Mathematics Must Do Problems of Angles and Trigonometric Ratios for NDA

Please Wait... While Loading Full Video

Must Do Problems of Angles and Trigonometric Ratios for NDA

Q 2630123012

If `cos x = tan y, cot = tan z` and `cot z = tan x`, then `sin x` is equal to
BCECE Mains 2015

(A)

` (sqrt5 + 1)/4`

(B)

` (sqrt5 - 1)/4`

(C)

` (sqrt5 + 1)/2`

(D)

` (sqrt5 - 1)/2`

Solution:

Given, `cos x = tan y, cot y = tan z`

and `cot z = tan x`

`:. cos x = tan y`

`=> cos x = 1/(tan z)`

` => cos x = cot z`

` => cos x = tan x`

` => cos x = (sin x)/(cos x)`

`=> cos^2 x = sin x`

`=> 1 - sin^2 x = sin x`

` => sin^2 x + sin x - 1 = 0`

` :. sin x = (-1 pm sqrt(1 - 4 xx (-1)))/(2 xx 1)`

` => sin x = ( -1 pm sqrt5)/2`

` => sin x = ( sqrt5 - 1)/2 [ ∵ (-1 - sqrt5)/2 < - 1 ]`

Correct Answer is `=>` (D) ` (sqrt5 - 1)/2`

Q 2600612518

If ` (sin^4 theta)/a + (cos^4 theta)/b = 1/(a + b)` , then which one of the following is incorrect?
BCECE Mains 2015

(A)

` (sin^4 theta)/a^2 = (cos^4 theta)/b^2`

(B)

` (sin^4 theta)/b^2 = (cos^4 theta)/a^2`

(C)

` (sin^8 theta)/a^3 + (cos^8 theta)/b^3 = 1/(a+b)^3`

(D)

`sin^4 theta = a^2/(a + b)^2`

Solution:

We have, ` (sin^4 theta)/a + (cos^4 theta)/b = 1/(a + b)`

` => (a + b) ( (sin^4 theta)/a + (cos^4 theta)/b ) = ( sin^2 theta + cos^2 theta)^2`

` => sin^4 theta + cos^4 theta + b/a sin^4 theta + a/b cos^4 theta`

` = sin^4 theta + cos^4 theta + 2 sin^2 theta cos^2 theta`

` => ( sqrt(b/a) sin^2 theta - sqrt(a/b) cos^2 theta )^2 = 0`

` => sqrt(b/a) sin^2 theta = sqrt(a/b) cos^2 theta`

` => b sin^2 theta = a cos^2 theta`

` => ( sin^2 theta)/a = (cos^2 theta)/b = ( sin^2 theta + cos^2 theta )/(a + b)`

` => sin^2 theta = a/(a + b) ` and ` cos^2 theta = a/(a + b)`

Clearly, these values satisfy options (a), (c) and (d).

Q 2620512411

Let ` (3pi)/4 < theta < pi ` and `sqrt (2 cot theta + 1/(sin^2 theta)) = k - cot theta`, then ` k ` is equal to
BCECE Mains 2015

(A)

`1`

(B)

`-1`

(C)

`0`

(D)

`1//2`

Solution:

We have, ` sqrt (2 cot theta + 1/(sin^2 theta)) = k - cot theta`

` => sqrt( 2 cot theta + cosec^2 theta) = k - cot theta`

` => sqrt( 2 cot theta + 1 + cot^2 theta ) = k - cot theta`

` => sqrt ( (1 + cot theta)^2 ) = k - cot theta`

` => pm (1 + cot theta) = k - cot theta`

`=> - (1 + cot theta) = k - cot theta`

` [ ∵ (3 pi)/4 < theta < 0 < pi => - oo < cot theta < - 1 ]`

` [ => - oo < 1 + cot theta < 0]`

` => -1 - cot theta = k - cot theta`

` => k = -1`

Correct Answer is `=>` (B) `-1`

Q 2519691510

The number of values of `x in [0, n pi] , n in Z`, that satisfy the equation `log_(| sin x |) (1 + cos 2x) = 2`, is
BCECE Mains 2015

(A)

`0`

(B)

`n`

(C)

`2n`

(D)

None of the above

Solution:

We observe that `log_(| sin |) (1 + cos x)` is defined if

`x != n pi , (2n +1) pi/2 , n in Z`.

