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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.
| 501. |
If `3tanA+4=0`, then the value of `2cotA-5cosA+sinA` is equal toA. `23/10ifpi/2ltAltpi`B. `23/10if(3pi)/2ltAlt2pi`C. `(-53)/10ifpi/2ltAltpi`D. `-53/10if(3pi)/2ltAlt2pi` |
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Answer» Correct Answer - A::D We have `tanA=-3/4` If A lies in `2^(nd)` quadrant, then `sinA=4/5,cosA=(-3)/5,and cotA=(-3)/4` If lies `2^(nd)` quadrant, then `sinA=(-4)/5,cosA=(3)/5,and cotA=(-3)/4` |
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| 502. |
If A lies in the second quadrant and `3tanA + 4=0`, then find the value of `2cotA-5cosA+ sinA`.A. `(-53)/(10)`B. `(23)/(10)`C. `(37)/(10)`D. `(7)/(10)` |
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Answer» Correct Answer - B Given equation, `" "3tanA +4=0` `rArr" "3tanA =-4` `rArr" "tanA=(-4)/(3)` `rArr" "cotA=(-3)/(4)` `rArr" "secA =sqrt(1+(16)/(9))=sqrt((25)/(9))=pm(5)/(3)` `rArr" "secA =(-5)/(3)" "` [since, A lies in second quadrant] `" "cosA =(-3)/(5)` `" "sinA =sqrt(1-(9)/(25))=(sqrt(25-9))/(25)=pm(4)/(5)` `" "sinA =(4)/(5) " "` [since, A lies in second quadrant] `therefore" "2cotA -5cosA +sinA =2((-3)/(4))-5((-3)/(5))+(4)/(5)` `" "=(-6)/(4)+3+(4)/(5)` `" "=(-30+60+16)/(20)=(46)/(20)` `" "=(23)/(10)` |
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| 503. |
The polar co-ordinates of the point whose cartesian co-ordinates are `(sqrt(2), sqrt(2))`, areA. `(2, (7pi)/(4))`B. `(2, (5pi)/(4))`C. `(2, (3pi)/(4))`D. `(2, (pi)/(4))` |
| Answer» Correct Answer - D | |
| 504. |
The cartesian co-ordinates of point, whose polar co-ordinates are `(sqrt(2), (pi)/(4))` areA. `(1, -1)`B. `(-1, 0)`C. `(1, 1)`D. `(sqrt(2), sqrt(2))` |
| Answer» Correct Answer - C | |
| 505. |
If `(-sqrt(2),sqrt2)` are cartesian co-ordinates off the point, then its polar co-ordinates are . . .A. `(2, (7pi)/(4))`B. `(2, (5pi)/(4))`C. `(2, (3pi)/(4))`D. `(2, (pi)/(4))` |
| Answer» Correct Answer - C | |
| 506. |
The polar co-ordinates of the point whose cartesian co-ordinates are `((1)/(sqrt(2)), (-1)/(sqrt(2)))`, areA. `(1, (7pi)/(4))`B. `(1, (3pi)/(4))`C. `(1, (5pi)/(4))`D. `(1, (pi)/(4))` |
| Answer» Correct Answer - A | |
| 507. |
The cartesian co-ordinates of a point, whose polar co-ordinates are `((3)/(4), 135^(@))` areA. `((-3)/(4sqrt(2)), (-3)/(4sqrt(2)))`B. `((3)/(4sqrt(2)), (3)/(4sqrt(2)))`C. `((-3)/(4sqrt(2)), (3)/(4sqrt(2)))`D. `((3)/(4sqrt(2)), (-3)/(4sqrt(2)))` |
| Answer» Correct Answer - C | |
| 508. |
The polar co-ordinates of the point whose cartesian co-ordinates are `(0, (1)/(2))`, areA. `(2, (pi)/(2))`B. `(2, (3pi)/(2))`C. `((1)/(2), (pi)/(2))`D. `((1)/(2), (3pi)/(2))` |
| Answer» Correct Answer - D | |
| 509. |
The polar co-ordinates of the point whose cartesian co-ordinates are `(-3, 0)`, areA. `(-3, pi)`B. `(3, pi)`C. `(-3, (pi)/(2))`D. `(3, (pi)/(2))` |
| Answer» Correct Answer - B | |
| 510. |
