Analyzing the Setup
Welcome, fellow traveler of the JEE journey! Today, we face a problem that might look intimidating at first: maximizing the rational function:
Many students would immediately reach for the quotient rule of differentiation, but let us pause. In the world of competitive exams, speed and elegance are your best friends.
Look closely at the numerator and the denominator. They are nearly identical! This is a classic setup for a 'surgical strike' in algebra.
We can rewrite the numerator as (3x2+9x+7)+10. By splitting the fraction, we get:
Now, the problem transforms. We are no longer dealing with a complex rational function; we are dealing with a simple inverse relationship.
To make y as large as possible, we must make the fraction 3x2+9x+710 as large as possible. Since the numerator 10 is fixed, this means we must make the denominator 3x2+9x+7 as small as possible.
The Geometry of the Parabola
Now, let us visualize the denominator g(x)=3x2+9x+7. This is a quadratic expression.
Because the leading coefficient a=3 is positive, this parabola opens upwards, like a cup. It has a unique minimum point, the vertex.
We don't need calculus to find this! We can use the vertex formula:
Plugging in a=3, b=9, and c=7, we get:
gmin=4(3)4(3)(7)−(9)2=1284−81=123=41
This is the smallest value the denominator can ever take.
The Final Triumph
With the minimum denominator in hand, we return to our simplified expression:
Dividing by a fraction is the same as multiplying by its reciprocal, so 10÷41=10×4=40.
Adding the constant 1, we arrive at ymax=41.
It is a clean, beautiful integer. This is the power of algebraic insight—turning a daunting problem into a simple, elegant solution. Keep this mindset, and you will conquer any problem the JEE throws at you!