Question

The tangent to the curve $y=x{{e}^{{{x}^{2}}}}$ passing through the point (1, e) also passes through the point:

& A.\left( \dfrac{4}{3},2e \right) \\\ & B.\left( 2,3e \right) \\\ & C.\left( \dfrac{5}{3},2e \right) \\\ & D.\left( 3,6e \right) \\\ \end{aligned}$$

Answer 1

At first, differentiate the function y with respect to x using formula,
$\dfrac{d}{dx}\left( f\left( x \right)\cdot g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\cdot g\left( x \right)+f\left( x \right)\cdot \dfrac{d}{dx}\left( g\left( x \right) \right)\text{ and }\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)$
And find slope at (1, e). Then, find line using formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is point and m is slope. Then check for each point whether it satisfies or not.

Complete step-by-step solution:
In the question, we are given a function of the curve $y=x{{e}^{{{x}^{2}}}}$ and we have to find a point which its tangent passes, given that it also passes through (1, e).
So, for finding the point at first we have to find tangent, so for it, we have to find the slope of the function at the point (1, e).
So, at first, we have to differentiate y with respect to x.
As we know, $y=x{{e}^{{{x}^{2}}}}$
So on differentiation we can write,
$\dfrac{dy}{dx}=1\cdot {{e}^{{{x}^{2}}}}+x\cdot \dfrac{d}{dx}\left( {{e}^{{{x}^{2}}}} \right)$
By using formula,
$\dfrac{d}{dx}\left( f\left( x \right)\cdot g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\cdot g\left( x \right)+f\left( x \right)\cdot \dfrac{d}{dx}\left( g\left( x \right) \right)$
Now, we can write,
$\dfrac{dy}{dx}={{e}^{{{x}^{2}}}}+x\times 2x\times {{e}^{{{x}^{2}}}}$
Using formula $\dfrac{d}{dx}\left( g\left( f\left( x \right) \right) \right)$ equals to $g'\left( f\left( x \right) \right)\cdot f'\left( x \right)$
Hence,

& \dfrac{dy}{dx}=2{{x}^{2}}\cdot {{e}^{{{x}^{2}}}}+{{e}^{{{x}^{2}}}} \\\ & \Rightarrow \dfrac{dy}{dx}={{e}^{{{x}^{2}}}}\left( 2{{x}^{2}}+1 \right) \\\ \end{aligned}$$ So at point (1, e) the value of $\dfrac{dy}{dx}$ will be, $$\begin{aligned} & {{\left. \dfrac{dy}{dx} \right|}_{\left( 1,e \right)}}=e\left( 2+1 \right) \\\ & \Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{\left( 1,e \right)}}=3e \\\ \end{aligned}$$ So the tangent will be, $$y-e=3e\left( x-1 \right)$$ Using formula that the line is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ if slope is m and passes through point $\left( {{x}_{1}},{{y}_{1}} \right)$ Hence, the line of the tangent is, $$\begin{aligned} & y-e=3ex-3e \\\ & \Rightarrow y=3ex-2e \\\ & \Rightarrow y=e\left( 3x-2 \right) \\\ \end{aligned}$$ Now, we will check on every point whether it passes or not by substituting it in the line. If it satisfies then it is the answer. So for $\left( \dfrac{4}{3},2e \right)$ on substituting value of x as $\dfrac{4}{3}$ on right hand side, we get: $$e\left( 3\times \dfrac{4}{3}-2 \right)=e\left( 4-2 \right)=2e$$ Which is the value of y. Hence, it satisfies. So for (2, 3e) on substituting value of x as 2 on right hand side, we get: $$e\left( 3\times 2-2 \right)=e\left( 6-2 \right)=4e$$ Which is not equal to y. Hence, not satisfied. So for $\left( \dfrac{5}{3},2e \right)$ on substituting value of x as $\dfrac{5}{3}$ on right hand side, we get: $$e\left( 3\times \dfrac{5}{3}-2 \right)=e\left( 5-2 \right)=3e$$ Which is not equal to y. Hence, not satisfied. So for (3, 6e) on substituting value of x as 3 on right hand side, we get: $$e\left( 3\times 3-2 \right)=e\left( 9-2 \right)=7e$$ Which is not equal to the y. Hence, not satisfied. **Therefore, the correct option is A.** **Note:** We can do the same question by another method. At first, we will find the slope of curve $y=x{{e}^{{{x}^{2}}}}$ at point (1, e). Then, we find slope between two points (1, e) and each of the point using formula $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ if points are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$. Then, we will check whether they are same or not, whichever matches it is correct.