Now, `log_(|sin x|) (1 + cos x) =2`

`=> (1 + cos x) = | sin x |^2`

`=> cos^2 x + cos x = 0`

`=> cos x (1 + cos x) = 0`

But, `cos^2 x + cos x != 0` for any

`x in (0, n pi) - (2n - 1) pi//2, n in Z`

Hence, the given equation has no solution.

Q 2539591412

Find the value of the expression `cos^4 pi/8 + cos^4 (3 pi)/8 + cos^4 (5 pi)/8 + cos^4 (7 pi)/8`.
BCECE Mains 2015

(A)

`2/3`

(B)

`- 3/2`

(C)

`3/2`

(D)

`- 2/3`

Solution:

Given, ` cos^4 pi/8 + cos^4 (3 pi)/8 + cos^4 (5 pi)/8 + cos^4 (7 pi)/8`

` = cos^4 pi/8 + cos^4 (3 pi)/8 + cos^4(pi/2 + pi/8) + cos^4 ( pi/2 + (3 pi)/8)`

` = cos^4 pi/8 + cos^4 (3 pi)/8 + sin^4 pi/8 + sin^4 (3 pi)/8 `

` [ ∵ cos ( pi/2 + theta) = - sin theta]`

` = ( cos^4 pi/8 + sin^4 pi/8) + ( cos^4 (3pi)/8 + sin^4 (3pi)/8)`

` = [ cos^4 pi/8 + sin^4 pi/8 + 2 sin^2 pi/8 cos^2 pi/8 - 2 sin^2 pi/8 cos^2 pi/8 ]`

` + [ cos^4 (3pi)/8 + sin^4 (3pi)/8 + 2 sin^2 (3pi)/8 cos^2 (3pi)/8 - 2 sin^2 (3pi)/8 cos^2 (3pi)/8 ]`

` = ( cos^2 pi/8 + sin^2 pi/8)^2 - 2 sin^2 pi/8 cos^2 pi/8`

` + ( cos^2 (3pi)/8 + sin^2 (3pi)/8)^2 - 2 sin^2 (3 pi)/8 cos^2 (3pi)/8`

` = 1 - 1/2 ( 2 sin pi/8 cos pi/8)^2 + 1 - 1/2 ( 2 sin (3pi)/8 cos (3pi)/8)^2`

` = 2 - 1/2 (sin 2 xx pi/8)^2 - 1/2 ( sin 2 xx (3pi)/8)^2`

` = 2 - 1/2 sin^2 pi/4 - 1/2 sin^2 (3 pi)/4`

` = 2 - 1/2 xx (1/sqrt2)^2 - 1/2 xx (1/sqrt2)^2`

` [ ∵ sin (3pi)/4 = sin ( pi - pi/4) = sin pi/4 ]`

` = 2 - 1/2 xx 1/2 - 1/2 xx 1/2 = 2 - 1/4 - 1/4 = 2 - 1/2 = 3/2`

Correct Answer is `=>` (C) `3/2`

Q 2589491317

The arithmetic mean of the roots of the equation `4 cos^3 x - 4 cos^2 x - cos (315 pi + x) = 1` in the interval `(0, 315)` is
BCECE Mains 2015

(A)

`50 pi`

(B)

`51 pi`

(C)

`100 pi`

(D)

`315 pi`

Solution:

