InterviewSolution
Saved Bookmarks
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.
| 601. |
The equation `sin^4x+cos^4x+sin2x+alpha=0`is solvable for`-5/2lt=alphalt=1/2`(b) `-3lt=alpha |
|
Answer» `sin^4x+cos^4x+sin2x+alpha=0` `(sin^2x)^2+(cos^2x)^2+sin2x+alpha=0` `(sinn^2x+cos^2x)^2-2sin^2x*cos^2x+sin2x+alpha=0` `1-2sin^2x*cos^2x+sin2x+alpha=0` `1-2(sin^2 2x)/4+sin2x+alpha=0` `2-sin^2 2alpha+2sin2x+2alpha=0` `2alpha=sin^2 2alpha-2sin2x-2` `2alpha=y^2-2y-2` `2alpha=(y-1)^2-3` `alpha=(y-1)^2/2-3/2` `-1<=y<=1` `-2<=y-1<=0` `0<=(y-1)^2<=4` `-3<=(y-1)^2-3<=1` `-3<=2alpha<=1` `-3/2<=alpha<=1/2`. |
|
| 602. |
If `pcos e ctheta+qcottheta=2`and `p^2cos e c^2theta-q^2theta=5`then the value of `sqrt(81 p^(-2))`is________ |
|
Answer» Correct Answer - 16 We have `p cosec theta+qcottheta=2 …(1)` `And (pcosectheta)^2-(qcottheta)^2=5` `rArr pcosectheta-qcottheta=5/2 ...(2)` Form (1) and (2) `pcosectheta=9/4,qcot theta=(-1)/4` Now, `cosec^2theta-cot^2theta=1` `rArr 81/(16p^2)-1/(16p^2)=1` `rArr 81p^(-2)-q^(-2)=16` `rArr sqrt(81p^(-2)-q^(-2))=4` |
|
| 603. |
The general solution of `sin^2x - 2cosx +1/4=0` isA. `2npipm(pi)/(6), ninZ`B. `2npipm(pi)/(3), ninZ`C. `npipm(pi)/(6), ninZ`D. `npipm(pi)/(3), ninZ` |
| Answer» Correct Answer - B | |
| 604. |
If : `a * sec alpha+b * tan alpha=m` and : `atan alpha+b * sec alpha=n,` then : `a^(2)-b^(2) = `A. `m^(2)+n^(2)`B. `m^(2)-n^(2)`C. `m^(2)-a^(2)`D. `m^(2)-b^(2)` |
| Answer» Correct Answer - B | |
| 605. |
In any `DeltaABC`, prove that `(c-bcosA)/(b-c cos A)=(cos B)/(cosC)`A. `(cosC)/(cosB)`B. `-(cosC)/(cosB)`C. `(-cosB)/(cosC)`D. `(cosB)/(cosC)` |
| Answer» Correct Answer - D | |
| 606. |
If x takes negative permissible vlaue then `sin^(-1)x=`A. `cos^(-1)(sqrt(1-x^(2)))`B. `-cos^(-1)(sqrt(1-x^(2)))`C. `pi-cos^(-1)(sqrt(1-x^(2)))`D. `cos^(-1)(sqrt(x^(2)-1))` |
| Answer» Correct Answer - B | |
| 607. |
In triangle `A B C ,`let `/_c=pi/2dot`If `r`is the inradius and `R`is circumradius of the triangle, then `2(r+R)`is equal to`a+b`(b) `b+c``c+a`(d) `a+b+c` |
|
Answer» Here, `C = pi/2` Now, `c = 2RsinC = 2R` `=>R = c/2` `=>2R = c->(1)` `r = (s-c)tan(C/2) = (s-c)tan(pi/4)` `=>r = s-c` `=>2r = 2s-2c = a+b+c-2c = a+b-c` From (1), `=>2r = a+b-2R` `=>2(r+R) = a+b.` |
