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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.
| 151. |
General solution of `secx+sqrt(2)=0` isA. `2npipm(pi)/(4), ninZ`B. `2npipm(3pi)/(4), ninZ`C. `2npipm(5pi)/(4), ninZ`D. `2npipm(7pi)/(4), ninZ` |
| Answer» Correct Answer - B | |
| 152. |
If `sinA=sin^2B and 2cos^2A=3cos^2B` then the triangle ABC isA. right angledB. obtuse angledC. ospscelesD. equilateral |
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Answer» Correct Answer - B `sinA=sin^2Band2cos^2A=3cos^2B` `rArr2-2sin^2A=3-3sin^2B` `rArr2sin^2A-3sinA+1=0` `rArr(2sinA-1)(sinA-1)=0` `rArrA=30^@rArrA=90^@` If `A=30^@rArrB=45^@rArrC=105^@` If `A=90^@rArrB=90^@`, which is not possible |
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| 153. |
If `sinA=sin^2Ba n d2cos^2A=3cos^2B`then the triangle `A B C`isright angled (b) obtuse angled(c)isosceles (d) equilateral |
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Answer» `sinA = sin^2B ->(1)` `2cos^2A = 3cos^2B` `=>2cos^2A = 3(1-sin^2B)` From (1), `=>2(1-sin^2A) = 3(1-sinA)` `=>2sin^2A - 3sinA+1 = 0` `=>2sin^2A - 2sinA-sinA+1 = 0` `=>2sinA(sinA-1)-1(sinA-1) = 0` `=>(2sinA-1)(sinA-1) = 0` `=>sinA = 1 or sinA = 1/2` `=>A = 90^@ or A = 30^@` When `A = 90^@`, `sin^2B = 1 => B = 90^@`, which is not possible for a triangle. When `A = 30^@`, `sin^2B = 1/2 => B = 45^@` `=>C = (180-30-45)^@ = 75^@` `:. Delta ABC` is an obtuse angle triangle. So, option-`(b)` is the correct option. |
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| 154. |
If `sinx+sin^2x=1`, then find the value of `cos^(12)x +3cos^(10)x + 3 cos^8x + cos^6x-1 ` |
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Answer» Correct Answer - 3 We have `sinx+sin^2x=1` `or sinx=1-sin^2xorsinx=cos^2x` Now, `cos^12x+3sin^10x+3sin^4x+sin^3x-2` `=sin^6x+3sin^5x+3sin^4x-2` `=(sin^2x)^3+3(sin^2x)^2sinx+3(sin^2x)` `(sinx)^2+(sinx)^3-2` `=(sin^2x+sinx)^3-2=(1)^3-2=-1` |
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| 155. |
If `sin^2(theta-alpha)c a salpha=cos^2(theta-alpha)sinalpha=msinalphacosalpha,`then prove that `|m|geq1/(sqrt(2))` |
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Answer» `sin^2(theta-alpha)cosalpha=msinalphacosalpha` `sin^2(theta-alpha)=msinalpha-(1)` `cos^2(theta-alpha)sinalpha=msinalphacosalpha` `cos^2(theta-alpha)=mcosalpha-(2)` adding equation 1 and 2 `sin^2(theta-alpha)+cos^2(theta-alpha)=m(sinalpha+cosalpha)` `1=m(sinalpha+cosalpha)` `[sinalpha+cosalpha=1/m]1/sqrt2` `1/sqrt2sinnalpha+1/sqrt2cosalpha=1/(sqrt2m)` `sin(alpha+pi/4)=1/(sqrt2m)` `|sinx|<=1` `|sin(alpha+pi/4)|<=1` `|1/(sqrt2m)|<=1` `|1/m|<=sqrt2` `|m|>=1/sqrt2`. |
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| 156. |
General solution of `8tan^(2)((x)/(2))=1+secx` isA. `2npipmcos^(-1)((2)/(3)), ninZ`B. `2npipmcos^(-1)((1)/(3)), ninZ`C. `npipmcos^(-1)((2)/(3)), ninZ`D. `npipmcos^(-1)((1)/(3)), ninZ` |
| Answer» Correct Answer - B | |
| 157. |
Find the maximum value of `4sin^2x+3cos^2x+sin(x/2)+cos(x/2)dot` |
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Answer» `I=4sin^2x+3(1-sin^2x)+sin(x/2)+cos(x/2)` `I=4sin^2x+3-3sin^2x+sin(x/2)+cos(x/2)` `I=sin^2x+3+sin(x/2)+cos(x/2)` `I=sin^2x+3+sqrt2[1/sqrt2sin(x/2)+cos(x/2)*1/sqrt2]` `=3+sin^2x+sqrt2sin[pi/4+x/2]` `sin(pi/4+x/2)=1` When `x=pi/2` i.e.`sin(pi/4+pi/4)=sin(pi/2)=1` Max`I=x=pi/2` `I(atx=pi/2)=3+sin^2(pi/2)+sqrt2` `=3+1+sqrt2` `=4+sqrt2`. |
