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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.
| 201. |
If `2tan^(-1)(cosx)=tan^(-1)(2cosecx)`, then x=A. `(pi)/(4)`B. `(-pi)/(4)`C. `(pi)/(3)`D. `(pi)/(6)` |
| Answer» Correct Answer - A | |
| 202. |
Minimum value of `y=256 sin^2x+324cosec^2x AA x inR` isA. 432B. 504C. 576D. 776 |
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Answer» Correct Answer - C `y=256sin^2x+324 cosec^2x` `=(16sinx-18cosecx)^2+576ge576` |
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| 203. |
If `2tan^(-1)(cosx)=tan^(-1)(2cosecx)`, then sinx +cosx is equal toA. `2sqrt(2)`B. `sqrt2`C. `(1)/(sqrt(2)`D. `(1)/(2)` |
| Answer» Correct Answer - B | |
| 204. |
The variable `x`satisfying the equation `|sinxcosx|+sqrt(2+tan^2+cot^2x)=sqrt(3)`belongs to the interval`[0,pi/3]`(b) `(pi/3,pi/3)`(c) `[(3pi)/4,pi]`(d) none-existentA. `[0,pi/3]`B. `(pi/3pi/2)`C. `[(3pi)/4,pi)`D. None of these |
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Answer» Correct Answer - D `abs(sinxcosx)+abs(tanx+cotx)=sqrt3` `or abs(sinxcosx)+1/abs(sinxcosx)=sqrt3` But `abs(sinxcosx)+1/(abs(sinxcosx))ge2` hence, no solution. |
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| 205. |
If a and b are positive quantities, `(a gt b)` find minimum positive value of `(a sectheta- b tantheta)`A. 2abB. `sqrt(a^2-b^2)`C. a-bD. `sqrt(a^2+b^2)` |
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Answer» Correct Answer - B Let `s=asectheta-btantheta` `or btantheta+s=asectheta` `or (a^2-b^2)tan^2theta-2bstantheta+(a^2-s^2)=0` For `tantheta" to be real " 4b^2s^2-4(a^2-b^2)(a^2-s^2)ge0` `or a^2s^2gea^2(a^2-b^2)` `or sgesqrt(a^2-b^2)` Therefore, the minimum value of s is `sqrt(A^2-B^2)`. |
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| 206. |
General solution of `tanx+cotx=2` isA. `(npi)/(2)+(-1)^(n)(pi)/(4), ninZ`B. `(npi)/(2)+(-1)^(n)(pi)/(2), ninZ`C. `npi+(-1)^(n)(pi)/(4), ninZ`D. `npi+(-1)^(n)(pi)/(2), ninZ` |
| Answer» Correct Answer - A | |
| 207. |
Find the general solution of : `cotx+tanx=2cosecx`.A. `2npipm(2pi)/(3), ninZ`B. `2npipm(4pi)/(3), ninZ`C. `2npipm(5pi)/(3), ninZ`D. `2npipm(pi)/(3), ninZ` |
| Answer» Correct Answer - D | |
| 208. |
In triangle `A B C ,(sinA+sinB+sinC)/(sinA+sinB-sinC)`is equal to |
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Answer» `A+B+C=theta` `sin((A+B)/2)=cos(C/2)` `D->sinA+sinB-sinC` `=2sin((A+B)/2)cos((A-B)/2)-sinC` `=2cos(C/2)cos((A-B)/2)-2sin(C/2)cos(C/2)` `=2cos(C/2)[cos((A-B)/2)-sin(C/2)]` `=2cos(C/2)[cos((A-B)/2)-cos((A+B)/2)]` `=2cos(C/2)[2sin(A/2)sin(B/2)]` `=4sinn(A/2)sin(B/2)cos(C/2)` `N->sinA+sinB+sinC` `=2sin((A+B)/2)cos((A-B)/2)+2sin(C/2)cos(C/2)` `=2cos(C/2)cos((A-B)/2)+2sin(C/2)cos(C/2)` `=2cos(C/2)[cos((A-B)/2)+cos((A+B)/2)]` `=2cos(C/2)[2cos(A/2)cos(B/2)]` `=4cos(A/2)cos(B/2)cos(C/2)` `N/D=(4cos(A/2)cos(B/2)cos(C/2))/(4sin(A/2)sin(B/2)cos(C/2)` `=cot(A/2)cot(B/2)`. |
