Animated Solution for Mathematics - Differentiation: If $x=1$ is a critical point of the function $f(x)=(3x^2+ax-2–a)e^x$, then :

Welcome, fellow traveler on the path of JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a function to reveal its hidden geometry.

We are looking at the function $f(x)=(3x^2+ax-2-a)e^x$. At first glance, it looks like a standard calculus problem, but beneath the surface lies a beautiful interplay between algebra and the behavior of curves.

We are told that $x=1$ is a critical point. In the physical world, a critical point is that fleeting moment where you stop climbing or descending—the instant your vertical velocity is zero.

Mathematically, this means the first derivative, $f^{\prime }(x)$, must be zero at $x=1$. This is our North Star. We must find $f^{\prime }(x)$ and set it to zero at $x=1$ to uncover the mystery of the constant $a$.

To find the derivative, we observe that our function is a product of two distinct entities: a polynomial $u(x)=3x^2+ax-2-a$ and an exponential $v(x)=e^x$. When these two dance together, we must use the product rule:

Let us differentiate carefully. The derivative of our polynomial $u(x)$ is $6x+a$, and the derivative of $e^x$ is, elegantly, just $e^x$. Putting it all together, we get:

Now, watch the magic happen. We can factor out the $e^x$ term:

Notice how the $+a$ and $-a$ terms cancel out perfectly? It is as if the universe wants us to succeed. We are left with:

We know that at $x=1$, $f^{\prime }(1)=0$. Substituting $x=1$ into our derivative, we get:

Since $e^1$ is never zero, we can safely divide it away, leaving us with a simple linear equation:

We have found our constant! The derivative is now fully revealed as $f^{\prime }(x)=e^x(3x^2-x-2)$.

With $a=-7$, we can now find all the critical points by factorizing the quadratic $3x^2-x-2$. Splitting the middle term, we get $(3x+2)(x-1)$.

Thus, our critical points are $x=1$ and $x=-\frac{2}{3}$.

Now, we perform the sign analysis: For $x>1$, $f^{\prime }(x)>0$ (the function is rising). For $-\frac{2}{3}<x<1$, $f^{\prime }(x)<0$ (the function is falling). * For $x<-\frac{2}{3}$, $f^{\prime }(x)>0$ (the function is rising again).

At $x=-\frac{2}{3}$, the derivative changes from positive to negative—a peak, or a local maxima. At $x=1$, the derivative changes from negative to positive—a valley, or a local minima.

We have conquered the function. Keep this clarity, keep this focus, and you will master any problem the JEE throws your way.