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| 158. |
If `A`is the area and `2s`is the sum of the sides of a triangle, then`Alt=(s^2)/4`(b) `Alt=(s^2)/(3sqrt(3))``2RsinAsinBsinC`(d) `non eoft h e s e` |
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Answer» `25=a+b+c` `A^2=s(s-a)(s-b)(s-c)` `AM>=GM` `(s+(s-a)+(s-b)+(s-c))/4=(s(s-a)(s-b)(s-c))^(1/4)` `(4s-2s)/4=(A^2)^(1/4)` `s/2>=(A)^(1/2)` `AM>=GM` `((s-a)+(s-b)+(s-c))/3>=((s-a)(s-b)(s-c))^(1/3)` `s/3>=((A^2/S)^(1/3)` `S^3/27>=A^2/S` `A^2/S<=S^3/27` `A^2<=s^4/27` `A<=S^2/(3sqrt3)` Option B is correct. |
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| 159. |
If `A=pi/5,`then find the value of `sum_(r=1)^8tan(r A)*tan((r+1)A)dot` |
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Answer» `tan((r+1)A-(rA))=(tan(r+1)A-tan(rA))/(1+tan(r+1)A*tan(rA))` `1+tan(rA)*tan(r+1)A=(tan(r+1)A-tan(rA))/(tanA)` `tan(rA)*tan(r+1)A=(tan(r+1)A-tan(rA))/(tanA)-1` `S=sum_(r=1)^8(tan(r+1)A-tan(rA))/(tanA)-sum_(r=1)^8 1` `=1/(tanA)sum_(r=1)^8(tan(r+1)A-tan(rA)-8` `=1/(tanA)[(tan2A-tanA)+(tan3A-tan2A)+...(tan9A-tan8A]-8` `=1/(tanA)[tan9A-tanA]-8` `tan9A=tan(9/5pi)=tan(2pi-pi/5)` `Ss=1/(tanA)(-tanA-tanA)-8` `=(-2tanA)/(tanA)-8` `=-2-8` `S=-10`. |
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| 160. |
`sin(pi/18)*sin(5pi/18)*sin(7pi/18)` |
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Answer» by calculating`pi /18 = 10^@` we have to find `sin 10^@, sin 50^@, sin 70^@` `1/2 sin 10^@ [cos(70^@-50^@) + cos(70^@ +50^@)]` `1/2 sin 10^@[cos 20^@ - cos 120^@]` `1/2sin10^@[1-2sin^2 10^@ + 1/2]` `1/2[(3-4sin^2 10^@)Sin 10^@/2]` `(3*sin10^@ - 4sin^3 10^@)/4` `1/4sin 30^@ = 1/8` |
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| 161. |
The minimum value of`sqrt((3sin x-4cosx-10)(3sinx+4cosx-10))`is ________ |
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Answer» `f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)` `f(x)=[(3sinx-10)-4cosx][(3sinx-10)+4cosx]` `f(x)=(3sinx-10)^2-16cos^2x` `f(x)=9sin^2x+100-60sinx-16(1-sin^2x)` `f(x)=25sin^2x+84-60sinx` `f(x)=(5sinx-6)^2+48` at sinx=1 `f(x)=(5-6)^2+48=49` Minimum value of`sqrtf(x)=sqrt49=7`. |
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| 162. |
Let `f(theta)=1/(1+(ottheta)^x)and S=sum_(theta=1^@)^(89^@) f(theta)`, then the value of `sqrt5` is _________ . |
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Answer» Correct Answer - 44.5 We have ,`f(theta)=((sintheta)^x)/((costheta)^x+(sintheta)^x)` `rArr f(theta)+f(pi/2-theta)=1`. `:. S=underset(theta=1^@)overset(89^@)sumf(theta)=f(1^@)+f(2^@)+...+f(88^@)+f(89^@)` `=(ubrace(1+1+1+.........+1)_("44 times"))+1/2=44+1/2=89/2` |
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| 163. |
If `K=sin(pi/(18))sin((5pi)/(18))sin((7pi)/(18)),`then the numerical value of `K`is_____ |
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Answer» `k=1/2[2sinpi/18*sin5/18pi]*sin7/18pi` `k=1/2[cos(5/18pi-pi/18)-cos(5/18pi+pi/18)]*sin7/18pi` `k=1/2[cos4/18pi*sin7/18pi-cospi/3*sin7/18pi]` `k=1/2[1/2(sin(7/18pi+4/18pi)+sin(7/18pi-4/18pi))-1/2sin7/18pi]` `k=1/2[1/2(sin11/18pi+sinpi/6)-1/2sin7/18pi]` `k=1/2[1/2sin7/18pi+1/2*1/2-1/2sin7/18pi]` `k=1/2[1/4]` `k=1/8`. |