|
| 608. |
In a right-angled isosceles triangle, the ratio of the circumradius andinradius is`2(sqrt(2)+1):1`(b) `(sqrt(2)+1):1``2:1`(d) `sqrt(2):1` |
|
Answer» Sides of right angled isoceles triangle will be , `a,a and sqrt2a.` `:. Delta = 1/2*a*a = a^2/2` `s = (a+a+asqrt2)/2 = (a(2+sqrt2) )/2` `:.` Circumradius `(R) = (abc)/(4Delta) = (a*a*sqrt2a)/(4(a^2/2)) = a/sqrt2` Inradius `(r) = Delta/s = (a^2/2)/((a(2+sqrt2) )/2) = a/(2+sqrt2)` `:. R:r = a/sqrt2 : a/(2+sqrt2) = 2+sqrt2:sqrt2 = sqrt2+1:1` So, option `b` is the correct option. |
|
| 609. |
Solve the equation `2sinx+cosy=2`for the value of `xa n dydot` |
|
Answer» `2sinx+cosy = 2` `=>cosy = 2(1-sinx)->(1)` As `-1 le cosy le 1` `:. -1 le 2(1-sinx) le 1` `=> -1/2 le 1-sinx le 1/2` `=> 1/2 le sinx le 3/2` As maximum value of `sinx` can be `1`, `:. 1/2 le sinx le 1` `:. x in [pi/6,pi/2].` Now, when `x = pi/6`, `cosy = 2(1-sin(pi/6)) = 2(1/2) = 1` `=> y = 0` Now, when `x = pi/2`, `cosy = 2(1-sin(pi/2)) = 2(0) = 0` `=> y = pi/2` `:. x in [pi/6,pi/2].` `:. y in [0,pi/2].` |
|
| 610. |
In any triangle `A B C ,(a^2+b^2+c^2)/(R^2)`has the maximum value of3 (b)6 (c) 9(d) none of these |
|
Answer» `(a^2+b^2+c^2)/R^2=((2RsinA)^2+(2RsinB)^2+(2Rsinc)^2)/R^2` `4(sin^2A+sin^2B+sin^2C)` `4*2(1+cosAcosBcosC)` `8(1+cosAcosBcosC)` `8(1+1/8)` `8(9/8)` `9` Option C is correct. |
|
| 611. |
If `x=sin(theta+(7pi)/(12))+sin(theta+pi/(12))+sin(theta+(3pi)/(12))dot``Y=cos(theta+(7pi)/(12))+cos(theta+pi/(12))+cos(theta+(3pi)/(12))dot`then prove that `X/Y-Y/X=2tan2theta` |
|
Answer» `X=2sin(theta+pi/4)cos(pi/3)+sin(theta+3/12pi)` `X=2sin(theta+pi/4)` `Y=2cos(theta+pi/4)cos(pi/3)+cos(theta+pi/4)` `Y=2cos(theta+pi/4)` `X/Y-Y/X=(X^2-Y^2)/(XY)` `=(4sin^2(theta+pi/4)-4cos^2(theta+pi/4))/(4sin(theta+pi/4)cos(theta+pi/4))` `=(-4(cos^2(theta+pi/4)-sin^2(theta+pi/4)))/(2(2sin(theta+pi/4)cos(theta+pi/4))` `=(-2cos2(theta+pi/4))/(sin2(theta+pi/4))` `=-2(-sin2theta)/(cos2theta)` `=(2sin2theta)/(cos2theta)` `=tan2theta`. |
|
| 612. |
In any triangle, the minimum value of `r_1r_2r_3//r^3`is equal to1 (b)9 (c) 27(d) none of these |
|
