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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.
| 51. |
यदि `sin^(-1)""(1)/(2)=tan^(-1)x,` तो x का मान ज्ञात कीजिए । |
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Answer» दिया है - `sin^(-1)""(1)/(2)=tan^(-1)x ` ` implies tan^(-1)x=sin^(-1)""(1)/(2)=(pi)/(6)impliesx=tan""(pi)/(6)=(1)/(sqrt(3))` `:. x =(1)/(sqrt(3))` |
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| 52. |
मान ज्ञात कीजिए - `cot^(-1)(-sqrt(3))` |
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Answer» मान `cot^(-1)(-sqrt(3))=theta , pi//2 lt theta lt pi` `implies cot theta=-sqrt(3)=cot""(5pi)/(6)implies theta =(5pi)/(6)` ` :. cot^(-1)(-sqrt(3)) "का मुख्य मान "=(5pi)/(6)` |
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| 53. |
मान ज्ञात कीजिए - `cos^(-1)""((1)/(2))` |
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Answer» मान `cos^(-1)""(1)/(2)=theta , 0 le theta le (pi)/(2)` ` implies cos theta =(1)/(2)=cos""(pi)/(3)implies theta=(pi)/(3)` `:. cos^(-1)""(1)/(2) "का मुख्य मान "=(pi)/(3)` |
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| 54. |
मान ज्ञात कीजिए - `sin^(-1)""((1)/(sqrt(2)))` |
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Answer» मान `sin^(-1)((1)/(sqrt(2)))=theta , -(pi)/(2) le theta le(pi)/(2) ` `implies sin theta =(1)/(sqrt(2))=sin""(pi)/(2)implies theta =(pi)/(4)` `:. sin^(-1)((1)/(sqrt(2))) "का मुख्य मान "=(pi)/(4)` |
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| 55. |
मान ज्ञात कीजिए - `cot^(-1)(-1)` |
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Answer» मान `cot^(-1)(-1)=theta , 0 lt theta lt pi` ` implies cot theta=-1=cot""(3pi)/(4)implies theta=(3pi)/(4)` `:. cot^(-1)(-1) "का मुख्य मान "=(3pi)/(4)` |
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| 56. |
मान ज्ञात कीजिए - `sin^(-1)(-1)` |
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Answer» मान `sin^(-1)(-1)=theta, -pi//2 le theta le pi//2` `implies sintheta =-1 =sin(-(pi)/(2))implies theta =-(pi)/(2)` ` :. sin^(-1)(-1) "का मुख्य मान "=-(pi)/(2)` |
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| 57. |
`tan^(-1)sqrt(3)+cot^(-1)sqrt(3)` का मुख्य मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(pi)/(2)` | |
| 58. |
`tan^(-1)(sqrt(3))-cot^(-1)(-sqrt(3))` का मानA. `pi` हैB. `-(pi)/(2)` हैC. `0` हैD. `2sqrt(3)` |
| Answer» Correct Answer - B | |
| 59. |
मान ज्ञात कीजिए - ` "cosec"^(-1)(-2)` |
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Answer» मान `"cosec"^(-1)(-2)=theta , -pi//2 le theta lt 0 (theta lt 0)` `implies "cosec"theta =-2 implies "cosec" theta="cosec"(-(pi)/(6))` `implies theta=-(pi)/(6)` `:. "cosec"^(-1)(-2) "का मुख्य मान "=-(pi)/(6)` |
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| 60. |
मान ज्ञात कीजिए - ` cot^(-1)(sqrt(3))` |
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Answer» मान `cot^(-1)(sqrt(3))=theta, 0 lt theta lt pi` `implies cot theta =sqrt(3)=cot""(pi)/(6)implies theta =(pi)/(6)` ` :. cot^(-1)(sqrt(3)) "का मुख्य मान "=(pi)/(6)` |
