Let $f(x) = \sqrt{1 + x^2}$, then (A) $f(xy) = f(x) \cdot f(y)$ (B) $f(xy) \ge f(x) \cdot f(y)$ (C) $f(xy) \le f(x) \cdot f(y)$ (D) None of these

Let f(x) = √(1 + x²), then (A) f(xy) = f(x) · f(y) – Class 11 Mathematics

NCERT Exemplar · Q27

f(x) = \sqrt{1 + x^2}

Manipur CohsemMCQ· 1mImportance★★★★★est

85% · 85/100 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Function Operations

Function Operations: Combining Machines

Think of a function as a machine. You feed it an input (say, a number $x$ ), it does something, and out comes an output $f(x)$ . Now imagine you have two such machines, $f$ and $g$ . Function operations are simply ways to hook these machines together — to add, subtract, multiply, or divide their outputs, or to feed one machine's output into the other.

The core idea is simple: if you can do arithmetic with numbers, you can do arithmetic with functions. The only catch is that both functions must be "ready to work" on the same input at the same time.

The Four Arithmetic Operations

Let $f$ and $g$ be two functions. For any input $x$ that belongs to both their domains (the set of numbers each can accept), we define:

Operation	Notation	What it means
Sum	$(f+g)(x)$	$f(x) + g(x)$
Difference	$(f-g)(x)$	$f(x) - g(x)$
Product	$(f \cdot g)(x)$	$f(x) \cdot g(x)$
Quotient	$\left(\frac{f}{g}\right)(x)$	$\frac{f(x)}{g(x)}$ , provided $g(x) \neq 0$

Note

The domain of the new function is the intersection of the domains of $f$ and $g$ — the numbers both machines can handle. For the quotient, you must also exclude any $x$ where $g(x) = 0$ , because division by zero is undefined.

Example. Let $f(x) = \sqrt{x}$ (domain: $x \geq 0$ ) and $g(x) = x - 1$ (domain: all real numbers). Then:

$(f+g)(x) = \sqrt{x} + x - 1$ , domain: $x \geq 0$ .
$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x-1}$ , domain: $x \geq 0$ and $x \neq 1$ .

Composition: Feeding One Machine into Another

This is the most powerful operation. Instead of adding outputs side by side, you take the output of one function and feed it as the input to the other.

(f \circ g)(x) = f\big(g(x)\big)

Read " $f$ composed with $g$ ". You do $g$ first, then $f$ on the result.

Intuition. Suppose $g$ is a machine that converts Celsius to Fahrenheit, and $f$ is a machine that converts Fahrenheit to Kelvin. Then $f \circ g$ converts Celsius directly to Kelvin — one combined machine.

Domain trap. For $f(g(x))$ to make sense, two conditions must hold:

$x$ must be in the domain of $g$ (so $g(x)$ exists).
$g(x)$ must be in the domain of $f$ (so $f$ can accept it).

So the domain of $f \circ g$ is: all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$ .

Watch out

Composition is not commutative. $f \circ g$ is almost never the same as $g \circ f$ . For example, if $f(x) = x^2$ and $g(x) = x+1$ , then:

$(f \circ g)(x) = (x+1)^2 = x^2 + 2x + 1$
$(g \circ f)(x) = x^2 + 1$ …

Start your 14-day free trial to unlock the full solution →

The function $f(x) = \sqrt{1 + x^2}$ does NOT satisfy $f(xy) = f(x) f(y)$ in general. By comparing $f(xy)$ and $f(x) f(y)$ using the Cauchy-Schwarz inequality, we find $f(xy) \le f(x) f(y)$ for all real $x, y$ , so option (C) is correct.

We are given $f(x) = \sqrt{1 + x^2}$ . The question asks how $f(xy)$ compares with $f(x) \cdot f(y)$ . At first glance, this looks like a functional equation problem, but it's really about comparing two expressions: $\sqrt{1 + x^2 y^2}$ and $\sqrt{(1 + x^2)(1 + y^2)}$ .

The key insight: squaring both sides removes the square roots and lets us compare polynomials. Once squared, the comparison reduces to checking whether $1 + x^2 y^2 \le (1 + x^2)(1 + y^2)$ for all real $x, y$ . That inequality is always true, and equality holds only in special cases.

Let's work through it step by step.

Write down the expressions explicitly.

$f(xy) = \sqrt{1 + (xy)^2} = \sqrt{1 + x^2 y^2}$

$f(x) \cdot f(y) = \sqrt{1 + x^2} \cdot \sqrt{1 + y^2} = \sqrt{(1 + x^2)(1 + y^2)}$
Square both sides to remove the radicals.

Since both sides are non-negative for all real $x, y$ , comparing $f(xy)$ and $f(x) f(y)$ is equivalent to comparing their squares:

$[f(xy)]^2 = 1 + x^2 y^2$

$[f(x) f(y)]^2 = (1 + x^2)(1 + y^2) = 1 + x^2 + y^2 + x^2 y^2$
Subtract to see the difference.

$[f(x) f(y)]^2 - [f(xy)]^2 = (1 + x^2 + y^2 + x^2 y^2) - (1 + x^2 y^2) = x^2 + y^2$
Interpret the result.

For any real $x, y$ , we have $x^2 + y^2 \ge 0$ , and it equals $0$ only when $x = y = 0$ . Therefore:

$[f(x) f(y)]^2 \ge [f(xy)]^2$ for all $x, y$

Since both sides are non-negative, taking square roots preserves the inequality:

$f(x) f(y) \ge f(xy)$

This means $f(xy) \le f(x) \cdot f(y)$ for all real $x, y$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ26 — Range of f(x) = (1)/(1 - 2cos x) is (A) [(1)/(3),1] (B) [-1,(1)/(3)] (…→Next questionQ28 — Domain of √(a² - x²) (a > 0) is (A) (-a,a) (B) [-a,a] (C) [0,a] (D) (-…

Q.Let f(x)=1+x2f(x) = \sqrt{1 + x^2}f(x)=1+x2​, then (A) f(xy)=f(x)⋅f(y)f(xy) = f(x) \cdot f(y)f(xy)=f(x)⋅f(y) (B) f(xy)≥f(x)⋅f(y)f(xy) \ge f(x) \cdot f(y)f(xy)≥f(x)⋅f(y) (C) f(xy)≤f(x)⋅f(y)f(xy) \le f(x) \cdot f(y)f(xy)≤f(x)⋅f(y) (D) None of these

🎓Unlock everything free for 14 days

Q.Let $f(x) = \sqrt{1 + x^2}$ , then
(A) $f(xy) = f(x) \cdot f(y)$
(B) $f(xy) \ge f(x) \cdot f(y)$
(C) $f(xy) \le f(x) \cdot f(y)$
(D) None of these

Unlock everything free for 14 days