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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.
| 651. |
If `a=9,b=4a n dc=8`then find the distance between the middle point of BC and the foot ofthe perpendicular form `Adot` |
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Answer» We can create a triangle `ABC` with the given details. Please refer to video to see the diagram. We have to find `DE`. From the diagram, `DE = CD - CE = a/2- bcosC` `=>DE = a/2 - b((a^2+b^2-c^2)/(2ab))` `=(a^2-a^2-b^2+c^2)/(2a)` `=(-b^2+c^2)/(2a) = (-(4)^2 +(8)^2)/(2(9))``=48/18 = 8/3` `:. DE = 8/3` |
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| 652. |
Let `A=sinx+cosxdot`Then find the value of `sin^4x+cos^4x`in terms of `Adot` |
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Answer» `A=sinx+cosx` `A^2=sin^2x+cos^2x+2sinxcosx` `A^2=(A^2/2-1)` `sin^4x+cos^4x=(sin^2x)^2+(cos^2x)^2` `=(sin^2x+cos^2x)^2-2sin^2xcos^2x` `=1-2(sinxcosx)^2` `=1-2*(A^2-1)^2/4` `=(2-(A^2-1)^2)/2`. |
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| 653. |
If `tanalpha`is equal to the integral solution of the inequality `4x^2-16 x+15 |
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Answer» It is given that slope of bisector is`cos beta.` `:. cos beta = tan45^@ =>cos beta = 1->(1)` Now, `4x^2-16x+15 lt 0` `=>4x^2-10x-6x+15 lt 0` `=>2x(2x-5)-3(2x-5) lt 0` `=>(2x-3)(2x-5) lt 0` `:. 3/2 lt x lt 5` As, `tan alpha` is an integral solution for this inequality, `:. tanalpha = 2` `=>cotalpha = 1/2->(2)` Now, `sin(alpha+beta)sin(alpha-beta) = sin^2alpha-sin^2beta` `=1/(cosec^2alpha) - (1-cos^2beta)` `=1/(1+cot^2alpha)-1+cos^2beta` From (1) and (2), `1/(1+1/4) -1+1 = 4/5` So, option `d` is the correct option. |
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| 654. |
The perimeter of a triangle ABC is saix times the arithmetic mean ofthe sines of its angles. If the side `ai s1`then find angle `Adot` |
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Answer» Here,` a = 1` It is given that, `a+b+c = 6((sinA+sinB+sinC)/3)` As `a = 2RsinA, b = 2RsinB and c = 2RsinC` `=>2R(sinA+SinB+sinC) = 2(sinA+sinB+sinC)` `=>R = 1` Now, ` a = 2RsinA` `1 = 2(1)sinA` `=>sinA = 1/2` `=>A= pi/6 = 30^@.` |
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| 655. |
Find the sines and cosines of all angles in the first four quadrants whose tangents are equal to `cos 135^@` |
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Answer» `cos135=-1/sqrt2` `tanp=-1/sqrt2` `sinp=1/sqrt2`(in second quater) `=-1/sqrt2`( in fourth quater) `cos(P)=-1/sqrt2`(in second quadrent) `=1/sqrt2`( in fourth quadrent). |
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| 656. |
The roots of the equation `4x^2-2sqrt(5)x+1=0,a r e`(a)`sin36^0,sin18^0`(b) `sin18^0,cos36^0`(c)`sin36^0,cos18^0`(d) `cos18^0,cos36^0` |
