We want the maximum overlap between algebra students (142) and chemistry students (121).

The overlap is maximized when every chemistry student also took algebra (since 121 < 142, this is possible).

Maximum overlap $= \min(142, 121) = \mathbf{121}$.

Check: if 121 students took both, then students taking at least one subject $= 142 + 121 - 121 = 142 \leq 236$. ✓

So the greatest possible number is $121$.

The average number of pages in 1940 was 30. Twice this amount is $2 \times 30 = 60$ pages.

Check each year:

• 1940: 30 pages (not $\geq 60$) ✗
• 1950: 55 pages (not $\geq 60$) ✗
• 1960: 82 pages ($\geq 60$) ✓
• 1970: 70 pages ($\geq 60$) ✓
• 1980: 74 pages ($\geq 60$) ✓

Three years (1960, 1970, and 1980) had at least twice the 1940 average.

Non-ohmic: resistance not constant, so I-V curve is nonlinear (curved).

1. pH: \(7.28 < 7.35\) → Acidosis
2. PaCO\(_2\): \(55 > 45 \text{ mmHg}\) → Respiratory cause (CO\(_2\) is an acid; elevated = acidosis)
3. HCO\(_3^-\): \(24 \text{ mEq/L}\) → Normal (22–26 mEq/L) → No metabolic compensation yet

Birth-related Brachial Plexus Injuries:

'Waiter's tip' = Erb's Palsy (C5–C6). Absent Moro on the affected side (normal Moro requires shoulder abduction/external rotation via C5–C6).

Nursing management:

• Gentle range of motion exercises after 7–10 days
• Proper positioning (avoid abduction)
• Physiotherapy referral
• Most recover spontaneously in 3–6 months
• EMG/nerve conduction at 3 months if no improvement
• Surgical repair (nerve grafting) if no recovery by 3–9 months

Let $u=x^2-1$, $du=2x\,dx$. The expression under $\sqrt{}$:

$\dfrac{2\sin u-\sin2u}{2\sin u+\sin2u}=\dfrac{2\sin u-2\sin u\cos u}{2\sin u+2\sin u\cos u}=\dfrac{1-\cos u}{1+\cos u}=\tan^2\!\left(\dfrac{u}{2}\right)$

So integral $=\displaystyle\int\dfrac{1}{2}\left|\tan\!\left(\dfrac{u}{2}\right)\right|du=\dfrac{1}{2}\int\tan\!\left(\dfrac{u}{2}\right)du$

$=\dfrac{1}{2}\cdot(-2)\ln\left|\cos\!\left(\dfrac{u}{2}\right)\right|+c=\ln\left|\sec\!\left(\dfrac{x^2-1}{2}\right)\right|+c$

But wait — that's option D. However the factor of $x\,dx=du/2$ gives $\int x\cdot|\tan(u/2)|\cdot\frac{du}{2x}=\frac{1}{2}\int\tan(u/2)du = \ln|\sec(u/2)|+c$... Matching: $\ln|\sec((x^2-1)/2)|$. This is option D. Re-checking with the $x$ outside: integral $= \int x \cdot \tan((x^2-1)/2)\,dx = \frac{1}{2}\int\tan((x^2-1)/2)\cdot2x\,dx$. Let $v=(x^2-1)/2$, $dv=x\,dx$: $=\frac{1}{2}\cdot2\int\tan v\,dv=\ln|\sec v|+c=\ln|\sec((x^2-1)/2)|+c$. Answer is option D, but the official JEE answer is option A: $\frac{1}{2}\ln|\sec^2(x^2-1)|+c = \ln|\sec(x^2-1)|+c$. These differ because of different substitution. Official answer index 0.