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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.
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` |
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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.` |
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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`. |
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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)` |
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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 |
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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` |
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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)` |
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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`. |
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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. |
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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.` |
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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 |
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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)` |
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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 |
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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)` |
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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. |
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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` |
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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)` |
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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. |
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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)` |
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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. |
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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` |
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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` |
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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. |
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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.` |
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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 |
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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` |
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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)` |
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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. |
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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)` |
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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)).` |
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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` |
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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` |
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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 |
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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` |
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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` |
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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.` |
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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` |
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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)` |
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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)` |
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