Q.By using the properties of definite integrals, evaluate the integral $\int_{-\pi/2}^{\pi/2}\sin^2 x\,dx$

Concept and Intuition

Parity arguments let us evaluate integrals over symmetric intervals without doing any actual integration — an odd integrand always gives $0$ on $[-\pi,\pi]$.

Step-by-Step Solution

1. $a=2n$ is even, so $\sin^ax=(\sin x)^{2n}$; since $\sin(-x)=-\sin x$, $(\sin(-x))^{2n}=(\sin x)^{2n}$ — even function.
2. Hence $e^{\sin^ax}$ is also even (composition of an even function with $e^{(\cdot)}$ preserves evenness).
3. $b=2m+1$ is odd. $\cot((2n+1)x)$ satisfies $\cot((2n+1)(-x))=-\cot((2n+1)x)$ — an odd function of $x$.
4. Raising an odd function to an odd power keeps it odd: $\cot^b((2n+1)x)$ is odd in $x$. …

We begin by noticing that the expression inside the sine, $\log(x + \sqrt{x^2+1})$, is a well-known inverse hyperbolic function. This is the key to deciding parity.

Concept and intuition:

A function $g(x)$ is odd if $g(-x) = -g(x)$ for all $x$ in its domain. If we can show the inner function is odd, and the outer function ($\sin$) is also odd, then their composition is odd. That’s the path here.

1. Identify the inner function.

   Let $h(x) = \log\left(x + \sqrt{x^2+1}\right)$.

   Notice that $x + \sqrt{x^2+1} > 0$ for all real $x$, so the domain is all real numbers — no issues.

2. Check if $h(x)$ is odd.

   Compute $h(-x)$:

Multiply numerator and denominator by the conjugate:

Therefore,

So $h(x)$ is odd.

3. Now consider $f(x) = \sin(h(x))$. Since $\sin$ is an odd function ($\sin(-u) = -\sin u$), and $h$ is odd, we have: $$$$

Concept and Intuition

$f$ is odd if $f(-x)=-f(x)$ for all $x$ in the domain, and even if $f(-x)=f(x)$. A useful trick for expressions like $\log(x+\sqrt{x^2+1})$ is to rationalise: multiplying $x+\sqrt{x^2+1}$ by $-x+\sqrt{x^2+1}$ gives $(x^2+1)-x^2=1$, so the two factors are reciprocals of each other — this instantly reveals the odd symmetry via $\log(1/u)=-\log u$.

Step-by-Step Solution

I. $f(x)=x\left(\dfrac{e^x-1}{e^x+1}\right)$.

$f(-x) = -x\left(\dfrac{e^{-x}-1}{e^{-x}+1}\right)$. Multiply numerator and denominator by $e^x$: $\dfrac{e^{-x}-1}{e^{-x}+1}=\dfrac{1-e^x}{1+e^x}=-\dfrac{e^x-1}{e^x+1}$.

So $f(-x) = -x\cdot\left(-\dfrac{e^x-1}{e^x+1}\right) = x\cdot\dfrac{e^x-1}{e^x+1} = f(x)$. Even, not odd.

II. $f(x)=k^x+k^{-x}+\cos x$.

$f(-x) = k^{-x}+k^{x}+\cos(-x) = k^x+k^{-x}+\cos x = f(x)$. Even, not odd.

III. $f(x)=\log\left(x+\sqrt{x^2+1}\right)$.

$f(-x) = \log\left(-x+\sqrt{x^2+1}\right)$. Since $\left(x+\sqrt{x^2+1}\right)\left(-x+\sqrt{x^2+1}\right) = (x^2+1)-x^2 = 1$, we get $-x+\sqrt{x^2+1} = \dfrac{1}{x+\sqrt{x^2+1}}$. …

Concept and Intuition

To classify a function as odd/even, the standard method is to directly compute $f(-x)$ and algebraically manipulate it until it can be compared to $f(x)$ (or $-f(x)$). The tricky part with $\frac{x}{e^x - 1}$-type expressions is that they aren't obviously symmetric-looking, but a substitution trick (multiplying by $e^x/e^x$) reveals hidden structure — this exact function ($\frac{x}{e^x-1}+\frac{x}{2}$) is a famous even function related to $\frac{x}{2}\coth(x/2)$.

Step-by-Step Solution

1. Start with $f(x) = \dfrac{x}{e^x-1} + \dfrac{x}{2} + 1$.
2. Substitute $-x$: $f(-x) = \dfrac{-x}{e^{-x}-1} + \dfrac{-x}{2} + 1 = \dfrac{-x}{e^{-x}-1} - \dfrac{x}{2} + 1$.
3. Simplify $\dfrac{-x}{e^{-x}-1}$: multiply numerator and denominator by $e^x$: $\dfrac{-x \cdot e^x}{e^x(e^{-x}-1)} = \dfrac{-xe^x}{1 - e^x} = \dfrac{xe^x}{e^x - 1}$.
4. Rewrite $xe^x = x(e^x - 1) + x$, so $\dfrac{xe^x}{e^x-1} = \dfrac{x(e^x-1)}{e^x-1} + \dfrac{x}{e^x-1} = x + \dfrac{x}{e^x-1}$. …

Concept and Intuition

"Even" and "odd" are symmetry properties, not value constraints at a single point, except when the symmetry condition itself pins down that value. For an odd function the relation $f(-x)=-f(x)$ evaluated at $x=0$ directly forces $f(0)=0$ (it's the only value equal to its own negative, for real numbers). For an even function, the relation $f(-x)=f(x)$ evaluated at $x=0$ gives $f(0)=f(0)$, which carries no information at all — so $f(0)$ can be anything.

Step-by-Step Solution

1. Statement (A): "$f$ even $\Rightarrow f(0)=0$." Test with $f(x)=x^2+1$: this is even since $f(-x)=(-x)^2+1=x^2+1=f(x)$, but $f(0)=1\ne 0$. So statement (A) is false.
2. Statement (B): $f(x)=x-[x]$ is the fractional-part function. For any $x$, $f(x+1) = (x+1)-[x+1] = x+1-([x]+1) = x-[x]=f(x)$, so it repeats every 1 unit — it is genuinely periodic. Statement (B) is true.
3. Statement (C): if $f$ is odd, $f(-x)=-f(x)$ for all $x$, including $x=0$: $f(0)=-f(0) \Rightarrow 2f(0)=0 \Rightarrow f(0)=0$. This is a valid, always-true deduction. Statement (C) is true. …

Concept and Intuition

Functions of the form $\log(A(x) + B(x))$ where $B(x)^2 - A(x)^2 = 1$ (a constant) and $A$ is odd, $B$ is even, are always odd — because then $A+B$ and $B-A$ are reciprocals of each other, so $\log(B-A) = -\log(A+B)$.

Step-by-Step Solution

1. Let $A(x) = \dfrac{2x^2-3}{x}$ (this is odd: replacing $x\to-x$ flips its sign) and $B(x) = \sqrt{\dfrac{4x^4-11x^2+9}{x^2}}$ (this is even: depends only on $x^2$).
2. Compute $B^2 - A^2 = \dfrac{4x^4-11x^2+9}{x^2} - \dfrac{(2x^2-3)^2}{x^2} = \dfrac{(4x^4-11x^2+9)-(4x^4-12x^2+9)}{x^2} = \dfrac{x^2}{x^2}=1$.
3. So $B^2-A^2=1 \Rightarrow (B-A)(B+A)=1 \Rightarrow B-A = \dfrac{1}{A+B}$. …