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101.

If `x in [-1, 0)` then find the value of `cos^(-1) (2x^(2) -1) -2 sin^(-1) x`

Answer» Correct Answer - `pi`
`cos^(-1) (2x^(2) -1) = 2pi - 2 cos^(-1) x " " ("as " x lt 0)`
`rArr cos^(-1) (2x^(2) -1) -2 sin^(-1) x = 2pi - 2 cos^(-1) x - 2sin^(-1) x`
`= 2pi - 2 (cos^(-1) x + sin^(-1) x)`
`= 2pi - 2 (pi)/(2) = pi`
102.

Prove that `2sin^-1[3/5]-tan^-1[17/31]=pi/4`

Answer» Let `sin^-1[3/5] = theta`,
Then, `sin theta = 3/5``=>costheta = sqrt(1 - (3/5)^2) = 4/5`
`:. tan theta = sintheta/costheta = 3/4`
Now, `tan2theta = (2tantheta)/(1-tan^2theta) = (2**3/4)/(1-(3/4)^2) = (3/2)/(7/16) = 24/7`
`:. 2theta = tan^-1(24/7)`
Now, `L.H.S. = 2sin^-1(3/5) - tan^-1(17/31) = 2theta-tan^-1(17/31)`
`=tan^-1(24/7)-tan^-1(17/31)`
`=tan^-1((24/7-17/31)/(1+24/7**17/31))`
`=tan^-1(((744-119)/(31**7))/((217+408)/(31**7))))`
`=tan^-1(625/625) = tan^-1(1)`
`=tan^-1(tan(pi/4)) = pi/4 = R.H.S.`
103.

Prove that: `sin^-1(3/5)-cos^-1(12/13)=sin^-1(16/65)`

Answer» LHS
`sin^(-1)(3/5)-cos^(-1)(12/13)`
Let`cos^(-1)(12/13)=theta`
`costheta=12/13`
`sintheta=5/13`
`theta=sin^(-1)(5/13)`
`sin^(-1)(3/5)-theta`
`sin^(-1)(3/5)-sin^(-1)(5/13)`
`sin^(-1)x-sin^(-1)y`
`(xsqrt(1-y^2)-ysqrt(1-x^2))`
`sin^(-1)(3/5*sqrt(1-(5/13)^2)-5/13sqrt(1-(3/5)^2)`
`sin^(-1)(3/5*12/13-5/13*4/5)`
`sin^(-1)(36/65-20/65)`
`sin^(-1)(16/65)=RHS`.
104.

Which of the following is the solution set of the equation `2cos^(-1)x=cot^(-1)((2x^2-1)/(2xsqrt(1-x^2)))?`(0, 1) `(b)`(-1, 1) -{0}`(-1,0)`(d) `(-oo,-1)uu[1,oo]`A. `(0,1)`B. `(-1, 1) - {0}`C. `(-1, 0)`D. `[-1, 1]`

Answer» Correct Answer - A
`2 cos^(-1) x = cot^(-1) ((2x^(2) -1)/(2x sqrt(1 -x^(2))))`
Put `x = cos theta : L.H.S. = 2 theta, 0 le theta le pi`
and `-1 le x le 1`...(i)
`R.H.S. = cot^(-1) ((cos 2 theta)/(2 cos theta|sin theta|)) = cot^(-1) (cot 2 theta) = 2 theta`
if `0 lt 2 theta le pi`...(ii)
From Eqs. (i) and (ii) , we get `0 lt theta lt pi//2`. Thus,
`x in (0, 1)`
105.

Prove that `2 tan^(-1) sqrt(b/a) =cos^(-1) ((a-b)/(a+b))`

Answer» LHS=`2tan^(-1)sqrt(b/a)`
Let `tan^(-1)sqrt(b/a)=theta`
`sqrt(b/a)=tantheta`
`b/a=tan^2theta`
`2theta`
`cos2theta=(1-tan^2theta)/(1+tan^2theta)`
`2theta=cos^(-1)((1-tan^2theta)/(1+tan^2theta))`
`cos^(-1)((1-(b/a))/(1+(b/a)))`
`cos^(-11)((a-b)/(a+b))`=RHS
106.