Given,

`4 cos^3 x - 4cos^2 x - cos (315 pi + x) = 1`

`=> 4cos^3 x - 4cos^2 x + cos x - 1 = 0`

`[ ∵ cos (315 pi + x) = (-1)^(315) cos x = - cos x]`

` => (4 cos^2 x + 1) (cos x - 1) = 0`

`=> cos x - 1 = 0` or `4 cos^2 x + 1 != 0`

`=> cos x = 1`

`=> cos x = cos 0`

`=> x = 2 n pi , n in I`

`:. x = 2 pi , 4 pi , 6 pi , 8 pi , ... , 100 pi`

`[ ∵ 0 < x < 315]`

`:.` Required arithmetic mean

` = (2 pi + 4 pi + 6 pi + 8 pi + ... + 100 pi)/(50)`

` = ( 2 pi ( 1+2+3+ 4+ ...+ 50))/(50)`

` = ( 2 pi . (50)/2 . 51)/(50) = 51 pi`

Correct Answer is `=>` (B) `51 pi`

Q 2549491313

If `| z_1 -1| < 1, | z_2 -2 | < 2, | z_3 - 3 | < 3`, then `| z_1 + z_2 + z_3 |`
BCECE Mains 2015

(A)

is less than `6`

(B)

is more than `3`

(C)

is less than `12`

(D)

lies between `6` and `12`

Solution:

`∵ | z_1 - 1| < 1`

` => | z_1 | - | 1 | < 1` [by triangle inequality]

` => | z_1 | < 2` .........(i)

Similarly, `| z_2 - 2| < 2`

` => | z_2| - |2| < 2`

`=> |z_2| < 4` .........(ii)

and ` | z_3 - 3| < 3`

` => | z_3| - |3| < 3`

`=> | z_3 | < 6` ........(iii)

On adding Eqs. (i), (ii) and (iii), we get

` | z_1| + | z_2| + | z_3| < 2 + 4 + 6`

` => | z_1| + | z_2| + | z_3| < 12` .........(iv)

By using triangular inequality,

` | z_1 + z_2 + z_3 | <= | z_1| + | z_2| + | z_3|`

` => | z_1 + z_2 + z_3 | < 12` [from Eq (iv)]

Q 2539491312

If `a = cos ( (2pi)/7) + i sin ( (2 pi)/7 )` , then the quadratic equation whose roots are ` alpha = a + a^2 + a^4` and ` beta = a^3 + a^5 + a^6` , is
BCECE Mains 2015

(A)

`x^2 - x + 2 = 0`

(B)

`x^2 + x - 2 = 0`

(C)

`x^2 - x - 2 = 0`

(D)

`x^2 + x + 2 = 0`

Solution:

Given, `a = cos ( (2pi)/7) + i sin ( (2pi)/7)`

`=> a^7 = [ cos ( (2pi)/7) + i sin ( (2pi)/7) ]^2`

` => a^7 = cos 2 pi + i sin 2 pi`

` a^7 = 1` ..........(i)

Let ` S = alpha + beta = (a + a^2 + a^4)+ (a^3 +a^5 +a^6)`

`= a + a^2 +a^3 +a^4 + a^5 +a^6`

Since, it is GP.

`:. S = (a(1 - a^6))/(1 - a)`

` => S = (a - a^7)/(1 - a)` [from Eq. (i)]

` => S = (a - 1)/(1 - a)`

` => S = - (1 - a)/(1 - a) = -1 ` ..........(ii)

Again, `P = alpha beta = (a + a^2 + a^4 ) (a^3 + a^5 + a^6)`

`= a^4 +a^6 +a^7 + a^5 +a^7 +a^8+ a^7 + a^9 + a^(10)`

`= a^4 + a^6 + 1 + a^5 + 1 + a· a^7 + 1 + a^2 ·a^7 + a^3 · a^7`

`= a^4 + a^6 + 1 + a^5 + 1 + a^2 + 1 + a^2 + a^3`

`= 3 + (a+ a^2 + a^3 + a^4 + a^5 + a^6 )`

`= 3 + S = 3 - 1` [from Eq. (ii)]

`= 2`... (iii)

`:.` Required equation is

`x^2 - (alpha + beta ) x + alpha beta ) = 0`

`:. x^2 + x + 2 = 0` [from Eqs. (ii) and (iii)]

