IGDTUW 2026 — Mathematics PYQ

To find the value of the given expression, we can break it down into two main steps: solving the infinite geometric series inside the logarithm, and then simplifying the logarithmic exponent.

Step 1: Simplify the Infinite Geometric Series

Let the sum of the infinite series be $S$:

$$S=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$$

This is an Infinite Geometric Progression (GP) where:

• First term ($a$) = $\frac{1}{4}$

• Common ratio ($r$) = $\frac{1/8}{1/4}=\frac{1}{2}$

The formula for the sum of an infinite GP when |r| < 1 is:

$$S=\frac{a}{1-r}$$

Substituting the values into the formula:

$$S=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$$

Step 2: Substitute the Series Sum into the Main Expression

Now, substitute $S=\frac{1}{2}$ back into the original expression:

$$E=2^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}$$

Step 3: Simplify the Logarithmic Exponent

We can use the logarithmic property $a^{{\log }_{b}c}=c^{{\log }_{b}a}$ to swap the base and the argument:

$$2^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}={\left(\frac{1}{2}\right)}^{{\log }_{\sqrt{5}}2}$$

We know that $\sqrt{5}=5^{1/2}$. Using the base power rule of logarithms, ${\log }_{b^k}x=\frac{1}{k}{\log }_{b}x$:

$${\log }_{\sqrt{5}}2={\log }_{5^{1/2}}2=2{\log }_{5}2={\log }_5(2^2)={\log }_{5}4$$

Substituting this back gives:

$$E={\left(\frac{1}{2}\right)}^{{\log }_{5}4}$$

Alternatively, let's look at the options. Since none of the standard bases simplify ${\left(\frac{1}{2}\right)}^{{\log }_{5}4}$ or $2^{{\log }_{\sqrt{5}}(1/2)}$ into simple integers like $4$, $\log 4$, or $\log 2$, let's double-check the values. The base of the log is $\sqrt{5}$ and the argument is $\frac{1}{2}$, resulting in an irrational exponent power of $2$, which does not equal $4$, $\log 4$, or $\log 2$.

Therefore, the correct choice is (d) None.

Correct Answer: (d) None