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.
| 351. |
The expression `sum_(n=1)^(oo)cot^(-1)(n^(2)-3n+3)` simplifies toA. `(pi)/(4)`B. `(pi)/(2)`C. `(3pi)/(4)`D. `pi` |
|
Answer» Correct Answer - C `S=sum_(n=1)^(oo)cos^(-1)(n^(2)-3n+3)` `=sum_(n=1)^(oo)tan^(-1)((1)/(1+n^(2)-3n+2))` `=sum_(n=1)^(oo)tan^(-1)(((n-1)-(n-2))/(1+(n-1)(n-2)))` `=sum_(n=1)^(oo)(tan^(-1)(n-1)-tan^(-1)(n-2))` `=(pi)/(2)+(pi)/(4)=(3pi)/(4)` |
|
| 352. |
Let `tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2))`, where `|x| |
|
Answer» Correct Answer - B `(-1)/(sqrt3) lt x lt (1)/(sqrt3)` Let `x = tan theta` `:. (-pi)/(6) lt theta lt (pi)/(6)` `:. Tan^(-1) y = theta + tan^(-1) tan 2 theta = theta + 2theta = 3 theta` `:. Y = tan 3 theta = (3 tan theta - tan^(3) theta)/(1-3 tan^(2) theta)` `:. y = (3x -x^(3))/(1 -3x^(2))` |
|
| 353. |
Value of `tan^(- 1){sin(cos^(- 1)sqrt(2/3))}` isA. `pi//4`B. `pi//2`C. `pi//3`D. `pi//6` |
|
Answer» `tan^(-1){sin(cos^(-1)sqrt(2/3))}` `=tan^(-1){sin(sin^(-1)sqrt(1/3))}=tan^(-1)(1)/sqrt(3)=(pi)/(6)` |
|
| 354. |
Match the statements in List I with those in List II |
|
Answer» Correct Answer - `b rarr p, r` `tan^(-1) (x + 3) - tan^(-1) (x -3) = sin^(-1) (3//5)` `rArr tan^(-1).((x +3)-(x -3))/(1 + (x^(2) -9)) = tan^(-1).(3)/(4)` `rArr (6)/(x^(2) -8) = (3)/(4)` `:. X^(2) -8 = 8` or `x = +- 4` Clearly, both values satisfy the equation Note : Solution of the remaining parts are given in their respective chapters. |
|
| 355. |
If `sin^-1 x+sin^-1y=(2pi)/3,` then `cos^-1 x +cos^-1 y` is equal toA. `(2pi)/(3)`B. `(pi)/(3)`C. `(pi)/(6)`D. `pi` |
|
Answer» We have `sin^(-1)x+sin^(-1)y=(2pi)/(3)` `rarr (pi)/(2) cos^(-1)x+(pi)/(2)-cos^(-1)y=(2pi)/(3)` `rarr cos^(-1)x+cos^(-1)y=pi-(2pi)/(3)=(pi)/(3)` |
|
| 356. |
The value of `cot (sum_(n=1)^(23) cot^(-1) (1 + sum_(k=1)^(n) 2k))` isA. `(23)/(25)`B. `(25)/(23)`C. `(23)/(24)`D. `(24)/(23)` |
|
Answer» Correct Answer - B `cot(underset(n =1)overset(23)sum cot^(-1) (n^(2) + n + 1))` `= cot(underset(n =1)overset(23)sum tan^(-1) ((n + 1 -n)/(1+ n (n + 1))))` `= cot (underset(n=1)overset(23)sum (tan^(-1) (n +1) - tan^(-1)n))` `=cot (tan^(-1) 24 -tan^(-1) 1) rArr cot (tan^(-1) ((23)/(25)))` `= (25)/(23)` |
|
| 357. |
If `alpha=3sin^-1(6/11) `and `beta=3cos^-1(4/9)`, where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)A. `cos beta gt 0`B. `sin beta lt 0`C. `cos (alpha + beta) gt0`D. `cos alpha lt 0` |
|
Answer» We observe that `sqrt(3)/(2)gt6/11gt1/2 and 0le4/9le1/2` `therefore sin^(-1)sqrt(3)/(2)gtsin^(-1)6/11gtsin^(-1)(1/2)` and `cos^(-1)0gt^(-1)(4/9)gtcos^(-1)(1/2)` `rarr (pi)/(3)gtsin^(-1)(6/11)gt(pi)/(6)and (pi)/(2)gtcos^(-1)(4/9)gt(pi)/(3)` `rarr pigtalphagt(pi)/(2)and (3pi)/(2)gtbetapi` `rarr (pi)/(2)ltalpha,alphale(3pi)/(1)rarr(3pi)/(2)rarr(3pi)/(2)ltalpha+betalt(5pi)/(2)` `rarr cos alphalt0 sin beta lt 0 cos beta lt 0` Hence option b,c and d are correct |
|
| 358. |
If `alpha=3sin^-1(6/11) `and `beta=3cos^-1(4/9)`, where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)A. `cos beta gt 0`B. `sin beta lt 0`C. `cos (alpha + beta) gt 0`D. `cos alpha lt 0` |