Answer» Here, we will use the property G.M. is always greater than or equal to H.M. `:. (r_1/r* r_2/r* r_3/r )^(1/3) ge 3/(r/r_1+ r/r_2+ r/r_3)->(1)` Now, `r = Delta/s, r_1 = Delta/(s-a),r_2 = Delta/(s-b),r_3 = Delta/(s-c)` `:. r/r_1 = (s-a)/s, r/r_2 = (s-b)/s,r/r_3 = (s-c)/s` `:. r/r_1+r/r_2+r/r_3 = 1/s[s-a+s-b+s-c] = 1/s[3s-(a+b+c)]` `= 1/s[3s-2s] = 1` `:. r/r_1+r/r_2+r/r_3 = 1` So, `(1)` becomes, `=> (r_1/r* r_2/r* r_3/r )^(1/3) ge 3/1` `=>(r_1/r* r_2/r* r_3/r ) ge 3^3` `=>(r_1r_2r_3)/r^3 ge 27` So, minimum value of `(r_1r_2r_3)/r^3` will be `27`. |
|
| 613. |
The number of solutions of the equation `cos6x+tan^2x+cos(6x)tan^2x=1`in the interval `[0,2pi]`is(a)`4`(b)` 5`(c) `6` (d)`7` |
|
Answer» `cos6x+tan^2x+cos6x*tan^2x=1` `cos6x[1+tan^2x]=1-tan^2x` `cos6x=(1-tan^2x)/(1+tan^2x)` `cos6x=cos2x` `cos6x-cos2x=0` `-2sin((6x+2x)/2)*sin((6x-2x)/2)=0` `-2sin4xsin2x=0` `sin4x=0` `0<=x<=2pi` `0<=4x<=8pi` `4x=0,pi,2pi,3pi,4pi,5pi,6pi,7pi,8pi` `x=0,pi/4,pi/2,3/4pi,pi,5/4pi,3/2pi,7/4pi,2pi` `sin2x=0` `0<=x<=2pi` `0<=2x<=4pi` `2x=0,pi,2pi,3pi,4pi` `x=0,pi/2,pi/3/2pi,2pi` `x=9`. |
|
| 614. |
`tan^(- 1)((sqrt(1+x)+sqrt(1-x))/(sqrt(1+x)-sqrt(1-x)))`A. `(pi)/(2)-cos^(-1)x`B. `(pi)/(2)+cos^(-1)x`C. `(pi)/(4)-(1)/(2)cos^(-1)x`D. `(pi)/(4)+(1)/(2)cos^(-1)x` |
| Answer» Correct Answer - C | |
| 615. |
If `x in[0, 1]`, then `(1)/(2)cos^(-1)((1-x)/(1+x))=`A. `tan^(-1)x`B. `tan^(-1)(sqrt(x))`C. `(1)/(2)tan^(-1)x`D. `(1)/(2)tan^(-1)(sqrt(x))` |
| Answer» Correct Answer - B | |
| 616. |
Solve `y=tan^(-1)((sqrt(1+x^2)-1)/x)`A. `(x)/(2)`B. `(1)/(2)tan^(-1)x`C. `tan^(-1)x`D. `2tan^(-1)x` |
| Answer» Correct Answer - B | |
| 617. |
Prove that : `tan1/2{sin^(-1)(2x)/(1+x^2)+cos^(-1)(1-y^2)/(1+y^2)}=(x+y)/(1-x y)`, if `|x|0`and `x y |
| Answer» Correct Answer - C | |
| 618. |
In triangle ABC, prove that the maximum value of `tanA/2tanB/2tanC/2i s R/(2s)` |
|
Answer» `tan(A/2)tan(B/2)tan(C/2)` `sqrt(((s-b)(s-c))/(s(s-a)))sqrt(((s-a)(s-c))/(s(s-b)))sqrt((s-a)(s-b))/(s(s-c)` `sqrt((s(s-a)(s-b)(s-c))/s^4` `/_/s^2=r/s<=R/(2s)`. |
|
| 619. |
A triangle has sides 6,7, and 8. The linethrough its incenter parallel to the shortest side is drawn to meet the othertwo sides at P and Q. Then find the length of the segment PQ. |
|