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| 209. |
If `y=(sinx+cosecx)^2+(cosx+secx)^2` then the minimum value of `y,AAx in R`, isA. 7B. 3C. 9D. 0 |
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Answer» Correct Answer - C `y=(sin^2x-cos^2x0+2(sinxcosecx+cosxsecx)+sec^2x+cosec^2x` `=5+2+tan^2x+cot^2x` `=7+(tanx-cotx)^2+2` `:. t_("min")=9` |
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| 210. |
The minimum value of the function `f(x)=sinx/(sqrt(1-cos^2x))+cosx/sqrt(1-sin^2x)+tanx/sqrt(1-sec^2x-1)+cotx/sqrt(1-cosec^2x-1)` whenever it is defined isA. 4B. -2C. 0D. 2 |
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Answer» Correct Answer - B `f(x) =sinx/sqrt(1-cos^2x)+(cosecx)/sqrt(1-sin^2x)` `+.tanx/sqrt(sec^2x-1)+cotx/sqrt(cosec^2x-1)` `=sinx/abssinx+cosx/abscosx+tanx/abstanx+cotx/abscotx` `={{:(4","" "x in"1st quadrant"),(-2","" "x in "2nd quadrant"),(0","" "x in " 3rd quadrant"),(-2"," " "x in "4th quadrant"):}` `f(x)_("min")=-2` |
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| 211. |
Prove that `(sinx-cosx+1)/(sinx+cos-1)=secx+tanx`. |
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Answer» `(sinx-cosx+1)/(sinx+cosx-1)=((sin^2x)-(cosx-1)^2)/((sinx+cosx-1)^2)` `=(2cosx-2cos^2x)/(2+2sinxcosx-2cosx-2sinx)` `=(2cosx(1-cosx))/(2(1-sinx)(1-cosx))` `=(cosx)/(1-sinx)=((1+sinx))/cosx=secx+tanx` |
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| 212. |
If `tanx=b/a ,`then `sqrt((a+b)/(a-b))+sqrt((a-b)/(a+b))`is equal to(a)`2sinx//sqrt(sin2x)`(b) `2cosx//sqrt(cos2x)`(c)`2cosx//sqrt(sin2x)`(d) `2sinx//sqrt(cos2x)` |
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Answer» `sqrt((a+b)/(a-b)*(a+b)/(a+b))+sqrt((a-b)/(a+b)*(a-b)/(a+b))` `sqrt((a+b)^2/(a^2-b^2))+sqrt((a-b)^2/(a^2-b^2))` `(a+b+a-b)/sqrt(a^2-b^2` `(2a)/sqrt(a^2-b^2)` `(2a)/(asqrt(1-b^2/a^2))` `2/sqrt(1-b^2/a^2)` `2/sqrt(1-tan^2x)` `2/(1-(sin^2x)/cos^2x)` `2/(((cos^2x-sin^2x)^(1/2))/cosx)` `(2cosx)/sqrt(cos2x)`. |
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| 213. |
The value of `(2sinx)/(sin3x)+(tanx)/(tan3x)`is________. |
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Answer» `(2sinx)/(sin3x) +tanx/(tan3x)` `=(2sinx)/(3sinx-4sin^3x) +tanx/((3tanx - tan^3x)/(1-3tan^2x))` `=2/(3-4sin^2x) +(1-3tan^2x)/(3-tan^2x)` `=2/(3-4sin^2x) +(1-3sin^2x/cos^2x)/(3-sin^2x/cos^2x)` `=2/(3-4sin^2x) +(1-3sin^2x/(1-sin^2x))/(3-sin^2x/(1-sin^2x))` `=2/(3-4sin^2x) +(1-sin^2x - 3sin^2x)/(3-3sin^2x-sin^2x)` `=2/(3-4sin^2x) +(1-4sin^2x)/(3-4sin^2x)` `=(2+1-4sin^2x) /(3-4sin^2x) ` `=(3-4sin^2x) /(3-4sin^2x) ` `=1` `:. (2sinx)/(sin3x) +tanx/(tan3x) = 1.` |
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| 214. |
