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BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Mathematics
Chapter: Indefinite Integration

BITSAT 2009 · Mathematics · Indefinite Integration

The equation of tangent to the curve (y= x) at the point ((π, 0)) is

The equation of tangent to the curve (y=\sin x) at the point ((\pi, 0)) is

Options

A $x+y=0$
B(x+y=\pi)
C(x-y=\pi)
D $x-y=0$

Correct answer

B. (x+y=\pi)

Step-by-step solution

(y=\sin x \Rightarrow

\frac{d y}{d x}=\cos

x \Rightarrow\left(\frac{d y}{d x}\right)_{(\pi, 0)}=-

1) Therefore the equation of tangent at ((

\pi,

0)) is given by (y-0=-1(x-

\pi) \Rightarrow

x+y=

\pi

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