|
Answer» Correct Answer - B::C::D `alpha = 3 sin^(-1).(6)/(11) gt 3 sin^(-1).(6)/(12) = 3 sin^(-1).(1)/(2) = (pi)/(2)` and `beta = 3 cos^(-1).(4)/(9) gt 3 cos^(-1).(4)/(8) = pi` So, `alpha gt (pi)/(2) and beta gt pi` `:. alpha + beta gt (3pi)/(2)` |
|
| 359. |
The value of `cos (2Cos^-1 0.8)` isA. 0.48B. 0.96C. 0.6D. 0.28 |
| Answer» Correct Answer - D | |
| 360. |
If `x^2+y^2+z^2=r^2,t h e ntan^(-1)((x y)/(z r))+tan^(-1)((y z)/(x r))+tan^(-1)((x z)/(y r))`is equal to`pi`(b) `pi/2`(c) 0(d) none of these |
|
Answer» `tan^-1((xy)/(zr))+tan^-1((yz)/(xr))+tan^-1((xz)/(yr))` `=tan^-1(((xy)/(zr)+(yz)/(xr))/(1-(xy)/(zr)(yz)/(xr)))+tan^-1((xz)/(yr))` `=tan^-1(((xy)/(zr)+(yz)/(xr))/(1-(xy)/(zr)(yz)/(xr)))+tan^-1((xz)/(yr))` `=tan^-1((x^2yr+yz^2r)/(r^2zx-y^2xz))+tan^-1((xz)/(yr))` `=tan^-1((yr(x^2+z^2))/(zx(r^2-y^2)))+tan^-1((xz)/(yr))` As, `x^2+y^2 +z^2 = r^2 =>x^2+z^2 = r^2-y^2` so, it becomes, `=tan^-1((yr(x^2+z^2))/(zx(x^2+z^2)))+tan^-1((xz)/(yr))` `=tan^-1((yr)/(zx))+tan^-1((xz)/(yr))` `=tan^-1((yr)/(zx))+cot^-1((yr)/(zx))` `=pi/2` So, option `(b)` is the correct option. |
|
| 361. |
If `-1 lt x lt 0`, then `cos^(-1) x` is equal toA. `sec^(-1).(1)/(x)`B. `pi - sin^(-1) sqrt(1 -x^(2))`C. `pi + tan^(-1).(sqrt(1 -x^(2)))/(x)`D. `cot^(-1).(x)/(sqrt(1 -x^(2)))` |
|
Answer» Correct Answer - A::B::C::D `cos^(-1) x = sec^(-1).(1)(x) " for all " x in [-1, 0)` Leet `x = -y` `:. Cos^(-1) x = cos^(-1) (-y) = pi - cos^(-1) y` `= pi - sin^(-1) sqrt(1 -y^(2))`...(i) `=pi - tan^(-1).(sqrt(1 - y^(2)))/(y)` ..(ii) From (i), `cos^(-1) = pi - sin^(-1) sqrt(1 - y^(2)) = pi - sin^(-1) sqrt(1 -x^(2))` From (ii), `cos^(-1) x = pi - tan^(-1).(sqrt(1 -y^(2)))/(y)` `= pi - tan^(-1).(sqrt(1 -x^(2)))/(-x)` `= pi + tan^(-1).(sqrt(1 -x^(2)))/(x)` `= pi + cot^(-1).(x)/(sqrt(1 -x^(2))) -pi " " ("as " x lt 0)` `= cot^(-1).(x)/(sqrt(1 -x^(2)))` |
|
| 362. |
The value of`sin^(-1)("cot"(sin^(-1)((2-sqrt(3))/4+cos^(-1)(sqrt(12))/4+sec^(-1)sqrt(2))))i s`(a)`0`(b) `pi/2`(c) `pi/3`(d) none of these |
|
Answer» `sin^(-1)(cot(sin^(-1)(sqrt3-1)/(2sqrt3)+cos^(-1)(sqrt3/2)+sec^(-1)sqrt2)` `sin^(-1)(cot(15+30+45))` `sin^(-1)(cot(90))` `sin^(-1)0=0`. |
|
| 363. |
Match the following lists : A. `{:(a,b,c,d),(s,r,q,p):}`B. `{:(a,b,c,d),(q,s,r,p):}`C. `{:(a,b,c,d),(s,r,p,q):}`D. `{:(a,b,c,d),(r,p,s,q):}` |
|
Answer» Correct Answer - B (a) `sin^(-1) (3x -4x^(3)) in [-(pi)/(2), (pi)/(2)]` `rArr pi - 3 sin^(-1) x in [-(pi)/(2), (pi)/(2)]` `rArr 3 sin^(-1) x in [(pi)/(2), (3pi)/(2)]` `rArr sin^(-1) x in [(pi)/(6), (pi)/(2)]` `rArr x in [(1)/(2), 1]` (b) `sin^(-1).(2x)/(1 + x^(2)) in [-(pi)/(2), (pi)/(2)]` `rArr pi - 2 tan^(-1) x in [-(pi)/(2), (pi)/(2)]` `rArr 2 tan^(-1) xin [(pi)/(2), pi)` `rArr tan^(-1) x in [(pi)/(4), (pi)/(2))` `rArr x in [1, oo)` (c) `tan^(-1).(2x)/(1 -x^(2)) in (-(pi)/(2), (pi)/(2))` `rArr 2 tan^(-1)x - pi in (-(pi)/(2), (pi)/(2))` `rArr 2 tan^(-1) x in ((pi)/(2), pi)` `rArr tan^(-1) x in ((pi)/(4), (pi)/(2))` `rArr x in (1, oo)` (d) `sin^(-1) (2x sqrt(1 -x^(2))) in [-(pi)/(2), (pi)/(2)]` `rArr pi - 2 sin^(-1) x in [-(pi)/(2), (pi)/(2)]` `rArr 2sin^(-1) x in [(pi)/(2), pi]` `rArr x in [(1)/(sqrt2) ,1]` |
