The equation of the tangent to the curve $6 y = 7 - x^{3}$ at point $(1, 1)$ is

2 x + y = 3

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x + 2 y = 3

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x + y = 1

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x + y + 2 = 0

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Solution

The correct option is C $x + 2 y = 3$
Given curve is, $6 y = 7 - x^{3}$
$\Rightarrow 6 \frac{d y}{d x} = - 3 x^{2} \Rightarrow \frac{d y}{d x} = - \frac{1}{2} x^{2}$
Thus slope of tangent to this curve at $(1, 1,)$ is $m = - \frac{1}{2}$
The equation of required tangent is,
$(y - 1) = - \frac{1}{2} (x - 1) \Rightarrow x + 2 y = 3$
Hence, option 'B' is correct.

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Similar questions

The equation of the tangent to the curve $y = 2 x^{2} + 3 s i n x a t x = 0$ is

P (x, y)

(1, 1)

X

Y

A

B

A P : B P = 3 : 1

List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

The normal at the point (1, 1) on the curve 2y + x² = 3 is

(A) x + y = 0 (B) x − y = 0

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