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| 61. |
मान ज्ञात कीजिए - ` sin^(-1)""(1)/(2)` |
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Answer» मान ` sin^(-1)""(1)/(2)=theta , -pi //2 le theta le pi//2` `implies sin theta =(1)/(2)=sin""(pi)/(6)implies theta=(pi)/(6)` `:. sin^(-1)""(1)/(2) "का मुख्य मान "=(pi)/(6)` |
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| 62. |
मान ज्ञात कीजिए - `sin[cos^(-1)""((3)/(5))]` |
| Answer» Correct Answer - `(4)/(5)` | |
| 63. |
मान ज्ञात कीजिए - `cos[tan^(-1)""(3)/(4)]` |
| Answer» Correct Answer - `(4)/(5)` | |
| 64. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `tan^(-1)(tan.(pi)/(4))`. |
| Answer» `tan^(-1)(tan.(pi)/(4))=(pi)/(4)," "[because (pi)/(4) in (-(pi)/(2),(pi)/(2))]` | |
| 65. |
मान निकालें : `sin^(-1) (sin ""(3pi)/(5))` |
| Answer» Correct Answer - `(2pi)/(5)` | |
| 66. |
मान ज्ञात कीजिए - `tan^(-1)[tan""(3pi)/(4)]` |
| Answer» Correct Answer - `-(pi)/(4)` | |
| 67. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `sin^(-1)(sin.(3pi)/(5))` |
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Answer» चूँकि `sin^(-1)` की मुख्य शाखा का परिसर `[-(pi)/(2),(pi)/(2)]` होता है। `therefore" "sin^(-1)(sin.(3pi)/(5))ne (3pi)/(5)," "[because (3pi)/(5) cancel(in)[-(pi)/(2),(pi)/(2)]]` अब , `sin^(-1)(sin.(3pi)/(5))` `=sin^(-1)[sin(pi-(2pi)/(5))]," "[because (3pi)/(5)=pi-(25)/(5)]` `=sin^(-1)(sin.(2pi)/(5))," "[because sin(pi-theta)=sin theta]` `=(2pi)/(5)in[-(pi)/(2),(pi)/(3)]," "[because sin^(-1)(sin theta)=theta]` `therefore" "sin^(-1)(sin.(3pi)/(5))=(2pi)/(5).` |
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| 68. |
प्रश्न संख्या 16 से 18में दिए प्रत्येक व्यंजन का मान ज्ञात कीजिए : `tan^(-1)(tan""(3pi)/(4))` |
| Answer» Correct Answer - `(-pi)/(4)` | |
| 69. |
मान निकालें : `tan^(-1) (tan""(3pi)/(4))` |
| Answer» Correct Answer - `-(pi)/(4)` | |
| 70. |
` sin^(-1)(sin""(pi)/(3))` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(pi)/(3)` | |
| 71. |
मान ज्ञात कीजिए - `cos^(-1)[cos""((4pi)/(3))]` |
| Answer» Correct Answer - `(2pi)/(3)` | |
| 72. |
`2 sin^(-1)""(1)/(2)+cos^(-1)""(-(1)/(2))` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `pi` | |
| 73. |
`tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))=(pi)/(4)-(1)/(2)cos^(-1)x,-(1)/(sqrt2)lexle1` |
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Answer» माना `x=costhetaimpliestheta=cos^(-1)x` `L.H.S.=tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))` `=tan^(-1)((sqrt(1+costheta)-sqrt(1-costheta))/(sqrt1+costheta+sqrt(1-costheta)))` `=tan^(-1)""(sqrt(2cos^(2)""(theta)/2)-sqrt(2sin^(2)""(theta)/(2)))/(sqrt(2cos^(2)""(theta)/(2))+sqrt(2sin^(2)""(theta)/(2)))` `=tan^(-1)((cos""(theta)/(2)-sin""(theta)/(2))/(cos""(theta)/(2)+sin""(theta)/(2)))` `=tan^(-1)((1-tan""(theta)/(2))/(1+tan""(theta)/(2)))` `=tan^(-1)tan((pi)/(4)-(theta)/(2))=(pi)/(4)-(theta)/(2)` `=(pi)/(4)-(1)/(2)cos^(-1)x=R.H.S.` |
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| 74. |
`sin^(-1)(sin""(3pi)/(5))` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(3pi)/(5)` | |