The value of `tan1^@tan2^@tan3^@...tan89^@` isA. `0`B. `1`C. `(1)/(2)`D. Not defined |
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Answer» Correct Answer - B Given expression, `tan1^(@)tan2^(@)tan3^(@) ... tan89^(@)` `=tan1^(@)tan2^(@)...tan45^(@)*tan(90^(@)-44^(@))tan(90^(@)-43^(@))...tan(90^(@)-1^(@))` `=tan1^(@)*cot1^(@)*tan2^(@)*cot2^(@)...tan89^(@)*cot89^(@)` `1*1...1*1=1` |
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| 511. |
If `tan theta = p/q` and `theta = 3 phi( 0 < theta < pi/2)`, then `p/(sin phi) - q/ (cos phi)` = |
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Answer» Here, ` tantheta = p/q=>p = qtantheta->(1)` `=>sec theta = sqrt(1+p^2/q^2) = sqrt(p^2+q^2)/q` `=>costheta = q/(sqrt(p^2+q^2))->(2)` `:. p/sinphi - q/cosphi = (qtantheta)/sinphi - q/cosphi` `=q(tantheta/sinphi - 1/cosphi)` `=q(sintheta/(costheta sinphi) - 1/cosphi)` `=q((sinthetacosphi -costhetasinphi)/(costheta sinphicosphi))` `=2q(sin(theta-phi)/(2sinphicosphi(costheta)))` `=2q(sin(3phi-phi)/(sin2phi(costheta)))` .....(As `theta = 3phi`) `=(2q)/costheta` `=2q(sqrt(p^2+q^2)/q)` .....From(2) `=2sqrt(p^2+q^2)` `:. p/sinphi - q/cosphi = 2sqrt(p^2+q^2)` |
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| 512. |
The principal solution `sqrt(3)secx-2=0` areA. `(5pi)/(6), (7pi)/(6)`B. `(pi)/(6), (11pi)/(6)`C. `(pi)/(6), (5pi)/(6)`D. `(7pi)/(6), (11pi)/(6)` |
| Answer» Correct Answer - B | |
| 513. |
If `sin 18^@=(sqrt(5)-1)/4` then what is the value of `sin 81^@` |
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Answer» `Sin 18^@ = (sqrt5 - 1)/4` `cos 18^@ = sqrt(1- sin^2 18)` `= sqrt(1- ((sqrt5-1)/4)^2)` `= sqrt((16-5-1+ 2sqrt5)/16)= sqrt((10+ 2 sqrt5)/16)` `cos 18^@ = sqrt((10 + 2 sqrt5)/4)` `sin 81^@ = sin(90- 9)= cos9^@` `cos 9^@ = sqrt((1+ cos18^@)/2)` `cos 9^@ = sqrt(((4+ sqrt(10+2sqrt5))/4)/2)` `cos 9^@ = sqrt((4+ sqrt(10 + 2 sqrt5))/8)` `sin 81^@ = cos 9^@ = sqrt(1/2 + sqrt(10+2 sqrt5)/8)` Answer |
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| 514. |
If `15 sin^(4)alpha+10cos^(4)alpha=6`, then the value of `8 cosec^(6)alpha+27 sec^(6)alpha` isA. 150B. 175C. 225D. 250 |
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Answer» Correct Answer - D `15 sin^(4)alpha+10 cos^(4)alpha=6` Dividing by `cos^(4)alpha`, we get `15 tan^(4)alpha+10=6 sec^(4)alpha` `rArr 15 tan^(4)alpha+10=6(1+tan^(2)alpha)^(2)` `rArr 9tan^(4)alpha-12tan^(2)alpha+4=0` `rArr (3 tan^(2)alpha-2)^(2)=0` `rArr tan^(2)alpha=(2)/(3)` Now `8 cosec^(6)alpha+27sec^(6)alpha` `8(1+cot^(2)alpha)^(3)+27(1+tan^(2)alpha)^(3)` `=8(1+(3)/(2))^(3)+27(1+(2)/(3))^(3)` `=125+125=250` |
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| 515. |
`sin theta + sin phi = a`,`cos theta + cos phi=b`then prove `tan [(theta -phi)/2]=+-sqrt[(4-a^2-b^2)/(a^2+b^2)]` |