Correct Answer is `=>` (D) `x^2 + x + 2 = 0`

Q 2581145927

`tan 10° + tan 3.5^o + tan 10° tan 35°` is equal to:
BCECE Stage 1 2015

(A)

(B)

`1/2`

(C)

`-1`

(D)

`1`

Solution:

Now , `tan(10^o + 35^o) = (tan10^o + tan 35^o)/(1 -tan 10^o tan 35^o)`

`=> tan 45^o =1 = (tan 10^o + tan 35^o)/(1 -tan 10^o tan 35^o)`

`=> tan 10^o + tan 35^o + tan 10^o tan 35^o =1`

Correct Answer is `=>` (D) `1`

Q 2501045828

If sin A= sin B and cos A cos B, then A is equal to:
BCECE Stage 1 2015

(A)

`2npi + B`

(B)

`2npi - B`

(C)

`npi + B`

(D)

`npi+ (-1)^nB`

Solution:

Since sin A = sin B

`=> A = B` and `A = pi - B` ... (i)

and `cos A = cos B`

`=> A= B` and `A= 2pi - B` ... (ii)

From Eqs. (i) and (ii)

`=> A = 2n pi + B`

Correct Answer is `=>` (A) `2npi + B`

Q 2511734629

If `alpha, beta, gamma` are the angles when a directed line makes with the positive directions the coordinate axes, then `sin ^2 alpha + sin^2 beta + sin^2 gamma` is equal to
BCECE Stage 1 2015

(A)

(B)

(C)

(D)

None of theses

Solution:

If the line makes em angle `alpha, beta, gamma` to the coordinate axes, then

`cos^ alpha + cos^2 beta + cos^2 gamma =1`

We know that

`cos^2alpha + cos^2 beta + cos^2 gamma =1`

`=> 1 -sin^2 alpha +1 - sin^2 beta +1 - sin^2 gamma =1`

`=> sin^2alpha +sin^2 beta +sin^2 gamma = 3-1`

`=2`

Correct Answer is `=>` (B) 2

Q 2551534424

If `cos(theta + phi) = m cos (theta - phi)`. then `tantheta` is equal to:
BCECE Stage 1 2015

(A)

`[(1 + m)/(1- m)] tan phi`

(B)

`[(1 - m)/(1 + m)] tan phi`

(C)

`[(1- m)/(1 + m)] cot phi`

(D)

`[(1 + m)/(1 - m)] sec theta`

Solution:

We have

`cos(theta + phi) = m cos(theta- phi)`

`=> cos theta cos phi - sin theta sin phi = m (cos theta cos phi +sin theta sin phi)`

`=> cos theta cos phi (1- m ) = sin theta sin phi (1 +m)`

`=> tan theta = ((1 -m)/(1+m)) cot phi`

Correct Answer is `=>` (C) `[(1- m)/(1 + m)] cot phi`

Q 2541334223

The maximum value of `12 sin 8- 9 sin^2 theta` is :
BCECE Stage 1 2015

(A)

(B)

(C)

(D)

None of these

Solution:

Let `f(theta) = 12 sintheta - 9 sin^ 2 theta`

On differentiating w.r.t. e, we get

`f' (theta) = 12 cos theta - 18 sin theta cos theta`

`= 12 cos theta - 9 sin^ 2 theta`

For maximum or minimum, put `f' (theta) = 0`.

`12 cos theta- 18 sin theta cos theta = 0`

`=> 6 cos theta (2- 3 sin theta) = 0`

`=> cos theta = 0 ` or `sin theta =2/3`

`=> theta = pi/2` or `theta = sin ^-1 \ \ 2/3`

`f'' (theta) = -12 sin theta - 18 cos 2 theta`

at `theta = pi/2`

`f'' (theta) = -12 -18 (-1) =6 > 0`, minimum

At `sin theta =2/3`

`f''(theta) = -12 (2/3) - 18 (1 - 2 (2/3)^2 )`

`=-8 - 18 ((9-8)/9) =-10 < 0` , maximum

`:. ` The maximum value of `f(theta)` at `theta = sin^-1\ 2/3` is

`f(theta) = 12 (2/3) -9 (2/3)^2`

`=8-4=4`

Correct Answer is `=>` (B) 4

Q 2580345217

If `tan x · tan y = a` and `x + y = pi/6`, then `tan x` and `tan y` satisfy the equation
BCECE Stage 1 2013