| 75. |
मान ज्ञात कीजिए - `cos^(-1)""(-(1)/(2))` |
| Answer» Correct Answer - `(2pi)/(3)` | |
| 76. |
मान ज्ञात कीजिए - `sec^(-1)""(-(2)/(sqrt(3)))` |
| Answer» Correct Answer - `(5pi)/(6)` | |
| 77. |
मान ज्ञात कीजिए - `cos^(-1)""((1)/(sqrt(2)))` |
| Answer» Correct Answer - `(pi)/(4)` | |
| 78. |
मान ज्ञात कीजिए - `sin^(-1)""(-(sqrt(3))/(2))` |
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Answer» Correct Answer - `-(pi)/(3)` माना `sin^(-1)""(-(sqrt(3))/(2))=x ` `implies sinx=-(sqrt(3))/(2)=-sin""(pi)/(3)=sin(-(pi)/(3))implies x=-(pi)/(3)` |
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| 79. |
मान ज्ञात कीजिए - `tan^(-1)""(-(1)/(sqrt(3)))` |
| Answer» Correct Answer - `-(pi)/(6)` | |
| 80. |
`cot(tan^(-1)a+cot^(-1)a)` |
| Answer» `cot(tan^(-1)a+cot^(-1)a)=cot""(pi)/(2)=0` | |
| 81. |
यदि `sin(sin^(-1)""(1)/(5)+cos^(-1)x)=1` तो x का मान ज्ञात कीजिए। |
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Answer» `sin(sin^(-1)""(1)/(5)+cos^(-1)x)=1` `impliessin(sin^(-1)""(1)/(5)+cos^(-1)x)=sin""(pi)/(2)` `impliessin^(-1)""(1)/(5)+cos^(-1)x+(pi)/(2)` `impliessin^(-1)""(1)/(5)=(pi)/(2)-cos^(-1)x` `impliessin^(-1)""(1)/(5)=sin^(-1)ximpliesx=(1)/(5)` |
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| 82. |
`cot(tan^(-1)3)` का मान ज्ञात कीजिए। |
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Answer» माना `tan^(-1)3=x` `impliestanx=3" implies "cotx=(1)/(3)` `impliescot(tan^(-1)3)=(1)/(3)` |
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| 83. |
यदि `sin^(-1)x=(pi)/(4)" तो "cos^(-1)x` का मान ज्ञात कीजिए। |
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Answer» `sin^(-1)x=(pi)/(4)` `impliesx=sin"(pi)/(4)=(1)/(sqrt2)` `:.cos^(-1)x=cos^(-1)""(1)/(sqrt2)` `=cos^(-1)(cos""(pi)/(4))=(pi)/(4)` |
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| 84. |
निम्नलिखित समीकरण को हल कीजिए - `sec^(-1)((x)/(a))-sec^(-1)((x)/(b))=sec^(-1)(b)-sec^(-1)(a).` |
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Answer» दिया गया समीकरण है - `sec^(-1)((x)/(a))-sec^(-1)((x)/(b))=sec^(-1)(b)-sec^(-1)(a)` `rArr cos^(-1)((a)/(x))-cos^(-1)((b)/(x))=cos^(-1)((1)/(b))-cos^(-1)((1)/(a))," "[because sec^(-1)x=cos^(-1)((1)/(x))]` `rArr cos^(-1)((a)/(x))+cos^(-1)((1)/(a))=cos^(-1)((1)/(b))+cos^(-1)((b)/(x))` `rArr cos^(-1)[(a)/(x).(1)/(a)-sqrt(1-(a^(2))/(x^(2)))sqrt(1-(1)/(a^(2)))]` `" "=cos^(-1)[(1)/(b).(b)/(x)-sqrt(1-(1)/(b^(2)))sqrt(1-(b^(2))/(x^(2)))]` `rArr cos^(-1)[(1)/(x)-sqrt((x^(2)-a^(2))/(x^(2)))sqrt((a^(2)-1)/(a^(2)))]` `=cos^(-1)[(1)/(x)-sqrt((b^(2)-1)/(b^(2)))sqrt((x^(2)-b^(2))/(x^(2)))]` `rArr (1)/(x)-sqrt((x^(2)-a^(2))/(x^(2)))sqrt((a^(2)-1)/(a^(2)))` `rArr sqrt((x^(2)-a^(2))/(x^(2)))sqrt((a^(2)-1)/(a^(2)))=sqrt((b^(2)-1)/(b^(2)))sqrt((x^(2)-b^(2))/(b^(2)))` `rArr ((x^(2)-a^(2))(a^(2)-1))/(a^(2)x^(2))=((x^(2)-b^(2))(b^(2)-1))/(b^(2)x^(2)),` |
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| 85. |
समीकरण `tan^(-1)(1+x)+tan^(-1)(1-x)=(pi)/(2)` का एक हल है- |
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Answer» Correct Answer - A |