|
| 364. |
Prove that : `2 tan^(-1) (cosec tan^(-1) x - tan cot^(-1) x) = tan^(-1) x`A. `cot^(-1) x`B. `cot^(-1).(1)/(x)`C. `tan^(-1) x`D. none of these |
|
Answer» Correct Answer - C `2 tan^(-1 (cosec tan^(-1) x - tan cot^(-1) x)` `= 2 tan^(-1) [cosec {cosec^(-1) (sqrt(1 + x^(2)))/(x)} - tan^(-1) {tan^(-1) ((1)/(x))}]` `= 2 tan^(-1) [sqrt((1 + x^(2))/(x)) - (1)/(x)] = 2 tan^(-1) [(sqrt(1 + x^(2)) - 1)/(x)]` `= 2 tan^(-1) [(sec theta -1)/(tan theta)]` [Putting `x = tan theta`] `= 2 tan^(-1) [(1 - cos theta)/(sin theta)] = 2 tan^(-1) [(2 sin^(2).(theta)/(2))/(2 sin.(theta)/(2) cos.(theta)/(2))]` `= 2 tan^(-1) tan.(theta)/(2) = 2 xx (theta)/(2) = theta = tan^(-1) x` |
|
| 365. |
If `x |
|
Answer» Correct Answer - A::B::C `tanx, x lt 0` Let `x = -y, y gt 0` `:. Tan^(-1) x = - tan^(-1) y` `= -cos^(-1).(1)/(sqrt(1 + y^(2)))` ...(i) `= -sin^(-1).(y)/(sqrt(1 + y^(2)))`..(ii) `= -cosec^(-1).(sqrt(1 + y^(2)))/(y)`...(iii) `=-cot^(-1).(1)/(y)` ...(iv) From (i) `tan^(-1) x = - cos^(-1).(1)/(sqrt(1 + x^(2)))` From (ii), `tan^(-1) x = -sin^(-1).(-x)/(sqrt(1 + x^(2))) = sin^(-1).(x)/(sqrt(1 + x^(2)))` From (iii), `tan^(-1) x = - cosec^(-1).(sqrt(1 + x^(2)))/(-x) = cosec^(-1).(sqrt(1 + x^(2)))/(x)` From (iv), `tan^(-1) x = - cot^(-1).(1)/(y)` `= -cot^(-1).(1)/(-x) = -(pi - cot^(-1).(1)/(x)) = -pi + cot^(-1).(1)/(x)` |
|
| 366. |
The value of `cos(1/2cos^(-1)(1/8))`is(a)`3/4`(b) `-3/4`(c) `1/(16)`(d) `1/4` |
|
Answer» `cos(1/2cos^(-1)(1/8))` `cos(1/2theta)` `cos(theta/2)=3/4` `cos^(-1)(1/8)=theta` `costheta=1/8` `2cos^2(theta/2)-1=1/8` `2cos^2(theta/2)=1+1/8` `2cos^2(theta/2)=9/8` `cos(theta/2)=3/4`. |
|
| 367. |
Solve `sin^(-1) (1 - x) - 2 sin^(-1) x = (pi)/(2)` |
|
Answer» Given , `sin^(-1) (1 - x) - 2 sin^(-1) x = pi//2` Let `x sin y` Therefore, the given equation reduces sto `sin^(-1) (1 - sin y) - 2 sin^(-1) sin y= (pi)/(2)` or `sin^(-1) (1 - sin y) - 2 y = (pi)/(2)` or `sin^(-1) (1 - sin y) = (pi)/(2) + 2y` or `1 - sin y = sin ((pi)/(2) + 2y)` or `1 - sin y = cos 2 y` or `1 - cos 2y = sin y` or `2 sin^(2) y = sin y` or `2 sin^(2) y - sin y = 0` or `sin y (2 sin y - 1) = 0` `rArr sin y = 0 " or " 1//2` `rArr x = 0 " or " 1//2` But, when `x = (1)/(2)`, it can be observed that `L.H.S. = sin^(-1). (1 - (1)/(2)) - 2 sin ^(-1). (1)/(2)` `= sin^(-1). ((1)/(2)) - 2 sin^(-1). (1)/(2)` `= - sin^(-1). (1)/(2)` `= - (pi)/(6) != (pi)/(2)` Therefore, `x = (1)/(2)` is not the solution of the given equation Thus, `x = 0` |
|
| 368. |
The solution set of inequality `(cot^(-1)x)(tan^(-1)x)+(2-pi/2),cot^(-1)x-3tan^(-1)x-3(2-pi/2)>0`is `(a , b),`then the value of `cot^(-1)a+cot^(-1)b`is____ |
|
