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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.
| 401. |
If `sin x + cos x = 1/5`, then tan 2x is |
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Answer» `sinx+cosx=1/5` we can replace `cosx=sqrt(1-sin^2(x))` =>`sqrt(1-sin^2(x))=1/5-sinx` squaring on both sides `1-sin^(2)x=1/25+sin^(2)x-2/5sinx` on solving we get =>`50sin^(2)x-10sinx-24=0` =>`50sin^(2)x-40sinx+30sinx-24=0` =>`(5sinx-4)(10sinx+6)=0` => thus, `sinx=4/5 or -3/5` thus `cosx=-3/5 or 4/5` thus `tanx=-4/3 or -3/4` `tan2x=(2tanx)/(1-tan^(2)x)` substituting values of `tanx` we get `tan2x=24/7 or -24/7` |
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| 402. |
`(1/(sec^2A-cos^2A)+1/(cosec^2A-sin^2A))sin^2Acos^2A=(1-sin^2Acos^2A)/(2+sin^2Acos^2A)` |
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Answer» LHS `=(1/(1/cos^2A -cos^2A) + 1/(1/sin^2A - Sin^2A))sin^2A.cos^2A` `=(cos^2A/(1-cos^4A) + sin^2A/(1- Sin^4A))sin^2Acos^2A` `=((cos^2A)/((1-cos^2A)(1+cos^2A)) + (sin^2A)/((1-sin^2A)(1+sin^2A)))sin^2Acos^2A` `= (cos^2Asin^2Acos^2A)/(sin^2A(1+cos^2A)) + (sin^2A sin^2Acos^2A)/(cos^A(1+sin^2A))` `= cos^4A/(1+ cos^2A) + sin^4A/(1+sin^2A)` `= (cos^4A(1+sin^2A) + sin^4A(1+ cos^2A))/((1+cos^2A)(1+sin^2A))` `= (cos^4A + sin^2A.cos^4A + sin^4A + sin^4Acos^2A)/(1 + cos^2A + sin^2A + sin^2Acos^2A)` `= ((cos^2A)^2 + (sin^2A)^2 + sin^2Acos^2A(cos^2A+sin^2A))/(2 + sin^2Acos^2A)` by using identity, `(a+b)^2 = a^2 + b^2 + 2ab` `= ((cos^2A + sin^2A)^2 - 2cos^2Asin^2A + sin^2Acos^2A)/(2 + sin^2Acos^2A)` `= (1- sin^2Acos^2A)/(2+sin^2Acos^2A) ` = RHS |
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| 403. |
Find the range of `y=sin^3x-6sin^2x+11sinx-6.`A. `[-24, 2]`B. `[-24,0]`C. `[0,24]`D. `[-24,24]` |
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Answer» Correct Answer - B `y=sin^(3)x-6sin^(2)x+11 sin x-6` Put sin x = 1 `therefore y=t^(3)-6t^(2)+11t-6,-1le t le 1` `therefore (dy)/(dt)=3t^(2)-12t+11=3(t-2)^(2)-1gt 0` for `-1le t le 1` `therefore` Function is increasing Hence, range is `[f(-1),f(1)]` or `[-24,0]` |
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| 404. |
IF ` tan ^(-1) 2x + tan ^(-1) 3x =(pi)/(4) , ` then x=A. `1`B. `-1`C. `(1)/(3)`D. `(1)/(6)` |
| Answer» Correct Answer - D | |
| 405. |
If `sin^(-1)(1-x)-2sin^(-1)x=cos^(-1)x`, then x=A. `0, (1)/(2)`B. `1`C. `0`D. `(1)/(2)` |
| Answer» Correct Answer - A | |
| 406. |
`2sin^(-1)((3)/(5))=`A. `tan^(-1)((4)/(3))`B. `tan^(-1)((7)/(24))`C. `tan^(-1)((3)/(4))`D. `tan^(-1)((24)/(7))` |
| Answer» Correct Answer - D | |
| 407. |
In ` A B C ,`if `A B=c`is fixed, and `cosA+cosB+2cos C=2`then the locus of vertex `C`isellipse (b) hyperbola (c)circle (d) parabola |
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Answer» `cosA+cosB+2cosC=2` `cosA+cosB=2(1-cosC)` `cosA+cosB=4sin^2(C/2)` `2cos((A+B)/2)cos((A-B)/2)=4cis^2(C/2)` `cos((A-B)/2)=2sinn(C/2)` `2cos(C/2)cos((A-B)/2)=4sin(C/2)cos(C/2)` `2sin((A+B)/2)cos((A-B)/2)=sinC` `sinA+sinB=2sinC` `a+b=2c` Locus os vertex C isan ecllipse with vertex A and B as focii. |