Which of the following is/are correct ?(a) `(tanx)^(In (cosx))lt(cotx)^(In(cosx))AAx in(pi/4,pi/2 )`(b)`(sinx)^(In (secx))gt(cosx)^(In(secx))AA x in(0,pi/4)`(c ) `(sec. pi/3)^(In (tanx))gt(sec. pi/3)^(In(cosx))AA x in (pi/4,pi/2)`(d) `(1/2)^(In(sinx))gt(3/4)^(In(sinx))AA x in(0,pi/2)` |
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Answer» (a) For `pi/4lt xlt pi/2,tanxgtcotx` but In `(cosx)lt0` `:. (tanx)^(In(secx))lt(cosx)^(In(secx))` Hence, the statement is correct. (b) For `x in (0,pi/4),cosxgt sinx` but In (secx) gt 0 (as sec x gt1) `:. (sinx)^(In(secx))lt(cosx)^(In(sexx))` Hence, the statement is not correct (c ) For `x in (pi/4,pi/2),tanxgt1` `:. In (tanx)gt0` but In`(cosx)lt0` `:. (sec. pi/3)^(In(tanx))gt (sec. pi/3)^(In(cosx))` Hence, the statement is correct. (d) For `x in(0,pi/2)` In `(sinx)lt0` as `1/2lt3/4` `:. (1/2)^(In(sinx))gt(3/4)^(In(sinx))` Hence, the statement is correct. |
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| 215. |
Given a triangle `A B C`with sides a=7, b=8 and c=5. Find the value of expression `(sinA+sinB+sinC)(cot A/2+cotB/2+cotC/2)` |
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Answer» `(a/(2R)+b/(2R)+c/(2R))((s(s-a))/ /_+(s(s-b))/ /_+(s(s-c))/ /_)` `((a+b+c)/(2R))((s(3s-a+b+c))/ /_)` `((2S)/(2R))((S*S)/ /_)` `((2S)/(2R))(s^2/ /_)` `s^3/(R/_)` `4/(abc)((a+b+c)/2)^3` `4/(7*8*5)((7+8+5)/2)^3` `100/7`. |
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| 216. |
If the inequality `sin^2x+acosx+a^2>1+cosx`holds for any `x in R ,`then the largest negative integral value of a is`-4`(b) -3(c) `-2`(d) `-1` |
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Answer» `sin^2x+acosx +a^2 gt 1+cosx` for `x in R` As `x in R`, we can put, `x = 0` `=>(sin0)^2+acos0 +a^2 gt 1+cos0` `=>a+a^2 gt 2` `=>a^2+a -2 gt 0` `=> (a+2)(a-1) gt 0` `:. a in (-oo,-2) uu (1,oo)` So, largest negative integral value of `a` will be `-3` as `-2` is not included. |
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| 217. |
The smallest positive value of `x`(in radians) satisfying the equation `(log)_(cosx)((sqrt(3))/2sinx)=2-(log)_(secx)(tanx)`is(a)`pi/(12)`(b) `pi/6`(c) `pi/4`(d) `pi/3` |
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Answer» `log_(cosx)(sqrt3/2sinx)=-log_(secx)(tanx)` `log_(secx)(tanx)=(lntanx)/(lnsecx)=(lntanx)/(-lncosx)=-log_(cosx) tanx` `log_cosxx(sqrt3/2sinx)=2+log_cosx(tanx)=log_cosx tanx` `log_cosx(sqrt3/2cosx)=2` `sqrt3/2cosx=cos^2x` `cosx[cosx-sqrt3/2]=0` `cosx=sqrt3/2` `x=pi/6`. |
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| 218. |
`sinxtanx-1=tanx-sinx`A. `2npipm(3pi)/(4), npi+(-1)^(n)(pi)/(2), ninZ`B. `npipm(3pi)/(4), npi+(-1)^(n)(pi)/(2), ninZ`C. `2npi+(pi)/(4), npi+(-1)^(n)(pi)/(2), ninZ`D. `npi+(pi)/(4), npi+(-1)^(n)(pi)/(2), ninZ` |
| Answer» Correct Answer - B | |
| 219. |
In `DeltaABC, sin(A-B)/sin(A+B)=`A. `(a^(2)-b^(2))/(2c^(2))`B. `(b^(2)-c^(2))/(2a^(2))`C. `(a^(2)-b^(2))/(c^(2))`D. `(b^(2)-a^(2))/(c^(2))` |