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Answer» `sqrt((4-(a^2 + b^2))/(a^2 + b^2))` `= sqrt(4/(a^2 + b^2) -1)` let `a^2 + b^2 = (sin theta + sin phi)^2 + (cos theta + cos phi)^2` `= sin^2 theta + sin^2 phi + 2 sin theta sin phi + cos^2 theta + cos^2 phi + 2 cos theta cos phi` `1 + 1 + 2[cos theta cos phi + sin theta sin phi]` `= 2 + 2 cos ( theta - phi)` `= 2[ 1+ cos( theta - phi)]` `= 2[ 1 + (1- tan^2((theta - phi)/2))/(1 + tan^2((theta - phi)/2))]` `= 2[(1 + tan^2 ((theta - phi)/2) + 1 - tan^2((theta - phi)/2))/(1 + tan^2((theta - phi)/2))]` `a^2 + b^2 = 4/(1 + tan^2((theta - phi)/2))` `1 + tan^2 ( (theta - phi)/2) = 4/(a^2 + b^2)` `tan( (theta - phi)/2) = +- sqrt(4/(a^2 + b^2) - 1)` `= +- sqrt((4 - a^2 - b^2)/(a^2 + b^2))` Answer |
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| 516. |
If `sin^(-1)(x/5) + cosec^(-1) (5/4) = pi/2 `, then the value of x isA. `1`B. `3`C. `4`D. `5` |
| Answer» Correct Answer - B | |
| 517. |
Find the tange of `f(x)=cos^4x+sin^2x-1`. |
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Answer» Correct Answer - [-1/4, 0] `f(x)=cos^4x+sin^2x-1` `=cos^4x+1-cos^2x-1` `=cos^4-cos^2x` `=(cos^2x-1//2)^2-1//4` Now, `0lecos^2xle1` `rArr -1//2cos^2x-1//2le1//2` `rArr-1//4le(cos^2x-1//2)^2-1//4le0` |
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| 518. |
If `cos^2x+cosx=a+2`, then find the value of a for which equation has solution. |
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Answer» Correct Answer - [-9/4, 0] `cos^2x+cosx=a+2` `rArr (cosx+1/2)^2=a+9/4` Now, `-1/2lecosx+1/2le3/2` `:. (cosx+1/2)^2in[0,9/4]` So, `0lea+9/4le9/4` `:. -9/4leale0` |
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| 519. |
Prove that `(sintheta+cosectheta)^2+(costheta+sectheta)^2ge9`. |
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Answer» `(sintheta+cosectheta)^2+(costheta+sectheta)^2` `=sin^2theta+cosec^2 theta+2+cos^2theta+sec^2theta+2` `=(sin^2theta+cos^2theta)+(cosec^2theta+sec^2theta)+4` `=5+1+tan^2thetta+1+cot^2theta` `=7+(tantheta-cottheta)^2+2` `=9+(tantheta-cottheta)^2ge9` |
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| 520. |
The principal solution `sqrt(3)cotx-1=0` areA. `(pi)/(3), (4pi)/(3)`B. `(2pi)/(3), (5pi)/(3)`C. `(pi)/(3), (2pi)/(3)`D. `(4pi)/(3), (5pi)/(3)` |
| Answer» Correct Answer - A | |
| 521. |
The greatest value of `sin^4theta+cos^4theta` isA. `1//2`B. 1C. 2D. 3 |
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Answer» Correct Answer - B `sin^4theta+cos^4=(sin^2theta+cos^2theta)^2-2sin^2thetacos^2theta` `=1-2sin^2thetacos^2thetale1` |
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| 522. |
Find the range of `f(x)=1/(2abscosx-3)` |
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Answer» Correct Answer - [-1, 1/3] `0leabscosxle1` `rArr 0le2abscosxle2` `rArr-3le2abscosx-3le-1` `rArr 1/(2abs(cosx)-3)in[-1,-1//3]` |
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| 523. |
Find the range of `f(x)=cosec^2x+25sec^2x`. |
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Answer» Correct Answer - [-1, 1/3] `f(x)=(1+cot^2x)+25(1+tan^2x)` `=26+cot^2x+25tan^2x` `=26+10+(cot^2x+25tan^2x-10)` `=36+(cotx-5tanx)^2ge36` |