The number of solution `(s)` of the equation `sin^(- 1)(1-x)-2sin^(- 1)x=pi/2` is/are

Answer» Correct Answer - C
`sin^(-1)(1-x)+2sin^(-1)x=(pi)/(2)`
`rArr sin^(-1)(1-x)=((pi)/(2)-2sin^(-1)x)`
`rArr 1-x =sin((pi)/(2)-2 sin^(-1)x)`
`rArr 1-x sin.(pi)/(2)cos(2 sin^(-1)x)-cos.(pi)/(2)sin(2sin^(-1)x)`
`rArr 1-x cos (2sin^(-1)x)^(2)`
`rArr 1-x=cos(cos^(-1)(1-2x^(2)))rArr 2x^(2)-x=0`
`rArr x =0` or `x = 1//2`
107.

if `x in (-oo,-1) then tan^(-1)(2x)/(1-x^(2))` equalsA. `2tan^(-1)x`B. `-pi+2 tan^(-1)x`C. `pi+2 tan^(-1)x`D. none of these

Answer» Correct Answer - C
108.

If `x in (0, 1)`, then find the value of `tan^(-1) ((1 -x^(2))/(2x)) + cos^(-1) ((1 -x^(2))/(1 + x^(2)))`

Answer» Correct Answer - `(pi)/(2)`
`tan^(-1) ((1 -x^(2))/(2x)) + cos^(-1) ((1 -x^(2))/(1 + x^(2)))`
`= (pi)/(2) - tan^(-1) ((2x)/(1 - x^(2))) + cos^(-1) ((1 -x^(2))/(1 + x^(2)))`
`= (pi)/(2) - 2tan^(-1) x + 2tan^(-1) x " " ("as " x in (0, 1))`
`= (pi)/(2)`
109.

`cos^(-1)x+sin^(-1)(x/2)= pi/6`

Answer» `(pi/2-sin^(-1)x)+sin^(-1)(x/2)=pi/6`
`pi/22-pi/6=sin^(-1)x-sin^(-1)(x/2)`
`sin^(-1)x-sin^(-1)(x/2)=pi/3=sin^(-1)(sqrt3/2)`
`sin^(-1)x=sin^(-1)(sqrt3/2)+sin^(-1)(x/2)`
`sin^(-1)(x)=sin^(-1)(sqrt3/2sqrt(1-x^2/4)+x/2sqrt(1-3/4))`
`sin^(-1)x=sin^(-1)(sqrt3/2sqrt(4-x^2)/2+x/2*1/2]`
`x=(sqrt3sqrt(4-x^2))/4+x/4`
`sqrt3/4x=sqrt3/4sqrt(4-x^2`
`3x^2=4-x^2`
`x^2=1`
`x=pm1`.
110.

Prove that `tan^(-1)(2/11)+tan^(-1)(7/24)=tan^(-1)(1/2)`

Answer» LHS
`tan^(-1)(2/11)+tan^(-1)(7/24)`
`tan^(-1)(((2/11)+(7/24))/(1-(2/11)(7/24)))`
`tan^(-1)((2*24+7*11)/(11*24-14))`
`tan^(-1)(125/250)`
`tan^(-1)(1/2)`
RHS.
111.

Prove that: `tan^(-1)(2/11)+tan^(-1)(7/24)=tan^(-1)(1/2)`

Answer» We know, `tan^-1x+tan-^-1y = tan^-1((x+y)/(1-xy))`
So,`L.H.S. = tan^-1(2/11)+tan^-1(7/24)`
`=tan^-1((2/11+7/24)/(1-2/11**7/24))`
`=tan^-1(125/250)`
`=tan^-1(1/2)=R.H.S.`
112.

Prove that:`2sin^(-1)(3/5)=tan^(-1)((24)/7)`

Answer» We have =`2 sin^(-1)(3/5)`
=`2 tan^(-1)(3/4)`
=`tan^(-1)(3/4)+tan^(-1)(3/4)`
=`tan^(-1)(((3/4)+(3/4))/(1-(3/4)(3/4)))`
=`tan^(-1)((24/16)/(7/16))`
=`tan^(-1)(24/7)`
hence proved
113.

