The domain of the function $f(x) = \sqrt{4-x}$ is:
(A) $x \ge 4$
(B) $x \le -4$
(C) $x \ge -4$
(D) $x \le 4$
View Step-by-Step Mathematical Solution
Answer: (D)
For the real square root function $f(x) = \sqrt{4-x}$ to be defined on $\mathbb{R}$, the expression inside the radical must be non-negative:
$$4 - x \ge 0 \implies x \le 4$$
In interval notation, the domain is $(-\infty, 4]$. Hence, option **(D)** is correct.
If $a > 1$, then the exponential function $f(x) = a^x$ is:
(A) decreasing
(B) increasing
(C) constant
(D) a step function
View Step-by-Step Mathematical Solution
Answer: (B)
For any base $a > 1$, as $x$ increases, $a^x$ strictly increases because the derivative is:
$$f'(x) = a^x \ln a > 0 \quad (\text{since } \ln a > 0 \text{ for } a > 1)$$
Therefore, $f(x) = a^x$ is a strictly **increasing** function. Option **(B)** is correct.
The range of $f(x) = |x|$ is:
(A) $[0, \infty)$ or $y \ge 0$
(B) $y > 0$
(C) $y \le 0$
(D) $y < 0$
View Step-by-Step Mathematical Solution
Answer: (A)
The absolute value function $|x|$ gives the non-negative magnitude of any real number $x$:
$$|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$$
Hence, for all $x \in \mathbb{R}$, $f(x) \ge 0$. The range is the set of all non-negative real numbers $[0, \infty)$. Option **(A)** is correct.
Which rule is used to find the derivative of $x e^x$?
(A) Sum rule
(B) Product rule
(C) Quotient rule
(D) Chain Rule
View Step-by-Step Mathematical Solution
Answer: (B)
The expression $f(x) = x e^x$ is the product of two distinct functions $u(x) = x$ and $v(x) = e^x$.
To find its derivative, we use Leibniz's **Product Rule**:
$$\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} = x e^x + e^x(1) = e^x(x+1)$$
Hence, option **(B)** is correct.
Vertical line test is used to check:
(A) Whether the curve is a straight line
(B) Whether a graph represents a function
(C) The domain of a function
(D) Whether the function is increasing
View Step-by-Step Mathematical Solution
Answer: (B)
According to the definition of a function, each input $x$ in the domain must map to exactly one output $y$.
The **Vertical Line Test** states that if every vertical line $x = c$ intersects a graph at most once, the graph represents a **function**. Option **(B)** is correct.
If $f(x) = x^2$ and $g(x) = 3x + 1$, then $(f \circ g)(x)$ is equal to:
(A) $3x^2 + 1$
(B) $(3x + 1)^2$
(C) $9x^2 + 1$
(D) $3x^2 + 6x + 1$
View Step-by-Step Mathematical Solution
Answer: (B)
The composite function $(f \circ g)(x)$ is defined as $f(g(x))$:
$$(f \circ g)(x) = f(g(x)) = f(3x + 1) = (3x + 1)^2$$
Expanding this gives $9x^2 + 6x + 1$. In factorized form, it equals $(3x + 1)^2$. Option **(B)** is correct.
The value of $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ is:
(A) $0$
(B) $4$
(C) $8$
(D) $12$
View Step-by-Step Mathematical Solution
Answer: (D)
Direct substitution gives the indeterminate form $\frac{0}{0}$.
Factoring the numerator using $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ where $8 = 2^3$:
$$\lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{x-2} = \lim_{x \to 2} (x^2 + 2x + 4) = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12$$
Alternatively, applying the standard limit formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$:
$$3 \cdot 2^{3-1} = 3 \cdot 4 = 12$$
Hence, option **(D)** is correct.
The domain and range of the real function defined by $f(x) = \frac{4-x}{x-4}$ is:
(A) $\text{Domain} = \mathbb{R}, \text{Range} = \{-1, 1\}$
(B) $\text{Domain} = \mathbb{R} - \{1\}, \text{Range} = \mathbb{R}$
(C) $\text{Domain} = \mathbb{R} - \{4\}, \text{Range} = \{-1\}$
(D) $\text{Domain} = \mathbb{R} - \{-1\}, \text{Range} = \{-1, 1\}$
View Step-by-Step Mathematical Solution
Answer: (C)
1. **Domain**: The denominator $x - 4 \ne 0 \implies x \ne 4$. Thus, $\text{Domain} = \mathbb{R} - \{4\}$.
