Analyzing the Setup
The integral provided is:
I=∫((2x)x+(x2)x)log2xdx
The presence of x in both the base and the exponent is a significant indicator that standard integration techniques will not suffice. In the context of JEE Advanced, we often look for elegance over brute force, such as differentiating potential options to recover the integrand.
Defining the Functions
Let us define our two primary functions:
f(x)=(2x)x
g(x)=(x2)x
To differentiate these, we employ
Logarithmic Differentiation. For any function
y=u(x)v(x), we use the identity:
y1dxdy=v′(x)ln(u(x))+v(x)u(x)u′(x)
Differentiating f(x)
Taking the natural log of
f(x):
ln(f(x))=xln(2x)
Differentiating both sides with respect to
x:
f(x)f′(x)=1⋅ln(2x)+x⋅2x1⋅21
Simplifying the expression, we obtain:
f(x)f′(x)=ln(2x)+1
f′(x)=(2x)x(lnx−ln2+1)
Differentiating g(x)
Following the same logic for
g(x)=(x2)x:
ln(g(x))=xln(x2)
Differentiating yields:
g(x)g′(x)=ln(x2)+x⋅x21⋅(−x22)
This simplifies to:
g(x)g′(x)=ln(x2)−1
g′(x)=(x2)x(ln2−lnx−1)
The Moment of Truth
Observe the relationship between the derivatives. We can rewrite
g′(x) as:
g′(x)=−(x2)x(lnx−ln2+1)
When we consider the sum
f′(x)+g′(x), we find:
f′(x)+g′(x)=[(2x)x−(x2)x](lnx−ln2+1)
We compare this to our original integrand, which contains log2x=ln2lnx. Because the derived expression contains (lnx+1−ln2) rather than a simple log2x term, the integral does not resolve into a standard elementary form.
Conclusion: This specific problem was identified as having a flaw in the provided options. By maintaining a rigorous mathematical approach, you have successfully verified the inconsistency, demonstrating the analytical confidence required to excel in competitive examinations.