Find the value of x which satisfy equation `2 tan^(-1) 2x = sin^(-1).(4x)/(1 + 4x^(2))`

Answer» Correct Answer - `-(1)/(2) le x le (1)/(2)`
`2 tan^(-1) 2x = sin^(-1).(4x)/(1 + 4x^(2)) = sin^(-1).(2(2x))/(1 + (2x)^(2))`
`rArr -(pi)/(2) le 2 tan^(-1) 2x le (pi)/(2)`
`rArr -(pi)/(4) le tan^(-1) 2x le (pi)/(4)`
`rArr -1 le 2x le 1`
`rArr -(1)/(2) le x le (1)/(2)`
114.

`cos^-1(4/5)+sin^-1(5/13) =tan^-1(56/33)`

Answer» `cos^-1 4/5 = theta`
`cos theta = 4/5`
`sin theta = 3/5`
`theta = sin^-1 ( 3/5)`
`sin^-1 (3/5) + sin^-1(5/13)`
`sin^-1 x + sin^-1 y = sin^-1 ( x sqrt(1-y^2) + ysqrt(1 - x^2))`
`sin^-1 ( 3/5sqrt(1 - (5/13)^2 ) + 5/3 sqrt(1 - (3/5)^2 ) )`
`sin^-1(3/5 sqrt(144/169) + 5/13sqrt(16/25))`
`sin^-1(3/5 xx 12/13 + 5/13 xx 4/5`
`sin^-1((36+20)/65)= sin^-1(56/65) = phi`
`sin phi = 56/65`
as `ab^2 = ac^2 - bc^2`
`= 65^2 - 56^2 `
`= (121 xx 9)= 1089= 33^2`
so`ab = 33`
`tan phi = 56/33`
`phi = tan^-1 (56/33)`
Answer
115.

If `sin^-1 x+sin^-1=(2pi)/3,` then `cos^-1 x cos^-1 y` is equal to

Answer» Correct Answer - `(pi)/(3)`
`sin^(-1) x + sin^(-1) y = (2pi)/(3)`
`rArr (pi)/(2) - cos^(-1) x + (pi)/(2) - cos^(-1) y = (2pi)/(3)`
`rArr cos^(-1) x + cos^(-1) y = pi - (2pi)/(3) = (pi)/(3)`
116.

If `|cos^(-1)((1-x^2)/(1+x^2))|

Answer» `|cos^-1((1-x^2)/(1+x^2))| lt pi/3`
`=> - pi/3 lt cos^-1((1-x^2)/(1+x^2)) lt pi/3`
we know, range of `cos^-1y` is from `0` to `pi`.
`:. 0 le cos^-1((1-x^2)/(1+x^2)) lt pi/3`
`=> 1/2 lt (1-x^2)/(1+x^2) le 1`
`=> 1+x^2 lt 2-2x^2 le 2+2x^2`
`=> 0 le x^2 lt 1/3`
`=> :. x in (-1/sqrt3,1/sqrt3).`
So, option `c` is the correct option.
117.

The number of solution of the equation `tan^(-1) (1 + x) + tan^(-1) (1 -x) = (pi)/(2)` isA. 2B. 3C. 1D. 0

Answer» Correct Answer - C
`tan^(-1) (1 + x) + tan^(-1) (1- x) = (pi)/(2)`
or `tan^(-1) (1 + x) = (pi)/(2) - tan^(-1) (1 -x)`
`= cot^(-1) (1 -x)`
`= tan^(-1) ((1)/(1 -x))`
or `1 + x = (1)/(1 -x)`
or `1 - x^(2) = 1`
or `x = 0`
118.

If `tan(x+y)=33`, and `x= tan^(-1)3`, then: y=A. 0.3B. `tan^(-1)(1.3)`C. `tan^(-1)(0.3)`D. ` tan^(-1)(1/18)`

Answer» Correct Answer - C
119.

If `tan^-1(1/y)=-pi+cot^-1 y,` where `y=x^2-3x+2,` then find the value of `x`

Answer» Correct Answer - `x in (1, 2)`
`tan^(-1) ((1)/(y)) = -pi + cot^(-1) y`
`rArr y lt 0`
`rArr x^(2) - 3x + 2 lt 0`
`rArr (x - 1) (x -2) lt 0`
`rArr x in (1, 2)`
120.