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| 86. |
निम्नलिखित समीकरण को हल कीजिए - `tan^(-1)(cotx)+cot^(-1)(tanx)=(pi)/(4)` |
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Answer» दिया गया समीकरण है - `tan^(-1)(cotx)+cot^(-1)(tanx)=(pi)/(4)` `tan^(-1)[tan((pi)/(2)-x)]+cot^(-1)[cot((pi)/(2)-x)]=(pi)/(4),` `[because tan((pi)/(2)-x)=cotx , cot((pi)/(2)-x)=tanx]` `(pi)/(2)-x+(pi)/(2)-x=(pi)/(4)` `=(pi)/(2)+(pi)/(2)-(pi)/(4)=2x` `(pi)/(2)+(pi)/(2)-(pi)/(4)=2x` `2x=(3pi)/(4) rArr x =(3pi)/(8).` |
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| 87. |
निम्नलिखित समीकरण को हल कीजिए - `tan^(-1).(1-x)/(1+x)=sin^(-1).(2x)/(1+x^(2)).` |
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Answer» दिया गया समीकरण है - `tan^(-1).(1-x)/(1+x)=sin^(-1).(2x)/(1+x^(2))` `tan^(-1)1-tan^(-1)x=2tan^(-1)x` `(pi)/(2)=2tan^(-1)x+tan^(-1)x` `rArr" "3tan^(-1)x=(pi)/(4)` `rArr" "3tan^(-1)x=(pi)/(4)` `rArr" "tan^(-1)=(pi)/(12)` `rArr" "x=tan.(pi)/(12)=tan.(180^(@))/(12)` `rArr" "x=tan15^(@)=tan(45^(@)-30^(@))` `rArr" "x=(tan45^(@)-tan30^(@))/(1+tan45^(@)tan30^(@))` `rArr" "x=(1-(1)/(sqrt3))/(1+(1)/(sqrt3))=((sqrt3-1)/(sqrt3))/((sqrt3+1)/(sqrt3))` `rArr" "x=(sqrt3-1)/(sqrt3+1)=((sqrt3-1)(sqrt3-1))/((sqrt3+1)(sqrt3-1))` `rArr" "x=((sqrt3-1)^(2))/((sqrt3)^(2)-(1)^(2))=((sqrt3)^(2)+1-2sqrt3)/(3-1)` `rArr" "x=(3+1-2sqrt3)/(3-1)=(4-2sqrt3)/(2)` `rArr" "x=2-sqrt3.` |
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| 88. |
`tan^(-1)((3a^(2)x-x^(3))/(a^(3)-3ax^(2))),agt0,(-a)/(sqrt3)ltxlt(a)/(sqrt3)` |
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Answer» `tan^(-1)((3a^(2)x-x^(3))/(a^(3)-3ax^(2)))` `=tan^(-1)((3a^(2).atantheta-a^(3)tan^(3)theta)/(a^(3)-3a.a^(2)tan^(2)theta))` `=tan^(-1)((3tantheta-tan^(3)theta)/(1-3tan^(2)theta))" "" माना "{:(x=atantheta),(implies(x)/(a)=tantheta):}` `=tan^(-1)(tan3theta)=3theta" "theta=tan^(-1)((x)/(a))` `=3tan^(-1)""(x)/(a)` |
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| 89. |
`tan^(-1)sqrt((1-cosx)/(1+cosx))*0 le x ltpi` को सरलतम रूप में लिखे | |
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Answer» `tan^(-1) sqrt((1-cosx)/(1+cosx)) = tan^(-1) ""sqrt((2sin^(2) ""(x)/(2))/(2 cos^(2)""(x)/(2)))=tan^(-1)|tan ""(x)/(2)|` ` = tan^(-1)(tan""(x)/(2)) = (x)/(2) " "[because 0 le x ltpi therefore 0 le(x)/(2)lt(pi)/(2)]` |
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| 90. |
सरलतम रूप में लिखिए - `tan^(-1)((3a^(2)x-x^(3))/(a^(3)-3ax^(2))), a gt 0 :(-a)/(sqrt3) le x le (a)/(sqrt3).` |
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Answer» माना `x=a tan theta` `rArr" "tan theta=(x)/(a)` `rArr" "theta=tan^(-1)((x)/(a))` चूँकि`" "-(a)/(sqrt3) lt x lt (a)/(sqrt3)` `rArr" "-(a)/(sqrt3) lt a tan theta lt (a)/(sqrt3)` `rArr" "-(1)/(sqrt3) lt tan theta lt (1)/(sqrt3)` अब, `tan^(-1)[(3a^(2)x-x^(3))/(a^(3)-3ax^(2))]` `=tan^(-1)[(3a^(2).a tan theta-a^(3) tan^(3)theta)/(a^(3)-3a.a^(2)tan^(2)theta)]` `=tan^(-1)[(a^(3)(3tan theta-tan^(3)theta))/(a^(3)(1-3tan^(2)theta))]` `=tan^(-1)[(3tan theta-tan^(3)theta)/(1-3tan^(2)theta)]` `=tan^(-1)(tan3 theta)` `=3theta` `=3tan^(-1).(x)/(a)` जो कि अभीष्ट सरलतम रूप है। |
