Q.(ix) The solution of the differential equation $\frac{dy}{dx}=\frac{x+2y}{x}$ is $x+y=kx^2$. (State True or False.)

Differentiate the relation $x+y=\tan^{-1}y$ with respect to $x$:

Multiply both sides by $(1+y^2)$:

$1+y^2+y^2y'=0$ …

Step 1 — Write $y$ with negative exponents (so the power rule applies to both terms).

Step 2 — First derivative (power rule $\frac{d}{dx}x^{m} = m x^{m-1}$):

Step 3 — Second derivative:

Step 4 — Multiply by $x^{2}$ and factor.

$x^{2}\frac{d^{2}y}{dx^{2}} = 2n(n+1)x^{n+1} + 3n(n+1)x^{-n} = n(n+1)\Big[\,2x^{n+1} + 3x^{-n}\,\Big] = n(n+1)\,y.$ …

Step 1 — The concept: equality of complex numbers.

If $a+ib=c+id$ with $a,b,c,d\in\mathbb{R}$, then $a=c$ and $b=d$. This works because $\{1,i\}$ is a basis of $\mathbb{C}$ over $\mathbb{R}$: a real number can never equal a non-zero purely imaginary number, so the two components cannot compensate for one another.

Step 2 — Write both sides in the standard $a+ib$ form.

Here $x,y$ are real, so $3x$ is the real part on the left and $(4x-y)$ is the imaginary part.

Step 3 — Compare real parts. …

Step 1 — Solve the first inequality.

Step 2 — Solve the second inequality.

So Statement (I) — "the solution sets are $x<3$ and $x\ge2$ respectively" — is TRUE.

Step 3 — Find the common solution set.

Intersecting $x<3$ with $x\ge2$:

Step 4 — Why Statement (II) fails — two independent reasons.

Statement (II) claims the common set is $\{2,3\}$. …

Step 1 — Simplify the base by rationalising the denominator.

The standard move for a complex fraction is to multiply top and bottom by the conjugate of the denominator. The conjugate of $1-i$ is $1+i$:

Step 2 — Expand numerator and denominator.

Numerator (using $i^2=-1$):

Denominator (difference of squares — this is why we use the conjugate: it clears $i$ from the bottom):

Therefore

Step 3 — Rewrite the equation.

Step 4 — The concept: powers of $i$ are periodic with period 4.

and then the cycle repeats: $i^5=i$, $i^6=-1$, and so on. In general $i^{x}$ depends only on $x \bmod 4$, and

$i^{x} = 1 \iff x \equiv 0 \pmod 4 \iff x = 4n.$ …

Step 1 — The three terms

The terms are

Step 2 — Add a constant (A.P. is preserved)

If $T_1,T_2,T_3$ are in A.P., so are $T_1+2,\ T_2+2,\ T_3+2$ (adding the same constant to every term does not change the common difference).

Write $2 = 1 + 1$ and absorb one of the $1$s as $\frac{p}{p}$:

By the same symmetry,

Step 3 — Strip the constants

Subtract $1$ from each, then divide each by the common non-zero factor $k=\left(\frac1p+\frac1q+\frac1r\right)$. Both operations preserve an A.P. We are left with

in A.P.

So the given terms are in A.P. iff $p,q,r$ are in A.P.

Step 4 — Numerical verification …

Step 1 — Why this works.

The modulus is multiplicative ($|z_1z_2|=|z_1||z_2|$, $|z_1/z_2|=|z_1|/|z_2|$), so we never need to expand the complex arithmetic.

Step 2 — Numerator moduli.

Step 3 — Denominator moduli.

$\text{Denominator}=2\sqrt{10}\cdot 2\sqrt2=4\sqrt{20}=8\sqrt5$ …

Step 1 — Set up the data

The observations are $-1,\;0,\;1,\;k$, so $n=4$.

Step 2 — Use the computational formula for variance

The formula $\sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}$ (mean of squares minus square of the mean) is the efficient route here, because $\bar x$ is not a whole number and the deviation form would be messy.

Step 3 — Impose the given standard deviation

Given $\sigma=\sqrt5$, so $\sigma^{2}=5$:

Multiply throughout by $16$ (the LCM of the denominators):

$8+4k^{2}-k^{2}=80$ …

Step 1 — Model the situation.

A handshake is completely determined by which two people shake — the order does not matter (A shaking B is the same handshake as B shaking A), and a person cannot shake their own hand. So the count of handshakes among $n$ people is a combination, not a permutation:

Step 2 — Form and solve the equation.

$\Rightarrow\; (n-10)(n+9) = 0 \;\Rightarrow\; n = 10 \ \text{or}\ n = -9$ …