| Answer» Correct Answer - C | |
| 220. |
In `triangleABC, a^(2)sin(B-C)=`A. `(b^(2)-c^(2))sinA`B. `(c^(2)-b^(2))sinA`C. `(2b^(2)-c^(2))sinA`D. `2(c^(2)-b^(2))sinA` |
| Answer» Correct Answer - A | |
| 221. |
In any triangle `A B C ,`prove that: ` a^3sin(B-C)+b^3sin(C-A)+c^3sin(A-B)=0`A. `a^(2)b^(2)+b^(2)c^(2)+a^(2)c^(2)`B. `a+b+c`C. `a^(2)+b^(2)+c^(2)`D. `0` |
| Answer» Correct Answer - D | |
| 222. |
General solution of `3sin^(2)x+10cosx-6=0` isA. `2npipmcos^(-1)((1)/(3)), ninZ`B. `2npipmcos^(-1)((2)/(3)), ninZ`C. `npipmcos^(-1)((1)/(3)), ninZ`D. `npipmcos^(-1)((2)/(3)), ninZ` |
| Answer» Correct Answer - A | |
| 223. |
Let A and B denote the statementsA: `"cos"a+"cos"b+"cos"g=""0`B : `"sin"a+"sin"b+"sin"g=""0`If `cos(b g)""+""cos(g a)""+""cos(a b)""=""""3//2`,then(1) A is true and B is false(2) A is false and B is true(3) both A and B are true(4) both A and Bare false |
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Answer» `cos(beta- gamma) + cos(gamma- alpha) + cos(alpha- beta) = -3/2` `2cos(beta- gamma) + 2 cos ( gamma- alpha) + 2 cos( alpha - beta) + 3= 0` `2 cos( beta- gamma) + 2cos( gamma - alpha) + 2cos(alpha- beta) + sin^2 alpha + cos^2 alpha + sin^2 beta + cos^2 beta + sin^2 gamma + cos^2 gamma = 0` now, using identity, `(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc+ ac)` `(cos alpha + cos beta + cos gamma)^2 + (sin alpha + sin beta + sin gamma)^2 = 0` `= 0+ 0 = 0` so, option 3 is correct |
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| 224. |
`int_0^pi[cotx]dx`, where [.] denotes the greatest integerfunction, is equal to(1) `pi//2`(2) 1(3) `1`(4) `pi//2` |
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Answer» `I = int_0^pi [ cot x] dx` `= int_0^pi [ cot(pi- x)] dx` `= int_0^pi [cot(pi-x)] dx` `= int _0^ pi [ - cot x] dx` `(i) + (ii)` `2 I= int_0^pi [ cot x] + int_0^pi [-cot x] dx` `[x] + [-x] = -1 x !in z` `= 0 x !in z` `2I= int_0^pi -1 dx` `2I = -1 xx pi ` `I = - pi/2` option 4 is correct |
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| 225. |
For a regular polygon, let r and R be the radii of the inscribed andthe circumscribed circles. A false statement among the following isThere is a regular polygon with `r/R=1/(sqrt(2))`(17)There is aregular polygon with `r/R=2/3`(30)There is aregular polygon with `r/R=(sqrt(3))/2`(47)There is aregular polygon with `r/R=1/2`(60) |
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Answer» `cos pi/n = r/R` `n=1; r/R= -1` `n=2 ; r/R=0` `n=3 ; r/R=1/2` `n=4 ; r/R = 1/sqrt2` `n=5 ; r/R= cos 36^@` `n=6 ; r/R = sqrt3/2` option 2 is correct |
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| 226. |
The polar co-ordinates of the point whose cartesian co-ordinates are `(0, -2)`, areA. `(-2, (pi)/(2))`B. `(-2, (3pi)/(2))`C. `(2, (pi)/(2))`D. `(2, (3pi)/(2))` |