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| 408. |
For any triangle ABC, prove that`(a-b)/c=(sin((A-B)/2))/(cosC/2)` |
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Answer» From sine law, we have,`a/sinA = b/sinB = c/sinC = k` So, `a = ksinA, b = ksinB, c = ksinC` `:. L.H.S. = (a-b)/c = (k(sinA - sinB))/(ksinC)` `= (sinA- sinB)/(sinC)` `=(2sin((A-B)/2)cos((A+B)/2))/(2sin(C/2)cos(C/2))` Here, `A+B+C = 180. So, A+B = 180- C` So, our expression becomes, `= (sin((A-B)/2)cos((180-C)/2))/(sin(C/2)cos(C/2))` `= (sin((A-B)/2)cos(90-C/2))/(sin(C/2)cos(C/2))` `= (sin((A-B)/2)sin(C/2))/(sin(C/2)cos(C/2))` `= sin((A-B)/2)/cos(C/2) = R.H.S.` |
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| 409. |
In any triangle ABC, if `a= 18 , b= 24 , c= 30`, findsinA, sinB, sinC |
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Answer» Here, `30^2 = 18^2+24^2` That means triangle ABC is aright angle triangle with `/_C = 90^@` With the given details,`sinA = (BC)/(AB) = a/c = 18/30 = 3/5` `sinB= (AC)/(AB) = b/c = 24/30 = 4/5` `sinC = sin90^@ = 1` |
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| 410. |
In `triangleABC`, If `angleA=25^(@), angleB=85^(@) and c=3,4`, then a=A. `3.0604`B. `3.604`C. `1.0529`D. `1.529` |
| Answer» Correct Answer - D | |
| 411. |
In `triangleABC`, if `a=72, angleB=108^(@), angleA=25^(@)`, then b=A. `162.04`B. `162.14`C. `124.61`D. `124.16` |
| Answer» Correct Answer - A | |
| 412. |
The value of `sin(sin^(-1)1/2+cos^(-1)1/2)=?`A. `2`B. `-2`C. `1`D. `-1` |
| Answer» Correct Answer - C | |
| 413. |
The value of `sin^(-1)((2sqrt(2))/(3))+sin^(-1)((1)/(3))` isA. `(pi)/(2)`B. `(3pi)/(2)`C. `(9pi)/(4)`D. `(9pi)/(8)` |
| Answer» Correct Answer - A | |
| 414. |
For any triangle ABC, prove that`(a+b)/c=(cos((A-B)/2))/(sinC/2)` |
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Answer» let `a/sinA = b/sin B = c/Sin C = k` (say) `(a+b)/c = (ksinA + ksinB)/(ksinC) ` `= (sinA + sinB)/(sinC)` `= (2sin((A+B)/2)cos((A-B)/2))/(2sin(C/2)cos(C/2))` `` as A+B+C = `180^@` `= (sin(90^@ - C/2)cos((A-B)/2))/(sin(C/2)cos(C/2))` `= (cos(C/2).cos((A-B)/2))/(sin(C/2)cos(C/2)) ` `= (Cos((A-B)/2))/Sin(c/2)` |
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| 415. |
`tan^(-1)(1)+cos^(-1)((1)/(2))+sin^(-1)((1)/(2))=`A. `(pi)/(4)`B. `(3pi)/(4)`C. `(pi)/(12)`D. `(2pi)/(3)` |
| Answer» Correct Answer - B | |
| 416. |
If : `x=sin(B-C) * cos A, ` `y=sin(C-A) * cos B, ` `z=sin(A-B) * cos C, ` then : x+y+z=(a) sin(A-B-C) (b) cos(A-B-C) (c)0 (d)-1A. sin (A-B-C)B. cos (A-B-C)C. 0D. -1 |
| Answer» Correct Answer - C | |
| 417. |
If `a=9, b =8 and c=x` satisfies `3 cos C=2,` thenA. `4`B. `5`C. `6`D. `7` |
| Answer» Correct Answer - D | |
| 418. |
In any triangle ABC, if `a=18 , b=24 , c=30`, findcosA, cosB, cosC |
