Q.Let $A = \{1, 2, 3\}$, $B = \{4, 5, 6, 7\}$ and let $f = \{(1, 4), (2, 5), (3, 6)\}$ be a function from A to B. Show that $f$ is one-one.

Step 1 — Why we compress into one sine.

A sum $a\sin\theta+b\cos\theta$ is not obviously monotonic, but $R\sin(\theta-\alpha)$ is: a sine increases steadily on any interval where its argument sweeps from $-\pi/2$ to $\pi/2$. So the whole question becomes "where does the argument live?"

Here $a=\sqrt{3},\ b=-1$, so

Step 2 — Find the phase.

and by $\sin(P-Q)=\sin P\cos Q-\cos P\sin Q$,

The $+4$ is a vertical shift; it changes no injectivity.

Step 3 — Impose one-one-ness.

$\sin t$ is strictly increasing (hence one-one) for $t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Put $t=2x-\frac{\pi}{6}$:

Step 4 — Solve for $x$.

Add $\frac{\pi}{6}$: …

Step 1 — Statement (I): is $f(x) = \sin x$ one-one on $[0, \pi/2]$?

so $f$ is strictly increasing on $[0, \pi/2]$, and a strictly monotonic function is injective. ✓

Is $g(x) = \cos x$ one-one on $[0, \pi/2]$?

so $g$ is strictly decreasing, hence also injective. ✓

Statement (I) is TRUE.

Step 2 — Statement (II): is $f + g$ one-one? Write

As $x$ runs over $[0, \pi/2]$, the argument $x + \pi/4$ runs over $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$ — an interval that straddles $\pi/2$, where $\sin$ peaks. So $f+g$ rises to $\sqrt2$ at $x = \pi/4$ and then falls: it is not monotonic.

Step 3 — Explicit counterexample. …

Concept & Intuition

Injectivity means: if $f(a)=f(b)$ then $a=b$. Equivalently, no two different inputs give the same output. For polynomials and rational functions, a quick way is to check whether the function is strictly monotonic (always increasing or always decreasing) on its domain — if it is, it’s injective. If it has a turning point (a local max or min), it fails because the horizontal line through that extremum will hit twice. Absolute value functions often have a V-shape, so they can be injective if we restrict to one side of the vertex.

Step-by-step reasoning

1. Option (A): $f(x)=x^2+2$ on $\mathbb{R}$.

   This is a parabola opening upward, with vertex at $x=0$. It decreases on $(-\infty,0]$ and increases on $[0,\infty)$.

   For example, $f(-1)=3$ and $f(1)=3$ — two different inputs give the same output.

   Hence not injective.

2. Option (B): $f(x)=|x+2|$ on $[-2,\infty)$.

   The absolute value function $|x+2|$ has its vertex at $x=-2$. On the given domain $[-2,\infty)$, we are on the right half of the V-shape, where $x+2 \ge 0$, so $f(x)=x+2$.

   That’s a straight line with slope $1$ — strictly increasing.

   Therefore, if $f(a)=f(b)$ then $a+2 = b+2$ so $a=b$.

   This is injective.

3. Option (C): $f(x)=\frac{4x^2+3x-5}{4+3x-5x^2}$ on $\mathbb{R}$.

   A rational function like this can be messy, but we can test simple values.

   At $x=0$: $f(0)=\frac{-5}{4} = -1.25$.

   At $x=1$: numerator $4+3-5=2$, denominator $4+3-5=2$, so $f(1)=1$.

   At $x=-1$: numerator $4-3-5=-4$, denominator $4-3-5=-4$, so $f(-1)=1$. …

Step 1 — (a) $f(x)=x|x|$.

Both one-sided derivatives at $0$ are $0$, so $f'(x)=2|x|$ exists for every $x\in(-1,1)$. Hence $x|x|$ is differentiable in $(-1,1)$ → (ii).

Step 2 — (b) $f(x)=\sqrt{|x|}$.

For $x>0$, $f'(x)=\dfrac{1}{2\sqrt{x}}\to +\infty$ as $x\to 0^+$; for $x<0$ it $\to -\infty$. The one-sided derivatives at $x=0$ are infinite and opposite — a cusp. So it is not differentiable at at least one point in $(-1,1)$ → (iv).

Step 3 — (c) $f(x)=x+[x]$.