| Answer» Correct Answer - D | |
| 227. |
The polar co-ordinates of the point whose cartesian co-ordinates are `(5, 0)`, areA. `(-5, 0)`B. `(-5, pi)`C. `(5, 0)`D. `(5, pi)` |
| Answer» Correct Answer - C | |
| 228. |
The polar co-ordinates of the point whose cartesian co-ordinates are `(1, sqrt(3))`, areA. `(2,, -(pi)/(3))`B. `(2, (pi)/(3))`C. `(2, (4pi)/(3))`D. `(2, (2pi)/(3))` |
| Answer» Correct Answer - B | |
| 229. |
The cartesian co-ordinates of a point, whose polar co-ordinates are `(4, (pi)/(2))` areA. `(0, 2)`B. `(0, 4)`C. `(2, 0)`D. `(4, 0)` |
| Answer» Correct Answer - B | |
| 230. |
The cartesian co-ordinates of a point, whose polar co-ordinates are `((1)/(2), 210^(@))` areA. `((-sqrt(3))/(4), (1)/(4))`B. `((sqrt(3))/(4), (-1)/(4))`C. `((-sqrt(3))/(4), (-1)/(4))`D. `((sqrt(3))/(4), (1)/(4))` |
| Answer» Correct Answer - C | |
| 231. |
The polar coordinates of the point whose cartesian coodinates are `(-1/sqrt(2),-1/sqrt(2))` areA. `(1, (pi)/(4))`B. `(1, (3pi)/(4))`C. `(1, (7pi)/(4))`D. `(1, (5pi)/(4))` |
| Answer» Correct Answer - D | |
| 232. |
The cartesian co-ordinates of point, which polar co-cordinats=es are `(2, (pi)/(4))` areA. `(sqrt(2), sqrt(2))`B. `(2, 2)`C. `((1)/(sqrt(2)), (1)/(sqrt(2)))`D. `((1)/(2), (1)/(2))` |
| Answer» Correct Answer - A | |
| 233. |
If ` (2sinalpha)/(1+cosalpha +sinalpha)=y` , then prove that `(1-cosalpha+sinalpha )/(1+sinalpha)` is also equal to y. |
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Answer» Given that, ` (2sinalpha)/(1+cosalpha +sinalpha)=y` Now, `(1-cosalpha+sinalpha )/(1+sinalpha)` = `((1-cosalpha+sinalpha) )/((1+sinalpha))*((1+cosalpha+ sinalpha))/((1+cosalpha+sinalpha))` `" "=({(1+sinalpha)-cosalpha})/((1+sinalpha))*({(1+sinalpha)+cosalpha})/((1+cosalpha+sinalpha))` `" "=((1+sinalpha)^(2)-cos^(2)alpha)/((1+sinalpha)(1+sinalpha+ cosalpha))` `" "=((1+sin^(2)alpha+2sinalpha )-cos^(2)alpha)/((1+sinalpha )(1+sinalpha+cosalpha))` `" "=(1+sin^(2)alpha+2sinalpha-1+sin^(2 )alpha)/((1+sinalpha)(1+sinalpha+cosalpha))` `" "=(2sin^(2)alpha+2sinalpha)/((1+sinalpha)(1+sinalpha+cosalpha))` `" "=(2sinalpha(1+sinalpha))/((1+sinalpha)(1+sinalpha+cosalpha))` `" "=(2sinalpha)/(1+sinalpha+cosalpha)=y" "` Hence proved. |
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| 234. |
Prove that `1/(secA-tanA)-1/(cosA)=1/cosA-1/(secA-tanA)`. |
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Answer» Given `1/(secA-tanA)-1/(cosA)=1/cosA-1/(secA-tanA)` or `1/(secA-tanA)+1/(secA-tanA)=1/(cosA)+1/cosA` Here `R.H.S.=2/cosA` Now `L.H.S.=1/(secA-tanA)+1/(secA+tanA)` `=(secA+tanA+secA-tanA)/((secA-tanA)(secA+tanA))` `=2/cosA` Thus, L.H.S.=R.H.S. |
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| 235. |
Prove that `(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/(cosA)`. |