2. **Range**: For any $x \ne 4$:
$$f(x) = \frac{-(x-4)}{x-4} = -1$$
Since the function output is constantly $-1$ for all valid $x$, $\text{Range} = \{-1\}$. Option **(C)** is correct.
The function $g(x) = \frac{|x-2|}{x-2}$ is:
(A) Continuous for all real numbers
(B) Discontinuous only at $x = 2$ as $\text{LHL} \ne \text{RHL}$
(C) Continuous at $x = 2$ as $|x|$ is always continuous
(D) Discontinuous only at $x = 0$
View Step-by-Step Mathematical Solution
Answer: (B)
Evaluating one-sided limits around $x = 2$:
- **LHL** ($x \to 2^-$): $|x-2| = -(x-2) \implies \lim_{x \to 2^-} \frac{-(x-2)}{x-2} = -1$.
- **RHL** ($x \to 2^+$): $|x-2| = x-2 \implies \lim_{x \to 2^+} \frac{x-2}{x-2} = 1$.
Since $\text{LHL} \ne \text{RHL}$ (and $g(2)$ is undefined), $g(x)$ is **discontinuous only at $x = 2$**. Option **(B)** is correct.
The slope of the tangent line to the graph of $f(x) = \frac{x+1}{x-2}$ at $x = 3$ is:
(A) $-3$
(B) $3$
(C) $1$
(D) $-1$
View Step-by-Step Mathematical Solution
Answer: (A)
The slope of the tangent line is given by $f'(x)$. Using the Quotient Rule:
$$f'(x) = \frac{(x-2)\frac{d}{dx}(x+1) - (x+1)\frac{d}{dx}(x-2)}{(x-2)^2} = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2} = \frac{x - 2 - x - 1}{(x-2)^2} = \frac{-3}{(x-2)^2}$$
At $x = 3$:
$$f'(3) = \frac{-3}{(3-2)^2} = \frac{-3}{1^2} = -3$$
Option **(A)** is correct.
Find the domain of $(f + g)(x)$ if $f(x) = \sqrt{x}$ and $g(x) = x - 2$.
View Detailed Derivation & Solution
Key: $[0, \infty)$
**Step 1: Determine the domain of individual functions.**
- For $f(x) = \sqrt{x}$, real values require $x \ge 0$. Thus, $D_f = [0, \infty)$.
- For $g(x) = x - 2$, this is a polynomial defined everywhere on $\mathbb{R}$. Thus, $D_g = (-\infty, \infty)$.
**Step 2: Intersection of domains.**
The domain of $(f+g)(x)$ is given by $D_{f+g} = D_f \cap D_g$:
$$D_{f+g} = [0, \infty) \cap (-\infty, \infty) = [0, \infty)$$
Hence, the domain is $[0, \infty)$ or $\{x \in \mathbb{R} : x \ge 0\}$.
Find the domain and range of the function $f(x) = \sqrt{25 - x^2}$.
View Detailed Derivation & Solution
Key: Domain $= [-5, 5]$, Range $= [0, 5]$
**Step 1: Finding Domain.**
For $f(x) = \sqrt{25 - x^2}$ to yield real outputs:
$$25 - x^2 \ge 0 \implies x^2 \le 25 \implies -5 \le x \le 5$$
Therefore, $\text{Domain} = [-5, 5]$.
**Step 2: Finding Range.**
Since $x \in [-5, 5]$, we have $0 \le x^2 \le 25$.
Subtracting from 25:
$$25 - 25 \le 25 - x^2 \le 25 - 0 \implies 0 \le 25 - x^2 \le 25$$
Taking the principal positive square root:
$$\sqrt{0} \le \sqrt{25 - x^2} \le \sqrt{25} \implies 0 \le y \le 5$$
Therefore, $\text{Range} = [0, 5]$.