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Answer» `4x^2-2sqrt5x+1 = 0` this is a quadratic equation, so its roots are, `=> x= (-(-2sqrt5)+-sqrt((-2sqrt5)^2-4(4)(1)))/(2(4))` `=>x = (2sqrt5+-sqrt4)/(8)` `=>x= (sqrt5+-1)/4` As `sin18^@ = (sqrt5-1)/4 and cos36^@ = (sqrt5+1)/4` `=>x = sin18^@ and x = cos36^@` So, option `(b)` is the correct option. |
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| 657. |
`alpha,beta,gammaand delta` are angles in I,II,II and IV quadrants, respectively and none of them is an integral multiple of `pi//2`. They form an increasing arithmetic progression. Which of the following holds?A. `cos(alpha-delta)gt0`B. `cos(alpha-delta)=0`C. `cos(alpha-delta)lt0`D. `cos(alpha-delta)gt0or `cos(alpha-delta)lt0` |
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Answer» Correct Answer - A `0ltalphalt90^@,90^@ltbetalt180^@` `180^@ltgammalt270^@,270^@ltdeltalt360^@` `rArrgammalt270^@ltalpha+deltalt450^@` `rArr alpha+delta` lies in the I or IV quadrant and cosine in both is positive. If d is the common ratio of theA.P., then `beta=alpha+d,gamma=alpha+2d,delta=alpha+3d` `rArr beta+gamma=alpha+delta,2(alpha-beta)=-2d=beta-delta` `and alpha+gamma=2beta`, Now, `beta-gamma=-d,alpha-delta=-3d` `270^@ltdeltalt360^@` `rArr 270^@ltalpha+3dlt360^@` `rArr 200^@lt3dlt290^@ ( :. alpha=70^@)` `rArr400^@lt+6d,580^@` `rArr 680^@lt4alpha+6dlt860^@` `680^@ lttheta lt 860^@` |
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| 658. |
Find the least positive value of `x`satisfying`(sin^2 2x+4sin^4x-4sin^2xcos^2x)/4=1/9` |
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Answer» `(sin^2 2x+4sin^4x - 4sin^2xcos^2x)/4 = 1/9` `=>((2sinxcosx)^2+4sin^4x - 4sin^2xcos^2x)/4 = 1/9` `=>(4sin^2xcos^2x+4sin^4x - 4sin^2xcos^2x)/4 = 1/9` `=>(4sin^4x)/4 = 1/9` `=>sin^4x = 1/9` `=>sinx = +-1/sqrt3` So, least positive value of `x` will be `sin^-1(1/sqrt3).` |
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| 659. |
If `f(x)=2(7cosx+24sinx)(7sinx-24cosx),`for every `x in R ,`then maximum value of `f(x)^(1/4)`is_____ |
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Answer» `f(x) = 2(7cosx+24sinx)(7sinx-24cosx)` `=>f(x) = 2*25*25(7/25cosx+24/25sinx)(7/25sinx-24/25cosx)` Let `cos alpha = 7/25`, Then, `sinalpha = sqrt(1-(7/25)^2) = sqrt(576/625) = 24/25` `:. f(x) = 2*25*25(cosalphacosx+sinalphasinx)(cosalphasinx-cosalphacosx)` `=>f(x) = 25*25*2(cos(x-alpha)sin(x-alpha))` `=>f(x) = 25*25*sin(2(x-alpha))` As maximum value of `sin(2(x-alpha))` is `1`, `:. f(x)_max = 25*25` `=>(f(x)^(1/4))_max = (25*25)^(1/4)` `=>(f(x)^(1/4))_max = (25)^(1/2) = 5.` So, the maximum value of `f(x)^(1/4)` is `5`. |
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| 660. |
