Q.Verify that the function $y = c_1 e^{ax}\cos bx + c_2 e^{ax}\sin bx$, where $c_1, c_2$ are arbitrary constants is a solution of the differential equation $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$.

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

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

Rearranging:

This is exactly option (C).

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.

The bracket is exactly the original $y$ — that is the whole point of the question: $y$ is a solution of the Euler–Cauchy equation $x^{2}y'' = n(n+1)y$.

Quick check with $n = 1$: $y = 2x^{2} + 3x^{-1}$, $y'' = 4 + 6x^{-3}$, so $x^{2}y'' = 4x^{2} + 6x^{-1} = 2\left(2x^{2}+3x^{-1}\right) = 2y = n(n+1)y$ since $n(n+1)=2$. ✓

ANSWER: B

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 4 — Compare imaginary parts.

Substituting $x=2$:

Step 5 — Check.

$3(2)+i\big(4(2)-9\big)=6+i(8-9)=6-i$ ✓ — exactly the right-hand side.

ANSWER: B

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

(Geometrically: $i = e^{i\pi/2}$ is a quarter-turn about the origin. Four quarter-turns make a full revolution and land back at $1$ — so the exponent must be a multiple of 4.)

Step 5 — Test the options against the cycle.

• (A) $x=4n+1$ $\Rightarrow i^{4n+1} = i^{4n}\cdot i = 1\cdot i = i \ne 1$ ✗
• (B) $x=2n+1$ (odd) $\Rightarrow i^{\text{odd}} = \pm i \ne 1$ ✗
• (C) $x=2n$ $\Rightarrow$ e.g. $n=1$ gives $i^2 = -1 \ne 1$ ✗ (works only when $n$ is itself even — not for all $n$)
• (D) $x=4n$ $\Rightarrow i^{4n} = (i^4)^n = 1^n = 1$ ✓ for every $n$

ANSWER: D

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.

Step 4 — Divide.

Step 5 — Rationalise.

ANSWER: C

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

Take $p=1,q=2,r=3$ (an A.P.):

Common difference: $\tfrac83-\tfrac56 = \tfrac{11}{6}$ and $\tfrac92-\tfrac83 = \tfrac{11}{6}$. ✓ A.P.

Now test a G.P., $p=1,q=2,r=4$: $T_1=\tfrac34,\ T_2=\tfrac52,\ T_3=6$; differences $1.75$ and $3.5$ — not an A.P. So a G.P. does not satisfy the hypothesis, ruling out (A).

ANSWER: B

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 $\{2,3\}$.

1. It is not a two-element set. No domain restriction to integers is given, so $x$ ranges over the reals. $x=2.5$ satisfies both ($3(2.5)+8=15.5<17$ ✓ and $2(2.5)+8=13\ge12$ ✓) yet is not in $\{2,3\}$. The solution is an uncountable interval.
2. It wrongly includes 3. At $x=3$: $3(3)+8=17$, and $17<17$ is false. So $3$ is excluded — the inequality is strict.

Hence Statement (II) is FALSE.

Step 5 — Combine.

(I) true, (II) false ⇒ option (A).

ANSWER: A

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):

Step 4 — Take the required root

The condition $k>0$ selects

Step 5 — Verify

With $k=2\sqrt6$: $\bar x = \dfrac{2\sqrt6}{4}=\dfrac{\sqrt6}{2}$, $\sum x_i^2 = 2+24=26$.

ANSWER: C

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.

Step 3 — Reject the impossible root.

A number of persons cannot be negative, so $n = -9$ is rejected. Hence $n = 10$.

Step 4 — Verify.

(Checking (A): ${}^{9}C_{2} = 36 \neq 45$; (C): ${}^{5}C_{2} = 10$; (D): ${}^{15}C_{2} = 105$.)

ANSWER: B