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1.

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
Here, `alpha = 3sin ^(-1) ((6)/(11)) and beta = 3 cos ^(-1) ((4)/(9)) "as " (6)/(11) gt (1)/(2)`
`rArr sin^(-1)((6)/(11)) gt sin ^(-1)((1)/(2)) = (pi)/(6)`
`therefore " " alpha = 3 sin ^(-1) ((6)/(11)) gt (pi)/(2)`
`rArr " " cos alpha lt 0`
Now, `" " beta = 3 cos ^(-1)((4)/(9))`
As ` " " (4)/(9) lt (1)/(2) rArr cos ^(-1) ((4)/(9)) gt cos ^(-1) ((1)/(2)) = (pi)/(3)`
`therefore " " beta = 3 cos ^(-1) ((4)/(9)) gt pi`
`therefore cos beta lt 0 and sin beta lt 0`
Now, `alpha + beta ` is slightly greater that ` ( 3pi )/(2)`.
`therefore " " cos ( alpha + beta ) gt 0`
Discussion
2.

if `cos^(-1)((2)/(3x))+cos^(-1)((3)/(4x))=(pi)/2, x>3/4` then `x=` (a) `(sqrt(145))/(11)` (b) `(sqrt(145))/(12)` (c) `(sqrt(146))/(10)` (d) `(sqrt(146))/(11)`A. `(sqrt ( 145))/( 10 )`B. `( sqrt ( 146))/( 12)`C. ` (sqrt ( 145))/( 12)`D. `(sqrt ( 145))/( 11)`

Answer» Correct Answer - C
We have ` cos ^(-1) ((2)/(3x)) + cos ^(-1) ((3)/(4x)) = (pi)/(2)`
`rArr cos ^(-1) ((2)/(3x). ( 3)/(4x) - sqrt (1 - (4)/( 9x ^(2))) sqrt (1 - ( 9)/( 16 x ^(2))) = (pi)/(2)`
`[ because cos ^(-1) x + cos ^(-1) y = cos ^(-1) ( xy - sqrt (1 - x ^(2)) sqrt (1 -y ^(2)))]`
`rArr cos^(-1) ((1)/(2x^2)) - (sqrt ( 9x ^(2) - 4) sqrt ( 16 x ^(2) -9 ))/( 12 x ^(2))) = (pi)/(2)`
`rArr = ( 6 - sqrt(9x ^(2) - 4 ) sqrt ( 16x ^(2) -9))/( 12 x ^(2) ) = cos ""(pi )/(2) = 0 `
`rArr sqrt (9x^(2) - 4) sqrt ( 16x ^(2) - 9 ) = 6`
On squaring both sides,
`rArr (9x ^(2) - 4) ( 16 x ^(2) -9) = 36`
`rArr 144 x ^(4) - 81 x ^(2) - 64 x ^(2) + 36 = 36`
`rArr " " x ^(2)( 144 x ^(2) - 145) =0`
`rArr " " x =0 or x= pm sqrt ((145)/( 144)) = pm =( sqrt ( 145))/( 12)`
But ` x gt (3)/(4), " " x = ( sqrt ( 145))/( 12)`
Discussion
3.

The number of real solutions of `tan^(-1)sqrt(x(x+1))+sin^(-1)sqrt(x^2+x+1)=pi/2`is`z`ero b. one``c. twod. infiniteA. zeroB. oneC. twoD. infinite

Answer» Correct Answer - C
Given function is
`tan^(-1)sqrt(x(x+1)) + sin ^(-1)sqrt(x^(2) + x + 1) = (pi)/(2)`
Function is defined, if
(i) `x (x + 1) ge 0`, since domain of square root function.
(ii) `x^(2) + x + 1 ge 0`, since domain of square root function.
(iii) ` sqrt(x^(2) + x + 1) le 1`, since domain of `sin ^(-1)` function.
From (ii) and (iii)`, 0 le x^(2) + x + 1 le 1 nn x^(2) + x ge 0`
`rArr " " 0 le x ^(2) + x + 1 le 1 nn x^(2) + x + 1 ge 1`
`rArr " " x ^(2) + x + 1 = 1 `
`rArr " " x^(2) + x = 0`
`rArr " " x(x+1) =0`
`rArr " " x=0, x =-1`
Discussion
4.

The numerical value of `"tan"(2tan^(-1)(1/5)-pi/4`is equal to____

Answer» Correct Answer - `(- (7)/(17))`
` tan [ 2 tan ^(-2) ((1)/(5)) - (pi)/(4)] = tan [ tan ^(-1) (( 2* (1)/(5))/( 1 - ( 1)/( 25))) - (pi)/(4)]`
`" "= tan [ tan ^(-1) ((5)/( 12)) - (pi)/( 4)] `
` " " = ( tan [ tan ^(-1) ((5)/( 12)) ] - tan (( pi)/( 4)))/( 1 + tan [ tan ^(-1) ((5)/( 12)) ] tan ""(pi)/( 4))`
` " " = ((5)/( 12) - 1 )/( 1 + ( 5)/( 12) * 1) = - ( 7)/( 17)`
Discussion
5.