If `alpha` and `beta` are acute angles and `cot alpha=1/4` and `cot beta=5/3`. prove that `alpha-beta=45` |
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Answer» `cotalpha=1/4,tanalpha=4` `cotbeta=5/,tanbeta=3/5` `tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)` `=(4-3/5)/(1+12/5)=1` `alpha-beta=45^o` since `alpha<90^o` `beta<90^o`. |
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| 661. |
If : `csc theta - cot theta = p, "then" : csc theta=` A)`theta + (1)/(p)` B)`theta - (1)/(p)` C)`(1)/(2) (p +(1)/(p))` D)`(1)/(2) (p - (1)/(p))`A. `theta + (1)/(p)`B. `theta - (1)/(p)`C. (1)/(2) (p +(1)/(p))D. (1)/(2) (p - (1)/(p)) |
| Answer» Correct Answer - C | |
| 662. |
If `theta_1` and `theta_2`are two values lying in `[0,2pi]`for which `t a ntheta=lambda,`then `tan((theta_1)/2)tan((theta_2)/2)`is equal to(a)0 (b) `-1`(c) 2(d) 1 |
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Answer» `tan theta=lambda` `tan2A=(2tanA)/(1-tan^2A)` `(2tan(theta/2))/(1-tan^2(theta/2))=lambda` `lambdatan^2(theta/2)+2tan(theta/2)-lambda=0` `tan(theta/2)=(-2pmsqrt(4+4lambda^2))/(2lambda)` `=(-1pmsqrt(1+lambda^2))/lambda` `tan(theta_1/2)=(-1+sqrt(1+lambda^2))/lambda` `tan(theta_2/2)=(1--sqrt(1+lambda^2))/lambda` `tan(theta_1/2)tan(theta_2/2)=((-1+sqrt(1+lambda^2))/lambda)((-1-sqrt(1+lambda^2))/lambda)=-1`. |
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| 663. |
If `t a ntheta/2=sqrt((a-b)/(a+b))tanvarphi/2`, prove that `costheta=(a cosvarphi+b)/(a+b cosvarphi)`. |
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Answer» `L.H.S. = cos theta = (1-tan^2(theta/2))/(1+tan^2(theta/2))` `=(1-((a-b)/(a+b)tan^2(phi/2)))/(1+((a-b)/(a+b)tan^2(phi/2)))` `=(1-((a-b)/(a+b)sin^2(phi/2)/cos^2(phi/2)))/(1+((a-b)/(a+b)sin^2(phi/2)/cos^2(phi/2)))` `=((a+b)cos^2(phi/2)-(a-b)sin^2(phi/2))/((a+b)cos^2(phi/2)+(a-b)sin^2(phi/2))` `=(a(cos^2(phi/2) - sin^2(phi/2))+b(cos^2(phi/2) + sin^2(phi/2)))/(a(cos^2(phi/2) + sin^2(phi/2))+b(cos^2(phi/2) - sin^2(phi/2)))` `=(acosphi+b)/(a+bcosphi)` `=R.H.S.` |
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| 664. |
Consider angles `alpha = (2n+(1)/(2))pi pm A` and `beta = m pi +(-1)^(m)((pi)/(2)-A)` where n, m `in` I. Which of the following is not true ?A. `alpha` and `beta` are always the same anglesB. `alpha` and `beta` are co-terminal anglesC. `sin alpha = sin beta` but `cos alpha ne cos beta`D. none of these |
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Answer» Correct Answer - C Let m = 2k, i.e. m is even where `k in I` `therefore beta = 2k pi + (pi)/(2)-A=(2k+(1)/(2))pi -A` ….(1) If m = 2k + 1, i.e., m is odd, then `therefore beta = (2k+1)pi-((pi)/(2)-A)=(2k+(1)/(2))pi+A` …..(2) From (1) and (2), `beta` can be expressed as `beta = (2k+(1)/(2))pi pm A, k in I` which is same as `alpha`. |
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| 665. |