(A)

`x^2 - sqrt3 (1- a) x + a = 0`

(B)

`sqrt3 x^2 - (1- a)x + a sqrt(3) = 0`

(C)

`x^2 + sqrt(3) (1 + a ) x - a = 0`

(D)

`sqrt3 x^2 + (1+ a)x - a sqrt(3) = 0`

Solution:

`∵ tan x ·tan y =a`

and `tan(x + y) = tan ( pi/6)`

` => (tan x + tan y )/(1- tan x · tan y) = 1/sqrt3`

`=> tan x + tan y = 1/sqrt3 (1- a)`

Equation whose roots are `tan x` and `tan y` is

`x^2 - ( 1 - a)/sqrt3 . x + a = 0`

`=> sqrt(3)x^2 - (1 - a) x + a sqrt3 = 0`

Correct Answer is `=>` (B) `sqrt3 x^2 - (1- a)x + a sqrt(3) = 0`

Q 2560245115

If `n` be a positive integer such that `sin (pi/(2n)) + cos (pi/(2n)) = sqrtn/2`, then
BCECE Stage 1 2013

(A)

`n = 6`

(B)

`n = 2`

(C)

`n = 1`

(D)

`n = 3,4,5`

Solution:

`sin (pi/(2n)) +cos (pi/(2n)) = sqrtn/2`

`=> sqrt2 { 1/sqrt2 . cos (pi/(2n)) + 1/sqrt2 . sin (pi/(2n))}`

` = sqrt n/2`

`=> sqrt2 {cos ( pi/4 - pi/(2n) ) } = sqrt n/2`

` => cos ( pi/4 - pi/(2n) ) = pi/(2sqrt2)` .....(i)

` => sqrt2 {cos ( pi/4 - pi/(2n) )} = sqrt n/2`

` => cos ( pi/4 - pi/(2n) ) = pi/(2sqrt2)` .......(ii)

when ` n = 6 , LHS = cos ( pi/6) = sqrt3/2`

and `RHS = sqrt6/(2sqrt2) = sqrt3/2`

Eq. (i) is not satisfied for `n = 1, 2, 3, 4, 5`.

Correct Answer is `=>` (A) `n = 6`

Q 2520401311

`tan [ pi/4 + 1/2 cos^-1 (a/b) ] + tan [ pi/2 - 1/2 cos^-1(a/b) ]` is equal to
BCECE Stage 1 2014

(A)

`(2a)/b`

(B)

`(2b)/a`

(C)

`a/b`

(D)

`b/a`

Solution:

`tan [ pi/4 +1/2 cos^-1 (a/b) ] +tan[ pi/4 - 1/2 cos^-1 (a/b) ]`

`= tan [ pi/4 + phi] + tan[ pi/4 -phi]` put (`1/2 cos^-1 (a/b) = phi`)

`= (! + tan phi)/(1 -tan phi ) + (1 - tanphi)/(1 + tan phi)`

`=( (1 + tan phi ) ^2 + (1 - tan phi)^2)/(1 -tan^2 phi)`

`= (2 (1 + tan ^2 phi))/(1 - tan^2 phi)`

`= (2 (cos^2 phi + sin^2 phi))/(cos^2 phi - sin^2 phi)`

`=2/(cos 2 phi)`

`= (2b)/a \ \ \ \ \ \ [ because 1/2 cos ^-1(a/b) = phi => cos 2 theta = a/b]`

Correct Answer is `=>` (B) `(2b)/a`

Q 2570301216

The number of values of x in `[0, 2pi]` satisfying the equat.ion `3 cos 2x- 10 cos x + 7 = 0` is
BCECE Stage 1 2014

(A)

(B)

(C)

(D)

Solution:

Given, `3cos 2x -10 cos x + 7 = 0`

`=> 3 (2 cos^2 x - 1) - 10 cosx + 7 = 0`
`=> 6 cos^2 x - 10 cos x + 4 = 0`

`=> 2 [(3cos x -2)(cos x -1)]= 0`
`=> cos x =1 ` and `cos x =2/3`

Since, cos x is positive in 1st and IV th quadrants. Hence, total number of solutions is 4.

