Tag: proving properties of curves

Questions Related to proving properties of curves

If line $PQ$, whose equation is $y = 2x + k,$  is a normal to the parabola whose vertex is   $(-2,3)$ and the axis parallel to the $x$-axis with latus rectum equal to $2$, then the possible value of k is

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $\dfrac{{58}}{8}$

  2. $\dfrac{{50}}{8}$

  3. $1$

  4. $-1$

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Correct Option: C
Explanation:
Normal to parabola $\rightarrow y=2x+k$
parabola $\rightarrow (y-3)^{2}=4a(x+2)$
Latus rectum$=2$
$4a=2$
$a=\dfrac{1}{2}$
parabola $\rightarrow (y-3)^{2}=2(x+2)$
$y=3+\sqrt{2}\sqrt{x+2}$
slope of normal =$\dfrac{1}{-y'}$
$y'=\dfrac{\sqrt{2}}{2\sqrt{x+2}}=\dfrac{1}{\sqrt{2x+4}}$
$m=-\sqrt{-2x+4}$
if $2=-\sqrt{2x+4}$
then $x=0, y=5, 1$
$(0,5$) should also lies on $y=2x+k$
then
$k=5,1$
$C$ is correct

If $y = \dfrac { 1 } { 1 + x ^ { n - m } + x ^ { p - m } } + \dfrac { 1 } { 1 + x ^ { m - n } + x ^ { p - n } } + \dfrac { 1 } { 1 + x ^ { m - p } + x ^ { n - p } }$ then $\dfrac { d y } { d x }$ at $x = e ^ { m ^ { n p } }$ is equal to

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $e ^ { m n p }$

  2. $e ^ { m n / p }$

  3. $e ^ { n p / m }$

  4. $0$

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Correct Option: D
Explanation:
$\\y=(\dfrac{1}{1+(\dfrac{x^n}{x^m})+(\dfrac{x^p}{x^m})})+(\dfrac{1}{1+(\dfrac{x^m}{x^n})+(\dfrac{x^p}{x^n})})+(\dfrac{1}{1+(\dfrac{x^m}{x^p})+(\dfrac{x^n}{x^p})})$

$\\=(\dfrac{x^m}{x^m+x^n+x^p})+(\dfrac{x^n}{x^n+x^m+x^p})+(\dfrac{x^p}{x^p+x^m+x^n})$

$\\=(\dfrac{x^m+x^n+x^p}{x^m+x^n+x^p})$

$\\=1$
$\\\therefore\>(\dfrac{dy}{dx})=0$

Let f(x) be a differentiable function and $f\left( \alpha  \right) = f\left( \beta  \right) = 0\,\left( {\alpha  < \beta } \right)$, then in the interval $\left( {\alpha ,\beta } \right)$

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $f\left( x \right) + f'\left( x \right) = 0$ has at least one root.

  2. $f\left( x \right) - f'\left( x \right) = 0$ has at least one root.

  3. $f\left( x \right) . f'\left( x \right) = 0$ has at least one root.

  4. None of these

Show answer
Correct Option: C
Explanation:

$\therefore f(\alpha) = f(\beta) = 0$


Thus, By Rolle's theorem
($\because f(x)$ is a differential function)
$f'(c) = 0$     where $\in (\alpha, \beta)$

$\therefore $ In the interval $(\alpha, \beta)$ 
$f(x) . f'(x) = 0$ has at least one root.

If $y = \log \left( \frac { 1 + x } { 1 - x } \right) ^ { 1 / 4 } - \frac { 1 } { 2 } \tan ^ { - 1 } x ,$ then $\frac { d y } { d x } =$

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $\frac { x ^ { 2 } } { 1 - x ^ { 4 } }$

  2. $\frac {2 x ^ { 2 } } { 1 - x ^ { 4 } }$

  3. $\frac { x ^ { 2 } } { 2 \left( 1 - x ^ { 4 } \right) }$

  4. None of these

Show answer
Correct Option: A
Explanation:

$\dfrac{dy}{dx}=\dfrac{1}{(\dfrac{1+x}{1-x})^{\frac{1}{4}}}\times$ $\dfrac{1}{4\times{(\dfrac{1+x}{1-x})^{\frac{3}{4}}}}\times $ $\dfrac{(1-x)(1)-(1+x)(-1)}{(1-x)^{2}}-$ $\dfrac{1}{2}\dfrac{1}{x^2+1}$ 


$\dfrac{1-x}{4(1+x)}\times\dfrac{2}{(1-x)^2}$ $-\dfrac{1}{2(x^2+1)}$  $=\dfrac{1}{2(1-x^2)}-$ $\dfrac{1}{2(x^2+1)}$ =$\dfrac{x^2}{1-x^4}$

If f'$\left( x \right) =\sqrt { { 2x }^{ 2 }-1 } $ and y=f$\left( { x }^{ 2 } \right) $ then $\dfrac { dy }{ dx } $ at x=1 is

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. 2

  2. 1

  3. -2

  4. none of these

Show answer
Correct Option: D
Explanation:

We have,

$f\left( x \right)=\sqrt{2{{x}^{2}}-1}$

And

$ y=f\left( {{x}^{2}} \right) $

$ y={{\left( \sqrt{2{{x}^{2}}-1} \right)}^{2}} $

$ y=2{{x}^{2}}-1 $


On differentiating and we get,

$ \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 2{{x}^{2}}-1 \right) $

$ \dfrac{dy}{dx}=4x-0 $

$ \dfrac{dy}{dx}=4x $

At point $\left( x=1 \right)$

So,

$ \dfrac{dy}{dx}=4x=4\left( 1 \right) $

$ \dfrac{dy}{dx}=4 $

Hence, this is the answer.