Answer» We can draw the diagram with the given details. Please refer to video to see the diagram. Now, `Delta = r**s = r**(a+b+c)/2 = r**(6+7+8)/2 = 21/2r` `:. Delta = 21/2 r->(1)` From the diagram, we can see that, `Delta = 1/2**h**BC = 1/2**h**6 = 3h` `:. Delta = 3h->(2)` From (1) and (2), `=>21/2r = 3h => r/h = 2/7` Now, `Delta APQ and Delta ABC` are similar triangles. `:. (h-r)/h = (PQ)/(BC)` `=>1-r/h = (PQ)/6` `=>6(1-2/7) = PQ` `=>30/7 = PQ` Therefore, length of `PQ` is `30/7.` |
|
| 620. |
For triangle `ABC,R=5/2 and r=1.` Let `I` be the incenter of the triangle and `D,E and F` be the feet of the perpendiculars from `I->BC,CA and AB,` respectively. The value of `(IDxIExIF)/(IAxIBxIC)` is equal to (a) `5/2` (b) `5/4` (c) `10` (d) `1/5` |
|
Answer» In`/_AIF` and `/_AIE` `(IF)/(sin(A/2))=AI=(IE)/(sin(A/2))` `(AI)^2=(IE*IF)/(sin^2(A/2))` `(BI)^2=(IF*ID)/(sin^2(B/2))` `(CCI)^2=(IE*ID)/(sin(C/2))` `(AI)^2*(BI)^2*(CI)^2=((IE)^2(IF)^2(ID)^2)/(sin^2(A/2)sin^2(B/2)sin^2(C/2))` `(IE*IF*ID)/(AI*BI*CI)=sin(A/2)sin(B/2)sin(C/2)=r/(4R)` `r=4Rsin(A/2)sin(B/2)sin(C/2)` `r=1/10` Option C is correct. |
|
| 621. |
prove that `a^2sin2B+b^2sin2A=4Delta` |
|
Answer» `L.H.S. = a^2sin2B+b^2sin2A` `= (2RsinA)^2(2sinBcosB)+(2RsinB)^2(2sinAcosA)` `=8R^2sinAsinB(sinAcosB+sinBcosA)` `=8R^2sinAsinB(sin(A+B))` `=8R^2sinAsinB(sin(pi-C))` `=8R^2sinAsinBsinC` Now, we know, `Delta = 2R^2sinAsinBsinC ` `:. 8R^2sinAsinBsinC = 4Delta = R.H.S.` |
|
| 622. |
If the lengths of the sides of a triangle are `3, 5, 7`, then its largest angle of the triangle isA. `90^(@)`B. `150^(@)`C. `120^(@)`D. `135^(@)` |
| Answer» Correct Answer - C | |
| 623. |
In ` A B C ,`show that `a^2(s-a)+b^2(s-b)+c^(2)(s-c))=4R (a+r sin (A/2)sin( B/2)sin( C/2))` |
|
Answer» `L.H.S. = a^2(s-a)+b^2(s-b)+c^2(s-c)` Putting `s = (a+b+c)/2` in the above, we get, `=1/2a^2(b+c-a)+1/2b^2(c+a-b)+1/2c^2(a+b-c)` `=1/2(a(b^2+c^2-a^2)+b(c^2+a^2-b^2)+c(a^2+b^2-c^2))` `=1/2(2abc cosA +2abc cosB+2abc cosC)` `=abc (cosA+cosB+cos C)` `=abc(1+4sin(A/2)sin(B/2)sin(C/2))` As `Delta = (abc)/(4R), => abc = 4RDelta` So, it becomes, `= 4RDelta(1+4sin(A/2)sin(B/2)sin(C/2)) = R.H.S.` |
|
| 624. |
If area of a triangle is 2 sq. units, then find the value of theproduct of the arithmetic mean of the lengths of the sides of a triangle andharmonic mean of the lengths of the altitudes of the triangle. |
|