Find the value of `sin(1/2cot^(-1)(-3/4))`

Answer» `cot^(-1)(-3/4)=theta`
`costheta=-3/5`
`sinn(1/2theta)`
`sin(theta/2)`
`costheta=-3/5`
`1-2sin^2(theta/2)=-3/5`
`1+3/5=2sin^2(theta/2)`
`2si^2(theta/2)=8/5`
`sin^2(theta/2)=4/5`
`sintheta=2/sqrt5,-2/sqrt5``-2/sqrt5` is not possible.
121.

If `0 le x le 1 then cos^(-1)(2x^(2)-1)` equalsA. `2cos^(-1)x`B. `pi-2 cos^(-1)x`C. `2pi -2 cos^(-1)x`D. none of these

Answer» Correct Answer - A
122.

The value of `tan{(cos^(- 1)(-2/7)-pi/2)]` is

Answer» Correct Answer - `(2)/(3sqrt5)`
`tan {(cos^(-1)(-(2)/(7)) - (pi)/(2))}`
`= tan {pi - cos^(-1) ((2)/(7)) - (pi)/(2)}`
`= tan {(pi)/(2) - cos^(-1) ((2)/(7))}`
`= tan {sin^(-1) ((2)/(7))}`
`= tan tan^(-1) ((2)/(3sqrt5)) = (2)/(3sqrt5)`
123.

Find the number of solution of `2tan^(-1)(tanx)=6-xdot`

Answer» `tan^(-1)(tanx)=(6-x)/2`
`y=tan^(-1)(tanx)`
`y=(6-x)/2`
`2y=6-x-(1)`
`x+2y-6=0-(2)`
There are 3 number of solutions of equation 1 and 2.
124.

Arithmetic mean of the non-zero solutions of the equation `tan^-1 (1/(2x + 1)) + tan^-1 (1/(4x + 1)) = tan^-1 (2/x^2)`A. 2B. 3C. 4D. none of these

Answer» Correct Answer - B
`tan^(-1).(1)/(1 + 2x) + tan^(-1).(1)/(1 + 4x) = tan^(-1).(2)/(x^(2))`
or `tan^(-1) [((1)/(1 + 2x) + (1)/(1+ 4x))/(1 - (1)/(1 + 2x) (1)/(1 + 4x))] = tan^(-1).(2)/(x^(2))`
or `(2 + 6x)/(6x + 8x^(2)) = (2)/(x^(2))`
or `6x^(3) - 14x^(2) - 12x = 0`
or `x(x -3) (3x + 2) = 0`
or `x = 3 " or " x = -2//3`(as `x != 0`)
But for `x = -2//3`, L.H.S. `lt 0 and R.H.S. gt 0`
Hence, the only solution is `x = 3`
125.

The value of tan `(2 "tan"^(-1)(1)/(5)-(pi)/(4))` isA. 1B. 0C. `7/17`D. `-7/17`

Answer» Correct Answer - D
126.

If `-1 le x le 0 then cos^(-1)(2x^(2)-1)` equalsA. `2 cos^(-1)x`B. `pi-2 cos^(-1)x`C. `2pi-2cos^(-1)x`D. `-2 so^(-1)x`

Answer» Correct Answer - C
127.

Find the value of `tan^(-1) (-tan.(13pi)/(8)) + cot^(-1) (-cot((9pi)/(8)))`

Answer» Correct Answer - `pi`
`tan^(-1) (-tan.(13pi)/(8)) + cot^(-1) (-cot ((19pi)/(8)))`
`= - tan^(-1) (tan.(13pi)/(8)) + pi - cot^(-1) (cot ((19pi)/(8)))`
`= -((13pi)/(8) - 2pi) + pi - ((19 pi)/(8) - 2pi)`
`= (3pi)/(8) + (5pi)/(8) = pi`
128.

The equation `3^(-1)x-pix-pi/2=0`hasone negative solutionone positive solutionno solutionmore than one solution

Answer» `3cos^-1x-pix -pi/2 = 0`
`=>3cos^-1x = pix - pi/2`
`=>cos^-1x = (pix)/3 - pi/6`
Now, if we draw graph for `cos^-1x` and `(pix)/3 - pi/6`, we can see there is only one point of intersection.
Please refer to video to see the graph.
Hence, the equation has one positive solution.
129.