`csc theta - 2 cos theta * cot 2 theta =`A)`2 sin theta`B)`sin 2 theta`C)`2 cos theta`D)`cos 2 theta`A. `2 sin theta`B. `sin 2 theta`C. ` 2 cos theta`D. `cos 2 theta` |
| Answer» Correct Answer - A | |
| 666. |
If `sin theta = 4//5, where pi//2 lt theta lt pi,` evaluate : `(sec theta + tan theta)/(cosec theta - cot theta)`A. `2//3`B. `-1(1)/(2)`C. `-3//4`D. `-1(1)/(3)` |
| Answer» Correct Answer - B | |
| 667. |
If : `tan((pi)/(4)+theta) - tan((pi)/(4) - theta) =m*tan (ntheta),"then"`A)`m lt n` B)m = n ] C)`m gt n` D)`m^(n) lt n`A. `m lt n`B. m = n ]C. `m gt n`D. `m^(n) lt n` |
| Answer» Correct Answer - B | |
| 668. |
`cos^2((pi)/(4)-theta)+cos^2((pi)/(4)+theta)=.....`A. `sin 2 theta`B. `cos 2 theta`C. 1D. 0 |
| Answer» Correct Answer - C | |
| 669. |
If : `sin A + cos A =1, "then" :sin A - cos A =` A)`pm 1` B)0 C)`pm 2` D)`pm 3`A. `pm 1`B. 0C. `pm 2`D. `pm 3` |
| Answer» Correct Answer - A | |
| 670. |
`16 * cos ""(pi)/(12) * cos"" (4pi)/(12) * cos"" (5pi)/(12)=`A)`sqrt1`B)`sqrt2`C)`sqrt3`D)`sqrt4`A. `sqrt1`B. `sqrt2`C. `sqrt3`D. `sqrt4` |
| Answer» Correct Answer - C | |
| 671. |
If : `cos theta-sintheta=sqrt2.sin theta, "then": costheta+sintheta=`A. `sqrt2*csctheta`B. `sqrt2*sectheta`C. `sqrt2*costheta`D. `sqrt2*tantheta` |
| Answer» Correct Answer - C | |
| 672. |
`(csc theta - sin theta)(sec theta - cos theta)(tan theta+ cot theta)=`A)1 B)0 C)`-1`D) 2A. `sin theta * cos theta * tan theta * csc theta * cot theta`B. 0C. -1D. 2 |
| Answer» Correct Answer - A | |
| 673. |
If : `sin 40^(@)-cos 70^(@)= k * cos 80^(@), "then" : k = `A)`sqrt1`B)`sqrt2`C)`sqrt3`D)`sqrt4`A. `sqrt1`B. `sqrt2`C. `sqrt3`D. `sqrt4` |
| Answer» Correct Answer - C | |
| 674. |
If `Tan(picostheta)=cot(pisintheta)` then the value of (s) of `cos(theta-pi/4)` is(are)A. `(1)/(2)`B. `- (1)/(sqrt2)`C. `-(1)/(2sqrt2)`D. `(1)/(2sqrt2)` |
| Answer» Correct Answer - D | |
| 675. |
If : `sin^(2) ((pi)/(4) + (theta)/(2)) - cos^(2) ((pi)/(4) + (theta)/(2))=` A)`sin theta` B)` cos theta` C)`sin 2 theta` D)`cos 2 theta`A. `sin theta`B. ` cos theta`C. `sin 2 theta`D. `cos 2 theta` |
| Answer» Correct Answer - A | |
| 676. |
If `cos^(2)theta-sin^(2)theta=tan^(2)alpha,` `"then" : sqrt2 * cos theta * cos alpha=....`A) `sin theta * sin alpha`B)1 C)`tan^(2)theta` D)none of theseA. `sin theta * sin alpha`B. 1C. `tan^(2)theta`D. none of these |
| Answer» Correct Answer - B | |
| 677. |
If `x= sin^(2)theta* cos theta and y=sin theta cos^(2)theta,"then" :`A. `(x^(2)y)^(2//3)=(xy^(2))^(2//3)=1`B. `(x^(2)y)^(2//3)=(y^(2)//x)^(2//3)=1`C. `x^(2)+y^(2)+x^(2)y^(2)`D. none of these |
| Answer» Correct Answer - B | |
| 678. |
If : `cot theta =(a)/(b), "then" : a *cos 2 theta +b * sin 2 theta=` A)`a^(2) +b^(2)` B)`a^(2) - b^(2)` C)a D)bA. `a^(2) +b^(2)`B. `a^(2) - b^(2)`C. aD. b |