Answer» Let `a,b and c` are the sides and `h_1,h_2 and h_3` are the lengths of the altitudes. Then, `ah_1 = bh_2 = ch_3 = 2Delta` `=>1/h_1+1/h_2+1/h_3 = (a+b+c)/(2Delta)` `=>2Delta = (a+b+c)/(1/h_1+1/h_2+1/h_3)` `=>2Delta = (a+b+c)/3**3/(1/h_1+1/h_2+1/h_3)` Now, it is given that `Delta = 2` square units `:. 2*2 = (a+b+c)/3**3/(1/h_1+1/h_2+1/h_3)` `=>(a+b+c)/3**3/(1/h_1+1/h_2+1/h_3) = 4` Here, `(a+b+c)/3` is A.M. of the sides and `3/(1/h_1+1/h_2+1/h_3)` is the H.M. of the altitudes. Therefore, the required product is `4`. |
|
| 625. |
If two sides of a triangle are roots of the equation `x^2-7x+8=0`and the angle between these sides is `60^0`then the product of inradius and circumradius of the triangle is`8/7`(b) `5/3`(c) `(5sqrt(2))/3`(d) `8` |
|
Answer» `x^2-7x+8=0` a+b=7 ab=8 `/_=60^o` `cos60^@=(a^2+b^2-c^2)/(2ab)` `1/2=((a+b)^2-2ab-c^2)/(2ab)` `1/2=(49-2*8-c^2)/(2*8)` `c^2=25` `c=5` `r*R=D/S*(abc)/(4D)=(8*5)/(4(a+b+c)/2)` `r*R=5/3`. |
|
| 626. |
General solution of `sin^2x-5sinxcosx-6cos^2x=0`is`x=npi-pi/4,n in Z`onlyn`pi+tan^(-1)6,n in Zon l y`both (a) and (b)none of these |
|
Answer» `sin^2x/cos^2x-(5sinxcosx)/(cos^2x)-(6cos^2x)/(cos^2x)=0` `tan^2x-5tanx-6=0` `tan^2x-6tanx+tanx-6=0` `tanx(tanx-6)+1(tanx-6)=0` `(tanx+1)(tanx-6)=0` `tanx=-1` `x=npi-pi/4` `tanx=6` `v=npi+tan^(-1)6` `tantheta=tanalpha` `theta=npi+alpha` Option C is correct. |
|
| 627. |
Find the general solution of each of the equations : (i) `sin2x=-(1)/(2)` (ii) `tan3x=-1`A. `(npi)/(3)+(3pi)/(4), ninZ`B. `npi+(3pi)/(4), ninZ`C. `(npi)/(3)+(pi)/(4), ninZ`D. `npi+(pi)/(4), ninZ` |
| Answer» Correct Answer - C | |
| 628. |
If `tanmtheta=tanntheta`, then the general values of `theta` are inA. A.P.B. G.P.C. H.P.D. A.G.P. |
| Answer» Correct Answer - A | |
| 629. |
The general solution of the equation `tanx+ tan2x + tanx.tan2x= 1` isA. `npi+(pi)/(4), ninZ`B. `(npi)/(3)+(pi)/(12), ninZ`C. `npi+(pi)/(12), ninZ`D. `(npi)/(3)+(pi)/(4), ninZ` |
| Answer» Correct Answer - B | |
| 630. |
General solution of `(tan3x-1)/(tan3x+1)=sqrt(3)` isA. `(npi)/(3)-(7pi)/(12), ninZ`B. `(npi)/(3)+(7pi)/(12), ninZ`C. `(npi)/(3)-(7pi)/(36), ninZ`D. `(npi)/(3)+(7pi)/(36), ninZ` |
| Answer» Correct Answer - D | |
| 631. |
