1-(3)/(4))-(pm (sqrt(3))/(2)sqrt(1-x^(2)))]=sin^(-1)2x` `rArr (x)/(2)-(pm(sqrt(3))/(2)sqrt(1-x^(2)))=2x` `rArr -(pm sqrt(3)sqrt(1-x^(2)))=3x` On squaring both sides, we get `3(1-x^(2))=9x^(2)` `rArr 4x^(2)=1` `rArr x = pm (1)/(2)` |
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| 493. |
If `alpha` is the only real root of the equation `x^3 + bx^2 + cx + 1 = 0 (b < c)`, then the value of `tan^-1 alpha+ tan^-1 (alpha^-1)` is equal to : |
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Answer» Correct Answer - `-(pi)/(2)` Let `f(x) = x^(3) + bx^(2) + cx + 1` So, `f(0) = 1 gt 0` `f(-1) = b - c lt 0` So, `-1 lt alpha lt 0`, `:. Tan^(-1) (alpha) + tan^(-1) ((1)/(alpha)) = -(pi)/(2)` |
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| 494. |
If `alpha` is the only real root of the equation `x^3 + bx^2 + cx + 1 = 0 (b < c)`, then the value of `tan^-1 alpha+ tan^-1 (alpha^-1)` is equal to :A. `(pi)/(2)`B. `-(pi)/(2)`C. 0D. non existent |
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Answer» Let `f(x)=x^(3)+bx^(2)+cx+1` Then `f(0)=1 gt 0 and f(-1)=b-c lt 0` `rarr alpha lt 0` `rarr tan^(-1)((1)/(alpha))=-pi + cot^(-1)alpha` `rarr tan^(-1) alpha+tan^(-1)(1)/(alpha)=-pi + tan^(-1)alpha+cos^(-1)alpha` `rarr tan^(-1)alpha+tan^(-1)(1)/(alpha)=-pi++(pi)/(2)=-(pi)/(2)` |
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| 495. |
If `sin^(- 1)(x-(x^2)/2+(x^3)/4-.....)+cos^(- 1)(x^2-(x^4)/2+(x^6)/4-.....)=pi/2` for `0 lt |x| lt sqrt2` then `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» using `sin^(-1) x +cso^(-1) x=(pi)/(2)` in statement 1 we get `x-(x^(2))/(2)+(x^(2))/(4)…=x^(2)-(x^(4))/(2)+x^(6))/(4)` `rarr (x)/(1+(x)/(2))=(x^(2))/(1+x^(2))/(2)` `rarr x(2+x^(2))=x^(2)(2+x)` so statement -01 is true LHS of statement 2 si meaningful if `x^(2)+x ge 0 x^(32)+x+1ge0 and 0 le sqrt(x^(2)+x+1)le1` `rarr x^(2)+xge0 and x^(2)+xle0 rarr x^(2)+x=0 rarr x=0` `rarr x=-1` clearly x=-1 satisfies the statement -1 |
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| 496. |
if `-1/2 le x le 1/2 then cos^(-1)(4x^(3)-3x)` equalsA. `3cos^(-1)x`B. `2pi-3 cos^(-1)x`C. `-2 pi + 3 cos^(-1)x`D. `none of these |
| Answer» Correct Answer - B | |
| 497. |
If `1/2 le x le 1 then cos^(-1)(4x^(3)-3x)` equalsA. `3 cos^(-1)x`B. `2[I -3 cos^(-1)]x`C. `-2 pi + 3cos^(-1)x`D. none of these |
| Answer» Correct Answer - A | |
| 498. |
Let `x_i in [-1,1]" for "i=1,2,3,…24`, such that `sin^(-1)x_1+sin^(-1)x_2+…+sin^(-1)x_24=12pi` then the value of `x_1+2x_2+3x_3+…+24x_24` isA. 276B. 300C. 325D. 351 |
| Answer» Correct Answer - B | |
| 499. |
If `-1 le x le -1/2`, then `sin^(-1)(3x-4x^3)` equalsA. `3 sin^(-1)x`B. `pi-3 sin^(-1)x`C. `-pi-3 sin^(-1)x`D. none of these |
| Answer» Correct Answer - C | |
| 500. |
Statement -1: If `a^(2)+b^(2)=c^(2),c ne 0` then the non zero solution of the equation `sin^(-1)((ax)/(c ))+sin^(-1)((bx)/(c))=sin^(-1)x` is `pm`1,. Statement-2: `sin^(-1)x+sin^(-1)y= sin^(-1)(x+y)`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» Clearly statement 2 is not true we have `c^(2)=a^(2)+b^(2)` so let `a=c cos alpha and b =c sin alpha` then `sin^(-1)(ax)/(c )+sin^(-1)(bx)/(c )=sin^(-1)x` `rarr sin^(-1)(x cos alpha) s+sin^(-1) (x sin alpha)=sin^(-1)x` `rarr sin^(-1)(xcos alphasqrt(1=-x^(2))sin^(2)alpha)+x sin alpha sqrt(1-x^(2) cos^(2) alpha)` `rarr cos alpha sqrt(1-x^(2)) alpha+sin alpha sqrt(1-x^(2) cos^(2) alpha=1)` clearly `x=pm1` satisfies this equaiton |
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