🗣️ Q.1 · Indirect Speech
Suresh said, “I did it yesterday.” → choose correct reported speech
Direct speech → indirect: past tense reporting verb usually forces one tense back.
“said” (past) → tense inside must go one step further into past:
"I did" (simple past) ➔ "he had done" (past perfect)
✅ Past inside past → past perfect
“I” (Suresh) → he (never keep “I” when reporting someone else)
❌ “Suresh said that I did…” – makes no sense (Suresh talking about himself in first person?)
yesterday → the day before (never leave “yesterday” in indirect past)
Other examples: today → that day, tomorrow → the next day.
🔎 option dissection
📅 Q.2 · Day of the week
2nd June = Thursday → find day for 3rd July (same year)
Week repeats every 7 days → count total days difference, remainder = days to jump.
(days difference) mod 7 = shift from given day
Method A (two segments):
• June has 30 days. From June 2 to June 30 = 28 days (30 − 2)
• 28 mod 7 = 0 → June 30 is also Thursday.
• Then July 1 (Fri), July 2 (Sat), July 3 (Sun) ➔ +3 days → Sunday.
Method B (total difference):
• days left in June after 2nd = 28
• add first 3 days of July = 3
• Total = 31 days
• 31 ÷ 7 = 4 weeks + 3 days remainder
• Thursday + 3 = Sunday
- Counting inclusive: 2 June to 3 July = 32 days? Wrong, we count difference.
- Forgetting June has 30 days (June = 31? big mistake)
- Leap year confusion — Feb not involved, so irrelevant.
🔎 checking options
🎯 Thursday + 3 days = Sunday.
🧠 Q.3 · Analogy (word pairs)
Exacerbate : Mitigate :: ___________ ?
Exacerbate = make worse Mitigate = reduce severity
👉 They are direct antonyms (worsen ↔ reduce).
Exacerbate ≈ aggravate (both = worsen); Mitigate ≈ alleviate (both = lighten).
antonym equation original = worsen : reduce → (A) = same.
😄 Analogy is about relationship, not word‑décor.
“Aggravate” makes it grave, “alleviate” lifts the burden. Perfect antonym twins.
📊 Q.4 · Patients with cholesterol, BP, diabetes
Find number with both diabetes and high BP
|Cholesterol| = 10, |BP| = 45, only BP = 20, no cholesterol+diabetes, cholesterol ⊆ BP
- ✅ C ⊆ B (every cholesterol patient also has BP)
- ✅ C ∩ D = ∅ (no patient has both cholesterol & diabetes)
- ✅ |B| = 45, |C| = 10, |only B| = 20
- ✅ |B ∪ C ∪ D| = 75 (since C⊆B, this is just |B ∪ D| = 75)
So 20 + 10 + |B ∩ D| = 45 → |B ∩ D| = 45 – 30 = 15
✅ Already we have answer = 15.
|B ∪ D| = 75, and |B| = 45, |B ∩ D| = 15.
Union formula: 75 = 45 + |D| – 15 → |D| = 45.
That means diabetes total = 45 (15 with BP, 30 diabetes only). Consistent with all conditions ✔
And diabetes refuses to be friends with cholesterol (C∩D = ∅). Simple arithmetic then gives 15.
🧩 Q.5 · Who is the youngest? (P, Q, R, S)
Four people, different ages – statements decoded
Q: “I am neither the youngest nor the oldest.”
R: “P is older than me.”
Based on these, youngest person = ?
P's line: “I am younger than S” → P < S (P younger → age(P) less than age(S))
Q's line: “neither youngest nor oldest” → Q not rank 1, not rank 4 → Q is 2nd or 3rd.
R's line: “P is older than me” → R < P (R younger than P).
From R < P and P < S → R < P < S
So among these three, R is the youngest in that subset. Q is still floating, but Q cannot be youngest (Q said so). Therefore R must remain youngest overall.
Possible orders (Q in middle):
- Case 1: R < Q < P < S (Q 2nd, P 3rd, S oldest) ✓ Q not youngest/oldest
- Case 2: R < P < Q < S (P 2nd, Q 3rd) ✓ Q not youngest/oldest
In both, R is still the youngest. No other placement makes Q youngest (would violate Q's statement).
Q says “I’m neither youngest nor oldest” – middle child energy. So R stays youngest.
R = the undisputed baby.
🧮 Q.6 · Quotient space M₃(ℝ) / T₃(ℝ) (upper triangular)
Which vector space is it isomorphic to?
Which option is isomorphic to M₃ / T₃ ?
dim(M₃) = 3×3 = 9 (free entries).
Upper triangular: diagonal (3) + above diagonal (3) = 6 free entries.
(positions: (1,1),(2,2),(3,3) + (1,2),(1,3),(2,3))
dim(T₃) = 6
dim( M₃ / T₃ ) = dim(M₃) – dim(T₃) = 9 – 6 = 3.
So we need a space of 3×3 matrices with dimension exactly 3.
🔹 (A) Skew‑symmetric (Aᵀ = –A): diagonal = 0, above‑diagonal determines below → free = choose 3 above‑diag = dim = 3 ✅
🔹 (B) Symmetric (Aᵀ = A): diagonal (3) + above‑diag (3) = 6 ❌
🔹 (C) Lower triangular: diagonal (3) + below (3) = 6 ❌
🔹 (D) Trace zero: one linear constraint on 9 entries → dim = 9 – 1 = 8 ❌
Only skew‑symmetric has dimension 3 → matches quotient.