The general solution of the equation `8cosxcos2xcos4x=(sin6x)/(sin x)`is`x=((npi)/7)+(pi/(21)),AAn in Z``x=((2pi)/7)+(pi/(14)),AAn in Z``x=((npi)/7)+(pi/(14)),AAn in Z``x=(npi)+(pi/(14)),AAn in Z` |
|
Answer» `4(2sinxcosx)cos2xcos4x=sin6x` `4sin2xcos2xcos4x=sin6x` `2(2sinxcos2x)cos4x=sin6x` `2sin4xcos4x=sin6x` `sin8x-sin6x=0` `2cos((8x+6x)/2)*sin((8x-6x)/2)=0` `2cos7xsinx=0` `cos7x=0` `7x=npi+pi/2` `x=npi/7+pi/14` Option B is correct. |
|
| 632. |
General solution of `tantheta+tan((pi)/(2)-theta)=2` isA. `npi+(pi)/(8), ninZ`B. `2npi+(pi)/(8), ninZ`C. `npi+(pi)/(4), ninZ`D. `2npi+(pi)/(4), ninZ` |
| Answer» Correct Answer - C | |
| 633. |
The general solution of the equation `tan x+tan(x+pi/3)+tan(x+(2pi)/3)=3` isA. `(2n+1)(pi)/(6), ninZ`B. `(4n+1)(pi)/(6), ninZ`C. `(2n+1)(pi)/(12), ninZ`D. `(4n+1)(pi)/(12), ninZ` |
| Answer» Correct Answer - D | |
| 634. |
General solution of `sin(x+(pi)/(5))=0` isA. `npi+(pi)/(5), ninZ`B. `npi-(pi)/(5), ninZ`C. `2npi+(pi)/(5), ninZ`D. `2npi-(pi)/(5), ninZ` |
| Answer» Correct Answer - B | |
| 635. |
General solution of `cos((5x)/2)=0` isA. `(4n+1)(5pi)/(2), ninZ`B. `(4n+1)(2pi)/(5), ninZ`C. `(4n+1)(pi)/(2), ninZ`D. `(4n+1)(pi)/(5), ninZ` |
| Answer» Correct Answer - D | |
| 636. |
Find the general solution : `cos 3x + cos x cos 2x = 0`A. `2npipm(pi)/(2), 2npipm(2pi)/(3), ninZ`B. `2npipm(pi)/(2), npipm(2pi)/(3), ninZ`C. `npipm(pi)/(4), 2npipm(2pi)/(3), ninZ`D. `npipm(pi)/(4), npipm(pi)/(3), ninZ` |
| Answer» Correct Answer - C | |
| 637. |
General solution of `sin2x+sin4x=sin3x` isA. `(npi)/(6), 2npipm(pi)/(3), ninZ`B. `(npi)/(3), 2npipm(pi)/(3), ninZ`C. `(npi)/(3), 2npipm(pi)/(6), ninZ`D. `(npi)/(6), 2npipm(pi)/(6), ninZ` |
| Answer» Correct Answer - B | |
| 638. |
If `(sin^-1x+sin^-1w)(sin^-1y+sin^-1z)=pi^2`, then |
|
Answer» given `pi^2 = pi xx pi = - pi xx - pi` `(sin^-1 x + sin^-1 w)(sin^-1 y + sin^-1 z) = pi xx pi ` max value of `sin^-1 x + sin^-1 w = sin^-1 1 + sin^-1 1 = pi/2 + pi/2 = pi` max `(sin^-1 y + sin^-1z) = -pi/2 - pi/2 = -pi` `pi^2, x=1, w=1, y=1, z=1` `D= |(1,1),(1,1)| = 0` when `-pi^2 , x=y=z=w=-1` `D= |(-1^(n1), -1^(n2)), (-1^(n3), -1^(n4))|` `= -1^(n1+n4) - (-1)^(n2+n3)` max [D]=`max[(-1)^(n1+n4) - (-1)^(n2+n3)]` `= 1-(-1)=2` min(D) = `-1+ (-1)= -2` option a & d is correctanswer |
