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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{x}=\text{a}\cos\ \text{nt}-\text{b}\sin\ \text{nt}$ then $\frac{\text{d}^2\text{x}}{\text{dt}^2}$ is:
  • A
    $n^2x$
  • ✓
    $-n^2x$
  • C
    $-nx$
  • D
    $nx$
✓

Answer

Correct option: B.
$-n^2x$
Here
$\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
Differentiating $\text{w.r.t.t},$ we get
$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$
Differentiating $\text{w.r.t.t},$ we get
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{an}^2\cos\text{nt}+\text{bn}^2\sin\text{nt}$
$=-\text{n}^2\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
$=-\text{n}^2\text{x}$

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