If (a^2-1) x+a y+(3-a)=0 is a normal to the c | vedclass

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(B) Given the curve $x y=1$,we have $y = \frac{1}{x}$.Calculating the derivative: $\frac{d y}{d x} = -\frac{1}{x^2}$.The slope of the tangent at any p

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$(-\infty,-1] \cup(0,1]$

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(B) Given the curve $x y=1$,we have $y = \frac{1}{x}$.Calculating the derivative: $\frac{d y}{d x} = -\frac{1}{x^2}$.The slope of the tangent at any point $(x, y)$ is $m_t = -\frac{1}{x^2}$.The slope of the normal $m_n$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{m_t} = x^2$.Since $x^2 \geq 0$ for all real $x$,the slope of the normal must be non-negative $(m_n \geq 0)$.The given equation of the normal is $(a^2-1) x+a y+(3-a)=0$,which can be rewritten as $a y = -(a^2-1) x - (3-a)$,or $y = -\frac{a^2-1}{a} x - \frac{3-a}{a}$.The slope of this line is $m = -\frac{a^2-1}{a} = \frac{1-a^2}{a}$.Since $m \geq 0$,we have $\frac{1-a^2}{a} \geq 0$.Multiplying by $-1$ reverses the inequality: $\frac{a^2-1}{a} \leq 0$,which is $\frac{(a-1)(a+1)}{a} \leq 0$.Using the sign scheme (Wavy Curve Method),the values of $a$ satisfying this are $a \in (-\infty, -1] \cup (0, 1]$.

If $(a^2-1) x+a y+(3-a)=0$ is a normal to the curve $x y=1$,then the interval in which '$a$' lies is

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