Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}\ \text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}(\text{t})\cos\text{t},$ then $\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=$
  • A
    $\text{f}(\text{t})-\text{f}(\text{t})$
  • B
    $\{\text{f}(\text{t})-\text{f}(\text{t})\}^2$
  • ✓
    $\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$
  • D
    $\text{None of these}$
✓

Answer

Correct option: C.
$\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$
Here,

$\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}'(\text{t})\sin\text{t}$

$\text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}'(\text{t})\cos\text{t}$

$ \Rightarrow\frac{\text{dx}}{\text{dt}}=\text{f}'(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}-\text{f}'(\text{t})\cos\text{}\text{t}$

$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}'(\text{t})\sin\text)\cos\text{t}+\text{f}'{t}+\text{f}(\text{t})$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}$

$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}$

Thus

$\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=\{-\text{f}(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\Big\}^2$

$=\{\text{f}(\text{t})\sin\text{t}+\text{f}''(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\}^2$

$=\sin^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2+\cos^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2(\sin^2\text{t}+\cos^2\text{t})$

$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

A fair die is thrown until $2$ appears. Then the probability, that $2$ appears in even number of throws, is
→
If $\text{A}=\begin{bmatrix}a^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}$and $\text{a}^2+\text{b}^2+\text{c}^3=1$ then $\text{A}^2=$
→
$\int {\frac{{{{\sin }^{ - 1}}x - {{\cos }^{ - 1}}x}}{{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x}}} dx = $
→
Diffrential coefficient of ${\left( {{x^{\frac{{\ell \, + \,m}}{{m\, - \,n}}}}} \right)^{\frac{1}{{n\, - \,\ell }}}}\,\,\,\,.\,\,\,\,{\left( {{x^{\frac{{\,m + \,n}}{{n\, - \,\ell }}}}} \right)^{\frac{1}{{\,\ell \, - \,m}}}}\,\,\,.\,\,\,{\left( {{x^{\,\frac{{n\, + \,\ell \,}}{{\ell \,\, - \,\,m}}}}} \right)^{\frac{1}{{m\, - \,n\,}}}}\,$ w.r.t. $x$ is
→
If $y = \sec {x^0}$, then ${{dy} \over {dx}} = $
→
$\int_{}^{} {\frac{{\cot x\tan x}}{{{{\sec }^2}x - 1}}} \;dx = $
→
The solution of the equation $(x + 2{y^3})\frac{{dy}}{{dx}} - y = 0$ is
→
Which one of the following statements is not true:
→
If the straight lines $x = 1 + s,$ $y = - 3 - \lambda s,$ $z = 1 + \lambda s$ and $x = t/2,y = 1 + t,z = 2 - t$, with parameters $s$ and $t$ respectively, are co-planar, then $\lambda $ equals
→
If $f(x)$ = $\left\{ \begin{gathered}
  \frac{{a + 3\cos x}}{{{x^2}}},\,\,\,\,\,\,\,\,\,x < 0 \hfill \\
  b\,\tan \left( {\frac{\pi }{{\left[ {x + 3} \right]}}} \right),\,x \geqslant 0 \hfill \\ 
\end{gathered}  \right.$ is continuous at $x = 0$ , then
→