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| 524. |
Find the range of `f(x)=cos^2x+sec^2x`. |
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Answer» We have `f(x)=cos^2x=sec^2x` `=(cos-secx)^2+2cosxsecx` `=2+(cosx-secx)^2ge2` |
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| 525. |
If `f(x)=sin^6x+cos^6x ,`then range of `f(x)`is`[1/4,1]`(b) `[1/4,3/4]`(c) `[3/4,1]`(d) none of theseA. `[1/4,1]`B. `[1/4,3/4]`C. `[3/4,1]`D. None of these |
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Answer» Correct Answer - A `f(x)=cos^6x+sin^6x` `=(cos^2x+sin^2x)^3-3cos^2xsin^2x(cos^2x+sin^2x)` `=1-3cos^2x(1-cos^2x)` `=3cos^4x-3cos^2x+1` `=3(cos^4x-cos^2x+1/2)` `=3((cos^2x=1/2)^2+1/12)` least value f(x) is `1/4," when "cos^2x-1/2=0` Greatest value of f(x) is 1, when `cos^2x=0or1` |
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| 526. |
For all real values of `theta`, choose the correct options. |
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Answer» Correct Answer - 4 a. `A=sin^2theta+cos^4theta` `=1-cos^2theta+cos^4theta` `=(cos^2theta-1//2)^2+3//4` `0le cos^2thetale1` `rArr-1//2lecos^2theta-1//2le1//2` `rArr 0le (cos^2theta-1//2)^2le1//4` `rArr 0le (cos^2theta-1//2)^2le1//4` `rArr 3//4 le (cos^2theta-1//2)^2+3//4le1` b. `3cos^2theta+sin^4theta=3-3sin^2theta+sin^4theta` `=(sin^2theta-3//2)^2+3//4` `0le sin^2thetale1` `rArr -3//2 lesin^2theta-3//2le-1//2` `rArr 1//4 le(sin^2theta-3//2)^2+3//4le3` `rArr 1le(sin^2theta-3//2)^2+3//4le3` c. `A=sin^2theta-cos^4theta` `=1-cos^2theta-cos^4theta` `=5//4-(cos^2theta+1//2)^2` `0lecos^2thetale1` `rArr 1//2lecos^2theta+1//2le3//2` `rArr 1//4le(cos^2+1//2)^2le9//4` `rArr -9//4le-(cos^2theta+1//2)^2le-1//4` `rArr-1le5//4-(cos^2theta+1//2)^2le1` d. `A=tan^2theta+2cot^2theta` `=(tantheta-sqrt2cottheta)^2+2sqrt2ge2sqrt2` |
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| 527. |
Let `f(x) = sin^6x + cos^6x + k(sin^4 x + cos^4 x)` for some real number k. Determine(a) all real numbers k for which `f(x)` is constant for all values of x.A. [-1,0]B. `[0,1/2]`C. `[-1,-1/2]`D. None of these |
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Answer» Correct Answer - C f(x)=0 `rArr (1-3 ) sin^2 x cos^2 x)+k[1-2 sin^2 x cos^2 x]=0` `rArr K+1 =(sin ^2 x cos^2 x)/(1-2 sin^2x cos^2x)` `rArrk=(3sin^2xcos^2x-1)/(1-2sin^2xcos^2x)` `=-3/2(1-2sin^2xcos^2x-1/3)/(1-2sin^2xcos^2x)` `=-3/2(1-(1/3)/(1-2sin^2xcos^2x))` minimum of `sin^2xcos^2x " is 0 at " x=0,pi//2` Maximum of `sin^2xcos^2x" is "1//4" at " x=pi//2` Hence, k in[1,-1/2] |
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| 528. |
Find the range of `f(x)=1/(5sinx-6)` |
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Answer» `-1lesinxle1` or `-5le 5sinxle5` or -11le5sinx-6le-1` or -1le1/(5sinx-6)le-1//11` or `1/(5sinx-6)in[-1,-1//11]` |
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| 529. |