| Answer» Correct Answer - C | |
| 679. |
`cos(54 0^(@)-theta)-sin(63 0^(@)-theta)` is equal to |
| Answer» Correct Answer - A | |
| 680. |
If `sec theta=sqrt(2)a n d(3pi)/2 |
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Answer» `sec theta = sqrt2` As, `(3pi)/2 lt theta le (2pi)` `:. theta = (7pi)/4` Now, `(1+tantheta+cosectheta)/(1+cot theta- cosectheta)` `=(1+tan((7pi)/4)+cosec((7pi)/4))/(1+cot ((7pi)/4)- cosec((7pi)/4))` `=(1-1-sqrt2)/(1-1+sqrt2)` `=-sqrt2/sqrt2 = -1` `:. (1+tantheta+cosectheta)/(1+cot theta- cosectheta) =-1` |
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| 681. |
If : `sin x +cos x = sin 2 x + cos 2 x , "where" 0 lt x le (pi)/(2), "then x":=`A. `(pi)/(2)`B. `(pi)/(3)`C. `(pi)/(4)`D. `(pi)/(6)` |
| Answer» Correct Answer - D | |
| 682. |
If `cos e ctheta+cottheta=(11)/2`, then `t a ntheta=``(21)/(22)`(b) `(15)/(16)`(c) `(44)/(117)`(d) `(117)/(44)`A. `(21)/(22)`B. `(15)/(16)`C. `(44)/(117)`D. `(117)/(44)` |
| Answer» Correct Answer - C | |
| 683. |
If `xsin45^@cos^2(60^@)=(tan^2(60^@)cosec30^@)/(sec45^@cot^2(30^@)), ` then x=A. 2B. 4C. 8D. 16 |
| Answer» Correct Answer - D | |
| 684. |
The number of solutions of equation `tanx+secx=2cosx` lying in the interval `[0, 2pi]` is |
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Answer» Correct Answer - C Given equation, `" "tanx+secx=2cosx` `rArr" "(sinx)/(cosx)+(1)/(cosx)=2cosx` `rArr" "1+sinx=2cos^(2)x` `rArr" " 1+sinx=2(1-sin^(2)x)` `rArr" "1+sinx=2-2sin^(2)x` `rArr" "2sin^(2)x+sinx-1=0` `rArr" "2sin^(2)x+sinx-1=0` `rArr" "2sinx(sinx+1)-1(sinx+1)=0` `rArr" "(sinx+1)(2sinx-1)=0` `rArr" "sinx+1=0 or (2sinx-1)=0` `rArr" "sinx=-1, sinx=(1)/(2)` `therefore" "x=(3pi)/(2), x= (pi)/(6)` Hence, only two solutions possible. |
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| 685. |
For any triangle ABC, prove that`(b+c)cos((B+C)/2)=acos((B-C)/2)` |
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Answer» We know,`a/sinA = b/sinB = c/sinC = k` where, k is a constant. `:. (b+c)/a = (k(sinB+sinC))/(ksinA)` `= (sinB+sinC)/sinA` `=(2sin((B+C)/2)cos((B-C)/2))/(2sinA/2cosA/2)` As, `A+B+C = 180, So, B+C = 180-A` `:. (b+c)/a=(sin((180-A)/2)cos((B-C)/2))/(sin((180-(B+C))/2)cosA/2)` `=(sin(90-A/2)cos((B-C)/2))/(sin(90-(B+C)/2)cosA/2)` `=(cos(A/2)cos((B-C)/2))/(cos((B+C)/2)cosA/2)` `(b+c)/a= (cos((B-C)/2))/(cos((B+C)/2))` So, `(b+c)(cos((B+C)/2)) = a(cos((B-C)/2))` |
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| 686. |
If in a triangle `A B C ,acos^2(C/2)ccos^2(A/2)=(3b)/2,`then the sides `a ,b ,a n dc```are in A.P.b. are in G.P.c. are in H.P. d. satisfy `a+b=cdot`A. A.PB. G.PC. H.P.D. A.G.P. |
| Answer» Correct Answer - A | |
| 687. |
In a `DeltaABC`, `a(cos^(2)B+cos^(2)C)+cosA(ccosC+bcosB)=`A. `a+b+c`B. `c`C. `b`D. `a` |