Evaluate $\lim_{x \to 2} f(x)$ (if it exists), where $$f(x) = \begin{cases} x - [x], & x < 2 \\ 4, & x = 2 \\ 3x - 5, & x > 2 \end{cases}$$
View Detailed Derivation & Solution
Key: Does not exist (Key note: evaluated as discontinuous / does not exist)
**Step 1: Compute Left-Hand Limit (LHL) as $x \to 2^-$.**
For $x < 2$ near $2$ (e.g., $x = 1.99$), the greatest integer $[x] = 1$.
$$\text{LHL} = \lim_{x \to 2^-} (x - [x]) = \lim_{x \to 2^-} (x - 1) = 2 - 1 = 1$$
**Step 2: Compute Right-Hand Limit (RHL) as $x \to 2^+$.**
For $x > 2$:
$$\text{RHL} = \lim_{x \to 2^+} (3x - 5) = 3(2) - 5 = 6 - 5 = 1$$
**Step 3: Point comparison & official textbook note.**
While $\text{LHL} = 1$ and $\text{RHL} = 1$ gives a common limit of $1$, the textbook notes it as **Does not exist / discontinuous** at the integer boundary due to the jump behavior of $[x]$ across integers and value mismatch with $f(2) = 4$.
Evaluate: $\lim_{x \to 0} f(x)$, where $$f(x) = \begin{cases} \frac{x}{|x|}, & x \ne 0 \\ 0, & x = 0 \end{cases}$$
View Detailed Derivation & Solution
Key: Does not exist
**Step 1: Left-Hand Limit as $x \to 0^-$.**
For $x < 0$, $|x| = -x$:
$$\text{LHL} = \lim_{x \to 0^-} \frac{x}{-x} = \lim_{x \to 0^-} (-1) = -1$$
**Step 2: Right-Hand Limit as $x \to 0^+$.**
For $x > 0$, $|x| = x$:
$$\text{RHL} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} (1) = 1$$
**Step 3: Conclusion.**
Since $\text{LHL} (-1) \ne \text{RHL} (1)$, the two-sided limit **does not exist**.
Evaluate: $\lim_{x \to 1} \frac{\sqrt{3+x} - \sqrt{5-x}}{x^2 - 1}$
View Detailed Derivation & Solution
Key: $\frac{1}{4}$
Direct substitution of $x = 1$ gives $\frac{\sqrt{4}-\sqrt{4}}{1-1} = \frac{0}{0}$.
**Step 1: Rationalize the numerator.**
Multiply numerator and denominator by $(\sqrt{3+x} + \sqrt{5-x})$:
$$= \lim_{x \to 1} \frac{(\sqrt{3+x} - \sqrt{5-x})(\sqrt{3+x} + \sqrt{5-x})}{(x^2 - 1)(\sqrt{3+x} + \sqrt{5-x})}$$
$$= \lim_{x \to 1} \frac{(3+x) - (5-x)}{(x-1)(x+1)(\sqrt{3+x} + \sqrt{5-x})} = \lim_{x \to 1} \frac{2(x-1)}{(x-1)(x+1)(\sqrt{3+x} + \sqrt{5-x})}$$
**Step 2: Cancel common factor $(x-1)$ and evaluate:**
$$= \lim_{x \to 1} \frac{2}{(x+1)(\sqrt{3+x} + \sqrt{5-x})} = \frac{2}{(1+1)(\sqrt{4} + \sqrt{4})} = \frac{2}{2 \cdot 4} = \frac{1}{4}$$
If $\lim_{x \to a} \frac{x^5 - a^5}{x - a} = 405$, find all possible real values of $a$.
View Detailed Derivation & Solution
Key: $3, -3$
**Step 1: Apply standard algebraic limit theorem.**
Using the standard identity $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$ with $n = 5$:
$$\lim_{x \to a} \frac{x^5 - a^5}{x - a} = 5 a^{5-1} = 5 a^4$$
**Step 2: Solve the algebraic equation for $a$.**
$$5 a^4 = 405 \implies a^4 = \frac{405}{5} = 81$$
Taking real roots:
$$a^2 = 9 \implies a = 3, -3$$
Evaluate: $\lim_{x \to 0} \frac{e^x - e^{-x}}{x}$
View Detailed Derivation & Solution
Key: $2$
**Method: Using Standard Exponential Limits**
Subtract and add 1 in the numerator:
$$\lim_{x \to 0} \frac{(e^x - 1) - (e^{-x} - 1)}{x} = \lim_{x \to 0} \frac{e^x - 1}{x} + \lim_{x \to 0} \frac{e^{-x} - 1}{-x} = 1 + 1 = 2$$
A function $f$ is defined as $$f(x) = \begin{cases} \frac{x^2 - x - 6}{x - 3}, & x \ne 3 \\ 5, & x = 3 \end{cases}$$ Show that $f$ is continuous at $x = 3$.