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| 164. |
Find the values of `xa n dy`for which cosec `theta=(x^2-y^2)/(x^2+y^2)`is satisfied. |
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Answer» `x^2-y^2<=x^2+y^2` `cosectheta>=1` `cosectheta<=-1` `(x^2-y^2)/(x^2+y^2)=1` `x^2-y^2=a^2+y^2` `y=0` `(x^2-y^2)/(x^2+y^2)=-1` `x^2-y^2=-(x^2+y^2)` `x=0`. |
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| 165. |
The minimum value of`sqrt((3s in x-4cosx-10(3sinx+4cosx-1))`is ________ |
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Answer» Correct Answer - 7 `f(x) =9sin^2x-16cos^2x-10(3sinx-4cosx)-10(3sinx+4cosx)+100` `25 sin^2x-16cos^2x+84` `=(5sinx-6)^2+48` The minimum value of f(x) occurs when sin x = 1 Therefore , the minimum value of `sqrt(f(x)) is 7`. |
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| 166. |
The set of all real numbers `a`such that `a^2+2a ,2a+3,a n da^2+3a+8`are the sides of a triangle is_____ |
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Answer» `x=a^2+2a` `y=2a+3` `z=a^2+3a+8` `x+y>Z` `a^2+4a+3>a^2+3a+8` `a>5` `y+z>x` `a^2+5a+11>a^2+2a` `a>-11/3` `z+x>y` `2a^2+sqrta+8>2a+3` `2a^2+3a+5>0` D is negative. `a>5`. |
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| 167. |
Suppose that for some angles `xa n dy ,`the equations `sin^2x+cos^2y=(3a)/2a n dcos^2x+sin^2y=(a^2)/2`hold simultaneously. the possible value of `a`is ___________ |
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Answer» Correct Answer - 1 `sin^2x+cos^2y=(3a)/2 ...(i)` `andcos^2x+sin^2y =a^2/2 ...(ii)` Adding (i) and (ii), we get `2=(3a)/2 +a^2/2` `or a^2+3a-4=0` `or (a+4)(a-1)=0` `:. A=-4("rehected"),a=1` |
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| 168. |
Prove that `1+cot theta |
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Answer» `1+cot theta - cot(theta/2) = 1+(cot^2(theta/2) - 1)/(2cot(theta/2)) - cot(theta/2)` `=(2cot(theta/2)+cot^2(theta/2)-1 -2cot^2(theta/2))/(2cot(theta/2))` `=-(cot^2(theta/2)-2cot(theta/2)+1)^2/(2cot(theta/2))` `=-(cot theta-1)^2/(2cot(theta/2))` As `-(cot (theta/2)-1)^2/(2cot(theta/2))` is always less than or equal to `0`, `:. 1+cot theta - cot(theta/2) le 0` `=>1+cot theta le cot(theta/2).` Now, when equality sign holds, `(cot(theta/2) - 1)^2 = 0` `=>cot (theta/2) = 1` `=>theta/2 = pi/4` `=>theta = pi/2.` |
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| 169. |
If sin 2x + sin x = 0, then x is(A)`npi + pi/3, 2npi + pi/6`(B) `npi , npi + (-1)^n pi/3`(C) `npi, 2npi + 2pi/3, 2npi - 2pi/3`(D) `2npi, npi + 2pi/3` |
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Answer» `sin2x+sinx = 0` `=>2sinxcosx+sinx = 0` `=>sinx(2cosx+1) = 0` So, `sinx = 0` and `2cosx+1 = 0` `=>sinx = 0` and `cosx = -1/2` `=> x = npi ` and `x = 2npi+-(2pi)/3` So, option `C` is the correct option. |
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| 170. |
In triangle ABC, angle A is greater than angle B. If the measure of angles A and B satisfy the equation `3sinx-4sin^3x-k=0`. (A) `pi/3` (B) `pi/2` (C) `(2pi)/3` (D) `(5pi)/6` |