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Answer» Here, `30^2 = 18^2+24^2` That means triangle ABC is aright angle triangle with `/_C = 90^@` With the given details,`cosA = (AC)/(AB) = b/c = 24/30 = 4/5` `cosB = (BC)/(AB) = a/c = 18/30 = 3/5` `cosC = cos90^@ = 0` |
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| 419. |
`[(1-tan^(2)30^(@))div(1+tan^(2)30^(@))]-sin30^(@)=`A)-1 B)0 C)1 D)2A. -1B. 0C. 1D. 2 |
| Answer» Correct Answer - B | |
| 420. |
In `triangleABC, sin((B-C)/(2))=`A. `2((b-c)/(a))cos((A)/(2))`B. `((b-c)/(a))cos((A)/(2))`C. `2((c-b)/(a))cos((A)/(2))`D. `((c-b)/(a))cos((A)/(2))` |
| Answer» Correct Answer - B | |
| 421. |
If : `x*cos A *cos B= sin (A-B), ` `y*cos B *cos C= sin (B-C), ` `z*cos C *cos A= sin (C-A), ` then : x+y+z= (a)-1 (b)0 (c)sin(A+B+C) (d) cos(A-B-C)A. -1B. 0C. sin(A + B + C)D. cos (A - B - C) |
| Answer» Correct Answer - B | |
| 422. |
General solution of `3(sec^2 x+tan^2x)=5` isA. `2npipm(pi)/(3), ninZ`B. `2npipm(pi)/(6), ninZ`C. `npipm(pi)/(3), ninZ`D. `2npipm(pi)/(3), ninZ` |
| Answer» Correct Answer - D | |
| 423. |
In `triangleABC, cos((A-B)/(2))=`A. `((a+b)/(c))sin((C)/(2))`B. `((a+b)/(2c))sin((C)/(2))`C. `((a+b)/(c))cos((C)/(2))`D. `((a+b)/(2c))cos((C)/(2))` |
| Answer» Correct Answer - A | |
| 424. |
`tan^(-1)1+tan^(-1)2+tan^(-1)3=`A. `0`B. `pi`C. `(pi)/(4)`D. `(pi)/(2)` |
| Answer» Correct Answer - B | |
| 425. |
`cos(tan^(-1)((1)/(3))+tan^(-1)((1)/(2)))=`A. `(-1)/(sqrt(2))`B. `(1)/(sqrt(2))`C. `(1)/(2)`D. `(-1)/(2)` |
| Answer» Correct Answer - B | |
| 426. |
`tan^(-1)((1)/(2))+tan^(-1)((1)/(3))=`A. `0`B. `(pi)/(6)`C. `(pi)/(4)`D. `(-pi)/(4)` |
| Answer» Correct Answer - C | |
| 427. |
If `(sec^4theta)/a+(tan^4theta)/b=1/(a+b),`then prove that `|b|lt=|a|`. |
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Answer» `(sec^4theta)/a+(tan^4b)/b=1/(a+b)=(sec^2theta-tan^2theta)/(a+b)` `rArr(sec^4theta)/(a(a+b))[(a+b)sec^2theta-a]+tan^2theta/(b(a+b))[(a+b)tan^2theta+b]=0` `rArra tan^2theta+bsec^2theta=0` `rArrsin^2theta=-b/a` Since`sin^2thetale1` `rArrabs(b/a)le1orabsbleabsa` |
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| 428. |
If `tan^(-1)(1+x)+tan^(-1)(1-x)=pi/2` then `x=?`A. `-1`B. `0`C. `1`D. `pi` |
| Answer» Correct Answer - B | |
| 429. |
`tan^(-1)((1)/(2))+tan^(-1)((2)/(11))=`A. `tan^(-1)((4)/(3))`B. `tan^(-1)((3)/(4))`C. `pi+tan^(-1)((4)/(3))`D. `pi+tan^(-1)((3)/(4))` |
| Answer» Correct Answer - B | |
| 430. |
If `xge0 and tan^(-1)((1-x)/(1+x))=(1)/(2)tan^(-1)x,` then x=A. `3`B. `2`C. `(1)/(sqrt(3))`D. `(1)/(sqrt(2))` |
| Answer» Correct Answer - C | |
| 431. |
Evaluate: `tan{2t a n-1 1/5-pi/4}`A. `(-17)/(7)`B. `(-7)/(17)`C. `(17)/(7)`D. `(7)/(17)` |
| Answer» Correct Answer - B | |
| 432. |
Find the value of `cos (-171 0^(@))` |
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Answer» We know, for any natural number n`cos(2npi +x) = cos x`. So, `cos(-1710^@) =cos(5**2pi + (-1710^@))` `cos(1800^@-1710^@) = cos90^@ = 0` |
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| 433. |
If `sin^(-1)(1-x) sin^(-1)x=(pi)/(2)` then x equalA. `(-1)/(2)`B. `1`C. `0`D. `(1)/(2)` |