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| 91. |
`tan^(-1)""((x)/(y))-tan^(-1)""(x-y)/(x+y) ` का मान है :A. `(pi)/(2) `हैB. `(pi)/(3) ` हैC. `(pi)/(4)` हैD. `(3pi)/(4)` |
| Answer» Correct Answer - C | |
| 92. |
हल कीजिए : `tan^(-1)((1)/(1+2x))+tan^(-1)((1)/(1+4x))=tan^(-1)""(2)/(x^(2))`. |
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Answer» Correct Answer - 0, 3, `-(2)/(3)` |
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| 93. |
`3cos^(-1)x=cos^(-1)(4x^(3)-3x),x""in[(1)/(2),1]` |
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Answer» माना `cos^(-1)x=theta` `impliesx=costheta` `R.H.S=cos^(-1)(4x^(3)-3x)` `=cos^(-1)(4cos^(3)theta-3costheta)` `=cos^(-1)(cos3theta)` `=3theta=3cos^(-1)x=L.H.S` |
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| 94. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए - `cos^(-1)(-(1)/(sqrt2))` |
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Answer» माना `cos^(-1)(-(1)/(sqrt2))=theta` `rArr" "cos theta=-(1)/(sqrt2)," "theta in [0,pi]` `rArr" "cos theta=-cos.(pi)/(4)` `rArr" "cos theta=cos(pi-(pi)/(4))` `" "[because cos (pi-theta)=-cos theta]` `rArr" "cos theta = cos.(3pi)/(4)` `rArr" "theta=(3pi)/(4)` अतः `cos^(-1)(-(1)/(sqrt2))` का मुख्य मान `(3pi)/(4)` है। |
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| 95. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए : `cos^(-1)(-(1)/(sqrt2))` |
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Answer» `cos^(-1)(-(1)/(sqrt2))=pi-cos^(-1)((1)/(sqrt2))` `=pi-cos^(-1)(cos""(pi)/(4))` `=pi-(pi)/(4)=(3pi)/(4)` |
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| 96. |
`sin^(-1)(-(1)/(2))` का मुख्य मान निकालें |
| Answer» Correct Answer - `(pi)/(6)` | |
| 97. |
`tan^(-1) 1 + cos^(-1)""(-(1)/(2))+sin^(-1)""((1)/(2))` का मुख्य मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(3pi)/(4) ` | |
| 98. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए - `sin^(-1)((1)/(sqrt2))` |
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Answer» हम जानते हैं कि `sin^(-1)` की मुख्य शाखा का परिसर `[-(pi)/(2),(pi)/(2)]` होता है। माना `sin^(-1)((1)/(sqrt2))=theta` `rArr" "sin theta=(1)/(sqrt2), theta in [-(pi)/(2),(pi)/(2)]` `rArr" "sin theta= sin.(pi)/(4)` `rArr" "theta=(pi)/(4)` अतः `sin^(-1)((1)/(sqrt2))` का मुख्य मान `(pi)/(4)` है। |
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| 99. |
निम्नलिखित के मुख्य मानों को ज्ञात कीजिए - `"cosec"^(1)(2)` |
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Answer» हम जानते हैं कि `"cosec"^(-1)` की मुख्य शाखा का परिसर `[-(pi)/(2),(pi)/(2)]-{0}.` होता है। माना `"cosec"^(-1)(2=0` `rArr" cosec"theta2," "theta in [-(pi)/(2),(pi)/(2)]-{0}` `rArr" cosec"theta="cosec"(pi)/(6)` `rArr" "theta=(pi)/(6).` अतः `"cosec"^(-1)(2)` का मुख्य मान `(pi)/(6)` है। |
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| 100. |
`cos^(-1)(-(1)/(2))` का मुख्य मान निकालें |
| Answer» Correct Answer - `(2pi)/(3)` | |