|
| 639. |
If `sinx+cosecx=2," then "sin^nx+cosec^nx` is equal toA. 2B. `2^n`C. `2^(n-1)`D. `2^(n-2)` |
|
Answer» Correct Answer - A `sinx+cosecx=2` `or (sinx-1)^2=0` `or sinx=1` `rArr sin^nx+cosec^nx=1+1=2` |
|
| 640. |
If `tantheta+sintheta=mandtantheta-sintheta=n`,thenA. `m^2-n^2=4mn`B. `m^2+n^2=4mn`C. `m^2-n^2=m^2+n^2`D. `m^2-n^2=4sqrt(mn)` |
|
Answer» Correct Answer - D From the given relations, `m+n=2tantheta,m-n=2sintheta`. Thus,`m^2-n^2=4tanthetasintheta` Also `sqrt(mn)=sqrt(tan^2theta-sin^2theta)=sinthetatantheta` From Eqs. (i) and (ii), we get `m^2-n^2=4sqrt(mn)`. |
|
| 641. |
If `cosectheta-cottheta=q`, then the value of `cosectheta` isA. `q+1/q`B. `q-1/q`C. `1/2(q+1/q)`D. None of these |
|
Answer» Correct Answer - C `cosectheta-cottheta=q` `:. cosectheta+cottheta=1/q` `:. cosectheta=1/2(q+1/q)`(on addition) |
|
| 642. |
Prove that:`(sin5A-sin3A)/(cos5A+cos3A)=tanA``(sinA+sin3A)/(cosA+cos3A)=tan2A` |
|
Answer» 1)`=(sin5A-sin3A)/(cos5A+cos3A)` `=(2sin((5A-3A)/2)cos((5A+3A)/2))/(2cos((5A+3A)/2)cos((5A-3A)/2)` `=sinA/cosA` `=tanA` RHS 2)`=(sinA+sin3A)/(cosA+cos3A)` `=(2cos((A-3A)/2)sin((A+3A)/2))/(2cos((A+3A)/2)cos((A-3A)/2)` `=(2cos(-A)sin2A)/(2cos2Acos(-A)` `=(sin2A)/(cos2A)` `=tan2A` RHS |
|
| 643. |
In `triangleABC`, if `c^(2)sinAsinB=ab,` then the triangle isA. an equilateralB. an isoscelesC. a right angledD. a scalene |
| Answer» Correct Answer - C | |
| 644. |
In `DeltaABC,BC=1,sin.(A)/2=x_1,sin.(B)/2=x_2,cos.(A)/2=x_3andcos.(B)/2=x_4" with "(x_1/x_2)^2007-(x_3/x_4)^2006=0` If `angleA=90^@`, then area of `DeltaABC` isA. 1/2 sq. unitsB. 1/3 sq. unitsC. 1 sq. unitsD. 2sq. Units |
|
Answer» Correct Answer - A In given `DeltaABC" both " A/2and B/2"lie strictly in "(0,pi/2)and sin x " always increasing in "(0,pi/2)` whereas cos x is always decreasing in `(0,pi/2)`. So, if `A/2ltB/2` `rArr sin(A)/2gt sin(B)/2` `or x_1gtx_2` `and x_3ltx_4` `or 1/x^3gt1/x^4` So, `x_1^2007x_4^2006=x_2^2007x_3^2006` is not valid. Similarly for `A/2ltB/2` `rArr sin(A)/2sin(B)/2` `rArr x_1ltx_2` `and 1/x^3lt1/x^4` For this also `x_1^2007x_4^2006=x_2^2007x_3^2006` is not valid. So,`(x_1/x_2)^2007-(x_3/x_4)^2006=0" is possible only when " A/2=B/2`. `rArr x_1=x_2and 1/x_3=1/x_4` Hence, `DeltaABC` is isosceles with `angleABC=angleCAB`. `rArr BC=AC=1"unit"` if `angleA=90^@` Area, `A=1/2BCxxAC=1/2" sq. units"` |
