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Calculation 

HHH → 0

HHT → 0

HTH → 1

HTT → 0

THH → 1

THT → 1

TTH → 1

TTT → 0

Probability distribution

\(\begin{array}{c|c|c} \mathrm{x}{\mathrm{i}} & 0 & 1 \ \hline \mathrm{P}\left(\mathrm{x}{\mathrm{i}}\right) & 1 / 2 & 1 / 2 \end{array}\)

\(\mu=\sum \mathrm{x}{\mathrm{i}} \mathrm{p}{\mathrm{i}}=\frac{1}{2}\)

$\sigma^{2}=\sum x_{i}^{2} p_{i}-\mu^{2}$

= $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$64\left(\mu+\sigma^{2}\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48$

Hence option 2 is correct

Calculation

a₁ .a₅ = 28 ⇒ a.ar⁴ = 28 ⇒ a² r⁴ = 28 …(1)

a₂ + a₄ = 29 ⇒ ar + ar³ = 29 

⇒ ar(1 + r²) = 29 

a² r² (1 + r²)² = (29)² …(2) 

By Eq. (1) & (2)

$\frac{\mathrm{r}^{2}}{\left(1+\mathrm{r}^{2}\right)^{2}}=\frac{28}{29 \times 29}$

⇒ $\frac{\mathrm{r}}{1+\mathrm{r}^{2}}=\frac{\sqrt{28}}{29} \Rightarrow \mathrm{r}=\sqrt{28}$

∵ a² r⁴ = 28 ⇒ a² × (28)² = 28 

⇒ $a=\frac{1}{\sqrt{28}}$

∴ $a_{6}=a r^{5}=\frac{1}{\sqrt{28}} \times(28)^{2} \sqrt{28}=784$

Hence option 3 is correct

Calculation

P(2λ + 1, 3λ + 2, 4λ + 3) on L₁

Q(3µ + 2, 4µ + 4, 5µ + 5) on L₂

Dr’s of PQ = 3µ – 2λ + 1, 4µ – 3λ + 2, 5µ – 4λ + 2

PQ ⊥ L₁ 

⇒ (3µ – 2λ + 1)2 + (4µ – 3λ + 2)3 + (5µ – 4λ + 2)4 = 0

38µ – 29λ + 16 = 0 …(1)

PQ ⊥ L₂ 

⇒ (3µ – 2λ + 1)3 + (4µ – 3λ + 2)4 + (5µ – 4λ + 2)5 = 0

50µ – 38λ + 21 = 0 …(2)

By (1) & (2) 

$λ=\frac{1}{3} ; \mu=\frac{-1}{6}$

∴ \mathrm{P}\left(\frac{5}{3}, 3, \frac{13}{3}\right) & \ \mathrm{Q}\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)

Line PQ 

$\begin{array}{ccc} \frac{x-\frac{5}{3}}{\frac{1}{6}} & \frac{y-3}{\frac{-1}{3}} & \frac{z-\frac{13}{3}}{\frac{1}{6}} \end{array}$

$\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1}$

$\text { Point }\left(\frac{14}{3},-3, \frac{22}{3}\right)$

lies on the line PQ 

Hence option 4 is correct

Calculation

$\rm \mathrm{e}^{5(ℓ n x)^{2}+3}=\mathrm{x}^{8}$

⇒ $\ln \mathrm{e}^{5(ℓ \mathrm{n} x)^{2}+3}=\ln \mathrm{x}^{8}$

⇒ 5(ℓnx)² + 3 = 8ℓnx

(ℓnx = t)

⇒ 5t² – 8t + 3 = 0

\(\mathrm{t}{1}+\mathrm{t}{2}=\frac{8}{5}\)

\(\ell \operatorname{nx}{1} x{2}=\frac{8}{5}\)

x₁ x₂ = e^8/5

Hence option 1 is correct

Calculation

T_n = S_n – S_n–1

⇒ $\mathrm{T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)$

⇒ $\frac{1}{T_{n}}=\frac{8}{(2 n-1)(2 n+1)(2 n+3)}$

\(\lim {n \rightarrow \infty} \sum{r=1}^{n} \frac{1}{T_{r}}=\lim {n \rightarrow \infty} 8 \sum{r=1}^{n} \frac{1}{(2 n-1)(2 n+1)(2 n+3)}\)

