MHT CET · Maths · Application of Derivatives
The equation of normal to the curve \(\mathrm{y}=\sin \left(\frac{\pi x}{4}\right)\) at the point \((2,5)\) is
- A x+y=5
- B y=5
- C x=2
- D x+y=2
Answer & Solution
Correct Answer
(C) x=2
Step-by-step Solution
Detailed explanation
(D)
we have, \(y=\sin \frac{\pi x}{4}\)
\(\begin{array}{l}
\frac{d y}{d x}=\frac{\pi}{4} \cdot \cos \frac{\pi x}{4} \\
\left(\frac{d y}{d x}\right)_{(2,5)}=\frac{\pi}{4} \cos \frac{2 \pi}{4}=0
\end{array}\)
Since slope of tangent is zero, it is parallel to \(\mathrm{X}\) axis. So normal is parallel to \(\mathrm{Y}\) axis. Hence required equation of normal is \(\mathrm{x}=2\)
we have, \(y=\sin \frac{\pi x}{4}\)
\(\begin{array}{l}
\frac{d y}{d x}=\frac{\pi}{4} \cdot \cos \frac{\pi x}{4} \\
\left(\frac{d y}{d x}\right)_{(2,5)}=\frac{\pi}{4} \cos \frac{2 \pi}{4}=0
\end{array}\)
Since slope of tangent is zero, it is parallel to \(\mathrm{X}\) axis. So normal is parallel to \(\mathrm{Y}\) axis. Hence required equation of normal is \(\mathrm{x}=2\)
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