Let \[f\] be a real-valued differential function on \[R\](the set of all real numbers) such that \[f\left( 1 \right) = 1\]. If the y-intercept of the tangent at any point \[P\left( {x,y} \right)\] on the curve \[y = f\left( x \right)\] is equal to the cube of the abscissa of \[P\], then the value of \[f\left( { - 3} \right)\] is equal to:
A. 0
B. 1
C. 2
D. 9
Answer
657.3k+ views
Hint: First of all, find the slope of the given tangent and then find the equation of the tangent. Use Bernoulli’s differential equation to find the value of the given function and hence to obtain the required answer.
Complete step-by-step answer:
Given that \[f\] is a real-valued differential function on and \[f\left( 1 \right) = 1\]
Let \[m = \dfrac{{dy}}{{dx}}\] be the slope of the tangent at the point \[P\left( {x,y} \right)\] on the curve \[y = f\left( x \right)\].
So, the equation of the tangent is \[\left( {Y - y} \right) = m\left( {X - x} \right)\].
To get y-intercept, put \[x = 0\] in the equation of the tangent.
\[
\Rightarrow Y - y = m\left( { - x} \right) \\
\Rightarrow Y = y - mx \\
\]
Given that y-intercept of the tangent is equal to the cube of the abscissa of \[P\] i.e., \[Y = {x^3}\]
\[
\Rightarrow y - mx = {x^3} \\
\Rightarrow y - \dfrac{{dy}}{{dx}}x = {x^3} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - {x^3}}}{x} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x} - {x^2} \\
\Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{y}{x} = - {x^2} \\
\]
By Bernoulli’s differential equation, we have \[I.F = {e^{\int { - \dfrac{1}{x}dx} }} = {e^{ - \ln x}} = \dfrac{1}{x}\]
\[
\Rightarrow y \times \dfrac{1}{x} = \int { - {x^2} \times \dfrac{1}{x}dx} \\
\Rightarrow \dfrac{y}{x} = - \int {xdx} \\
\Rightarrow \dfrac{y}{x} = - \dfrac{{{x^2}}}{2} + c \\
\]
So, we have \[f\left( x \right) = - \dfrac{{{x^3}}}{2} + \dfrac{3}{2}x\]
Therefore, \[f\left( { - 3} \right) = - \dfrac{{{{\left( { - 3} \right)}^3}}}{2} + \dfrac{3}{2}\left( { - 3} \right) = \dfrac{{27}}{2} - \dfrac{9}{2} = \dfrac{{27 - 9}}{2} = \dfrac{{18}}{2} = 9\].
Thus, the correct option is D. 9
Note: The abscissa refers to the horizontal (x) axis and the ordinate refers to the vertical (y) axis of a standard two-dimensional graph. An ordinary differential equation is called a Bernoulli’s differential equation if it is of the form where it is a real number.
Complete step-by-step answer:
Given that \[f\] is a real-valued differential function on and \[f\left( 1 \right) = 1\]
Let \[m = \dfrac{{dy}}{{dx}}\] be the slope of the tangent at the point \[P\left( {x,y} \right)\] on the curve \[y = f\left( x \right)\].
So, the equation of the tangent is \[\left( {Y - y} \right) = m\left( {X - x} \right)\].
To get y-intercept, put \[x = 0\] in the equation of the tangent.
\[
\Rightarrow Y - y = m\left( { - x} \right) \\
\Rightarrow Y = y - mx \\
\]
Given that y-intercept of the tangent is equal to the cube of the abscissa of \[P\] i.e., \[Y = {x^3}\]
\[
\Rightarrow y - mx = {x^3} \\
\Rightarrow y - \dfrac{{dy}}{{dx}}x = {x^3} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - {x^3}}}{x} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x} - {x^2} \\
\Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{y}{x} = - {x^2} \\
\]
By Bernoulli’s differential equation, we have \[I.F = {e^{\int { - \dfrac{1}{x}dx} }} = {e^{ - \ln x}} = \dfrac{1}{x}\]
\[
\Rightarrow y \times \dfrac{1}{x} = \int { - {x^2} \times \dfrac{1}{x}dx} \\
\Rightarrow \dfrac{y}{x} = - \int {xdx} \\
\Rightarrow \dfrac{y}{x} = - \dfrac{{{x^2}}}{2} + c \\
\]
So, we have \[f\left( x \right) = - \dfrac{{{x^3}}}{2} + \dfrac{3}{2}x\]
Therefore, \[f\left( { - 3} \right) = - \dfrac{{{{\left( { - 3} \right)}^3}}}{2} + \dfrac{3}{2}\left( { - 3} \right) = \dfrac{{27}}{2} - \dfrac{9}{2} = \dfrac{{27 - 9}}{2} = \dfrac{{18}}{2} = 9\].
Thus, the correct option is D. 9
Note: The abscissa refers to the horizontal (x) axis and the ordinate refers to the vertical (y) axis of a standard two-dimensional graph. An ordinary differential equation is called a Bernoulli’s differential equation if it is of the form where it is a real number.
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