Correct Answer is `=>` (D) 4

Q 2540301213

The maximum value of `4 sin^2 x- 12sin x +7` is
BCECE Stage 1 2014

(A)

(B)

(C)

Does not exist

(D)

None of the above

Solution:

`4sin^2 x -12 sin x + 7 = 4 (sin^2 x - 3 sin x) + 7`

`= 4 [ (sinx - 3/2)^2 - 9/4] +7`

`= 4(sin x - 3/2)^2 -2`

Since, `- 1 <= sin x <= 1`

`:. -5/2 <= sin x - 3/2 <= - 1/2`

`=> 1/4 <= (sin x - 3/2 )^2 = 25/4`

`=> 1 <= 4 (sin x - 3/2)^2 <= 25`

`=> -1 <= 4 (sin x - 3/2)^2 - 2 <= 23`

Correct Answer is `=>` (D) None of the above

Q 2530301212

If `sin A - sqrt6 cos A = sqrt7 cos A`, then `cos A + sqrt6 sin A` is equal to
BCECE Stage 1 2014

(A)

`sqrt6 sinA`

(B)

`-sqrt7 sin A`

(C)

`sqrt6 cos A`

(D)

`sqrt 7 cos A`

Solution:

Given, `sin A- sqrt6 cos A = - sqrt7 cos A`

On squaring both sides, we get

`sin^2 A+ 6 cos^2 A-2sqrt 6 sinAcos A=7cos^2 A`

`=> sin^2 A+ 6(1-sin^2 A) = cos^2 A + 6cos^2 A + 2sqrt6 sinA cos A`

`=> sin^2A - 6cos^2 A + 6 = cos^2A + 6 sin^2 A + 2sqrt 6 sin A cos A`

`=> 7 sin^2A = (cos A + sqrt 6 sin A)^2`

`=> pm sqrt 7 sin A = cos A + sqrt 6 sin A`

Correct Answer is `=>` (B) `-sqrt7 sin A`

Q 2479878716

The value of `x` in `(0 , pi/2)` satifying the equation `sin x cos x = 1/4` is
BCECE Stage 1 2012

(A)

`pi/6`

(B)

`pi/12`

(C)

`pi/8`

(D)

`pi/4`

Solution:

`2sin xcosx = 1/2 = sin (pi/6)`

`=> 2x = npi +(-1)^n (pi/6)`

`=> x = (npi)/2+ (-1)^n (pi/12)`

For `x ϵ (0 , pi/2)`

`x = pi/12`

Correct Answer is `=>` (B) `pi/12`

Q 2419678519

The solution of the equation `cos^2 theta + sin theta + 1 = 0`, lies in the interval
BCECE Stage 1 2011

(A)

`(-pi/4 , pi/4)`

(B)

`(pi/4 ,(3pi)/4)`

(C)

`((3pi)/4, (5pi)/4)`

(D)

`((5pi)/4, (7 pi)/4)`

Solution:

We have, `cos^2 theta+sin theta=0`

`=> 1-sin^2 theta+sin theta+1=0`

`=> sin^2 theta-sin theta-2=0`

`=> (sin theta+1)(sin theta-2)=0`

`=> sin theta+1=0`

`=> sin theta=-1`

`=> theta=(3 pi)/2 in ((5pi)/4,(7 pi)/4)`

Correct Answer is `=>` (D) `((5pi)/4, (7 pi)/4)`

Q 2429756611

If `x= sec theta- cos theta` and `y = sec^n theta- cos^n theta`, then `((dy)/(dx))^2` is
BCECE Stage 1 2011

