Exams › JEE Main › Maths

If x = √(2^(cosec⁻¹ t)) and y = √(2^(sec⁻¹ t)) (|t| ≥ 1), then dy/dx is equal to -

1. y/x
2. −y/x
3. −x/y
4. x/y

Correct answer: −y/x

Solution

x^2 = 2^(cosec^-1 t) and y^2 = 2^(sec^-1 t), and cosec^-1 t + sec^-1 t = pi/2, so x^2 y^2 = 2^(pi/2) = constant. Differentiating ln(x^2) + ln(y^2) = const gives 2/x + (2/y)(dy/dx) = 0, so dy/dx = -y/x.

Related JEE Main Maths questions

• For the function f(x) = log(x + √(x² + 1)), which of the following describes it correctly?
• For the function f(x) = log((1 + x)/(1 − x)), which of the following relations is satisfied?
• Find the domain of the function f(x)=sin⁻¹{log2(1/2 x²)}.
• For which values of x does the inequality sin⁻¹(sin 5) > x² − 4x hold?
• Determine the range of the function f(x) = sin⁻¹x + tan⁻¹x + sec⁻¹x.
• If A = tan⁻¹ ((x√(3))/(2K-x)) and B = tan⁻¹ ((2x-K)/(K√(3))), then what is the value of A - B?

⚔️ Practice JEE Main Maths free + battle 1v1 →