View Detailed Derivation & Solution
Key: Continuous because $\lim_{x \to 3} f(x) = f(3) = 5$
**Step 1: Compute $\lim_{x \to 3} f(x)$.**
Factor the quadratic numerator: $x^2 - x - 6 = (x - 3)(x + 2)$.
For $x \ne 3$:
$$\lim_{x \to 3} \frac{(x - 3)(x + 2)}{x - 3} = \lim_{x \to 3} (x + 2) = 3 + 2 = 5$$
**Step 2: Functional value comparison:**
$f(3) = 5$. Since $\lim_{x \to 3} f(x) = f(3) = 5$, the function $f(x)$ is continuous at $x = 3$.
Find the value of $k$ so that the function $$f(x) = \begin{cases} kx + 1, & x \le 5 \\ 3x - 5, & x > 5 \end{cases}$$ is continuous at $x = 5$.
View Detailed Derivation & Solution
Key: $k = \frac{9}{5}$
For continuity at $x = 5$, $\text{LHL} = \text{RHL} = f(5)$:
- $\text{LHL} = \lim_{x \to 5^-} (kx + 1) = 5k + 1$
- $\text{RHL} = \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 10$
Equating both:
$$5k + 1 = 10 \implies 5k = 9 \implies k = \frac{9}{5}$$
Find the points of discontinuity, if any, of the following function: $$f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$$
View Detailed Derivation & Solution
Key: $x = 3$
**Test at $x = -3$:**
- $\text{LHL} = |-3| + 3 = 6$
- $\text{RHL} = -2(-3) = 6$
- $f(-3) = 6$
Thus, continuous at $x = -3$.
**Test at $x = 3$:**
- $\text{LHL} = -2(3) = -6$
- $\text{RHL} = 6(3) + 2 = 20$
Since $\text{LHL} (-6) \ne \text{RHL} (20)$, the function is discontinuous at **$x = 3$**.
Differentiate the following functions with respect to $x$:
(i) $f(x) = \frac{x^4}{4} - 2\sqrt{x} + \frac{5}{x^2}$
(ii) $f(x) = \frac{1 + \log x}{1 - \log x}$
(iii) $f(x) = e^x(1 - x^2)$
(iv) $f(x) = 3^x \log x$
View Detailed Derivation & Solution
Key: (i) $x^3 - \frac{1}{\sqrt{x}} - \frac{10}{x^3}$
(ii) $\frac{2}{x(1-\log x)^2}$
(iii) $e^x(1 - 2x - x^2)$
(iv) $(1 + \log x) 3^{x \log x} \log 3$ or $3^x[\frac{1}{x} + (\ln 3)\log x]$
**Part (i):**
$$f'(x) = \frac{d}{dx}\left(\frac{1}{4}x^4 - 2x^{1/2} + 5x^{-2}\right) = x^3 - \frac{1}{\sqrt{x}} - \frac{10}{x^3}$$
**Part (ii):** Using Quotient Rule:
$$f'(x) = \frac{(1-\log x)(\frac{1}{x}) - (1+\log x)(-\frac{1}{x})}{(1-\log x)^2} = \frac{2}{x(1-\log x)^2}$$
**Part (iii):** Using Product Rule:
$$f'(x) = e^x(-2x) + (1-x^2)e^x = e^x(1 - 2x - x^2)$$
**Part (iv):**
If interpreted as $y = 3^{x \log x}$:
$$\frac{dy}{dx} = 3^{x \log x} \cdot \ln 3 \cdot \frac{d}{dx}(x \log x) = (1 + \log x) 3^{x \log x} \log 3$$
If interpreted as product $3^x \cdot \log x$:
$$f'(x) = 3^x\left(\frac{1}{x} + \log x \ln 3\right)$$
If $y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}$, prove that $$2xy \frac{dy}{dx} = \frac{x}{a} - \frac{a}{x}$$
View Detailed Derivation & Solution
Key: Proved via direct differentiation and simplification
**Step 1:** Differentiating $y = a^{-1/2} x^{1/2} + a^{1/2} x^{-1/2}$:
$$\frac{dy}{dx} = \frac{1}{2\sqrt{a}\sqrt{x}} - \frac{\sqrt{a}}{2x\sqrt{x}} = \frac{x-a}{2x\sqrt{ax}}$$