A differentiable function function $y = h(x)$ satisfies $\displaystyle \overset{x}{\underset{0}{\int}} (x - t + 1)h(t)dt = x^4 + x^2; \forall x \ge 0$, then value of $h(0) + h'(0)$ is equal to

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $0$

  2. $1$

  3. $e^2$

  4. $2$

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Correct Option: A

Area of the triangle formed by the lines $x-y=0, x+y=0$ and ant tangent to the hyparabola $x^{2}-y^{2}=a^{2}$ is 

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $|a|$

  2. $\dfrac{1}{2}|a|$

  3. $a^{2}$

  4. $\dfrac{1}{2}a^{2}$

Show answer
Correct Option: C
Explanation:

Equation of line

$ x-y=0\,\,......\,\,\left( 1 \right) $

$ x+y=0\,\,.......\,\,\left( 2 \right) $


Equation of hyperbola is

${{x}^{2}}-{{y}^{2}}={{a}^{2}}\,\,......\,\,\left( 3 \right)$


Let the point $P(a\sec \theta, a\tan \theta)$ on the hyperbola.


Equation of tangent is,

$ x{{x} _{1}}-y{{y} _{1}}={{a}^{2}} $

$\Rightarrow a(x\sec \theta-y\tan\theta)=a^2$

$ \Rightarrow x\sec \theta -y\tan \theta =a\,\,......\,\,\left( 4 \right) $


Now,

Area of $\Delta AOB$ $=\dfrac{1}{2}$

$=\dfrac{1}{2}\left|a^2(\tan^2\theta-\sec^2\theta)-a^2(\sec^2\theta-\tan^2\theta)\right|$

$ =\dfrac{1}{2}\left| {{a}^{2}}\left( -1 \right)-{{a}^{2}}\left( 1 \right) \right| $

$ =\dfrac{1}{2}\left| -2{{a}^{2}} \right| $

$ =\left| -{{a}^{2}} \right| $

$ =\left| {{a}^{2}} \right| $


Hence, this is the answer.

If $x = \exp \left{ \tan ^ { - 1 } \left( \frac { y - x ^ { 2 } } { x ^ { 2 } } \right) \right}$ then $\frac { d y } { d x } =$

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $2 x [ 1 + \tan ( \log x ) ] + x \cdot \sec ^ { 2 } ( \log x )$

  2. $x [ 1 + \tan ( \log x ) ] + \sec ^ { 2 } ( \log x )$

  3. $2 x [ 1 + \tan ( \log x ) ] + x ^ { 2 } \sec ^ { 2 } ( \log x )$

  4. None of these

Show answer
Correct Option: D
Explanation:
$x=exp\left\{\tan^{-1}\left(\dfrac{y-x^2}{x^2}\right)\right\}$
$ln x=\tan^{-1}\left(\dfrac{y-x^2}{x^2}\right)$
$\Rightarrow \tan (ln x)=\dfrac{y}{x^2}-1$
$\Rightarrow y=x^2[\tan (ln x)+1]$
$=x^2\tan(ln x)+x^2$
$\dfrac{dy}{dx}=2x\tan (ln x)+x^2\sec^2(ln x)\dfrac{1}{x}+2x$
$=2x\tan (ln x)+x\sec^2(ln x)+2x$.

Consider the function $f(x)=\mathrm{s}\mathrm{g}\mathrm{n} x$ and $g(x)=x\left ( 1-x^{2} \right )$. Which of the following does NOT hold good?

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $(fog)(x)$ is neither odd nor even

  2. $(gof)(x)$ is odd

  3. $(fog)(x)$ is neither continuous nor differentiable for some $x$ on $\left ( -\infty, \infty \right )$

  4. $(fog)(x)$ is continuous and differentiable for every $x$ on $\left ( -\infty, \infty \right )$

Show answer
Correct Option: A,C
Explanation:

 $\text{sgn}(x) =\begin{cases}-1 & x <0\0 & x = 0\ 1 & x>0\end{cases} $
$g(x)=x(1-x^{2})$
$=x(1-x)(1+x)$
$g(x)=0$ implies, $x={-1,0,1}$.
$\text{sgn}(g(x))=\begin{cases}-1 & x\in(-\infty,-1)\cup(1,\infty) \0 & x =  \big{-1,0,1\big} \ 1 & x\in(-1,1)\end{cases} $

The set of all points of differentiability of the function $\displaystyle f(x) = \dfrac{\sqrt{x + 1} 1}{\sqrt{x}}$ for $x$  and $f(0)$ = 0 is

physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

  1. $(\infty , \infty)$

  2. $[0 , \infty)$

  3. $(0 , \infty)$

  4. $(\infty$ , $\infty)$ $-\left { 0 \right }$

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Correct Option: C