Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=\sin (a x+b)$, then what is $\frac{d^{2} y}{d x^{2}}$ at $x=-\frac{b}{a}$, where $a, b$ are constants and $\mathrm{a} \neq 0$?
Options:
Solution:
2951 Upvotes
Verified Answer
The correct answer is:
0
Let $y=\sin (a x+b)$
$\Rightarrow \frac{d y}{d x}=a \cos (a x+b)$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=-a^{2} \sin (a x+b)$
Now, $\frac{d^{2} y}{d x^{2}}$ at $x=-\frac{b}{a}$ is
$-a^{2} \sin \left(a\left(-\frac{b}{a}\right)+b\right)=-a^{2} \sin 0=0$
$\Rightarrow \frac{d y}{d x}=a \cos (a x+b)$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=-a^{2} \sin (a x+b)$
Now, $\frac{d^{2} y}{d x^{2}}$ at $x=-\frac{b}{a}$ is
$-a^{2} \sin \left(a\left(-\frac{b}{a}\right)+b\right)=-a^{2} \sin 0=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.