Answer» Correct Answer - 5 `(cot^(-1) x) (tan^(-1)x) + (2 -(pi)/(2)) cot^(-1) x - 3 tan^(-1) x 3 (2 - (pi)/(2)) gt 0` `rArr cot^(-1) x (tan^(-1) x -(pi)/(2)) + 2 cot^(-1) x - 6 - 3 (tan^(-1) x -(pi)/(2)) gt 0` `rArr -(cot^(-1) x)^(2) + 5 cot^(-1) x - 6 gt 0` `rArr (cot^(-1) x -3) (2 - cot^(-1) x) gt 0` `rArr (cot^(-1) x -3) (cot^(-1) x -2) lt 0` `rArr 2 lt cot^(-1) x lt3` `rArr cot 3 lt x lt cot2` [as `cot^(-1) x` is a decreasing function] `rArr` Hence, `x in (cot3, cot2)` `rArr cot^(-1) a + cot^(-1) b = cot^(-1) (cot 3) + cot^(-1) (cot 2) = 5` |
|
| 369. |
The solution set of the inequality `tan^(-1)x+sin^(-1)x ge (pi)/(2)` isA. `[-1,1]`B. `[sqrt((sqrt(5)-1)/(4)),1]`C. `[sqrt((sqrt(5)-1)/(2)),1]`D. `[(sqrt(5)-1)/(2),1]` |
|
Answer» Correct Answer - C `tan^(-1)x + sin^(-1)x ge (pi)/(2), x in [-1,1]` `rArr tan^(-1)x ge (pi)/(2) -sin^(-1)x` `rArr tan^(-1)x ge cos^(-1)x` `rArr x ge "tan"^(-1)(sqrt(1-x^(2)))/(x)` `rArr x ge (sqrt(1-x^(2)))/(x)` `x^(4)ge 1-x^(2)` `rArr x^(4)+x^(2)-1ge 0` `rArr x in [sqrt((sqrt(5)-1)/(2)),1]` |
|
| 370. |
If `sin^(-1)a+sin^(-1)b+sin^(-1)c=pi,`then the value of `asqrt((1-a^2))+bsqrt((1-b^2))+sqrt((1-c^2))`will be`2a b c`(b) `a b c`(c) `1/2a b c`(d) `1/3a b c`A. `2 abc`B. `abc`C. `(1)/(2) abc`D. `(1)/(3) abc` |
|
Answer» Correct Answer - A Let `sin^(-1) a = A, sin^(-1) b = B and sin^(-1) c = C` `rArr sin A = a, sin B = b, sin C = c` and `A + B + C = pi` `rArr sin 2A + sin 2B + sin 2C` `= 4 sin A sin B sin C`....(i) `rArr sin A cos A + sin B cos B + sin C cos C` `= 2 sin A sin B sin C` `rArr sin A sqrt(1 - sin^(2) A) + sin B sqrt(1 - sin^(2) B)` `+ sin C sqrt(1 - sin^(2) C) = 2 sin A sin B sin C` ..(ii) `rArr asqrt((1 - a^(2))) + bsqrt((1 -b^(2))) + c sqrt((1 - c)^(2)) = 2abc` Alternatively : Let `a = (1)/(sqrt2), b = (1)/(sqrt2), c =1` Then `asqrt(1 -a^(2)) + bsqrt(1 -b^(2)) + csqrt(1 -c^(2))` `= (1)/(sqrt2) sqrt(1 - (1)/(2)) + (1)/(sqrt2) sqrt(1 -(1)/(2)) + 1 sqrt(1 -1) = 1` `= 2. (1)/(sqrt2). (1)/(sqrt2) .1` |
|
| 371. |
Prove that `cot^(-1) ((sqrt(1 + sin x) + sqrt(1 - sin x))/(sqrt(1 + sin x) - sqrt(1 - sin x))) = (x)/(2), x in (0, (pi)/(4))` |
|
Answer» Consider `(sqrt(1 + sin x) + sqrt(1 - sin x))/(sqrt(1 + sin x) - sqrt(1 - sin x))` `= ((sqrt(1 + sin x) + sqrt(1 - sin x))^(2))/((sqrt(1 + sin x)^(2)) - (sqrt(1 - sin x))^(2))` (by rationalizing) `= ((1 + sin x) + (1 - sin x) + 2 sqrt((1 + sin x) (1 - sin x)))/(1 + sin x - 1 + sin x)` `= (2(1 + sqrt(1 - sin^(2)x)))/(2 sin x) = (1 + cos x)/(sin x)` `= (2 cos^(2) (x)/(2))/(2 sin (x)/(2) cos (x)/(2)) = cot. (x)/(2)` `rArr cot^(-1) ((sqrt(1 + sin x)+ sqrt(1 - sin x))/(sqrt(1 + sin x) - sqrt(1 - sin x))) = (x)/(2)` |
|
| 372. |
If `sin^(-1)a+sin^(-1)b+sin^(-1)c=pi,` then the value of `asqrt((1-a^2))+bsqrt((1-b^2))+sqrt((1-c^2))` will be(A) `2a b c`(B) `a b c`(C) `1/2a b c`(D) `1/3a b c` |
|
Answer» `sin^(-1)a=A,sinA=a` `sin^(-1)b=B,sinB=b` `sin^(-1)c=C,sinC=c` `A+B+C=pi` `sin2A+sin2B+sin2C=4sinAsinBsinC` `sinAcosA+sinBcosB+sinCcosC=2sinAsinBsinC` `sinAsqrt(1-sin2B)+sinBsqrt(1-sin2B)+sinCsqrt(1-sin2C)=2sinAsinBsinC` `asqrt(1-a^2)+bsqrt(1-b^2)+csqrt(1-c^2)=2abc` Option A is correct. |
|
| 373. |
lf `asin^-1 x -bcos^-1 x=c,`then `asin^-1 x +bcos^-1` equal to |
