Question

Find a real number ‘a’ such that the curve $f(x) = {e^x}$ is a tangent to the curve $g(x) = a{x^2}$
a)$\dfrac{{{e^2}}}{4}$
b) $\dfrac{{{e^2}}}{2}$
c) $\dfrac{e}{4}$
d) $\dfrac{e}{2}$

Answer 1

Hint: We are asked to find the value of ‘a’ such that the curve $f(x) = {e^x}$ is a tangent to $g(x) = a{x^2}$ . To find this we have to differentiate ‘g’ . Then we will take a general point say ${x_0}$ and equate both the functions there. This will result in finding ‘a’ .
Formula used:
1) The tangent to any curve at a point ‘c’ for ‘f’ is given by the derivative of ‘f’ at that point ‘c’
2) If $f(x) = a{x^n}$ then $f'(x) = n \times a{x^{n - 1}}$

Complete step-by-step answer:
The given curves are $f(x) = {e^x}$ and $g(x) = a{x^2}$
We are asked to find a real number ‘a’ such that ‘f’ is a tangent of the curve ‘g’.
Differentiating ‘g’ with respect to ‘x’ we get-
$g'(x) = 2a{x^{2 - 1}} = 2ax$
For a point ${x_0}$ we have the tangent as $g'({x_0}) = 2a{x_0}$
Now this represents the tangent of the curve $g(x) = a{x^2}$ at the point ${x_0}$ .
If ‘f’ is a tangent of ‘g’ at the point ${x_0}$ then we have,
$f({x_0}) = g'({x_0})$
Now putting these values together we get,
${e^{{x_0}}} = 2a{x_0}$
$ \Rightarrow \dfrac{{{e^{{x_0}}}}}{{2{x_0}}} = a$
So for ${x_0} = 1$ we have,
$a = \dfrac{e}{2}$
And for ${x_0} = 2$ we have,
$a = \dfrac{{{e^2}}}{4}$
Therefore, option ‘a’ and ‘d’ are correct.
So, the correct answer is “A and D”.

Note: The differentiation of any function with respect to the independent variable at a point always gives the equation of the tangent to that curve at that point. So always take a general point as it is necessary. Don’t just evaluate the function without taking a general point as it is not mathematically correct for every curve.