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Answer» `" "LHS=( tanA+secA-1)/( tanA-secA+1)` `" "=(tanA+secA-( sec^(2)A-tan^(2)A))/((tanA-secA+1 ))" "[becausesec ^(2)A-tan^(2)A=1 ]` `" "=((tanA+secA)-(secA+tanA)(secA-tanA))/((1-secA+tanA))` `" "=((secA+tanA )(1-secA+tanA))/(1 -secA+tanA)` `" "=secA+tanA=(1)/(cosA)=( sinA)/(cosA)` `=(1+sinA)/(cosA)=RHS " "` Hence proved. |
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| 236. |
prove that`1/(secA-tanA)-1/(cosA)=1/(cosA)-1/(secA+tanA)` |
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Answer» `L.H.S. = 1/(secA-tanA) - 1/cosA` `=1/(1/cosA-sinA/cosA) - 1/cosA` `=(cosA)/(1-sinA) - 1/cosA` `=(cos^2A-1+sinA)/(cosA(1-sinA))` `=(1-sin^2A-1+sinA)/(cosA(1-sinA))` `=(sinA(1-sinA))/(cosA(1-sinA))` `= tanA` `R.H.S. = 1/cosA - 1/(secA+tanA)``= 1/cosA - 1/(1/cosA+sinA/cosA)` `= 1/cosA - cosA/(1+sinA)` `=(1+sinA - cos^2A)/(cosA(1+sinA))` `=(1+sinA - 1+sin^2A)/(cosA(1+sinA))` `=(sinA(1+sinA))/(cosA(1+sinA))` `=tanA` `:. L.H.S. = R.H.S. = tanA` |
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| 237. |
Find the smallest positive root of the equation `sqrt(sin(1-x))=sqrt(cos"x")` |
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Answer» `sqrt(sin(1-x)) = sqrtcosx` `=>sin(1-x) = cosx` `=>cos(pi/2-(1-x)) = cosx` `=>pi/2-1+x = 2npi+-x`, where `n in Z.` `=>x = (2npi-pi/2+1)/2` As we have to find the smallest positive root of the equation, it will come when `n = 2`. `:. x = (4pi-pi/2 +1)/2` `=>x = (7pi)/4+1/2.` |
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| 238. |
Solve `sin^2x +1/4sin^2 3x=sinxsin^2 3x` |
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Answer» `sin^2x+1/4sin^2 2x=sinxsin^2 3x` `sin^2x-sinxsin^2 3x+1/4sin^2 3x=0` `(sinx-1/2sin^2 3x)^2-1/4sin^4 3x+1/4sin^2 3x=0` `(sinx-1/2sin^2 3x)^2+1/4sin^2 3x[-sin^2 3x+1]=0` `(sinx-1/2sin^2 3x)^2+1/4sin^2 3xcos^2 3x=0` `(sinx-1/2sin^2 3x)^2+1/4*(sin^2 6x)/4=0` `(sinx-1/2sin^2 3x)^2+1/16sin^2 6x=0` `(sinx-1/2sin^2 3x)^2+(1/4sin6x)^2=0` `sinx-1/2sin^2 3x=0` `2sinx=sin^2 3x` `2sinx=sin^2((npi)/2)` `x=0` `(sin(6x)/4)=0` `sin6x=0` `6x=npi` `x=(npi)/6` `sinx=1/2` `sinx=sinpi/6` `x=npi+(-1)^n pi/6`. |
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| 239. |
General solution of `1-cosx=sinxsin((x)/(2))` isA. `2npi, 4npi, ninZ`B. `npi, 2npi, ninZ`C. `npi, (npi)/(2), ninZ`D. `(npi)/(3), (2npi)/(3), ninZ` |
| Answer» Correct Answer - A | |
| 240. |
General solution of `sin4x=(sqrt(3))/(2)` isA. `(npi)/(4)+(-1)^(n)(pi)/(12), ninZ`B. `(npi)/(4)+(-1)^(n)(pi)/(3), ninZ`C. `npi+(-1)^(n)(pi)/(12), ninZ`D. `npi+(-1)^(n)(pi)/(3), ninZ` |
| Answer» Correct Answer - A | |
| 241. |
General solution of `cosec3x=(-2)/(sqrt(3))` isA. `(npi)/(3)+(-1)^(n)(5pi)/(3), ninZ`B. `(npi)/(3)+(-1)^(n)(5pi)/(9), ninZ`C. `(npi)/(3)+(-1)^(n)(4pi)/(3), ninZ`D. `(npi)/(3)+(-1)^(n)(4pi)/(9), ninZ` |
| Answer» Correct Answer - D | |