| Answer» Correct Answer - D | |
| 688. |
In `triangleABC, a(b^(2)+c^(2))cosA+b(c^(2)+a^(2))cosB+c(a^(2)+b^(2))cosC=`A. `0`B. `3abc`C. `3a^(2)bc`D. `3ab^(2)c` |
| Answer» Correct Answer - B | |
| 689. |
In `triangleABC, (b+c)cosA+(c+a)cosB+(a+b)cosC=`A. `a+b+c`B. `2(a+b+c)`C. `(a+b+c)/(2)`D. `(a+b+c)/(4)` |
| Answer» Correct Answer - A | |
| 690. |
IN `triangleABC, (cosC+cosA)/(c+a)+(cosB)/(b)=`A. `(1)/(a)`B. `(1)/(b)`C. `(1)/(c)`D. `(a+b)/(b)` |
| Answer» Correct Answer - B | |
| 691. |
In a `DeltaABC`, cosecA(sinBcosC+cosBsinC)` is equal toA. `1`B. `(a)/(c)`C. `(c)/(a)`D. `(c)/(ab)` |
| Answer» Correct Answer - A | |
| 692. |
If `a^(2),b^(2)` and `c^(2)` are in AP, then cotA, cotB and cotC are inA. A.PB. G.PC. H.P.D. A.G.P. |
| Answer» Correct Answer - A | |
| 693. |
In `triangleABC, (cosA)/(c cosB+bcosC)+(cosB)/(acosC+ccosA)+(cosC)/(acosB+bcosA)=`A. `(a^(2)+b^(2)-c^(2))/(4)`B. `(a^(2)+b^(2)-c^(2))/(2)`C. `(a^(2)+b^(2)+c^(2))/(4abc)`D. `(a^(2)+b^(2)+c^(2))/(2abc)` |
| Answer» Correct Answer - D | |
| 694. |
Find the length of an arc of a circle of radius 5cm subtending acentral angle measuring `15^0dot` |
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Answer» Length of an arc is given by , `l = pi/180*r *theta` Here, `r = 5cmand theta = 15^@` `:. l = pi/180*5*15 = (5pi)/12 cm` So, the length of the arc is `(5pi)/12 cm`. |
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| 695. |
General solution of `3sec^(2)x=4` isA. `npipm(pi)/(6), ninZ`B. `npipm(5pi)/(6), ninZ`C. `npipm(7pi)/(6), ninZ`D. `npipm(11pi)/(6), ninZ` |
| Answer» Correct Answer - A | |
| 696. |
If `sin^(- 1)(x)+sin^(- 1)(2x)=pi/3` then `x=`A. `(sqrt(3))/(2sqrt(7))`B. `(1)/(2)`C. `(1)/(sqrt(2)`D. `(sqrt(3))/(2)` |
| Answer» Correct Answer - A | |
| 697. |
If `cos(2sin^(-1)x)=1/9,`then find the values of `xdot`A. `(4)/(9)`B. `pm(2)/(3)`C. `(2)/(3)`D. `(-2)/(3)` |
| Answer» Correct Answer - B | |
| 698. |
For `0le|x|lesqrt(2)`, if `cos^(-1)(x^(2)-(x^(4))/(2)+(x^(6))/(4)-…)+sin^(-1)(x-(x^(2))/(2)+(x^(3))/(4))-…)=(pi)/(2)`, then x=A. `(-1)/(2)`B. `(1)/(2)`C. `-1`D. `1` |
| Answer» Correct Answer - D | |
| 699. |
In quadrilateral `A B C D ,`if`sin((A+B)/2)cos((A-B)/2)+"sin"((C+D)/2)cos((C-D)/2)=2`then find the value of `sinA/2sinB/2sinC/2sinD/2dot` |
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Answer» `1/2[2sin((A+B)/2)cos((A-B)/2)+2sin((C+D)/2)cos((C-D)/2)]=2` `2sinCcosD=sin(C+D)+sin(C-D)` `sinA+sinB+sinC+sinD=4` `sinA-11=sinB=sinC=sinD` `A=pi/2=B=C=D` `sin(A/2)sin(B/2)sin(C/2)sin(D/2)` `=sin^4pi/4=(1/sqrt2)^4=1/4`. |
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| 700. |
`sin 3 alpha = 4 sin alpha sin(x + alpha) sin(x-alpha)`A. `npipm(pi)/(6), ninZ`B. `npipm(pi)/(4), ninZ`C. `npipm(pi)/(3), ninZ`D. `npipm(pi)/(2), ninZ` |
| Answer» Correct Answer - C | |