Let `f(x) = sin^6x + cos^6x + k(sin^4 x + cos^4 x)` for some real number k. Determine(a) all real numbers k for which `f(x)` is constant for all values of x. |
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Answer» Correct Answer - A `(1-3sin^2xcos^2x)+k[1-2sin^2xcos^2x]=0` is an identity, i.e., `(1+k)-(3+2k)sin^2xcos^2=0` is an identity. `rArr 1+k=and 3+2k=0`, which do not hold simultaneously. |
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| 530. |
If `sinx=3/5,cosy=-(12)/(13),`where x and y both lie in second quadrant, find the value of `sin (x + y)dot` |
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Answer» Here, `sin x = 3/5` So, `cosx = sqrt(1-(3/5)^2)=+-4/5` As, x lies in 2nd quadrant, `cos x = -4/5` Now, `cos y = -12/13` So, `sin y = sqrt(1-(-12/13)^2)=+-5/13` As, y lies in 2nd quadrant, `sin y = 5/13` Now, we know, `sin(x+y) = sinxcosy+cosxsiny` `=3/5(-12/13)+(-4/5)(5/13) =-56/65` |
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| 531. |
Find the range of `f(x) =sin^2x-3sinx+2`. |
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Answer» `f(x)=sin^2x-3sinx+2` `=(sinx-3//2)^2+2-9//4` `=(sinx-3//2)^2-1//4` Now, `-1lesinxle1` or `-5//2lesinx-3//2le-1//2` or `1/4le(sinx-3//2)^2le25//4` or `0le(sinx-3//2)^2-1//4le6` |
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| 532. |
Show that `2(sin^6x+cos^6x)-3(sin^4x+cos^4x)+1=0`. |
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Answer» `2(sin^6x+cos^6x)-3(sin^4x+cos^4x)+1` `=2[(sin^2x)^3+(cos^2x)^(3)]-3(sin^4x+cos^4x)+1` `=2[(sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x)]-3[(sin^2x+cos^2x)^2-2sin^2xcos^2x]+1` `=2[1-3sin^2xcos^2x]-3[1-2sin^2xcos^2x]+1=0` |
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| 533. |
Solve `2cos^2x+3sinx=0`. |
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Answer» `2(1-sin^2x) + 3sinx = 0` `-2sin^2x + 3sinx + 2=0` `2sin^2x - 3sinx - 2=0` let `sinx = t` `2t^2 - 3t -2 = 0` `(2t +1)(t-2) = 0` `t = -1/2 or t=2` `sin x = -1/2 or sinx = 2` solution is not possible for `sinx = 2` `:. sinx = - sin(pi/6) = sin(-pi/6)` `x = npi + (-1)^n(-pi/6) ; n in z` answer |
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| 534. |
For any triangle ABC, If `B=3C`, show that `cosC=sqrt((b+c)/(4c)) and sinA/2=(b-c)/(2c)`A. `sinC`B. `cosC`C. `cotC`D. `tanC` |
| Answer» Correct Answer - B | |
| 535. |
Solve `sin 2x - sin 4x + sin 6x = 0`. |
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Answer» `sin2x- sin4x+ sin6x = 0` as we know that, `(sina + sinb = 2sin((a+b)/2)cos((a-b)/2))` so, `2sin4x.cos(-2x) - sin4x = 0` `= Sin4x(2cos2x-1) = 0` now it can possible that whether `sin4x = 0 or 2cos2x-1=0` when `sin 4x=0` `4x = n pi & n in z` `x= npi/4` when `2cos 2x-1=0` `cos2x= cos (pi/3) ` `2x = 2n pi +- pi/3` `x= n pi +- pi/6` answer = `(npi)/4 , npi +- pi/6` |
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| 536. |
Let `A=sinx+cosx`,Then find the value of `sin^4c+cos^4c`in terms of A. |