= $\lim _{n \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n+1)(2 n+3)}\right)$

= $\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right]$

= $\frac{2}{3}$

Hence option 3 is correct

Calculation 

= \(\underbrace{12 \mathrm{C}{2}}{\substack{\text { Sclection of two } \ \text { letters before } \mathrm{M}}} \times \underbrace{{ }^{13} \mathrm{C}{2}}{\substack{\text { Selection of two } \ \text { letters after } \mathrm{M}}}=5148\)

Hence option 4 is correct

Calculation

$\frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{1}{\mathrm{y}^{2}}\right) \mathrm{x}=\frac{1}{\mathrm{y}^{3}}$

$\text { I.F. }=\mathrm{e}^{\int \frac{1}{{y}^{2}} \mathrm{dy}}=\mathrm{e}^{-\frac{1}{y}}$

⇒ $x \cdot e^{-\frac{1}{y}}=\int\left(e^{-\frac{1}{y}}\right) \cdot \frac{1}{y^{3}} d y$

Put $-\frac{1}{y}=t$

$+\frac{1}{\mathrm{y}^{2}} \mathrm{dy}=\mathrm{dt}$

$x \cdot e^{-\frac{1}{y}}=-\int t \cdot e^{t} d t$

$x \cdot e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C$

$x \cdot e^{-\frac{1}{y}}=\frac{+1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C$

x = 1, y = 1 

$\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{e}}+\frac{1}{\mathrm{e}}+\mathrm{C}$

⇒ $\mathrm{C}=-\frac{1}{\mathrm{e}}$

Put y = $\frac{1}{2}$

$\frac{\mathrm{x}}{\mathrm{e}^{2}}=\frac{2}{\mathrm{e}^{2}}+\frac{1}{\mathrm{e}^{2}}-\frac{1}{\mathrm{e}}$

x = 3 – e

Hence option 3 is correct

Calculation

y = x² + px – 3

Let P(α, 0), Q(β, 0), R(0, –3)

Circle with centre (–1, –1) is (x + 1)² + (y + 1)² = r²

Passes through (0, –3)

1² + (–2)² = r²]

r² = 5

(x + 1)² + (y + 1)² = 5

Put y = 0

(x + 1)² = 5 – 1

(x + 1)² = 4

x + 1 = ±2

x = 1 or x = –3 

P(1, 0) and Q(–3, 0)

Area of ΔPQR = $\frac{1}{2}\left|\begin{array}{ccc} 1 & 0 & 1 \ -3 & 0 & 1 \ 0 & -3 & 1 \end{array}\right|=6$

Hence option 2 is correct

Calculation. 

S₁ : (x + 2)² + (y – 2)² = 2²

S₂ : (x – 2)² + (y – 5)² = r²

Both circle intersect at two points 

∴ |r₁ – r₂| < c₁ c₂ < r₁ + r₂

|r – 2| < 5 < 2 + r 

⇒ 3 < r < 7

r ∈ (3, 7)

α = 3, β = 7

3β – 2α = 15

Hence option 1 is correct

Calculation 

f(x) = (7tan⁶ x – 3tan² x)(sec² x)

$I_{1}=\int_{0}^{\pi / 4}\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right) d x$

Put tanx = t

$I_{1}=\int_{0}^{1}\left(7 t^{6}-3 t^{2}\right) d t=\left[t^{7}-t^{3}\right]_{0}^{1}=0$

$I_{2}=\int_{0}^{\pi / 4} x \underbrace{\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right)}_{\mathrm{II}} d x$

= \(\left[x\left(\tan ^{7} x-\tan ^{3} x\right)\right]{0}^{\pi / 4}-\int{0}^{\pi / 4}\left(\tan ^{7} x-\tan ^{3} x\right) d x\)

= $0-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{2} x-1\right)\left(1+\tan ^{2} x\right) d x$

Put tanx = t

= $-\int_{0}^{1}\left(t^{5}-t^{3}\right) d t=-\left[\frac{t^{6}}{6}-\frac{t^{4}}{4}\right]=\frac{1}{12}$

7I₁ + 12I₂ = 1 

Hence option 3 is correct