Prove that:`(9pi)/8-9/4sin^(-1)(1/3)=9/4sin^(-1)((2sqrt(2))/3)`

Answer» `L.H.S. = (9pi)/8 -9/4sin^-1(1/3)``=9/4(pi/2-sin^-1(1/3))`
As `pi/2-sin^-1x = cos^-1x`.So,our expression now is,
`=9/4cos^-1(1/3)`
If, we create a triangle with base 1 and hypotenuse 3, perpendicular comes `2sqrt2`. So,
`cos^-1(1/3) = sin^-1((2sqrt2)/3)`
Now, our expression becomes,
`9/4sin^-1((2sqrt2)/3) = R.H.S.`
130.

Statement -1: If a is twice the tangent of the arithmetic mean of `sin^(-1)x and cos^(-1)` x , b the geometric mean of tanx and cot x then `x^(2)-ax+b=0rarr x=1` statement-2: `tan((sin^(-1)x+cos^(-1)x)/(2))=1`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.

Answer» It is given that
`a=2tan(sin^(-1)x+cos^(-1)x)/(2)and b=sqrt(tanx xx cot x)` ltrbgt `rarr a=2 tan(pi)/(4)=2 and b=1`
`therefore x^(2)-ax+b=0 rarr x^(2)-2x+1=0 rarr (x-1)^(2)=0 rarr x=-1 `
so statement 1 is true
`tan(sin^(-1)x+cos^(-1))/(2)=tan (pi)/(4)=1`
so statement -2 is also true
131.

If `a^(2) + b^(2) = c^(2), c != 0`, then find the non-zero solution of the equation: `sin^(-1).(ax)/(c) + sin^(-1).(bx)/(c) = sin^(-1) x`

Answer» Correct Answer - `x = +- 1`
`sin^(-1).(ax)/(c) + sin^(-1).(bx)/(c) = sin^(-1) x`
`rArr sin^(-1) ((ax)/(c) sqrt(1 - (b^(2) x^(2))/(c^(2))) + (bx)/(c) sqrt(1 - (a^(2) x^(2))/(c^(2)))) = sin^(-1) x`
`rArr (ax)/(c) sqrt(1- (b^(2) x^(2))/(c^(2))) + (bx)/(c) sqrt(1 - (a^(2) x^(2))/(c^(2))) = x`
`rArr (a)/(c) sqrt(1 - (b^(2) x^(2))/(c^(2))) + (b)/(c) sqrt(1 - (a^(2) x^(2))/(c^(2))) =1`
`rArr (a^(2))/(c^(2)) (1 - (b^(2) x^(2))/(c^(2))) + (b^(2))/(c^(2)) (1- (a^(2) x^(2))/(c^(2))) + (2ab)/(c^(2)) sqrt(1 - (a^(2) x^(2))/(c^(2))) sqrt(1- (b^(2) x^(2))/(c^(2))) = 1`
`rArr (a^(2) + b^(2))/(c^(2)) - (2a^(2) b^(2) x^(2))/(c^(4)) + (2ab)/(c^(2)) sqrt(1 - (a^(2) x^(2))/(c^(2))) sqrt(1 - (b^(2) x^(2))/(c^(2))) = 1`
`rArr sqrt(1- (a^(2) x^(2))/(c^(2))) sqrt(1 - (b^(2) x^(2))/(c^(2))) = (abx^(2))/(c^(2))`
`rArr sqrt(c^(2) - a^(2) x^(2)) sqrt(c^(2) - b^(2) x^(2)) = abx^(2)`
`rArr c^(4) - c^(2) (a^(2) + b^(2)) x^(2) + a^(2) b^(2) x^(4) = a^(2) b^(2) x^(4)`
`rArr c^(4) -c^(2) (c^(2)) x^(2) = 0`
`rArr x = +- 1`
132.

If `x gt y gt 0`, then find the value of `tan^(-1).(x)/(y) + tan^(-1) [(x + y)/(x -y)]`

Answer» Correct Answer - `(3pi)/(4)`
Since `(x)/(y) xx (x + y)/(x - y) gt 1`, then the expression is equal to
`pi + tan^(-1) [((x)/(y) + (x + y)/(x -y))/(1 - (x)/(y) xx (x + y)/(x -y))] = pi + tan^(-1) (-1) = pi - (pi)/(4) = (3pi)/(4)`
133.