|
| 645. |
`sin^(2)A*tan^(2)A+cos^(2)A*cot^(2) A=`A)`1 + tan^(2)A + cot^(2) A` B)`tan^(2)A +cot^(2) A-1` C)`1 + sec^(2)A + tan^(2) A` D)`1 + csc^(2)A + cot^(2) A`A. `1 + tan^(2)A + cot^(2) A`B. `tan^(2)A +cot^(2) A-1`C. `1 + sec^(2)A + tan^(2) A`D. `1 + csc^(2)A + cot^(2) A` |
| Answer» Correct Answer - B | |
| 646. |
If `sinx+sin^2x=1` then `cos^2x+cos^4x`A. -1B. 0C. 1D. 2 |
| Answer» Correct Answer - C | |
| 647. |
The equation `asinx+Cos2x=2a-7` possesses a solution if |
|
Answer» Given equation is, `asinx + cos2x = 2a-7` `=>asinx+1-2sin^2x =2a - 7` `=>2sin^2x-asinx+2a-8 = 0` `=> sinx = (-(-a)+-sqrtD)/(2(2))` Here, `D = (-a)^2-4(2)(2a-8)` `=> D = a^2-16a+64 = (a-8)^2` `:. sinx = (a+-(a-8))/4` As `sin x` lies between `-1` and `1`. `:. sin x = (a+(a-8))/4` `=>sinx = (a-4)/2` `:. -1 le (a-4)/2 le 1`. `=>-2 le a-4 le 2` ` => 2 le a le 6` So, for `2 le a le 6`, given eqaution will possess a solution. |
|
| 648. |
If `asinx+bcos(x+theta)+bcos(x-theta)=d ,`then the minimum value of `|costheta|`is equal to(a)`1/(2|b|)sqrt(d^2-a^2)`(b) `1/(2|a|)sqrt(d^2-a^2)`(c)`1/(2|d|)sqrt(d^2-a^2)`(d) none of these |
|
Answer» `asinx+b[2cosxcostheta]=d` `asinx+(2bcostheta)cosx=d` `|d|<=sqrt(a^2+(2bcostheta)^2` `|d|<=sqrt(a^2+4b^2cos^2theta` `d^2<=a^2+4b^2cos^2theta` `(d^2-a^2)/(4b^2)<=cos^2theta` `|costheta|>=sqrt(d^2-a^2)/(2|B|)` Option A is correct. |
|
| 649. |
If `tanx=a/b` then show that `(asinx-bcosx)/(asinx+bcosx)=(a^2-b^2)/(a^2+b^2)`A. `(a^(2)+b^(2))/(a^(2)-b^(2))`B. `(2a^(2)-b^(2))/(a^(2)+2b^(2))`C. `(a^(2)-b^(2))/(a^(2)+b^(2))`D. none of these |
| Answer» Correct Answer - C | |
| 650. |
If the equation `asinx+cos2x=2a-7`possesses a solution, then find the value of `adot` |
|
Answer» `asinx +cos2x = 2a-7` `=>asinx+1-2sin^2x = 2a-7` `=>2sin^2x - asinx +2a-8 = 0` `=>sinx = (a+-sqrt(a^2 - 4(2)(2a-7)))/4` `=>sinx = (a+-sqrt(a^2 - 16a+64))/4` `=>sinx = (a+-sqrt((a-8)^2))/4` `=>sinx = (a+a-8)/4 and sin x = (a-a+8)/4` `=>sinx = (a-4)/2 and sinx = 2` We know, `sinx !=2`, `:. sinx = (a-4)/2` `:. -1 le (a-4)/2 le 1` `=>2 le a le 6` `:. a in [2,6]`, which is required solution. |
|