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Answer» `A>B` `3sinx-4sin^3x-k=0` `sin3x=k` `sin3A=k` `sin3B=k` `sin3A-sin3B=0` `2cos((3A+3B)/2)sin((3A-3B)/2)=0` `A+B=60` Now, we know `A+B+C=180` `C=120`. |
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| 171. |
The equality `sin A+ sin 2A + sin 3A = 3` holds for some real value of A. |
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Answer» False Given that, `sinA+sin2A+sin3A=3` It is possible only if `sinA, sin2A, sin3A` each has a value one because maximum value of `sinA` is a certain angle is 1. Which is not possible because angle are different. |
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| 172. |
Number of solution of the equation `cos^4 2x+2sin^2 2x=17(cosx+sinx)^8,0 |
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Answer» `cos^4 2x+2sin^2 2x=17(cosx+sinx)^8` `(cos^2 2x)^2+2(sin^2 2x)=17((cosx+sinx)^2)^4` `(1-sin^2 2x)^2+2sin^2 2x=17(cos^2x+sin^2x+2sinxcosx)^4` `1+sin^4 2x-2sin^2 2x+2sin^2 2x=17(1+sin2x)^4` `1+sin^4 2x=17(1+sin2x)^4` `1+y^2=1(1+y)^4` Let` sin2x=y` `1+y^4=17(1+y)^2(1+y)^2` `(1+y^4)=17(1+2y+y^2)` `1+y^4+17(1+4y^2+y^4+2y^2+4y^2+4y)` `16y^4+102y^2+68y^3+68y+16=0` `4y^2+17y^3+28y^2+17y+8=0` `y=-1/2` `sin2x=-1/2` `2x=3pi+pi/6` `x=7/12pi,11/12pi,19/12pi,23/12pi`. |
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| 173. |
If `"cosec"x=1+cotx`, then `x=2npi,2npi+(pi)/(2)` |
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Answer» True Given that, `" ""cosec"x=1+cotx` `rArr" "(1)/(sinx)=1+(cosx)/(sinx)rArr(1)/(sinx)=(sinx+cosx)/(sinx)` `rArr" "sinx+cosx=1` `rArr" "(1)/(sqrt(2))*sinx+(1)/(sqrt(2))*cosx=(1)/(sqrt(2))` `rArr" "sin""(pi)/(4)sinx+cosxcos""(pi)/(4)=(1)/(sqrt(2))` `rArr" "cos(x-(pi)/(4))=cos""(pi)/(4)` `therefore" "x-(pi)/(4)=2npipm(pi)/(4)` For positive sign, `" "x=2npi+(pi)/(4)+(pi)/(4)=2npi+(pi)/(2)` For negative sign, `" "x=2npi-(pi)/(4)+(pi)/(4)=2npi` |
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| 174. |
Prove that`cos(2pi)/(15)cos(4pi)/(15)cos(8pi)/(15)cos(14pi)/(15)=1/(16)` |
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Answer» `cos2/15picos4/15picos8/15picos14/15pi=1/16` `=-cospi/15cos2/15picos4/15picos8/15pi` `=cosAcos2Acos2^2A....cosA^nA=(sin2^nA)/(2^nsinA)` `A=pi/15` `=-cosAcos2Aco2^2Acos2^3A` `=(-sin2^4(16/15pi))/(2^4sin(pi/16))` `=(-sin(16/15pi))/(16sin(pi/15))` `=-[-sinpi/15]/(16sinpi/15)` `cos2/15picos4/15picos8/15picos14/15pi=1/16`. |
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| 175. |
`sin 10^@` is greater than `cos 10^@`. |
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Answer» False `" "sin10^(@)=sin(90^(@)-80^(@))` `" "sin10^(@)=cos80^(@)` `therefore" "cos80^(@)ltcos10^(@)` Hence, `" "sin10^(@)ltcos10^(@)` |
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| 176. |
One value of `theta` which satisfies the equation `sin^(4)theta-2sin^(2)theta-1` lies between 0 and `2pi`. |
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Answer» False Given equation, `" "sin^(4)theta-2sin^(2)theta-1=0` ltBrgt `rArr" "sin^(2)theta=(2pmsqrt(4+4))/(2)` `rArr" "sin^(2)theta=(2pm2sqrt(2))/(2)` `rArr" "sin^(2)theta=(1+sqrt(2))or (1-sqrt(2))rArr-1lesinthetale1` `rArr" "sin^(2)thetale1` `because " "sin^(2)theta=sqrt(2+1)or (1-sqrt(2))` which is not possible. |
