The domain of definition of the function y = 1 / log₁₀(1 - x) + √(x + 2) is (a) (-3, -2) excluding -

Q: Wait, why is the log argument strictly positive? Can't it be zero?

Great question! The logarithm function $\log_{b}(A)$ is only defined when $A > 0$. If $A = 0$, the function approaches negative infinity, which is not a real number. If $A 0$.

Q: Why do we exclude $x = 0$ from the final domain?

Excellent catch! While $x = 0$ satisfies the square root condition (since $0 + 2 = 2 > 0$) and the log argument condition (since $1 - 0 = 1 > 0$), it makes the denominator zero. Specifically, $\log_{10}(1 - 0) = \log_{10}(1) = 0$. Since division by zero is undefined, $x = 0$ must be excluded.

Q: How do I know when to use square brackets versus parentheses in the final answer?

Think of it as a 'closed' versus 'open' door. A square bracket $[a, b]$ means the endpoints $a$ and $b$ are included in the domain (the inequality is $\geq$ or $\leq$). A parenthesis $(a, b)$ means the endpoints are excluded (the inequality is $>$ or $<$). Since our condition $x \geq -2$ includes $-2$, we use a square bracket. Since $x < 1$ excludes $1$, we use a parenthesis.

Question

The domain of definition of the function $y = \frac{1}{\log_{10}(1 - x)} + \sqrt{x + 2}$ is

Options:

(A) $(-3, -2)$ excluding $-2.5$

(B) $[0, 1]$ excluding $0.5$

(C) $[-2, 1)$ excluding $0$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, aspiring mathematician! Today, we are embarking on a journey to uncover the 'domain of definition' for a function that might look intimidating at first glance: $$y = \frac{1}{\log_{10}(1 - x)} + \sqrt{x + 2}$$ In the world of JEE Advanced, finding the domain is like being a detective at a crime scene. You are looking for the 'forbidden' values of $x$ that would cause the function to break—to become undefined, imaginary, or mathematically impossible. Let's break this down into three manageable, logical phases. ### Phase 1: The Foundation (The Square Root) Our first suspect is the square root term, $\sqrt{x + 2}$. In the realm of real numbers, you cannot take the square root of a negative number. If the value inside the root is negative, the function ceases to exist in our real-number coordinate system. Therefore, we must enforce the constraint that the radicand is non-negative: $$x + 2 \geq 0$$ Solving this is simple algebra: $x \geq -2$. On our number line, this means we are only interested in the region starting from $-2$ and extending all the way to positive infinity. We mark this with a solid blue dot at $-2$ to show it is included. ### Phase 2: The Gatekeeper (The Logarithm) Next, we turn our attention to the logarithm, $\log_{10}(1 - x)$. Logarithms are notoriously picky; they refuse to accept zero or negative numbers as inputs. The argument of the log, which is $(1 - x)$, must be strictly greater than zero. So, we set up our second constraint: $$1 - x > 0$$ Rearranging this, we find $1 > x$, or $x < 1$. This is our second boundary. We place an open green circle at $1$ to indicate that $1$ itself is forbidden, and we shade everything to the left. ### Phase 3: The Hidden Trap (The Denominator) Now, here is where many students stumble. We have a fraction, and the logarithm is sitting right there in the denominator: $\frac{1}{\log_{10}(1 - x)}$. The golden rule of fractions is that the denominator can never be zero. If $\log_{10}(1 - x) = 0$, the entire expression becomes undefined. We must solve for the $x$ that causes this disaster: $$\log_{10}(1 - x) = 0$$ Converting this to exponential form, we get $1 - x = 10^0$, which simplifies to $1 - x = 1$. Solving for $x$, we find $x = 0$. This point is a pothole on our number line; even though it satisfies our previous two conditions, it must be strictly excluded from our final answer. ### The Grand Intersection We have our three pieces of the puzzle: $x \geq -2$, $x < 1$, and $x eq 0$. To find the final domain, we look for the overlap of these regions. The intersection of $x \geq -2$ and $x < 1$ gives us the interval $[-2, 1)$. Finally, we must 'punch a hole' at $x = 0$ by removing it from this interval. The result is the elegant set **$

Animated Solution for Mathematics - Functions: The domain of definition of the function $y = \frac{1}{l o g _{10} ( 1 - x )} + x + 2$ is

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Visualized Solution

Analyzing the Function

Constraint 1: Square Root

Setting up the Inequality

Solving Constraint 1

Constraint 2: Logarithm Argument

Setting up the Log Inequality

Solving Constraint 2

Constraint 3: The Denominator

Setting up the Non-Zero Condition

Solving the Log Equation

Finding the Excluded Value

Combining All Constraints

Final Domain

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Grand Intersection

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of definition of the function y=log10​(1−x)1​+x+2​ is