Let `tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2))`, where `|x|

Answer» Correct Answer - A
Given ` tan ^(-1) y = tan ^(-1) x + tan ^(-1) ((2x)/( 1- x ^(2)))`
where `|x| lt (1)/(sqrt 3) rArr tan ^(-1) y = tan ^(-1) {( x + (2x )/( 1-x ^(2)))/( 1- x (( 2x )/( 1 - x ^(2)))) }`
`[because tan ^(-1) x + tan ^(-1) y = tan ^(-1) ((x + y )/( 1- xy )), where x gt 0, y gt 0 and xy lt 1]`
`" " =tan ^(-1) ((x - x ^(3) + 2x )/( 1 - x ^(2) - 2x ^(2))) `
`tan ^(-1) y = tan ^(-1) ((3x - x ^(3))/( 1 - 3x ^(2)))`
`rArr " " y = ( 3x - x ^(3))/( 1 - 3x ^(2))`
` " " |x| lt (1)/(sqrt 3)`
`rArr " " - (1)/(sqrt (3)) lt x lt (1)/(sqrt 3)`
Let ` x = tan theta `
`rArr " " - ( pi )/(6) lt theta lt (pi)/(6)`
`therefore " " tan ^(-1) y = theta + tan ^(-1) ( tan 2 theta) = theta + 2theta = 3 theta `
`rArr " " y = tan 3 theta `
`rArr " " y = ( 3 tan theta - tan ^(3) theta)/( 1- 3 tan ^(2) theta)`
` rArr " " y = ( 3x - x ^(3))/( 1- 3x ^(2))`
Discussion
6.

Solve the equation `tan^(-1)2x+tan^(-1)3x=pi/4`

Answer» Correct Answer - `x = (1)/(6)`
Given, `tan ^(-1) 2 x + tan ^(-1) 3x = (pi)/(4)`
`rArr " " tan ^(-1) (( 2x + 3x )/( 1 - 6x^(2))) = (pi)/( 4)`
`rArr " " ( 5x)/( 1- 6x ^(2)) = 1`
`rArr " " 6x ^(2) + 5x - 1 = 0`
`rArr " " (x + 1 ) ( 6x - 1 ) = 0`
` rArr " " x = -1 or (1)/(6)`
But `x = -1 ` does not satisfy the given equation.
`therefore ` We take ` x = (1)/(6)`.
Discussion
7.

The value of `cot {sum_(23)^(n=1) cot^(-1)(1+ sum_(n=1)^(n) 2k)}` isA. `(23)/(25)`B. `(25)/(23)`C. `(23)/(24)`D. `( 24)/(23)`

Answer» Correct Answer - B
We have , ` cot [ overset ( 23) underset ( n = 1 ) ( sum) cot ^(-1) (1 + overset n underset (h = 1 ) (sum) 2k )]`
`rArr cot [ overset ( 23) underset(n = 1 ) (sum) cot ^(-1) (1 + 2 + 4 + 6 + 8 + ... + 2n )]`
`rArr cot [ overset ( 23) underset ( n = 1) ( sum) cot ^(-1) { 1 + n (n + 1 )}] `
` rArr cot [ overset ( 23) underset ( n = 1 ) (sum) tan ^(-1) ""(1)/( 1+ n (n + 1 ))]`
`rArr cot [ overset ( 23) underset ( n = 1) ( sum) tan ^(-1) { (( n + 1 ) - n )/( 1+ n ( n + 2 ))}]`
`rArr cot [ overset ( 23) underset ( n = 1 ) ( sum) ( tan ^(-1) ( n + 1 ) - tan ^(-1) In n ) ]`
`rArr cot [ (tan ^(-1) 2 - tan ^(-1) 1 ) + ( tan ^(-1) 3 - tan ^(-1) 2) + ( tan ^(-1) 4 - tan ^(-1 ) 3) ] + ... + ( tan ^(-1) 24 - tan ^(-1) 23)]`
`rArr cot ( tan ^(-1) 24 - tan ^(-1) 1)`
`rArr cot ( tan ^(-1) ""( 24 - 1 ) /( 1+ 24 *(1))) = cot ( tan ^(-1) ""( 23)/( 25))`
`" " = cot ( cot ^(-1) ""( 25)/( 23)) = ( 25)/( 23)`
Discussion