|
Answer» Correct Answer - D `a sin^(-1) x - b cos^(-1) x = c` We have `b sin^(-1) x + b cos^(-1) x = (bpi)/(2)` Adding `(a + b) sin^(-1) x = (bpi)/(2) + c` `rArr sin^(-1) x = (((bpi)/(2)) + c)/(a + b) = (b pi + 2c)/(2(a + b))` `:. cos^(-1) x = (pi)/(2) - (b pi + 2c)/(2(a + b)) = (pi a - 2c)/(2(a + b))` `rArr a sin^(-1) x + b cos^(-1) x = (pi ab + c(a - b))/(a + b)` |
|
| 374. |
Find the value of `sin((1)/(2) cot^(-1) (-(3)/(4)))` |
|
Answer» Let `cos^(-1) (-3//4) = theta` or `cos theta = -3//4` `rArr theta in (pi//2, pi)` `rArr cos theta = -3//5` `rArr 1 - 2 sin^(2) (theta//2) = - 3//5` `:. Sin^(2) theta//2 = 4//5 " or " sin theta//2 = 2 sqrt5` |
|
| 375. |
`cot^(-1)(-2)` is equal toA. `pi-tan^(-1)(1/2)`B. `tan^(-1)(1/2)`C. `-tan^(-1)1/2`D. `cot^(-1)2` |
| Answer» Correct Answer - A | |
| 376. |
Prove that `2 tan^(-1) (cosec tan^(-1) x - tan cot^(-1) x) = tan^(-1) x (x != 0)` |
|
Answer» Case I: `x gt 0` 2 tan^(-1) (cosec tan^(-1) x - tan cot^(-1) x)` `= 2 tan^(-1) `(cosec(cosec^(-1) (sqrt(1 + x^(2)))/(x)) - tan (tan^(-1). (1)/(x)))` `= 2 tan ^(-1) ((sqrt(1 + x^(2)))/(x) - (1)/(x))` `= 2 tan^(-1) ((sec theta - 1)/(sin theta))` (Putting `x = tan theta`) `= 2 tan^(-1) ((1 - cos theta)/(sin theta))` `= 2 tan^(-1) ((2 sin^(2) theta//2)/(2 sin theta//2 cos theta//2))` `= 2 tan^(-1) (tan.(theta)/(2)) = 2 xx (theta)/(2)` `= theta = tan^(-1) x` Case II: `x lt 0` Let `y = -x` `:. y gt 0` `:. 2 tan^(-1) (cosec tan^(-1) x - tan cot^(-1) x)` `= 2 tan^(-1) (cosec tan^(-1) (-y) - tan cot^(-1) (-y))` `=2 tan^(-1) (cosec (-tan^(-1) y) - tan (pi - cot^(-1) y))` `= 2 tan^(-1) (-cosec (tan^(-1) y) + tan (cot^(-1) y))` `= 2 tan^(-1) (-cosec (cosec^(-1) (sqrt(1 + y^(2)))/(y)) + tan (tan^(-1).(1)/(y)))` `= 2 tan^(-1) (-(sqrt1 + y^(2))/(y) + (1)/(y))` `= 2 tan^(-1) ((sqrt(1 + y^(2))-1)/(y))` `= -tan^(-1) y` `= tan^(-1) (-y)` `= tan^(-1) x` |
|
| 377. |
If `asin^(-1)x-bcos^(-1)x=c ,`then `asin^(-1)x+bcos^(-1)x`is equal to(a)`0 `(b) `(pia b+c(b-a))/(a+b)`(c)`pi/2`(d)`(pia b+c(a-b))/(a+b)` |
|
Answer» `asin^(-1)x-bcos^(-1)x=c-(1)` `b(cos^(-1)x+sin^(-1)x)=pi/2*b-(2)` Adding equation 1 and 2 `sin^(-1)x(a+b)=(bpi)/2+c` `sin^(-1)x=((bpi)/2+c)/(a+b)-(3)` `cosx=pi/2-((bx+2c)/(2(a+b)))` `=(pia-2c)/(2(a+b))` `asin^(-1)x+bcos^(-1)x=(xab+c(a-b))/(a+b)` Option D is correct. |
|
| 378. |
If `cos (2 sin^(-1) x) = (1)/(9)`, then find the value of x |
|
Answer» Let `sin^(-1) x = theta`. Then, `cos 2 theta = (1)/(9)` or `1 - 2 sin^(2) theta = 9` or `1 - 2x^(2) = (1)/(9)` or `x^(2) = (4)/(9)` or `x = +-(2)/(3)` |
|
| 379. |
`cosec^(-1)(-x),x in R-(-1,1),` is equal toA. `cosec^(-1)x`B. `-sin^(-1)x`C. `-sin^(-1)(1/x)`D. `pi-cosec^(-1)x` |
| Answer» Correct Answer - C | |
| 380. |
If `sin^(-1) x = pi//5`, for some `x in (-1, 1)`, then find the value of `cos^(-1) x` |
| Answer» `sin^(-1) x + cos^(-1) x = (pi)/(2) " or " cos^(-1) x = (pi)/(2) - sin^(-1) x = (pi)/(2) - (pi)/(5) = (3pi)/(10)` | |
| 381. |