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| 177. |
Prove that :`16 cos 2pi/15 cos 4pi/15 cos 8pi/15 cos16pi/15 = 1` |
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Answer» Here, we will use, `2sinxcosx = sin2x` Now, `L.H.S. = 16cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)cos((16pi)/15)` `=16cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)cos(pi+pi/15)` `=16cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)(-cos(pi/15))` `=-16cos(pi/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)` `=8/sin(pi/15)*2sin(pi/15)cos(pi/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)` `=8/sin(pi/15)*sin((2pi)/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)` `=4/sin(pi/15)*2sin((2pi)/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)` `=4/sin(pi/15)*sin((4pi)/15)cos((4pi)/15)cos((8pi)/15)` `=2/sin(pi/15)*2sin((4pi)/15)cos((4pi)/15)cos((8pi)/15)` `=2/sin(pi/15)*sin((8pi)/15)cos((8pi)/15)` `=sin((16pi)/15)/(sin((pi)/15))` `=sin(pi+pi/15)/(sin((pi)/15))` `=(sin((pi)/15))/(sin((pi)/15))` `=1 = R.H.S.` |
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| 178. |
`cos""(2pi)/(15)cos""(4pi)/(15)cos""(8pi)/(15)cos""(16pi)/(15)=(1)/(16)` |
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Answer» True `" "LHS=cos""(2pi)/(15)cos""(4pi)/(15)cos""(8pi)/(15)cos""(16pi)/(15)` `" "=cos24^(@)cos48^(@)cos96^(@)cos192^(@)` `" "=(1)/(16sin24^(@))[(2sin24^(@)cos24^(@))(2cos48^(@))(2cos96^(@))(2cos192^(@))]` ltBrgt `" "=(1)/(16sin24^(@))[2sin48^(@)cos48^(@)(2cos96^(@))(2cos192^(@))]` `" "=(1)/(16sin24^(@))` `[(2sin96^(@)cos96^(@))(2cos192^(@))]` ` " "=(1)/(16sin24^(@))[2sin192^(@)cos192^(@)]` `" "=(1)/(16sin24^(@))sin384^(@)=(sin(360^(@)+24^(@)))/(16sin24^(@))` `" "=(1)/(16)=RHS" "` Hence proved |
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| 179. |
`(cos21^(@) -sin21^(@))/(cos21^(@) + sin21^(@))`A)`tan21^(@)`B)`cot66^(@)`C)`tan42^(@)`D)`cot42^(@)`A. `tan21^(@)`B. `cot66^(@)`C. `tan42^(@)`D. `cot42^(@)` |
| Answer» Correct Answer - B | |
| 180. |
If `b >1,sint >0,cost >0a n d(log)_b(sint)=x ,t h e n(log)_b(cost)`is equal to`1/2(log)_b(a-b^(2x))`(b) `2log(1-b^(x/2))``(log)_bsqrt(1-b^(2x))`(d) `sqrt(1-x^2)` |
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Answer» `log_b(sint)=x` `sint=b^x` `sin^2t+cos^2t=1` `cos^2t=1-sin^2t` `cos^2t=1-(b^x)^2` `cost=sqrt(1-c^(2x))` `log_acost=log_bsqrt(1-b^(2x))` Option C is correct. |
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| 181. |
If `P`is a point on the altitude AD of the triangle ABC such the `/_C B P=B/3,`then AP is equal to`2asinC/3`(b) `2bsinC/3``2csinB/3`(d) `2csinC/3` |
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Answer» We can create a diagram using the given details. Please refer to video to see the diagram. From the diagram, `/_ABP = (2B)/3 and /_APB = 90+B/3` In `Delta APB`, `(AP)/(sin((2B)/3)) = c/(sin(90+B/3))` `=>(AP)/(2sin(B/3)cos(B/3)) = c/(cos(B/3))` `=>AP = 2csin(B/3)` |
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| 182. |
If `bgt1, sintgt0,costgt0andlog_b(sint)=x," then "log_b(cost)` is equal toA. `1/2log_b(1-b^(2x))`B. `2log(1-b^(x//2))`C. `log_bsqrt(1-b^(2x))`D. `sqrt(1-x^2)` |
