If the normal to the curve y=f(x) at (1,2) ma | vedclass

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(B) The slope of the tangent to the curve $y=f(x)$ at $(1,2)$ is given by $f^{\prime}(1)$.The slope of the normal to the curve at $(1,2)$ is given by

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(B) The slope of the tangent to the curve $y=f(x)$ at $(1,2)$ is given by $f^{\prime}(1)$.The slope of the normal to the curve at $(1,2)$ is given by $m_n = -\frac{1}{f^{\prime}(1)}$.Given that the normal makes an angle $	heta = \frac{3 \pi}{4}$ with the positive $X$-axis,the slope of the normal is $m_n = 	an\left(\frac{3 \pi}{4}ight)$.We know that $	an\left(\frac{3 \pi}{4}ight) = -1$.Equating the two expressions for the slope of the normal: $-\frac{1}{f^{\prime}(1)} = -1$.Multiplying both sides by $-1$,we get $\frac{1}{f^{\prime}(1)} = 1$.Therefore,$f^{\prime}(1) = 1$.

If the normal to the curve $y=f(x)$ at $(1,2)$ makes an angle $\frac{3 \pi}{4}$ with the positive $X$-axis,then $f^{\prime}(1)=$

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