The solution of the inequality `(log)_(1/2sin^(-1)x >(log)_(1//2)cos^(-1)x`is`x in [(0,1)/(sqrt(2))]`(b) `x in [1/(sqrt(2)),1]``x in ((0,1)/(sqrt(2)))`(d) none of these |
|
Answer» `log_(1/2)sin^(-1)x>log_(1/2)cos^(-1)x` `cos^(-1)x>sin^(-1)x` `cosx>pi/2-cosx` `2cosx>pi/2` `cosx>pi/4` `0ltx<1/sqrt2`. |
|
| 382. |
If `sin (sin^(-1).(1)/(5) + cos^(-1) x) = 1`, then find the value of x |
|
Answer» `sin(sin^(-1).(1)/(5) + cos^(-1)x) =1` or `sin^(-1).(1)/(5) + cos^(-1) x = (pi)/(2)` or `sin^(-1). (1)/(5) = (pi)/(2) - cos^(-1) x` or `sin^(-1).(1)/(5) = sin^(-1) x` or or `x = (1)/(5)` |
|
| 383. |
`cos^(-1)(-x),absx le 1`, is equal toA. `-cos^(-1)x`B. `cos^(-1)x`C. `pi/2-cos^(-1)x`D. `pi/2+sin^(-1)x` |
| Answer» Correct Answer - D | |
| 384. |
solve `sin^(-1) x le cos^(-1) x` |
|
Answer» `cos^(-1) x ge sin^(-1) x` or `(pi)/(2) - sin^(-1) x ge sin^(-1) x` `(pi)/(2) ge 2 sin^(-1) x` `(-pi)/(2) le sin^(-1) x le (pi)/(4)` or `-1 le x le sin ((pi)/(4))` `rArr x in [-1, (1)/(sqrt2)]` |
|
| 385. |
Find the range of `tan^(-1)((2x)/(1+x^2))`A. `[-(pi)/(4), (pi)/(4)]`B. `(-(pi)/(2), (pi)/(2))`C. `(-(pi)/(2), (pi)/(4)]`D. `[(pi)/(4), (pi)/(2)]` |
|
Answer» Correct Answer - A `1 + x^(2) ge 2 |x|` or `(2|x|)/(1 + x^(2)) le 1` or `-1 le (2x)/(1 + x^(2)) le 1` `rArr tan^(-1) ((2x)/(1 + x^(2))) in [-(pi)/(4), (pi)/(4)]` |
|
| 386. |
Find the range of `f(x) = sin^(-1) x + tan^(-1) x + cos^(-1) x` |
|
Answer» Clearly, the domain of the function is `[-1, 1]` Also, `tan^(-1) x in [-(pi)/(4), (pi)/(4)] " for " x in [-1, 1]` Now, `sin^(-1) x + cos^(-1) x = (pi)/(2) AA x in [-1, 1]` Thus, `f(x) = tan^(-1) x + (pi)/(2)`, where `x in [-1, 1]` Hence, the range is `[-(pi)/(4) + (pi)/(2), (pi)/(4) + (pi)/(2)] = [(pi)/(4), (3pi)/(4)]` |
|
| 387. |
The number of integer `x`satisfying `sin^(-1)|x-2|+cos^(-1)(1-|3-x|)=pi/2`is(a)` 1` (b)` 2` (c)` 3`(d)` 4` |
|
Answer» `sin^(-1)|x-2|+cos^(-1)(1-|3-x|)=pi/2` `alpha=beta` `|x-2|=1-|3-x|` `|x-2|+|3-x|=1|(x-2)+(3-x)|` `(x-2)(3-x)>=0` `2<=x<=3` Option B is correct. |
|
| 388. |
Find `x`satisfying `[tan^(-1)x]+[cos^(-1)x]=2,`where `[]`represents the greatest integer function. |
|
Answer» `0 lt cot^(-1) x lt pi and -pi//2 lt tan^(-1) x lt pi//2` `rArr [cot^(-1) x] in {0, 1, 2, 3} and [tan^(-1) x] in {-2, -1, 0, 1}` For `[tan^(-1) x] + cot^(-1) x] = 2`, following cases are possible Cose(i): `[cot^(-1)x] = tan^(-1) x] = 1` `rArr 1 le cot^(-1) x lt 2 and 1 le tan^(-1) x lt pi//2` `rArr x in (cot 2, cot 1] and x in [tan 1, oo)` `:. x in phi " as " cot 1 lt tan 1` Case (ii): `[cot^(-1) x] = 2, [tan^(-1) x] = 0` `rArr 2 le cot^(-1) x lt 3 and 0 le tan^(-1) x lt 1` `rArr x in (cot 3, cot 2] and x in [0, tan 1)` `:. x in phi " as " cot 2 lt 0` So no solution Case (iii) : `[cot^(-1) x] = 3, [tan^(-1) x] = -1` `rArr 3 le cot^(-1) x lt pi and - 1 le tan^(-1) x lt 0` `rArr x in (-oo, cot 3] and x in [- tan 1, 0)` `:. x in phi cot 3 lt - tan 1` Therefore, no such value of x exist |
|
| 389. |
The number of solutions of the equation `cos^(-1)((1+x^2)/(2x))-cos^(-1)x=pi/2+sin^(-1)x`is0 (b)1 (c) 2(d) 3 |
|