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Answer» Correct Answer - A::C `log_bsin_t=xorsint=b^x` Let `log_b(cost)=y,then b^y=cost` `or b^(2y)=cos^2t=1-sin^2t=1-b^(2x)` `or 2y=log_b(1-b^(2x))` `or y=1/3log_b(1-b^(2x))=log_bsqrt(1-b^(2x))` |
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| 183. |
Prove that: `2tan^(-1)1/2+tan^(-1)1/7=tan^(-1)(31)/(17)`A. `tan^(-1)((31)/(17))`B. `tan^(-1)((25)/(21))`C. `tan^(-1)((17)/(3))`D. `tan^(-1)((21)/(25))` |
| Answer» Correct Answer - A | |
| 184. |
If `(1+sint)(1+cost)=5/4`then find the value of `(1-sint)(1-cost)dot` |
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Answer» `(1+sint)(1+cost) = 5/4` `=>1+sint+cost+sintcost = 5/4` `=>(sint+cost) = 1/4-sintcost->(1)` `=>(sint+cost)^2 = (1/4-sintcost)^2` `=>sin^2t+cos^2t+2sintcost = 1/16+(sintcost)^2 -(sintcost)/2` `=>1+sin2t = 1/16+(sin^2 2t)/4 - (sin2t)/4` `=>16+16sin2t = 1+4sin^2 2t -4sin2t` `=>4sin^2 2t -20sin2t -15 = 0` `:. sin 2t = (20+-sqrt(400-4(-15)(4)))/8 = (5-2sqrt10)/2` Here, we have rejected the value `(5+2sqrt10)/2` as it becomes more than `1`. Now, `(1-sint)(1-cost) = 1-(sint+cost)+sintcost` `=1-(1/4-sintcost)+sintcost` `=3/4+sin2t` `=3/4+5/2-sqrt10` `=13/4-sqrt10` `:. (1-sint)(1-cost) = 13/4 - sqrt10.` |
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| 185. |
Let `P={theta:sintheta-costheta=sqrt2costheta}and Q={theta: sintheta+costheta=sqrt2sintheta}` be two sets. ThenA. `psubQ and Q-P=phi`B. `QcancelsubP`C. `PcancelsubQ`D. P=Q |
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Answer» Correct Answer - D In set P, `sintheta=(sqrt2+1)costheta` `or tantheta=sqrt2+1` In set Q, `(sqrt2-1)sintheta=costheta` `or tan thetaa=1/(sqrt2-1)=sqrt2+1` `rArr P=Q` |
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| 186. |
If `cos(alpha+beta)=(4)/(5) and sin(alpha-beta)=(5)/(13)` , where `alpha` lie between 0 and ` (pi)/(4)`, then find that value of `tan2alpha`. |
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Answer» Given that, ` cos(alpha+beta)=(4)/(5) and sin(alpha-beta)=(5)/(13)` `rArr" "sin(alpha+beta)=sqrt(1-(16)/(25))=sqrt((9)/(25))= pm (3)/(5)` `therefore" "sin(alpha+beta)=(3)/(5) ` and `" "cos(alpha-beta)=sqrt(1-(25)/(169))=sqrt((144)/(169))=pm(12)/(13) ` `" " cos(alpha-beta)=(12)/(13)` Now, `tan(alpha+beta)=(sin(alpha+beta))/(cos(alpha+beta))" "`[since, `alpha` lies between 0 and ` (pi)/(4)` `((3)/(5))/((4)/(5))=(3)/(4)` and `" "tan(alpha-beta)=(sin(alpha-beta))/(cos(alpha-beta))=((5)/(13))/((12)/(13))=(5)/(12)` `therefore" "tan2alpha= tan(alpha+beta+alpha-beta)` ` " "=(tan(alpha+beta)+ tan (alpha-beta))/(1- tan(alpha+ beta)*tan(alpha-beta))" "[becausetan(xpmy)=(tanxpmtany)/(1pmtanx*tany)] ` `" "=((3)/(4)+(5)/(12))/(1-(3)/(4)*(5)/(12))=((9+5)/(12))/((16-5)/(16))=(14xx16)/(12xx11)=(56)/(33)` |
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| 187. |
The set of values of `lambda in R`such that `sin^2theta+costheta=lambdacos^2theta`holds for some `theta,`is`(-oo,1]`(b) (`-oo,-1]``varphi`(d) `[-1,oo`) |