If `-1/2 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 - A
134.

If `1/2 le x le 1 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 - B
135.

If `alpha,beta(alpha < beta)` are the roots of equation `6x^2+11="" x+3="0` , then which following real? (a) `cos^(-1)alpha` (b) `sin^(-1)beta` (c) `cosec^(-1)alpha` (d) both `cot^(-1)alpha` and `cot^(-1)beta`

Answer» `6x^2+11x+3 = 0`
`=>6x^2+9x+2x+3 = 0`
`=>3x(2x+3)+1(2x+3) = 0`
`=>(2x+3)(3x+1) = 0`
`=> x = -3/2 or x = -1/3`
As `alpha` and `beta` ar the roots of the given equation,
`:. alpha = -3/2 or beta = -1/3`
Now, we will check the options.
(a) `cos^-1(alpha) = cos^-1(-3/2)`
As, domain of `cos^-1` is from `-1` to `1`, so `cos^-1(-3/2)` is not real.
(b) `sin^-1(beta) = sin^-1(-1/3) `
As, domain of `sin^-1` is from `-1` to `1`, so `sin^-1(-1/3)` is real.
(c) `cosec^-1(alpha) = cosec^-1(-3/2)`
As, domain of `cosec^-1` is from `-oo` to `-1`, so `cosec^-1(-3/2)` is real.
(d) `cot^-1(alpha) and cot^-1(beta)`
As, domain of `cot^-1` is from `-oo` to `oo`, so `cot^-1(-3/2) and cot^-1(-1/3)` is real.
136.

Find the value of `sin^(-1) ((3)/(5)) + tan^(-1) ((1)/(7))`

Answer» Correct Answer - `(pi)/(4)`
`sin^(-1).(3)/(5) + tan^(-1).(1)/(7) = tan^(-1).(3).(4) + tan^(-1).(1)/(7)`
`= tan^(-1) ((3 //4 + 1//7)/(1-3//4 xx 1//7))`
`= tan^(-1) ((25)/(25)) = tan^(-1) 1 = (pi)/(4)`
137.

Prove that `costan^(-1)sincot^(-1)x=sqrt((x^2+1)/(x^2+2))`

Answer» `cot^-1(x) = sin^-1(1/(sqrt(1+x^2)))` if `x gt 0`
`cot^-1(x) = sin^-1(pi-1/(sqrt(1+x^2)))` if `x lt 0`
In both cases,
`sin(cot^-1x) = 1/(sqrt(1+x^2)`
`=>tan^-1(sin(cot^-1x)) = tan^-1(1/(sqrt(1+x^2))) = cos^-1(sqrt(1+x^2)/sqrt(2+x^2))`
`:. costan^-1(sin(cot^-1x)) = coscos^-1(sqrt(1+x^2)/sqrt(2+x^2))`
`=>costan^-1(sin(cot^-1x)) = (sqrt(x^2+1)/sqrt(x^2+2)).`
138.

Prove that:`sin^(-1){(sqrt(1+x)+sqrt(1-x))/2}=pi/4+(sin^(-1)x)/2,""0 < x < 1`

Answer» `L.H.S. = sin^-1((sqrt(1+x)+sqrt(1-x))/2)`
Let `x = sintheta => theta = sin^-1x`
Then,
`L.H.S. = sin^-1((sqrt(1+sintheta)+sqrt(1-sintheta))/2)`
`= sin^-1((sqrt(1+cos(pi/2-theta))+sqrt(1-cos(pi/2-theta)))/2)`
`= sin^-1((sqrt(2cos^2(pi/4-theta/2))+sqrt(2sin^2(pi/4-theta/2)))/2)`
`= sin^-1((sqrt2cos(pi/4-theta/2)+sqrt2sin(pi/4-theta/2))/2)`
`= sin^-1(1/sqrt2cos(pi/4-theta/2)+1/sqrt2sin(pi/4-theta/2))`
`= sin^-1(sin(pi/4)cos(pi/4-theta/2)+cos(pi/4)sin(pi/4-theta/2))`
`= sin^-1(sin(pi/4+pi/4-theta/2))`
`= sin^-1(sin(pi/2-theta/2))`
`=pi/2-theta/2`
`=pi/2-sin^-1(x)/2`
`:. sin^-1((sqrt(1+x)+sqrt(1-x))/2) = pi/2-sin^-1(x)/2`
139.