(A)

`(n^2(y^2+4))/(x^2+4)`

(B)

`(n^2(y^2-4))/x^2`

(C)

`n ((y^2-4))/(x^2-4)`

(D)

`((ny)/x)^2-4`

Solution:

Differentiate given equations w.r.t. `theta`, we get

`(dy)/(d theta)=n sec^(n-1) theta * tan theta-n * cos^(n-1) theta(-sin theta)`

`=n tan theta (sec^n theta+cos^n theta)`

`(dx)/(d theta)=sec theta tan theta+ sin theta=tan theta(sec theta+ cos theta)`

`:. (dy)/(dx) =(n (sec^n theta+cos^n theta))/((sec theta+cos theta))`

`:. ((dy)/(dx))^2=(n^2{(sec ^n theta-cos^n theta)^2+4})/((sec theta -sec theta)^2 +4)`

`=(n^2(y^2+4))/((x^2+4))`

Correct Answer is `=>` (A) `(n^2(y^2+4))/(x^2+4)`

Q 2418180009

The number of solutions of the equation `3sin^2 x-7 sinx+2 = 0` in the interval `[0 , 5pi]` is
BCECE Stage 1 2015

(A)

`0`

(B)

`5`

(C)

`6`

(D)

`10`

Solution:

We have `3sin^2x-7sinx+2 = 0`

`=> (3sinx-1)(sinx-2) = 0`

`=> sinx = 1/3` `[because sinx -2 ne 0]`

`=> x = sin^(-1)(1/3) , pi-sin^(1/3) , 2pi+sin^(-1)(1/3)`

`3pi-sin^(-1)(1/3) , 4pi+sin^(-1)(1/3) , 5pi-sin^(-1)(1/3)`

Correct Answer is `=>` (C) `6`

Q 2436267172

The number of roots of the equation
`x + 2 tan x = pi/2` in the interval `[0, 2pi]` is
UPSEE 2013

(A)

`1`

(B)

`2`

(C)

`3`

(D)

infinite

Solution:

We have, `x + 2 tan x = pi/2`

`=> tan x = pi/4 - x/2`

Let `y = tan x ` and `y = pi/4 - x/2`

The curves `y =tan x` and `y = pi/4 - x/2` in interval

`[0, 2pi]`, intersect at three points. The abscissa
of these three points are the roots of the
equation.

Correct Answer is `=>` (C) `3`

Q 2476345276

The general solution of `tan (pi/2 sin theta)=cot(pi/2 cos theta)` is
UPSEE 2010

(A)

`theta=2 r pi +pi/2, r in z`

(B)

`theta=2 r pi, r in Z`

(C)

`theta=2 r pi, r in Z`

(D)

None of the above

Solution:

`tan(pi/2 sin theta)=cot (pi/2 cos theta)`

`=> tan (pi/2 sin theta)=tan (pi/2 -pi/2 cos theta)`

`=> pi/2 sin theta=r pi+pi/2 -pi/2 cos theta, r in Z`

`=> sin theta+cos theta=(2 r+1), r in Z`

`=> 1/sqrt 2 sin theta+1/sqrt 2 cos theta= (2r+1)/sqrt 2, r in Z`

`=> cos (theta-pi/4)=(2r+1)/sqrt 2, r in Z`

`=> cos(theta-pi/4)=1/sqrt 2` or `-1/sqrt 2` (for `r=0,-1`)

`=> theta-pi/4=2 r pi pm pi/4, r in Z`

`=> theta =2 r pi pm pi/4+pi/4, r in Z`

`=> theta=2 r pi, 2r pi+pi/2 , r in Z`

But `theta = 2r pi +pi/2, r in Z` does not satisfy the
given equation.