| Answer» Correct Answer - C | |
| 434. |
If `tan^(-1)(x-1)/(x-2)+tan^(-1)(x+1)/(x+2)=pi/4`, then find the value of x.A. `(1)/(2)`B. `pm(1)/(2)`C. `pm(1)/(sqrt(2))`D. `pmsqrt(2)` |
| Answer» Correct Answer - C | |
| 435. |
Find the value of sin `(31pi)/3`. |
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Answer» We know, `sin(2npi+x) = sinx` where n is a positive number. Now, `sin(31pi/3) = sin(10pi+pi/3)=sin(2(5pi)+pi/3)` `=sin(pi/3)=sqrt3/2` |
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| 436. |
The principal solution `cotx=1` areA. `(pi)/(4), (3pi)/(4)`B. `(pi)/(4), (5pi)/(4)`C. `(5pi)/(4), (7pi)/(4)`D. `(3pi)/(4), (7pi)/(4)` |
| Answer» Correct Answer - B | |
| 437. |
Find the principal and general solution of `cotx=-sqrt(3)` |
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Answer» Here, `cotx = -sqrt3` We know, `cot(pi/6) = sqrt3` and cot x is negative in 2nd and 4th quadrant. So, principal solution will be` (pi-pi/6)` and `(2pi-pi/6)` Principle solution `x = (5pi)/6 and (11pi)/6` General solution,`x = (npi-pi/6)` |
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| 438. |
Find the range of the function `5sinx -12cosx +7` |
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Answer» `f(x)=5sinx-12cosx+7` `sqrt(a^2+b^2)/sqrt(a^2+b^2)(asinx++bcosx)` `sqrt(a^2+b^2)(a/sqrt(a^2+b^2)sinx+b/sqrt(a^2+b^2)cosx)` `sinalpha=a/sqrt(a^2+b^2)` `cosalpha=b/sqrt(a^2+b^2)` `sqrt(5^2+12^2)=sqrt(25+144)=sqrt169=13` `13/13(5sinx-12cosx)+7` `13(5/13sinx-12/13cosx)+7` `cosalpha=5/13` `sinalpha=12/13` `13(cosalphasinx-sinalphacosx)+7` `13(sin(x-alpha))+7` `-1<=sin(x-alpha)<=1` `-13<=13sin(x-alpha)<=13` `-13+7<=13sin(x-alpha)+7<=13+7` `-6<=f(x)<=20` `f(x) in [-6,20]`. |
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| 439. |
` Cos { 2 pi/2^64 - 1) cos{2^2 pi/(2^64-1)}.........cos{2^64 pi/(2^64 - 1)}` = |
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Answer» Let `(2pi)/(2^64-1) = theta` Then, our expression becomes, `costhetacos2thetacos4theta......cos(2^63)theta` `=1/(2sintheta)(2sinthetacosthetacos2thetacos4theta.....cos(2^63)theta)` `=1/(2sintheta)(sin2thetacos2thetacos4theta.....cos(2^63)theta)` `=1/(4sintheta)(2sin2thetacos2thetacos4theta.....cos(2^63)theta)` `=1/(4sintheta)(sin4thetacos4theta.....cos(2^63)theta)` Similarly, if we continue, we will get the required expression as, `=1/((2^64)sintheta)(sin(2^64)theta)` Putting value of `theta`, our expression becomes, `=1/((2^64)sin((2pi)/(2^64-1) ))(sin(2^64)((2pi)/(2^64-1) ))` `=1/((2^64)sin((2pi)/(2^64-1) ))(sin((2pi)/(2^64-1) ))` `=1/(2^64)` `=1/((2^4)^16) = 1/16^16` So, value of the given expression will be `1/16^16`. |
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| 440. |
The principal solution `cotx=-1` areA. `(pi)/(4), (3pi)/(4)`B. `(pi)/(4), (5pi)/(4)`C. `(5pi)/(4), (7pi)/(4)`D. `(3pi)/(4), (7pi)/(4)` |
| Answer» Correct Answer - D | |
| 441. |
`cos (pi/11) cos(2 pi/11) cos(3 pi/11) cos(4 pi/11) cos(5pi/11)` |
