If the maximum value of the term independent of t in the expansion of (t² x^(1/5) + (1-x)^(1/10)/t)¹

Q: Why do we use differentiation to maximize $x(1-x)$? Can't we just guess the value?

While you might guess that $x=0.5$ is the peak for $x(1-x)$, differentiation is the rigorous mathematical proof required for JEE Advanced. It ensures you aren't just guessing, but finding the exact critical point where the slope of the tangent is zero. It is a robust habit to build.

Q: Is $\binom{15}{10}$ the same as $\binom{15}{5}$? Why does this property matter?

Yes, absolutely! The property $\binom{n}{r} = \binom{n}{n-r}$ is a lifesaver. Calculating $\binom{15}{10}$ involves large numbers, but $\binom{15}{5}$ is much easier to compute. Always look for symmetry to simplify your arithmetic.

Q: What if the exponent of $t$ resulted in a non-integer $r$? Does that mean there is no independent term?

Exactly! If the calculation for $r$ yields a fraction or a number outside the range $[0, 15]$, it would mean that no term in the expansion is independent of $t$. In such a case, the coefficient would be zero. Always verify that your $r$ is a valid integer.

Question

If the maximum value of the term independent of $t$ in the expansion of $\left(t^2 x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t}\right)^{15}, x \ge 0$, is $K$, then $8K$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, looks like a chaotic mess of variables and fractional exponents. We have the expression $\left(t^2 x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t}ight)^{15}$.

It is easy to feel overwhelmed by the $t$ in the denominator and the $x$ with its fractional powers. But remember, in JEE Advanced, complexity is often just a mask for elegance.

Our goal is to find the term independent of $t$. This means we are hunting for a term where $t$ effectively vanishes, leaving us with a function of $x$ that we can then maximize.

### The Hunt for Independence

To find this elusive term, we turn to our most powerful tool: the General Term formula. For any binomial expansion of the form $(a+b)^n$, the $(r+1)$-th term is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

Here, our $n$ is $15$, our $a$ is $t^2 x^{\frac{1}{5}}$, and our $b$ is $t^{-1} (1-x)^{\frac{1}{10}}$.

Let us set up the expression:

$$T_{r+1} = \binom{15}{r} \left(t^2 x^{\frac{1}{5}}ight)^{15-r} \left(t^{-1} (1-x)^{\frac{1}{10}}ight)^r$$

Now, pause. Do not expand everything yet. We only care about the $t$ terms. Let us isolate them:

$$t^{2(15-r)} \cdot t^{-r} = t^{30-2r} \cdot t^{-r} = t^{30-3r}$$

For the term to be independent of $t$, the power of $t$ must be zero. This is the filter through which we find our term.

We set $30 - 3r = 0$, which gives us $r = 10$. We have found our target; we are looking for the $11^{	ext{th}}$ term of the expansion.

### The Calculus Twist

With $r=10$, our term becomes:

$$T_{11} = \binom{15}{10} \left(x^{\frac{1}{5}}ight)^{15-10} \left((1-x)^{\frac{1}{10}}ight)^{10} = \binom{15}{10} x(1-x)$$

Now, the problem shifts from binomial expansion to function maximization. We have a function $f(x) = x(1-x) = x - x^2$.

This is a downward-opening parabola. To find its maximum, we use the power of calculus. We find the derivative $f'(x) = 1 - 2x$ and set it to zero.

This gives us $x = \frac{1}{2}$. Substituting this back, the maximum value of $f(x)$ is $\frac{1}{2}(1 - \frac{1}{2}) = \frac{1}{4}$.

### Final Calculation

We are in the home stretch. We need to calculate $\binom{15}{10}$. Using the symmetry property $\binom{n}{r} = \binom{n}{n-r}$, we know $\binom{15}{10} = \binom{15}{5}$.

Calculating this:

$$\binom{15}{5} = \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 3003$$

Thus, the maximum value $K$ is $3003 \cdot \frac{1}{4} = \frac{3003}{4}$. The question asks for $8K$.

So, $8 \cdot \frac{3003}{4} = 2 \cdot 3003 = \mathbf{6006}$.

Animated Solution for Mathematics - Binomial Theorem: If the maximum value of the term independent of $t$ in the expansion of $(t^{2} x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{15}, x \geq 0$ , is $K$ , then $8 K$ is equal to ______.

Enter Numerical Value:

Visualized Solution

The Binomial Expression

General Term $T_{r + 1}$

Raw Setup

Power of $t$

Term Independent of $t$

The $1 1^{th}$ Term

Defining $f (x)$

Maximizing $f (x)$

Finding the Critical Point

Max Value of $f (x)$

Calculating $(10 15)$

Finding $K$

Final Result $8 K$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Hunt for Independence

The Calculus Twist

Final Calculation

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Enter Numerical Value:

Visualized Solution

The Binomial Expression

General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​

Raw Setup

Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Term Independent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 11th Term

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Maximizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding the Critical Point

Max Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1015​)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Final Result .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8K