M₃ / T₃ kills all upper‑triangular info. What remains free? Exactly the three entries below diagonal: (2,1), (3,1), (3,2).
Those 3 degrees correspond to strictly lower‑triangular matrices (dim 3). Skew‑symmetric also has dim 3 → isomorphic as vector spaces (same finite dimension over ℝ).
✅ Isomorphism doesn’t mean “same set”, only same linear structure.
Skew‑symmetric also has exactly three free spirits. Same dimension → they're twins (isomorphic).”
📐 Q.7 · Change order of integration (dx dy → dy dx)
I = ∫₀¹∫₀^{√y} dx dy + ∫₁²∫_{√(y‑1)}¹ dx dy
From x ≤ √y → x² ≤ y. So region 1 = { (x,y): 0 ≤ x ≤ 1, x² ≤ y ≤ 1 }.
below y = 1, above parabola y = x², for x∈[0,1]
√(y‑1) ≤ x → y‑1 ≤ x² → y ≤ x²+1. And y ≥ 1, x ≤ 1, x ≥ 0.
So region 2 = { (x,y): 0 ≤ x ≤ 1, 1 ≤ y ≤ x²+1 }.
above y = 1, below parabola y = x²+1, x∈[0,1]
For a fixed x ∈ [0,1] :
• from region 1 : y ∈ [x² , 1]
• from region 2 : y ∈ [1 , x²+1]
Together: y runs from x² to x²+1 without a gap.
I = ∫₀¹ ∫_{x²}^{x²+1} dy dx
That’s exactly option (B).
🦠 Q.8 · Cantor set + Lebesgue integral
f defined on [0,1] with Cantor set C – find ∫₀¹ f(x) dx
Lebesgue integral ∫₀¹ f = ?
Lebesgue measure m(C) = 0. So any modification on C does not affect the integral.
∫_C x²⁰²⁶ dx = 0 (because set has zero length)
For integration we can ignore C:
• on [0,½] → f(x) = cos(πx) (since C has measure zero)
• on [½,1] → f(x) = sin(πx)
∫₀^{½} cos(πx) dx = [ (1/π) sin(πx) ]₀^{½} = (1/π)(sin(π/2) – 0) = 1/π
∫_{½}^{1} sin(πx) dx = [ –(1/π) cos(πx) ]_{½}^{1} = –(1/π)(cos π – cos(π/2)) = –(1/π)(–1 – 0) = 1/π
∫₀¹ f = 1/π + 1/π = 2/π
Lebesgue doesn’t care about x²⁰²⁶ on C. Only cos and sin matter → sum = 2/π.
🧩 Q.9 · Group of order 595
|G| = 595 = 5 × 7 × 17 – which statement is TRUE?
(A) G cannot have a proper normal subgroup.
(B) G must have a proper normal subgroup.
(C) G cannot have an element of order 17.
(D) The number of Sylow 5‑subgroups is 17.
|G| = 595 = 5 × 7 × 17. Prime divisors: 5, 7, 17.
17 divides |G| ⇒ ∃ an element of order 17. So option (C) is false.
n₅ satisfies:
• n₅ ≡ 1 (mod 5)
• n₅ divides 595/5 = 119 = 7×17 → divisors: 1, 7, 17, 119.
Check each mod 5: 1≡1 ✅, 7≡2 ❌, 17≡2 ❌, 119≡4 ❌ → only n₅ = 1.
∴ unique Sylow‑5 subgroup ⇒ normal & proper (order 5).
n₅=1 ⇒ exists proper normal
n₅=1 ⇒ normal subgroup of order 5
Cauchy says element order 17 exists
we found n₅ = 1, not 17
And Cauchy guarantees an element of order 17 — no escape. Option (C) dies immediately.
🔥 Q.10 · Heat equation on ℝ (Fourier method)
uₜ = k uₓₓ , u(x,0) = e^{-a|x|}, a>0 – find û(w,t)
û(w,t) = ∫_{-∞}^{∞} u(x,t) e^{iwx} dx → equals ?
ℱ[uₓₓ] = (-w²) û(w,t) (using e^{iwx} convention).
So ∂/∂t û(w,t) = k (-w²) û(w,t) → ∂û/∂t = -k w² û
û(w,t) = û(w,0) · e^{-k w² t}
Even integrand + Euler: odd part vanishes →
û(w,0) = 2 ∫₀^∞ e^{-ax} cos(wx) dx
Known formula: ∫₀^∞ e^{-ax} cos(wx) dx = a/(a²+w²) (a>0)
∴ û(w,0) = 2 · a/(a²+w²) = 2a/(a²+w²)
û(w,t) = 2a/(a²+w²) · e^{-k w² t}
This matches option (A).
missing factor 2 from evenness
exponent uses a² instead of w²t
that's the heat kernel in x‑space, not û
Initial shape e^{-a|x|} transforms to 2a/(a²+w²). Multiply = option A.
Forget the 2? You land in B. Use a² in exponent? That’s C — nonsense. Pick D? That’s the x‑space kernel, not the Fourier transform.”
Paper Solutions
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