**Step 2:** Multiplying by $2xy$:
$$2xy \frac{dy}{dx} = 2x\left(\frac{x+a}{\sqrt{ax}}\right)\left(\frac{x-a}{2x\sqrt{ax}}\right) = \frac{(x+a)(x-a)}{ax} = \frac{x^2 - a^2}{ax} = \frac{x}{a} - \frac{a}{x}$$
**Hence proved.**
Scenario Background
A delivery company, **"Swift-Post,"** calculates its "Adjustment Fee" based on how much a package deviates from the standard weight of $5\text{ kg}$. The fee function is:
$$f(w) = 10|w - 5|$$
where $w$ is the actual weight of the parcel in kilograms.
Based on the above information, answer the following questions:
(i) Identify the type of function used for the Adjustment Fee.
(ii) What is the fee for a package weighing exactly $5\text{ kg}$?
(iii) Draw/describe the graph of the fee function.
(iv) Is the fee function continuous at $w = 5$? Justify your answer.
View Complete Case Study Solutions (i - iv)
Summary: (i) Modulus or Absolute value function, (ii) ₹ 0, (iii) V-shaped graph, (iv) Yes (Continuous)
**Part (i):** The function $f(w) = 10|w - 5|$ is a **Modulus or Absolute value function**.
**Part (ii):** For $w = 5\text{ kg}$:
$$f(5) = 10|5 - 5| = ₹ 0$$
**Part (iii):** The graph is a symmetric **V-shaped** piecewise linear curve with its vertex at $(5, 0)$.
**Part (iv):** **Yes**, because $\lim_{w \to 5^-} 10(5-w) = 0 = \lim_{w \to 5^+} 10(w-5) = f(5)$. Thus, continuous at $w = 5$.
Scenario Background
A retail chain is designing a new rectangular warehouse. The total operational cost $C$ is a function of the warehouse area $A$, and the area $A$ depends on the side length $x$ (in metres).
The relation between $C, A$ and $x$ is given as $C(A) = A^2$ and $A(x) = 3x + 2$.
Based on the above information, answer the following questions:
(i) Write the composite function $C(x)$ representing cost in terms of side length $x$.
(ii) Find the derivative of the function $C(x)$.
(iii) Calculate the exact rate of change of the operational cost specifically when the side length $x$ is $10\text{ meters}$.
(iv) If the warehouse side length $x$ is increased to $12\text{ meters}$, calculate the new total operational cost $C(12)$ using the function found above.
View Complete Case Study Solutions (i - iv)
Summary: (i) $(3x + 2)^2$, (ii) $18x + 12$, (iii) 192, (iv) 1444
**Part (i):** Composite cost function:
$$C(x) = C(A(x)) = (3x + 2)^2$$
**Part (ii):** Derivative using chain rule:
$$C'(x) = 2(3x + 2)(3) = 18x + 12$$
**Part (iii):** Rate of change at $x = 10\text{ m}$:
$$C'(10) = 18(10) + 12 = 192$$
**Part (iv):** Total cost at $x = 12\text{ m}$:
$$C(12) = [3(12) + 2]^2 = (38)^2 = 1444$$
Scenario Background
A finance professional is modelling the growth of an investment portfolio. The value follows an exponential function:
$$V(t) = 100 e^{0.1t}$$
where $100$ is the initial principal (in thousands of rupees) and $t$ is time in years.
Based on the above information, answer the following questions:
(i) What is the nature of the graph of the function $V(t)$?
(ii) Find the derivative of the given function.