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Answer» `sin^2theta+costheta=lambdacos^2theta` `sin^2theta/cos^2theta+cos^2theta/cos^2theta=lambdacos^2theta/cos^2theta` `tan^2theta+sectheta=lambda` `sec^2theta-1+sectheta-lambda=0` `sec^2theta+sectheta-(lambda+1)=0` `sectheta=(-1pmsqrt(1+4(lambda+1)))/2` `=(-1pmsqrt(4lambda+5))/2` `4lambda+5>=0` `lambda>=-5/4` `(-1pmsqrt(4lambda+5))/2>=1` `sqrt(4lambda+5)>=3` `4lambda+5>=9` `4lambda>=4` `lambda>=1-(1)` `lambda<=-1-(2)` `lambda in[-1,oo)` Option D is correct. |
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| 188. |
Th range of k for which the inequaliity `kcos^2x-kcosx+1>=0 AA x in(-oo,oo) is`A. `klt- 1/2`B. `kgt4`C. `-1/2lekle4`D. `1/2lekle5` |
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Answer» Correct Answer - C `kcos^2x-kcos+1le0AAx in(-oo,oo)` `rArrk(cos^2x-cosx)+1ge0` But `cos^2x-cosx=(cosx-1/2)^2-1/4` `rArr -1/4lecos^2x-cosxle2` Now, from (i), we get `2k+1ge0rArr kge-1/2` Also, `-k/4+1ge0` `rArr kle4` `rArr -1/2lekle4` |
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| 189. |
If `cos^2x-(c-1)cosx+2cgeq6`for every `x in R ,`then the true set of values of `c`is`(2,oo)`(b) `(4,oo)`(c) `(-oo,-2)`(d) `(-oo,-4)` |
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Answer» `cos^2x-(c-1)x + 2c ge 6` `=>cos^2x-(c-1)cosx+2c-6 ge 0` `=>cos^2x-(c-3)cosx - 2cosx+(2c-6) ge 0` `=>cosx(cosx-(c-3))-2(cosx-(c-3)) ge 0` `=>(cosx-(c-3))(cosx-2) ge 0` As, `cosx - 2` can not be greater than or equal to `2`. `:. cosx le c-3` `=>1 le c-3` `=>4 le c` `:. c in [4,oo).` |
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| 190. |
The set of values of `lambda in R`such that `sin^2theta+costheta=lambdacos^2theta`holds for some `theta,`is`(-oo,1]`(b) (`-oo,-1]``varphi`(d) `[-1,oo`)A. `(-oo,1]`B. `(-oo,-1]`C. `phi`D. `[-1,oo)` |
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Answer» Correct Answer - D We have `tan^2theta+sectheta=lambda` `rArr sec^2theta+sectheta-(lambda+1)=0` `rArr sectheta=-(-1pmsqrt(4lambda+5))/2le-1` For real `sectheta`, `4lambda+5ge0rArrlambdage(-5)/4 ...(i)` Also,` secthetage or secthetale-1` `rArr(-1pmsqrt(4lambda+5))/2ge1or (-1pmsqrt(4lambda+5))/2ge-1` `rArr4alambda+5ge9or4lambda+5ge1` `rArrlambdage1orlambdage-1 ...(2)` `:. " From(1) and (2), we get "lambdain[-1,oo)`. |
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| 191. |
If the median AD of triangle ABC makes an angle `pi/4`with the side BC, then find the value of `|cotB-cotC|dot` |
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Answer» By m-n theoram `(m+n)cottheta=mcotalpha-cotbeta` `(m+n)cottheta=ncotbeta-mcotC` `(BD+DC)cot(x/4)=DC cot B-BD cotC` `BD+DC=DC cot B-BD cotC` `2BD=BD cot B- BD cot C` `|cotB-cotC|=2`. |
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| 192. |
If `cos^2x-(c-1)cosx+2cgeq6`for every `x in R ,`then the true set of values of `c`is`(2,oo)`(b) `(4,oo)`(c) `(-oo,-2)`(d) `(-oo,-4)`A. `[2,oo)`B. `[4,oo)`C. (-oo,-2]`D. (-oo,-4]` |
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Answer» Correct Answer - B `cos^2x-(c-1)cosx+2cge6AAx in R` `rArr (cosx-2)(cosx-AAx in R)` `rArrcosxlec-3` `rArrc-3ge1` `rArrc in[4,oo)` |
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| 193. |
If `a,b,c inR` then prove that `sec^2theta=(bc+ca+ab)/(a^2+b^2+c^2)`only if a = b = c. |
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Answer» We know that `sec^2thetage1` `:. (bc+ca+ab)/(a^2+b^2+c^2)ge 1` `rArr bc+ca+abgea^2+b^2+c^2` `rArra^2+b^2+c^2-ab-bc-cale0` `rArr(a-b)^2+(b-c)^2+(c-a)^2le0` `rArra-b=0,b-c=0" and "c-a=0` `rArra=b=c` |