Answer» Correct Answer - B `cos^(-1) ((1 + x^(2))/(2x)) = (pi)/(2) + (sin^(-1) x + cos^(-1) x)` `rArr cos^(-1) ((1 + x^(2))/(2x)) = pi` or `((1 + x^(2))/(2x)) = cos pi = -1` or `x^(2) + 1 + 2x = 0` or `x = -1` |
|
| 390. |
The number of integer `x`satisfying `sin^(-1)|x-2|+cos^(-1)(1-|3-x|)=pi/2`is1 (b)2 (c) 3(d) 4A. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - B `sin^(-1) |x -2| + cos^(-1) (1 - |3 -x|) = (pi)/(2)` or `|x -2| = 1 -|3 -x|` or `|x -2| + |3 -x| = |(x -2) + (3 -x)|` or `(x -2) (3-x) le 0` or `2 le x le 3` |
|
| 391. |
Range of `f(x)=sin^(-1)x+tan^(-1)x+sec^(-1)x`isA. `((pi)/(4), (3pi)/(4))`B. `[(pi)/(4), (3pi)/(4)]`C. `{(pi)/(4), (3pi)/(4)}`D. none of these |
|
Answer» Correct Answer - C `f(x) = sin^(-1) x + tan^(-1) x + sec^(-1) x`, clearly, domain of `f(x) " is " x = +- 1` Thus, the range is `{f(1), f(-1)}`, i.e., `{(pi)/(4), (3pi)/(4)}` |
|
| 392. |
If `(sin^(-1) x)^(2) - (cos^(-1) x)^(2) = a pi^(2)` then find the range of aA. `[-(3)/(4), (1)/(4)]`B. `[-(3)/(4), (3)/(4)]`C. `[-1, 1]`D. `[-1, (3)/(4)]` |
|
Answer» Correct Answer - A `a pi^(2) = (sin^(-1) x)^(2) - (cos^(-1) x)^(2)` `rArr a pi^(2) = (sin^(-1) x - cos^(-1) x) (sin^(-1) x + cos^(-1) x)` `rArr a pi^(2) = (sin^(-1) x - cos^(-1) x) (pi//2)` `rArr 2 a pi = (sin^(-1) x cos^(-1) x)` `rArr 2a pi = (2 sin^(-1) x - pi//2)` Now, `-pi le 2 sin^(-1) x le pi` `rArr -3pi//2 le 2 sin^(-1) x - pi//2 le pi//2` `:. -3 pi//2 le 2a pi le pi//2` or `-3//4 le a le 1//4` |
|
| 393. |
The cost prices of two tables a same. One is sold at a profit of and the other more than the first one. If the overall profit earned after selling the tables is 24%, what is the cost price of each table (B) R 3500 (A) R 4400 (C) R 4800 (D) 4250 (E) R 3820 |
|
Answer» Let price of each table is `x` rupees. Then, selling price of first table `= x**(100+20)/100 = 6/5x` Rs Now, selling price of second table `= 6/5x+340` Rs Cost price of each both tables `= x+x = 2x` Rs Selling price of both tables ` = 6/5x+6/5x+340 = 12/5x+340` Rs Now, total profit is `24%`. `:. 2x**(100+24)/100 = 12/5x+340` `=>62/25x -12/5x = 340` `=>2/25x = 340` `=>x = 170**25 = 4250` Rs So, cost of each table is `4250` Rs. |
|
| 394. |
Let `f(x)=sec^-1(x-10)+cos^-1(10-x).` Then range of `f(x)` isA. `{0,pi/2,pi}`B. `{0,pi/2}`C. `{pi/2}`D. `{pi}` |
| Answer» Correct Answer - D | |
| 395. |
Let f`(x) = cosec^-1[1 + sin^2x], where [*]` denotes the greatest integer function, then the range of fA. Is discrete and has two membersB. Is discrete and has four membersC. Is continuousD. Is `(pi/2,pi/6)` |
| Answer» Correct Answer - A | |
| 396. |
If `[cot^(-1)x]+[cos^(-1)x]=0`, where `[]`denotes the greatest integer functions, then the complete set of valuesof `x`is`(cos1,1)`(b) `cos1,cos1)``(cot1,1)`(d) none of these |
|
Answer» `[cot^-1x]+[cos^-1x] = 0` As, `cot^-1x and cos^-1 x ge 0`, So, only solution available for this equation is, `[cot^-1x] = [cos^-1x] = 0` When `[cot^-1x] = 0, x in (cot1,oo)` When `[cos^-1x] = 0, x in (cos1,1)` Intersection of above two cases will be `x in (cot 1, 1)`. So, option `c` is the correct option. |
