Match the following List:Column IA) The maximum value of sec − 1 ( 2 ( x 2 + 2 ) 7 − 5 ( x 2 + 3 ) ) is(B) The minimum value of cosec − 1 [ 3 x 2 + 4 5 ] + sec − 1 [ 3 x 2 + 4 1 ] is, where [.]

Question

Match the following List:Column IA) The maximum value of sec − 1 ( 2 ( x 2 + 2 ) 7 − 5 ( x 2 + 3 ) ​ ) is(B) The minimum value of cosec − 1 [ 3 x 2 + 4 5 ​ ] + sec − 1 [ 3 x 2 + 4 1 ​ ] is, where [.] denotes the greatest integer function(C) Points of non-differentiability of the function f ( x ) = min { tan ( x + 12 π ​ ) , cot ( x + 12 π ​ ) } for all x in (0, 3π/2) is/are(D) Tangent is drawn to hyperbola 8 x 2 ​ − y 2 = 1 at ( 2 2 ​ sec θ , tan θ ) , θ ∈ ( 0 , 2 π ​ ) . The value of θ such that the sum of intercepts on axes made by this tangent is maximum isColumn IIp) 6 π ​ (q) 4 π ​ (r) 3 2 π ​ (s) 6 7 π ​

TestHub · Accepted Answer

(A) (7-5(x^2+3))/(2(x^2+2)) = 1/(x^2+2) - 5/2. Since 1/(x^2+2) <= 1/2, the argument is at most -2, and sec^(-1) is increasing for arguments <= -1, so the maximum value is sec^(-1)(-2) = 2pi/3; (A) matches (r). (B) The sum is minimized when [3x^2 + 1/4] = 1, i.e. 3x^2 + 1/4 lies in [1, 2); then [3x^2 + 5/4] = 2, giving cosec^(-1)(2) + sec^(-1)(1) = pi/6 + 0 = pi/6; (B) matches (p). (C) The two branches cross where tan(x + pi/12) = cot(x + pi/12), i.e. tan^2(x + pi/12) = 1, so x + pi/12 = pi/4, 3pi/4 or 5pi/4 and x = pi/6, 2pi/3, 7pi/6 in (0, 3pi/2). At each crossing the minimum function has a corner and is non-differentiable, so (C) matches (p), (r), (s). (D) The tangent at (2 sqrt(2) sec t, tan t) is (x sec t)/(2 sqrt(2)) - y tan t = 1, with intercepts 2 sqrt(2) cos t and -cot t. The sum S(t) = 2 sqrt(2) cos t - cot t has S'(t) = -2 sqrt(2) sin t + cosec^2 t = 0, so sin^3 t = 1/(2 sqrt(2)) and sin t = 1/sqrt(2), i.e. t = pi/4; (D) matches (q). Answer: A-r, B-p, C-(p,r,s), D-q (option C).

Mathematics - Application of Derivative Question with Solution | TestHub

Options:

Doubts & Discussion