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Answer» Here, we will use, `2sinxcosx = sin2x` Now, `cos(pi/11)cos((2pi)/11)cos((3pi)/11)cos((4pi)/11)cos((5pi)/11)` `=1/(2sin(pi/11))(2sin(pi/11)cos(pi/11)cos((2pi)/11)cos((3pi)/11)cos((4pi)/11)cos((5pi)/11))` `=1/(2sin(pi/11))(sin((2pi)/11)cos((2pi)/11)cos((3pi)/11)cos((4pi)/11)cos((5pi)/11))` `=1/(4sin(pi/11))(2sin((2pi)/11)cos((2pi)/11)cos(pi-(8pi)/11))cos((4pi)/11)cos((5pi)/11))` `=1/(4sin(pi/11))(sin((4pi)/11)cos((4pi)/11)cos(-(8pi)/11)cos((5pi)/11))` `=-1/(8sin(pi/11))(2sin((4pi)/11)cos((4pi)/11)cos((8pi)/11)cos((5pi)/11))` `=-1/(8sin(pi/11))(sin((8pi)/11)cos((8pi)/11)cos((5pi)/11))` `=1/(16sin(pi/11))(2sin((8pi)/11)cos((8pi)/11)cos(pi+(5pi)/11))` `=1/(16sin(pi/11))(sin((16pi)/11)cos((16pi)/11))` `=1/(32sin(pi/11))(2sin((16pi)/11)cos((16pi)/11))` `=1/(32sin(pi/11))(sin((32pi)/11))` `=1/(32sin(pi/11))(sin(3pi-(pi)/11))` `=1/(32sin(pi/11))(sin(pi/11))` `=1/32` so, value of the given expression is `1/32.` |
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| 442. |
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 - B | |
| 443. |
If `a^2+2a+cosec^2(pi/2(a+x))=0`, then, find the values of a and x. |
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Answer» Correct Answer - `a = - 1 and x=2n, n in Z` We have `(a+1)^2+cosec^2(pi/2(a+x))-1=0` `or (a+1)^2+cot^2(pi/2(a+x))=0` `rArr a=-1" and " cot^2((-pi)/2+(pix)/2)=0` `rArr tan^2((pix)/2)=0or tan((pix)/2)=0` `rArr (pix)/2=npi,n in Z` `rArr x=2n, n in Z` |
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| 444. |
If `sin^(-1) x + sin^(-1) y = (2pi)/3", then " cos^(-1) x + cos^(-1) y `A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
| Answer» Correct Answer - C | |
| 445. |
If `sin(sin^(-1)1/5+cos^(-1)x)` =1 then x is equal toA. `(-1)/(5)`B. `(1)/(5)`C. `-5`D. `5` |
| Answer» Correct Answer - B | |
| 446. |
The principal solution `secx=sqrt(2)` areA. `(3pi)/(4), (5pi)/(4)`B. `(pi)/(4), (5pi)/(4)`C. `(3pi)/(4), (7pi)/(4)`D. `(pi)/(4), (7pi)/(4)` |
| Answer» Correct Answer - D | |
| 447. |
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 - A | |
| 448. |
The principal solution `secx=2` areA. `(pi)/(3), (2pi)/(3)`B. `(pi)/(3), (5pi)/(3)`C. `(2pi)/(3), (4pi)/(3)`D. `(4pi)/(3), (5pi)/(3)` |
| Answer» Correct Answer - B | |
| 449. |
The principal solution `secx=-2` areA. `(pi)/(3), (2pi)/(3)`B. `(pi)/(3), (5pi)/(3)`C. `(2pi)/(3), (4pi)/(3)`D. `(4pi)/(3), (5pi)/(3)` |
| Answer» Correct Answer - C | |
| 450. |
If ` cos x + 2 cos y + 3 cos z = sin x + 2 sin y + 3 sin z = 0` then the value of ` sin 3x + 8 sin 3y + 27 sin 3z` is : |
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Answer» `cosx+2cosy+3cosz=0` `isinnx+2isiny+3isinz=0` `e^(ix)+2e^(iy)+3e^(iz)=0` `a^3+b^3+c^3=3abc` `e^(13x)+2e^(i3y)+27e^(13z)=18e^(i(x+y+z))` `cos3x+isinx+8cos3y+i8sinn3y+27cos3z+isin3z=18[cos(x+y+z)+(isin(x+y+z)]` `sin3x+8sin3y+27sin3z=18sin(x+y+z)`. |
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