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Answer» `A=sinx+cosx` `:. A^2=1+2sinxcosx` Now, `sin^4+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x` `=1-2sin^2cos^2x` `=1-2((A^2-1)/2)^2` `=1-((A^2-1)^2)/2` `=(2-(A^4-2A^2+1))/2` `=(1+2A^2-A^4)/2` `=1/2+A^2-1/2A^4` |
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| 537. |
`sqrt((1-sintheta)/(1+sintheta))={sectheta-tantheta ,` if `-pi/2ltthetaltpi/2` and `-sectheta+tantheta`,` pi/2ltthetalt(3pi)/ 2` |
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Answer» `sqrt((1-sintheta)/(1+sintheta))sqrt((1-sintheta)/(1+sintheta))` `sqrt((1-sintheta)^2/(1-sin^2theta))` `sqrt((1-sintheta)^2/cos^2theta)` `sqrt((sectheta-tantheta)^2` `sectheta-tantheta,-pi/2ltxltpi/2` `tantheta-secctheta,pi/2ltxlt3/2pi`. |
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| 538. |
Leg `f(x)=log((log)_(1//3)((log)_(1/3)((log)_7(sinx+a)))`be defined for every real value of `x ,`then the possible value of `a`is3 (b) 4(c) 5 (d)6 |
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Answer» `f(x)=log(log_(1/3)(log_7(sinx+a)))` `log_(1/3)log_7(sinx+a))>0` `0`1<(sinx+a)<7` `1-sinx`2<a<6`. |
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| 539. |
If `x=(sin^3P)/cos^2P,y=cos^3P/sin^2P" and " sinP+cosP=1/2` then find the value of x + y. |
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Answer» `x+y=(sin^5P+cos^5P)/(cos^2P sin^2P)` Now, `sin^5P+cos^5P=(sinP+cosP)(sin^4P-sin^3PcosP+sin^2Pcos^2P-sinPcos^3P+cos^4P)` `=(sinP+cosP)[(sin^4P+cos^4P)-sinPcosP(sin^2P+cos^2P)+sin^2Pcos^2P]` `=(sinP+cosP)[1-2sin^2Pcos^2P-sinPcosP+sin^2Pcos^2P]` `=(sinP+cosP)[1-sin^2Pcos^2P-sinPcosP]` Now, `sinP+cosP=1/2` `rArr1+2sinPcosP=1/4` `rArr sinPcosP=-3/8` Using these values, we get `x+y=(1/2[1-9/64+3/8])/(9/64)` `=79/18` |
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| 540. |
Number of integral values of `a`for which the equation `cos^2x-sinx+a=0`has roots when `x in (0,pi/2)`is____________ |
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Answer» `cos^2x-sinx+a=0` `1-sin^2x-sinx+a=0` `sin^2x+sinx-a-1=0` `(-1)^2-4(1)(-a-x)>=0` `1+4(a+1)>=0` `a>=-5/4` `0<-a-1<1` `1<-a<2` `-1>a>-2` Number of integral value of a =0. |
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| 541. |
The equation `(cosp - 1) x^2 +cosp x + sinp = 0` where x is a variable, has real roots. then the interval of p may be any one of the following : |
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Answer» `D>=0` `b^2-4ac>=0` `cos^2P-4(cosP-1)sinP>=0` `c^2P>=4(cosP-1)sinP` `0<=cos^2P<=1` `sinP>=0` `P in[0,pi]` `(-1/sqrt2)^2>=4(-1/sqrt2=1)-1/sqrt2` `1/2>=40(1+sqrt2)/(sqrt2-sqrt2)=2(1+sqrt2)` `1/2<=2(1+sqrt2)` `P=3/4pi` is not possible. `P=-pi/4` is not possibble. option 4 is correct. |
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| 542. |
The number of integral values of `k` for which the equation `7cos x +5 sinx=2k+1` has a solution is (1) `4` (2) `8` (3) `10` (4) `12` |