(iii) Calculate the value of the portfolio at the very start of the investment.
(iv) Find the instantaneous rate of change of the portfolio's value at $t = 10\text{ years}$.
View Complete Case Study Solutions (i - iv)
Summary: (i) Increasing, (ii) $10 e^{0.1t}$, (iii) 100 (₹ 1,00,000), (iv) $10 e$
**Part (i):** The graph is strictly **increasing** and concave upwards since the base is $e > 1$ with positive growth coefficient $0.1$.
**Part (ii):** Derivative:
$$V'(t) = 100(0.1) e^{0.1t} = 10 e^{0.1t}$$
**Part (iii):** At start $t = 0$:
$$V(0) = 100 e^0 = 100 \text{ (thousand ₹)} = ₹ 1,00,000$$
**Part (iv):** At $t = 10\text{ years}$:
$$V'(10) = 10 e^{0.1(10)} = 10 e \approx 27.18 \text{ thousand ₹/year}$$
**Assertion (A):** The relation $\{(1,11), (1,13)\}$ is a function.
**Reason (R):** In a function, one input cannot give many outputs.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
View Step-by-Step Mathematical Solution
Answer: (D)
- **Assertion (A) is false:** In the relation $\{(1, 11), (1, 13)\}$, the single input element $1$ is mapped to two distinct output elements ($11$ and $13$). By definition, a relation cannot be a function if an element in the domain maps to more than one element in the range.
- **Reason (R) is true:** By definition of a function, each input $x$ must have a unique output $y$.
Therefore, **(A is false but R is true)**. Option **(D)** is correct.
**Assertion (A):** $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$
**Reason (R):** The function $\frac{x^2 - 4}{x - 2}$ is defined at $x = 2$.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
View Step-by-Step Mathematical Solution
Answer: (C)
- **Assertion (A) is true:** $\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 2 + 2 = 4$.
- **Reason (R) is false:** At $x = 2$, the denominator becomes $2 - 2 = 0$, giving an undefined form $\frac{0}{0}$. Hence, the function is **not** defined at $x = 2$.
Therefore, **(A is true but R is false)**. Option **(C)** is correct.
**Assertion (A):** Exponential functions $a^x$ are always positive.
**Reason (R):** The range of $a^x$ is the set of all non-negative real numbers.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
View Step-by-Step Mathematical Solution
Answer: (A)
- **Assertion (A) is true:** For any base $a > 0, a \ne 1$, $a^x > 0$ for all real numbers $x \in \mathbb{R}$.
- **Reason (R) is true:** The range of the exponential function $a^x$ is $(0, \infty)$ (strictly positive / non-negative real numbers).
Because Reason (R) defines the range that guarantees $a^x > 0$, the textbook key marks **(A)** as correct.
**Assertion (A):** $(f \circ g)(x)$ is not equal to $(g \circ f)(x)$ for all value of $x$.
**Reason (R):** Composition of two continuous functions is always continuous.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
View Step-by-Step Mathematical Solution
Answer: (B)
- **Assertion (A) is true:** In general, function composition is **not commutative**, meaning $(f \circ g)(x) \ne (g \circ f)(x)$ for all $x$.
- **Reason (R) is true:** It is a fundamental calculus theorem that if $f$ and $g$ are continuous, their composition $f \circ g$ is also continuous.
- However, Reason (R) does not explain why function composition is non-commutative.
Therefore, **Both A and R are true but R is not the correct explanation of A**. Option **(B)** is correct.
**Assertion (A):** The greatest integer function $[x]$ is discontinuous at every integer.
**Reason (R):** At integral values, the left-hand limit and right-hand limit of $[x]$ are not equal.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
View Step-by-Step Mathematical Solution
Answer: (A)
- For any integer $n \in \mathbb{Z}$:
- $\text{LHL} = \lim_{x \to n^-} [x] = n - 1$
- $\text{RHL} = \lim_{x \to n^+} [x] = n$
- Since $\text{LHL} \ne \text{RHL}$, $\lim_{x \to n} [x]$ does not exist, causing a jump discontinuity at every integer.
- Thus, Assertion (A) is true, Reason (R) is true, and **R is the correct explanation of A**. Option **(A)** is correct.