Q: Let = 1 60^ 61^ + 1 62^ 63^ + + 1 118^ 119^ Then the value of ( cosec 1^ )^2 is __________

aligned & = _ r -30 ^ 59 1 (2 r )^ (2 r +1)^ \\ & cosec 1^ = _ r -30 ^ 59 1^ (2 r )^ (2 r +1)^ \\ & = _ r -30 ^ 59 ( (2 r )^ - (2 r +1)^ ) \\ & = 60^ - 61^ \\ & + 62^ - 63^ \\ & \\ & + (118^ )- (119^ ) \\ & cosec 1^ = 60^ = 1 3 \\ & ( cosec 1^ )^2=3 aligned

Q: Let α and β be real numbers such that - π 4 < β < 0 < α < π 4 . If sin α + β = 1 3 and cos α - β = 2 3 , then the greatest integer les

sin α cos β + cos α sin β + cos β sin α + sin β cos α 2 = cos α - β sin β cos β + cos α - β sin α cos α 2 = 2 cos α - β 1 sin 2 β + 1 sin 2 α 2 = 16 9 2 sin α + β · cos α - β sin 2 α · sin 2 β 2 = 16 9 4 · 1 3 · 2 3 cos 2 α - 2 β - cos 2 α + 2 β 2 = 16 9 8 9 2 cos 2 α - β - 1 - 1 - 2 sin 2 α + β 2 = 16 9 8 9

Q: Let α = ∑ k = 1 ∞ sin 2 k π 6 . Let g : 0 , 1 → ℝ be the function defined by g x = 2 a x + 2 a 1 - x . Then, which of the following statements is/are TRUE?

Given, α = ∑ k = 1 ∞ sin 2 k π 6 and g x = 2 a x + 2 a 1 - x . Now solving, α = ∑ k = 1 ∞ 1 2 2 k = ∑ k = 1 ∞ 1 4 k = 1 4 1 - 1 4 = 1 3 Now putting the value of α in g x we get, g x = 2 x 3 + 2 1 - x 3 Now, g ' x = ln 2 3 2 2 x 3 - 2 1 3 2 x 3 Now finding the critical point by g ' x = 0 ⇒ x = 1 2 And, derivative changes sign from negative to positive at x = 1 2 , hence x = 1 2 is

Question 1

Let = (1 - 2 ( 11 ) ) (1 - 2 ( 3 11 ) ) (1 - 2 ( 9 11 ) ) (1 - 2 ( 27 11 ) ) (1 - 2 ( 81 11 ) ). Then the value of 5 - ^2 is _____________.

Accepted Answer

Let the given expression be . First, we simplify the angles in the product. ( 27 11 ) = (2 + 5 11 ) = ( 5 11 ) ( 81 11 ) = (7 + 4 11 ) = - ( 4 11 ) = ( - 4 11 ) = ( 7 11 ) Thus, the angles are 11 , 3 11 , 5 11 , 7 11 , 9 11 , which can be written as _k = (2k-1) 11 for k = 1, 2, 3, 4, 5. The expression becomes: = _ k=1 ^ 5 (1 - 2 _k ) Using the trigonometric identity 1 - 2 = 1 - 2(2 ^2( /2) - 1) = 3 - 4 ^2( /2). Multiplying and dividing by ( /2), we get: 1 - 2 = 3 ( /2) - 4 ^3( /2) ( /2) = - (3 /

Question 2

Let = 1 60^ 61^ + 1 62^ 63^ + + 1 118^ 119^ Then the value of ( cosec 1^ )^2 is __________

Accepted Answer

aligned & = _ r -30 ^ 59 1 (2 r )^ (2 r +1)^ \ & cosec 1^ = _ r -30 ^ 59 1^ (2 r )^ (2 r +1)^ \ & = _ r -30 ^ 59 ( (2 r )^ - (2 r +1)^ ) \ & = 60^ - 61^ \ & + 62^ - 63^ \ & \ & + (118^ )- (119^ ) \ & cosec 1^ = 60^ = 1 3 \ & ( cosec 1^ )^2=3 aligned

Question 3

Let &#945; and &#946; be real numbers such that - &#960; 4 &#60; &#946; &#60; 0 &#60; &#945; &#60; &#960; 4 . If sin &#945; + &#946; = 1 3 and cos &#945; - &#946; = 2 3 , then the greatest integer les

Accepted Answer

sin &#945; cos &#946; + cos &#945; sin &#946; + cos &#946; sin &#945; + sin &#946; cos &#945; 2 = cos &#945; - &#946; sin &#946; cos &#946; + cos &#945; - &#946; sin &#945; cos &#945; 2 = 2 cos &#945; - &#946; 1 sin 2 &#946; + 1 sin 2 &#945; 2 = 16 9 2 sin &#945; + &#946; &#183; cos &#945; - &#946; sin 2 &#945; &#183; sin 2 &#946; 2 = 16 9 4 &#183; 1 3 &#183; 2 3 cos 2 &#945; - 2 &#946; - cos 2 &#945; + 2 &#946; 2 = 16 9 8 9 2 cos 2 &#945; - &#946; - 1 - 1 - 2 sin 2 &#945; + &#946; 2 = 16 9 8 9

Question 4

Let  &#945; = &#8721; k = 1 &#8734; sin 2 k &#960; 6 . Let g : 0 , 1 &#8594; &#8477; be the function defined by g x = 2 a x + 2 a 1 - x . Then, which of the following statements is/are TRUE?

Accepted Answer

Given, &#945; = &#8721; k = 1 &#8734; sin 2 k &#960; 6  and  g x = 2 a x + 2 a 1 - x .  Now solving, &#945; = &#8721; k = 1 &#8734; 1 2 2 k = &#8721; k = 1 &#8734; 1 4 k = 1 4 1 - 1 4 = 1 3 Now putting the value of  &#945;  in  g x  we get, g x = 2 x 3 + 2 1 - x 3 Now, g ' x = ln 2 3 2 2 x 3 - 2 1 3 2 x 3 Now finding the critical point by  g ' x = 0 &#8658; x = 1 2 And, derivative changes sign from negative to positive at x = 1 2 , hence x = 1 2 is

Trigonometric Ratios & Identities — Previous Year Questions & Solutions