On $(-1,1)$, $[x]=-1$ for $-1<x<0$ and $[x]=0$ for $0\le x<1$:

$f(x)=\begin{cases}x-1,& -1<x<0\\ x,& 0\le x<1\end{cases}$ …

Step 1 — Order of composition.

So $f$ acts first. (A common slip is computing $f(g(x))$ instead.)

Step 2 — Substitute.

With $g(u)=\dfrac{u}{u^2+1}$ and $u=f(x)=3x^2-5$:

Step 3 — Expand the denominator.

Step 4 — Final form. …

Solution

Substitute $x=0$ into the matrix:

This matrix $M$ satisfies $M^{T}=-M$, i.e. it is skew-symmetric, and it is of odd order ($3\times3$). For any skew-symmetric matrix,

so $2\det(M)=0$ and hence $\det(M)=0$. Therefore $f(0)=0$. …

The key idea: the outer function $\cos^{-1}(t)$ is defined only when its argument $t$ is in $[-1, 1]$. Here the argument is $[x]$, the greatest integer less than or equal to $x$. So we need all $x$ such that $[x] \in [-1, 1]$.

1. Understand the greatest integer function $[x]$. It returns the largest integer not exceeding $x$. For example, $[1.2] = 1$, $[0.9] = 0$, $[-0.3] = -1$, $[-1] = -1$, $[-1.2] = -2$. It jumps at integer points.

2. Domain condition for $\cos^{-1}$: The inverse cosine function $\cos^{-1}(t)$ is defined only for $t \in [-1, 1]$. So we require $[x] \in \{-1, 0, 1\}$.

3. Find $x$ such that $[x] = -1$: This happens when $-1 \le x < 0$. (Check: $x = -1$ gives $[-1] = -1$, included; $x = -0.5$ gives $[-0.5] = -1$; $x = 0$ gives $[0] = 0$, not $-1$.)

4. Find $x$ such that $[x] = 0$: This happens when $0 \le x < 1$.

5. Find $x$ such that $[x] = 1$: This happens when $1 \le x < 2$. …

The key idea is that to decide where $x^x$ increases, we need its derivative. But $x^x$ is not a simple power or exponential — it’s both. The trick is to take the natural logarithm first, differentiate implicitly, and then analyse the sign.

Why logarithmic differentiation?

For a function like $f(x)=x^x$, the variable appears in both the base and the exponent. The standard differentiation rules (power rule, exponential rule) don’t apply directly. By writing $\log f(x) = x \log x$, we turn it into a product, which we can differentiate easily. Then we multiply by $f(x)$ to get $f'(x)$.

Now let’s work through it step by step.

1. Set up the function and take logs Let $f(x) = x^x$, with $x > 0$. Take natural logarithm of both sides:

2. Differentiate implicitly Differentiate both sides with respect to $x$:

Using the product rule on the right:

So we have:

3. Solve for $f'(x)$ Multiply both sides by $f(x) = x^x$:

4. Analyse the sign of $f'(x)$ For $x > 0$, $x^x$ is always positive. So the sign of $f'(x)$ is entirely determined by the factor $(1 + \log x)$. …

Step 1 — Simplify the base.

Multiply numerator and denominator by the conjugate of the denominator:

Using $i^{2} = -1$:

Step 2 — Raise to the 96th power using periodicity.

Powers of $-i$ cycle with period 4:

Since $96 = 4 \times 24$ is an exact multiple of 4:

Step 3 — Read off $a$ and $b$. …

Step 1 — Turn the sentence into an equation.

"The product of the slope of the tangent and the $x$-coordinate equals the $y$-coordinate":

Step 2 — Separate the variables.

The equation is variables-separable (all $y$'s on one side, all $x$'s on the other):

Step 3 — Integrate.

So the general solution is the family of straight lines through the origin.

Step 4 — Use the initial condition $(1, 1)$.

$1 = c(1) \;\Rightarrow\; c = 1 \qquad\Longrightarrow\qquad \boxed{y = x}$ …

Step 1 — Statement 1.

Both numerator and denominator are polynomials, hence continuous, so the limit is obtained by direct substitution provided the denominator does not vanish. At $x = 1$:

The condition $a + b + c \ne 0$ is exactly what guarantees the denominator is non-zero, so

Statement 1 is TRUE.

Step 2 — Statement 2.

As $x \to -2$, the first term tends to the finite value $\dfrac{1}{-2} = -\dfrac12$. But the second term has denominator $x + 2 \to 0$ while its numerator stays at $2$, so …