$x=t^2+t+1, y=t^2-t+1$; then $\frac{d y}

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

$x=t^2+t+1, y=t^2-t+1$; then $\frac{d y}{d x}$ is

Options:

$\frac{2 t+1}{2 t-1}$

$\frac{2 t-1}{2 t+1}$

$\frac{2 t+1}{(2 t-1)^2}$

none of these

Correct Answer:

$\frac{2 t-1}{2 t+1}$

Explanation:

$x=t^2+t+1, \frac{d x}{d t}=2 t+1, ~~\frac{d y}{d t}=2 t-1$

$\frac{d y}{d x}=\frac{2 t-1}{2 t+1}$

Hence (2) is correct answer.