Correct Answer is `=>` (B) `theta=2 r pi, r in Z`

Q 2456434374

The solution of the equation `sec theta -cosec theta=4/3` is
UPSEE 2010

(A)

`1/2 [ n pi +(-1)^n sin^(-1) 3//4]`

(B)

`(npi)/2 +(-1)^n sin^(-1) 3//4`

(C)

`n pi+(-1)^n sin^(-1) 3//4`

(D)

None of these

Solution:

`sec theta - cosec theta = 4/3`

`=> 3(sin theta-cos theta )=4 sin theta cos theta`

`=> 9(1-sin 2 theta)=4 sin^2 2 theta`

`=> 4 sin^2 2 theta+9 sin 2 theta-9=0`

`=> sin 2 theta= (-9 pm sqrt(81+144))/8=3/4 ,-3`

But `sin 2 theta ne -3`

`:. sin 2 theta=3/4`

`=> theta=1/2[n pi+(-1)^n sin^(-1) \ 3/4]`

Correct Answer is `=>` (A) `1/2 [ n pi +(-1)^n sin^(-1) 3//4]`

Q 2425491361

If `3 sin^2 theta + 2 sin^2 phi = 1` and `3 sin 2 theta = 2 sin 2 phi`,
`0 < theta < pi/2 ` and `0 < phi < pi/2`, then the value of
`theta + 2 phi` is
UPSEE 2014

(A)

`pi/4`

(B)

`pi/2`

(C)

`pi`

(D)

None of these

Solution:

Given that, `3 sin^2 theta + 2 sin^2 phi = 1`

`=> 3sin^2 theta = cos 2 phi` .....(i)

and `3 sin theta cos theta = sin 2 phi` ....(ii)

On squaring and adding Eqs. (i) and (ii), we get

`9 sin^2 theta (sin^2 theta + cos^2 theta ) = 1`

` => sin theta = 1/3` and `cos theta = (2sqrt(2))/3`

` :. cos 2 phi = 3 xx 1/9 = 1/3` and `sin 2 phi = (2sqrt(2))/3`

Now, `cos (theta + 2 phi) = cos theta cos 2 phi - sin theta sin 2 phi`

` = (2 sqrt(2))/3 . 1/3 - 1/3 . ( 2sqrt(2))/3 = 0`

and `theta + 2 phi < (3pi)/2`

` :. theta + 2 phi < pi/2`

Correct Answer is `=>` (B) `pi/2`

Q 2441723623

If the vectors a and b are linearly independent satisfying
`(sqrt 3tan theta + 1) a+ (sqrt 3 sec theta- 2) b = 0`, then the
value of `theta` is
UPSEE 2015

(A)

`pi/2`

(B)

`pi/6`

(C)

`(5 pi)/6`

(D)

`(11 pi)/6`

Solution:

Since, the vectors a and b are linearly
independent.

`:. sqrt 3 tan theta+1=0` and `sqrt 3 sec theta-2=0`

`=> tan theta=-1/sqrt 3` and `sec theta=2/sqrt 3`

`=> theta=(11 pi)/6`

Correct Answer is `=>` (D) `(11 pi)/6`

Q 2480091817

If `1/6 sin^2 theta, cos theta` and `sec theta` are in GP, then `theta` is equal to `(n in Z)`
UPSEE 2015

(A)

`2n pi pm pi/3`

(B)

`2 n pi pm pi/6`

(C)

`n pi + (-1)^n pi/3`

(D)

`n pi+pi/3`

Solution:

Since, `1/6 sin^2 theta, cos theta` and `sec theta` are in GP.

`:. cos^2 theta=1/6 sin^2 theta * sec theta`

`=> 6 cos^2 theta=(sin^2 theta)/(cos theta)`

`=> 6 cos^2 theta=(1-cos^2 theta)`

`=> 6 cos^3 theta+cos^2 theta-1=0`

`=> (2 cos theta-1) (3 cos^2 theta)+2 cos theta+1)=0`

`=> cos theta=1/2`

`=> theta=2 n pi pm pi/3 * n in Z`

[·: other factor gives imaginary roots]

Correct Answer is `=>` (A) `2n pi pm pi/3`

Q 1642634533