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Answer» `7cosx+5sinx=2k+1` `7/sqrt(7^2+5^2)cosx+5/sqrt(7^2+5^2)sinx=(2k+1)/sqrt(7^2+5^2)` `tanphi=7/5` `sin(x+phi)=(2k+1)/sqrt74` `-1<=sin(x+phi)<=1` `=-1<=(2k+1)/sqrt74<-1` after pitting the values. 4+4=8. option 2 is correct. |
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| 543. |
Solve `7cos^2theta+3sin^2theta=4` |
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Answer» `7cos^2theta+3sin^2theta=4` `4cos^2theta+3cos^2theta+2sin^2theta=4` `4cos^2theta+3=4` `4cos^2theta=1` `cos^2theta=1/4` `costheta=pm1/2` `theta=npipmpi/3`. |
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| 544. |
A variable triangle `A B C`is circumscribed about a fixed circle of unit radius. Side `B C`always touches the circle at D and has fixed direction. If B and C vary insuch a way that (BD) (CD)=2, then locus of vertex Awill be a straight line.parallel to side BCperpendicular to side BCmaking an angle `(pi/6)`with BCmaking an angle `sin^(-1)(2/3)`with `B C` |
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Answer» Here, `BD = s-b` `CD = s-c` `BD*CD = (s-b)(s-c)` `=>2 = (s-b)(s-c)` `=>2s(s-a) = s(s-a)(s-b)(s-c)` `=>2s(s-a) = Delta^2` `=>Delta^2/s^2 = (2(s-a))/s` `=>r^2 = 2-(2a)/s` As, `r` is constant, so `a/s` will be constant. Let `H_a` is the distance of `A` from `BC`. Then, `Delta = 1/2*a*H_a` `=>Delta/s = 1/(2s)aH_a` It is given that circle is of unit radius, ` :. 1 = a/(2s)H_a` `=>H_a = (2s)/a` It means `H_a` will be constant as `a/s` is a constant. So, locus of vertex `A` will be a straight line parallel to `BC`. So, option `(a)` is the correct option. |
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| 545. |
`tan(tan^(-1)((1)/(2))-tan^(-1)((1)/(3)))=`A. `(1)/(6)`B. `(1)/(7)`C. `(5)/(6)`D. `(7)/(6)` |
| Answer» Correct Answer - B | |
| 546. |
If `cos^(-1)x+cos^(-1)y+cos^(-1)z+cos^(-1)t=4pi`, then `x^(2)+y^(2)+z^(2)+t^(2)=`A. `6`B. `4`C. `xy+yz+zt`D. `1-2xyzt` |
| Answer» Correct Answer - B | |
| 547. |
If `tanA=(1)/(2) and tanB=(1)/(3)`, then `tan(2A+B)` is equal toA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C Given that, `tanA=(1)/(2) and tanB=(1)/(3)` Now, `" "tan(2A+B)=(tan2A+tanB)/(1-tan2A*tanB)" "...(i)` `Also, " "tan2A=(2tanA)/(1-tan^(2)A)=(2*(1)/(2))/(1-(1)/(4))=(4)/(3)` From Eq. (i), ` tan(2A+B)=((4)/(3)+(1)/(3))/(1-(4)/(3)*(1)/(3))=((4)/(3)+(1)/(3))/((9-4)/(9))=((5)/(3))/((5)/(9))=3` |
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| 548. |
Find the most general solution of the following `sec 4 x-sec 2x=2`A. `(2n+1)(pi)/(10), (2n+1)(pi)/(2), ninZ`B. `(2n+1)(pi)/(10), (2n+1)(pi)/(5), ninZ`C. `(2n+1)(pi)/(15), (2n+1)(pi)/(5), ninZ`D. `(2n+1)(pi)/(6), (2n+1)(pi)/(3), ninZ` |
| Answer» Correct Answer - A | |
| 549. |
If `x=2. y=3`, then `tan^(-1)x+tan^(-1)y=`A. `(pi)/(4)`B. `(pi)/(3)`C. `(-pi)/(4)`D. `pi` |
| Answer» Correct Answer - C | |
| 550. |
In `triangleABC,a(cosB+sinBcot((A)/(2)))=`A. `a+b+c`B. `c+a`C. `a+b`D. `b+c` |
| Answer» Correct Answer - D | |