| 242. |
Find the general value of x for which `sqrt(3)" cosec "x=2`.A. `npi+(-1)^(n-1)(pi)/(3), ninZ`B. `npi+(-1)^(n-1)(2pi)/(3), ninZ`C. `npi+(-1)^(n)(pi)/(3), ninZ`D. `npi+(-1)^(n)(2pi)/(3), ninZ` |
| Answer» Correct Answer - C | |
| 243. |
General solution of `sinx=(-sqrt(3))/(2)` isA. `npi+(-1)^(n)(pi)/(3), ninZ`B. `npi+(-1)^(n)(5pi)/(3), ninZ`C. `npi+(-1)^(n)(7pi)/(3), ninZ`D. `npi+(-1)^(n)(4pi)/(3), ninZ` |
| Answer» Correct Answer - D | |
| 244. |
General solution of `sin2x=(1)/(2)` isA. `npi+(-1)^(n)(pi)/(12), ninZ`B. `npi+(-1)^(n)(pi)/(6), ninZ`C. `(npi)/(2)+(-1)^(n)(pi)/(12), ninZ`D. `(npi)/(2)+(-1)^(n)(pi)/(6), ninZ` |
| Answer» Correct Answer - C | |
| 245. |
General solution of` sqrt(2)cosecx+cotx=sqrt(3)` isA. `2npi+(pi)/(6)pm(3pi)/(4), ninZ`B. `2npi-(pi)/(6)pm(3pi)/(4), ninZ`C. `2npi+(pi)/(3)pm(3pi)/(4), ninZ`D. `2npi-(pi)/(3)pm(3pi)/(4), ninZ` |
| Answer» Correct Answer - D | |
| 246. |
General solution of `cosecx=-sqrt(2)` isA. `2npi+(-1)^(n)(7pi)/(4), ninZ`B. `2npi+(-1)^(n)(5pi)/(4), ninZ`C. `npi+(-1)^(n)(7pi)/(4), ninZ`D. `npi+(-1)^(n)(5pi)/(4), ninZ` |
| Answer» Correct Answer - D | |
| 247. |
General solution of `cosecx=-sqrt(2)` isA. `(pi)/(2), (2pi)/(3)`B. `(-5pi)/(4), (-7pi)/(4)`C. `(5pi)/(4), (7pi)/(4)`D. `(pi)/(4), (3pi)/(4)` |
| Answer» Correct Answer - C | |
| 248. |
The principal solution `sqrt(3)cosecx+2=0` areA. `(pi)/(3), (3pi)/(3)`B. `(2pi)/(3), (5pi)/(3)`C. `(4pi)/(3), (5pi)/(3)`D. `(pi)/(3), (4pi)/(3)` |
| Answer» Correct Answer - C | |
| 249. |
In ` A B C ,`if `(sinA)/(csinB)+(sinB)/c+(sinC)/b=c/(a b)+b/(a c)+a/(b c),`then the value of angle `A`is`120^0`(b) `90^0`(c) `60^0`(d) `30^0` |
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Answer» `sinA/(csinB)+(sinB)/C+sinC/b=c/(ab)+b/(ac)+a/(bc)` Sine law `sinA/a=sinB/b=sinC/a=k` `SinA=ak` `SinB=bk` `SinC=ck` `(ak)/(cbk)+(bk)/(c)+(ck)/b=c/(ab)+b/(ac)+a/(bc)` `k[b/c+c/b]=1/a[c/b+b/c]` `k=1/a` `sinA=ak` `sinA=a*1/a` `SinA=1` `A=90`. |
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| 250. |
Let `f(x)=x^2a n dg(x)=sinxfora l lx in Rdot`Then the set of all `x`satisfying `(fogogof)(x)=(gogof)(x),w h e r e(fog)(x)=f(g(x)),`is`+-sqrt(npi),n in {0,1,2, dot}``+-sqrt(npi),n in {1,2, dot}``pi/2+2npi,n in { ,-2,-1,0,1,2}``2npi,n in { ,-2,-1,0,1,2, }` |
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Answer» `f(x)=x^2` `g(x)=sinx` `g(f(x))=sin(f(x))` `[email protected]=sinx^2` `g([email protected])=sin([email protected])` `=sin(sinx^2)` `f(ggf)=(ggf)^2` `=[sin(sinx^2)]^2` `=sin^2(sinx^2)` `(sin(sinx^2))^2=sin(sinx^2)` `sinx^2=y` `(siy)^2=siny` `(siny)^2-siny=0` `siny[siny-1]=0` `siny=0` `y=npi` `sinx^2=(4n+1)pi/2` This is not possible `siny=1` `y=(4n+1)pi/2` `sinx^2=npi` `sinx^2=0` `x^2=npi` `x=pmsqrt(npi)` `n in(0,1,2,3,4...)`. |
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