|
| 397. |
Complete solution set of `[cot^(-1)x]+2[tan^(-1)x]=0,`where `[]`denotes the greatest integer function, is equal to(a)`(0,cot1)`(b) `(0,tan1)``(tan1,oo)`(d) `(cot1,tan1)` |
|
Answer» `[cot^-1x]+2[tan^-1x] = 0` There can be two cases that satisfy the given equation. Case -1 : When `[cot^-1x] = 0 and [tan^-1x] = 0` `=> x in (cot1,oo) and x in (0,tan1)` `=> x in (cot1,tan1).` Case -2 : When `[cot^-1x] = 2 and [tan^-1x] = -1` `=> x in (cot1,cot2] and x in [-tan1,0)` `=> x in phi` as there is no intersection point of above two values of `x`. So, only solution is possible in Case -1, that is, `x in (cot1,tan1)`. So, option `(d)` is the correct option. |
|
| 398. |
Which of the following is/are true ?A. `"tan"^(-1)(1)/(3)=(1)/(2)"sin"^(-1)(3)/(5)`B. `"tan"^(-1)(1)/(3)=(pi)/(4)-cot^(-1)2`C. `"tan"^(-1)(1)/(3)=(pi)/(4)-(1)/(2)"cos"^(-1)(4)/(5)`D. `"tan"^(-1)(1)/(3)=(pi)/(2)-cot^(-1)3` |
|
Answer» Correct Answer - A::B::C `"tan"^(-1)(1)/(3)+"tan"^(-1)(1)/(2)=(pi)/(4)` `rArr "tan"^(-1)(1)/(3)=(pi)/(4)-"tan"^(-1)(1)/(2)=(pi)/(4)-cot^(-1)2` Also `2"tan"^(-1)(1)/(3)="tan"^(-1)((2)/(3))/(1-(1)/(9))="tan"^(-1)(3)/(4)` `= "sin"^(-1)(3)/(5)=(pi)/(2)-"cos"^(-1)(4)/(5)` |
|
| 399. |
Let `x_1,x_2,x_3,x_4` be four non zero numbers satisfying the equation `tan^-1 (a/x)+tan^-1(b/x)+tan^-1(c/x)+tan^-1(d/x)=pi/2` then which ofthe following relation(s) hold good?A. `x_(1)+x_(2)+x_(3)+x_(4)=a+b+c+d`B. `(1)/(x_(1))+(1)/(x_(2))+(1)/(x_(3))+(1)/(x_(4))=0`C. `x_(1)x_(2)x_(3)x_(4)=abcd`D. `(x_(2)+x_(3)+x_(4))(x_(3)+x_(4)+x_(1))(x_(1)+x_(2)+x_(3))=abcd` |
|
Answer» Correct Answer - B::C::D `"tan"^(-1)(a)/(x)+"tan"^(-1)(b)/(x)+"tan"^(-1)(c )/(x)+"Tan"^(-1)(d)/(x)=(pi)/(2)` `rArr "tan"^(-1)(a)/(x)"tan"^(-1)(b)/(x)=(pi)/(2)-("tan"^(-1)(c )/(x)+"tan"^(-1)(d)/(x))` `rArr "tan"^(-1)((a)/(x)+(b)/(x))/(1-(ab)/(x^(2)))=(pi)/(2)-"tan"^(-1)((c )/(x)+(d)/(x))/(1-(cd)/(x^(2)))` `rArr "tan"^(-1)((a+b)x)/(x^(2)-ab)="cot"^(-1)((c+d)x)/(x^(2)-cd)` `rArr ((a+b)x)/(x^(2)-ab)=(x^(2)-cd)/((c-d)x)` `rArr x^(4)-(ab+ac+ad+bc+bd+cd)x^(2)+abcd=0` This equation has roots `x_(1),x_(2),x_(3),x_(4)` `therefore Sigma x_(1)=0, Sigma x_(1)x_(2)=- Sigma ab, Sigma x_(1)x_(2)x_(3)=0` and `x_(1)x_(2)x_(3)x_(4)=abcd` `therefore sum(1)/(x_(1))=0` and `(x_(2)+x_(3)+x_(4))(x_(3)+x_(4)+x_(1))(x_(4)+x_(1)+x_(2))(x_(1)+x_(2)+x_(3))` `=x_(1)x_(2)x_(3)x_(4)` |
|
| 400. |
If `pgtqgt0` and `prlt-1ltqr`, then find the value of `tan^(-1)((p-q)/(1+qr))+tan^(-1)((q-r)/(1+qr))+tan^(-1)((r-p)/(1+qr))` |
|
Answer» As `p gt 0, q gt 0 => pq gt 0` `:. tan^-1((p-q)/(1+pq)) = tan^-1p-tan^-1q->(1)` As ` qr gt -1` `:. tan^-1((q-r)/(1+qr)) = tan^-1q-tan^-1r->(2)` As ` pr lt -1` `:. tan^-1((r-p)/(1+rp)) = pi +tan^-1r-tan^-1p->(3)` Adding (1),(2) and (3), `tan^-1((p-q)/(1+pq))+tan^-1((q-r)/(1+qr)) tan^-1((r-p)/(1+rp)) = tan^-1p-tan^-1q+ tan^-1q-tan^-1r + pi +tan^-1r-tan^-1p` `tan^-1((p-q)/(1+pq))+tan^-1((q-r)/(1+qr)) tan^-1((r-p)/(1+rp)) = pi.` |
|