Q95 / 200 · TS EAMCET 2001 · Mathematics

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

A
$2 x+y=3$
Correct answer
B
$x+2 y=3$
C
$x+y=-1$
D
$x+y+2=0$

Solution

Given curve is $6 y=7-x^3$ On differentiating w.r.t. $x$, we get $\begin{aligned} 6 y^{\prime} & =-3 x^2 \\ \Rightarrow \quad y_{(1,1)}^{\prime} & =-\frac{3}{6}=-\frac{1}{2}\end{aligned}$ The equation of the tangent is $\begin{array}{rlrl} & & y-1 & =-\frac{1}{2}(x-1) \\ \Rightarrow & & 2 y-2 & =-x+1 \\ \Rightarrow & x+2 y & =3\end{array}$

Hence, the correct answer is option: 'B'.

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