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| 194. |
Find the values of a for which `a^2-6sinx-5ale0,Aax inR`. |
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Answer» `a^2-6sinx-5a<=0` `(a^2-5a)/6<=sinx` `(a^2-5a)/6<=-1` `a^2-5a<=-6` `a^2-5a+6<=0` `(a-3)(a-2)<=0` `a in[2,3]`. |
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| 195. |
General solution of `sin^(3)x+cos^(2)x+sinxcosx=1` isA. `npi, npipm(pi)/(2), ninZ`B. `2npi, 2npipm(pi)/(2), ninZ`C. `npi, npipm(pi)/(4), ninZ`D. `2npi, 2npipm(pi)/(4), ninZ` |
| Answer» Correct Answer - B | |
| 196. |
The sides of a triangle are in the ratio `1: sqrt3:2.` Then the angles are in the ratioA. `1:sqrt(2):3`B. `1:sqrt(3):2`C. `sqrt(2):sqrt(3):3`D. `1:sqrt(3):3` |
| Answer» Correct Answer - B | |
| 197. |
The angles of a triangle are in the ratio `1:2:3` , then the sides of a triangle are in the ratioA. `1:sqrt(3):2`B. `2:sqrt(3):1`C. `sqrt(3):2:1`D. `3:2:1` |
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Answer» Correct Answer - A Given, `(angleA)/(1)=(angleB)/(2)=(angleC)/(3)=(A+B+C)/(1+2+3)=(180^(@))/(6)=30^(@)` `thereforeangleA=30^(@),angleB=60^(@),angleC=90^(@)` By sine rule , `a:b:c=sinA:sinB:sinC` `=sin30^(@):sin60^(@):sin90^(@)` `=(1)/(2):(sqrt(3))/(2):1=1:sqrt(3):2` |
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| 198. |
If the equation `cof^4x-2cos e c^2x+a^2=0`has at least one solution, then the sum of all possible integral valuesof a is equal toa. 4b. 3 c.2 d. 0A. 4B. 3C. 2D. 0 |
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Answer» Correct Answer - D `cot^4x-2(1+cot^2x)+a^2=0` `or cot^4x-2cot^2+a^2-2=0` `or (cot^2x-1)^2=3-a^2` To have at least one solution, `3-a^2ge0` `or a^2-3le0` `a in[-sqrt3,sqrt3]` Integral values are -1,0,1, therefore, the sum is 0. |
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| 199. |
`(sin(pi+x)cos(pi/2+x)tan((3pi)/2-x)cot(2pi-x))/(sin(2pi-x)cos(2pi+x)c o s e c(-x)sin((3pi)/2-x))=1` |
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Answer» `=(sin(pi+x)cos(pi/2+x)tan(3/2pi-x)cot(2pi-x))/(sin(2pi-x)cos(2pi+x)cosec(-x)sin(3/2pi-x))` `=(-sinx(-sinx)cotx(-cotx))/(-sinxcosx-cosecx-cosx)` `=(sin^2x*cos^2x)/(cos^2xsin^2x)` `=1`. |
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| 200. |
Let `alphaa n dbeta`be any two positive values of `x`for which `2cosx ,|cosx|,`and `1-3cos^2x`are in G.P. The minimum value of `|alpha-beta|`is`pi/3`(b) `pi/4`(c) `pi/2`(d) none of these |
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Answer» As `2cosx, |cosx| and 1-3cos^2x` are in G.P. `:. |cosx|^2 = 2cosx(1-3cos^2x)` `=>cos^2x = 2cosx-6cos^3x` `=>6cos^3x+cos^2x-2cosx = 0` `=>cosx(6cos^2x+cosx-2) = 0` `=>cosx(6cos^2x+4cosx-3cosx-2) = 0` `=>cosx(2cosx-1)(3cosx+2) = 0` `=>cosx = 0 or 2cosx-1 = 0 or 3cosx+2 = 0` `=> cosx = 0 or cos x = 1/2 or cosx = -2/3` `=> x = pi/2 or x = pi/3 or x = cos^-1(2/3)` But, as `alpha and beta` are positive, so we will take inly positive values. `:. alpha = pi/2 and beta = pi/3` `|alpha-beta| = pi/2-pi/3 = pi/6.` So, option `d` is the correct option. |
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