MHT CET · Maths · Application of Derivatives
The equation of tangent to the curve \(y=y=\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)\) at \(x=\frac{\pi}{4}\), is
- A \(2 x+y-\frac{\pi}{2}-1=0\)
- B \(2 x-y-\frac{\pi}{2}+1=0\)
- C \(x+y-\frac{\pi}{2}-1=0\)
- D \(x-y-\frac{\pi}{2}+1=0\)
Answer & Solution
Correct Answer
(A) \(2 x+y-\frac{\pi}{2}-1=0\)
Step-by-step Solution
Detailed explanation
\(\mathrm{y}=\sqrt{2} \sin \left(2 \mathrm{x}+\frac{\pi}{4}\right) \)
\( \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{2} \cos \left(2 \mathrm{x}+\frac{\pi}{4}\right)(2) \)
\( \therefore\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\frac{\pi}{4}}=(2 \sqrt{2}) \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=2 \sqrt{2}\) \(\left(-\sin \frac{\pi}{4}\right)\) \(=-2\)
When \(\mathrm{x}=\frac{\pi}{4}, \mathrm{y}=\sqrt{2} \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=1\)
Hence equation of required tangent is
\(
(y-1)=-2\left(x-\frac{\pi}{4}\right) \Rightarrow 2 x+y-\frac{\pi}{2}-1=0
\)
\( \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{2} \cos \left(2 \mathrm{x}+\frac{\pi}{4}\right)(2) \)
\( \therefore\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\frac{\pi}{4}}=(2 \sqrt{2}) \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=2 \sqrt{2}\) \(\left(-\sin \frac{\pi}{4}\right)\) \(=-2\)
When \(\mathrm{x}=\frac{\pi}{4}, \mathrm{y}=\sqrt{2} \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=1\)
Hence equation of required tangent is
\(
(y-1)=-2\left(x-\frac{\pi}{4}\right) \Rightarrow 2 x+y-\frac{\pi}{2}-1=0
\)
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