The value(s) of `x` satisfying `tan^(-1)(x+3)-tan^(-1)(x-3)=sin^(-1)(3/5)` may beA. `-2`B. `-1`C. 0D. 2

Answer» Correct Answer - C
`tan^(-1)(x+3)-tan^(-1)(x-3)="sin"^(-1)(3)/(5)`
`rArr "tan"^(-1)(6)/(1+x^(2)-9)="tan"^(-1)(3)/(4)`
`rArr 24=3x^(2)-24`
`rArr 3x^(2)=48 rArr x = pm 4`
140.

Let `P_1=(sin^(-1)x)^(sin^(-1)x),P_2=(sin^(-1)x)^(cos^(-1)x),P_3=(cos^(-1)x)^(sin^(-1)),P_4=(cos^(-1)x)^(cos^(-1)x)` then match the entries of Column I to the entries of Column II.

Answer» Correct Answer - A(q),B(s) , C (r ) , D(p)
141.

Statement-1:`sin^(-1)tan((tan^(-1))x+tan^(-1)(1-x))]` `=(pi)/(2)` has no non zero integral solution Statement-2: The greatest and least values of `(sin^(-1)x)^(3)+(cos^(-1)x)^(3) are (7pi)^(3)/(8) and (pi)^(3)/(32)` respectivelyA. 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.

Answer» `sin^(-1)[tan (tan^(-1)x+tan^(-1)(1-x))]=(pi)/(2)`
`rarr sin^(-1)[tan{tan^(-1)(x+1-x)/(1-x(1-x))}] =(pi)/(2)`
`rarr tan{ tan^(-1)(1)/(1-x+x^(2))}=1`
` rarr (1)/(1-x+x^(2))=1 rarr x^(2)-x+1=1 rarr x=0 0,1`
so statement 1 is not true
statement 2 si true
142.

If `tan^(-1)(a/x)+tan^(-1)(b/x)+tan^(-1)(c /x)+tan^(-1)(d/x)=(pi)/(2)` then `x^(4)-x^(2)(Sigma ab)+abcd=`A. `-1`B. 0C. 1D. 2

Answer» We have
`tan^(-1)(a)/(x)+tan^(-1)(b)/(x)+tan^(-1)(c )/(x)+tan^(-1)(d)/(x)=(pi)/(2)`
`rarr tan^(-1)(a)/(x)+tan^(-1)(b)/(x)=(pi)/(2)-{tan^(-1)(c )/(x)+tan^(-1)(d)/(x)}`
`rarr tan^(-1) ((a+b)x)/(x^(2)-ab)=(pi)/(2)-tan^(-1)((c+d)x)/(x^(2)-cd)`
`rarr tan^(-1)((a+b)x)/(x^(2)-ab)=co^(-1)((c+d)x)/(x^(2)-cd)`
`rarr tan^(-1)((a+b)x)/(x^(2)-ab)=(x^(2)-cd)/((c+d)x) rarr x^(4) -x^(2) (Sigma ab)+abcd=0`
143.

`cos^(- 1)sqrt((a-x)/(a-b))=sin^(- 1)sqrt((x-b)/(a-b))` is possible ,ifA. `a gt x gt b`B. `a lt x lt b`C. `a = x = b`D. `a gt b` and x takes any value

Answer» Correct Answer - A::B
`cos^(-1)(sqrt((a-x)/(a-b)))=sin^(-1)(sqrt(1-(a-x)/(a-b)))`
`=sin^(-1)(sqrt((x-b)/(a-b)))`
`therefore` We must have `(a-x)/(a-b)gt 0` and `(x-b)/(a-b)gt 0`
`0 le (a-x)/(a-b)lt 1` and `0 le (x-b)/(a-b)lt 1`
From both, we get `0 le (a-x)/(a-b)lt 1`
`therefore a gt x gt b` or `a lt x lt b`
144.

