
If the derivative of \[\left( {ax - 5} \right){e^{3x}}\] at \[x = 0\] is \[ - 13\], then the value of \[a\]is equal to
\[
(A){\text{ 8}} \\
(B){\text{ - 5}} \\
(C){\text{ 5}} \\
(D){\text{ - 2}} \\
(E){\text{ 2}} \\
\]
Answer
599.1k+ views
Hint:- Use the product rule to find derivatives.
Let, y\[ = \left( {ax - 5} \right){e^{3x}}\] (1)
As given in the question that the value of derivative of y with respect to x at \[x = 0\]is \[ - 13\].
As, y is a function of x. So, we can get the derivative of y easily by using product rule.
Which states that if u and v are two functions then,
\[ \Rightarrow \left( {\dfrac{{d(uv)}}{{dx}}} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
So, here u\[ = {e^{3x}}\] and v\[ = \left( {ax - 5} \right)\]
So, differentiating equation 1 with respect to x. We get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}\dfrac{{d(ax - 5)}}{{dx}} + (ax - 5)\dfrac{{d({e^{3x}})}}{{dx}}\] (By using product rule)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}(a) + (ax - 5)3{e^{3x}}\]
Now, putting \[x = 0\] in the above equation. We get,
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = {e^0}(a) + (a(0) - 5)3{e^0}\]
As, given in the question, the derivative of the given function i.e. y at \[x = 0\] is \[ - 13\]. So,
\[ \Rightarrow - 13 = a - 15\]
\[ \Rightarrow a = 2\]
Hence, the correct option is E.
Note:- Whenever we come up with this type of problem where we are given with a function and the value of the derivative of that function at a given point, we first calculate the derivative of that function at a known point, then equate it with the given value of the derivative of the function at that point to get the required value of the variable.
Let, y\[ = \left( {ax - 5} \right){e^{3x}}\] (1)
As given in the question that the value of derivative of y with respect to x at \[x = 0\]is \[ - 13\].
As, y is a function of x. So, we can get the derivative of y easily by using product rule.
Which states that if u and v are two functions then,
\[ \Rightarrow \left( {\dfrac{{d(uv)}}{{dx}}} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
So, here u\[ = {e^{3x}}\] and v\[ = \left( {ax - 5} \right)\]
So, differentiating equation 1 with respect to x. We get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}\dfrac{{d(ax - 5)}}{{dx}} + (ax - 5)\dfrac{{d({e^{3x}})}}{{dx}}\] (By using product rule)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{3x}}(a) + (ax - 5)3{e^{3x}}\]
Now, putting \[x = 0\] in the above equation. We get,
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = {e^0}(a) + (a(0) - 5)3{e^0}\]
As, given in the question, the derivative of the given function i.e. y at \[x = 0\] is \[ - 13\]. So,
\[ \Rightarrow - 13 = a - 15\]
\[ \Rightarrow a = 2\]
Hence, the correct option is E.
Note:- Whenever we come up with this type of problem where we are given with a function and the value of the derivative of that function at a given point, we first calculate the derivative of that function at a known point, then equate it with the given value of the derivative of the function at that point to get the required value of the variable.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

If overrightarrow a overrightarrow b overrightarrow class 12 maths CBSE

If a b and c are unit coplanar vectors then left 2a class 12 maths CBSE

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

When was the first election held in India a 194748 class 12 sst CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