Find the minimum value of the function`f(x)=(pi^2)/(16cot^(-1)(-x))-cot^(-1)x`

Answer» `f(x) = pi^2/(16cot^-1(-x)) - cot^-1x`
`=>f(x) = pi^2/(16cot^-1(-x)) - (pi-cot^-1(-x))`
`=>f(x) = cot^-1(-x)+pi^2/(16cot^-1(-x)) - pi`
`=>f(x) = (sqrt(cot^-1(-x)) - pi/(4sqrt(cot^-1(-x))))^2+pi/2 - pi`
Now, minimum value of `(sqrt(cot^-1(-x)) - pi/(4sqrt(cot^-1(-x))))^2` will be `0` as square of a number can not be less than `0`.
`:. f(x)_min = pi/2 - pi = -pi/2.`
145.

If equation `sin^(-1) (4 sin^(20 theta + sin theta) + cos^(-1) (6 sin theta - 1) = (pi)/(2)` has 10 solution for `theta in [0, n pi]`, then find the minimum value of n

Answer» Correct Answer - 9
We must have
`4 sin^(2) theta + sin theta = 6 sin theta -1`
`rArr 4 sin^(2) theta - 5 sin theta + 1 = 0`
`rArr (4 sin theta - 1) (sin theta - 1) = 0`
`rArr sin theta = (1)/(4) " or " sin theta = 1`
`sin theta = 1` is not possible as otherwise
`6 sin theta - 1 = 4 sin^(2) theta + sin theta = 5`
`:. sin theta = (1)/(4)`
For 10 solution, `theta in [0, 9pi] " or " [0, 10 pi]`
Thus, the least value of `n " is " 9`
146.

Match the following

Answer» Correct Answer - A(q, r, s,t), B(p,r,s,t) , C(p,q,r,s, t) , D(p, q , r,t)
147.

STATEMENT -1 : The function `f : [-1,1] to [0,pi],f(x)=cos^(-1)x` is not one-one. and STATEMENT -2 :The function `f : (-oo,oo)to[-1,1],f(x)=cosx is not one-one.A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -3B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -3C. Statement-1 is True, Statement -2 is FalseD. Statement -1 is False, Statement -2 is True

Answer» Correct Answer - D
148.

Match the following according to the number of solutions of the equation.

Answer» Correct Answer - A ( r) , B(s) , C(p) , D( r)
149.

If `cos^-1 (x^2-1)/(x^2+1)+ tan^-1 (2x)/(x^2-1) = (2pi)/3`, then x equal to (A) `sqrt(3)` (B) `2+sqrt(3)` (C) `2-sqrt(3)` (D) `-sqrt(3)`A. 2B. `sqrt(3)`C. 4D. 3

Answer» Correct Answer - B
`cos^(-1)((x^(2)-1)/(x^(2)+1))=pi - co^(-1)((1-x^(2))/(1+x^(2)))=pi-2 tan^(-1)x` (as x gt 1)
`tan^(-1)((2x)/(x^(2)-1))=-tan^(-1)((2x)/(1-x^(2)))=pi-2 tan^(-1)x` (as x gt 1)
`therefore` Given equation is `2pi-4 tan^(-1)x = 2pi//3`
`therefore tan^(-1)x = pi//3`
`therefore x = sqrt(3)`
150.

If `sin^(-1)((sqrt(x))/2)+sin^(-1)(sqrt(1-x/4))+tan^(-1)y=(2pi)/3`, thenA. maximum value of `x^(2)+y^(2)` is `(49)/(3)`B. maximum value of `x^(2)+y^(2)` is 4C. minimum value of `x^(2)+y^(2)` is `(1)/(3)`D. minimum value of `x^(2)+y^(2)` is 3

Answer» Correct Answer - A::C::D
`sin^(-1)((sqrt(x))/(2))+sin^(-1)(sqrt(1-(x)/(4)))+tan^(-1)y=(2pi)/(3),x in [0,4]`
`therefore sin^(-1)((sqrt(x))/(2))+cos^(-1)((sqrt(x))/(2))+tan^(-1)y=(2pi)/(3)`
`therefore (pi)/(2)+tan^(-1)y=(2pi)/(3)`
`therefore tan^(-1)y=(2pi)/(3)-(pi)/(2)=(pi)/(6)`
`rArr =(1)/(sqrt(3))`
`therefore` maximum value of `(x^(2)+y^(2))=16+(1)/(3)=(49)/(3)`
and minimum value of `(x^(2)+y^(2))=(0)^(2)+(1)/(3)=(1)/(3)`