Mathematics Test - 12

Given,

The sum of the 3^rd and the 7^th term of an A.P. is 30 and the sum of the 5^th and the 9^th term is 56.

Let the first term of the A.P. be 'a' and the common difference 'd'.

nth term of an A.P.,

\(a_n = a + (n-1)d\)

\(a_3 = a + (3-1)d\)

\(\Rightarrow a_3 = a + 2d\)

\(a_7 = a + (7-1)d\)

\(\Rightarrow a_7 = a + 6d\)

\(a_5 = a + (5-1)d\)

\(\Rightarrow a_7 = a + 4d\)

\(a_9 = a + (9-1)d\)

\(\Rightarrow a_9 = a + 8d\)

Therefore, we can write

\({a}_{3}+a_{7}\) = 30

a + 2d + a + 6d = 30

2a + 8d = 30 ...(i)

And, \({a}_{5}+a_{9}\) = 56

a + 4d + a + 8d = 56

2a + 12d = 56 ...(ii)

On solving equation (i) and equation (ii), we get

4a + 20d = 86

2a + 10d = 43.....(iii)

\({a}_{4}+a_{8}\) = a + (4-1)d + a + (8-1)d

= a + 3d +a + 7d

= 2a +10d

From equation (iii), we get

The sum of the fourth and eighth terms of the series,

\({a}_{4}+a_{8}\) = 43

Given, 

\(\mathrm{X}=8^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=(1+7)^{\mathrm{n}}-7 n-1\)

\(=1+7 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}\)

\(=49\left[\frac{\mathrm{n}(\mathrm{n}-1)}{2}+\ldots+7^{\mathrm{n}-2}\right]\)

So, the set \(\mathrm{X}\) will be some specific multiples of \(49\).

\(\Rightarrow\)\(\mathrm{Y}=49(\mathrm{n}-1)\)

Therefore, the set \(Y\) will be all multiples of \(49 .\) So, it will contain the elements of \(\mathrm{X}\) too.

So, \(\mathrm{X} \subset \mathrm{Y}\)

We know that:

\(\underset{{{x \rightarrow a}}}{\lim}\left[\frac{f(x)}{g(x)}\right]=\frac{\underset{{{x \rightarrow a}}}{\lim}f(x)}{\underset{{{x \rightarrow a}}}{\lim}g(x)}\), provided \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\)

Given:

\(\lim _{x \rightarrow 0} \frac{\tan 6 x+4 x}{4 x+\tan x}\)

Dividing by \(6x\) in numerator and denominator, we get:

\(=\underset{{{{x} \rightarrow 0} }}{\lim}\frac{\frac{\tan 6 x}{6 x}+\frac{4 x}{6 x}}{\frac{4 x}{6 x}+\left(\frac{1}{6}\right) \frac{\tan x}{x}} \quad\left[\because \underset{{{{x} \rightarrow 0} }}{\lim} \frac{\tan x}{x}=1,\underset{{{{x} \rightarrow 0} }}{\lim} \frac{\tan 6 x}{6 x}=1\right]\)

\(=\underset{{{{x} \rightarrow 0} }}{\lim} \frac{1+\frac{4}{6}}{\frac{4}{6}+\frac{1}{6}}\)

\(=\frac{1+\frac{4}{6}}{\frac{4}{6}+\frac{1}{6}}\)

\(=\frac{\frac{6+4}{6}}{\frac{4+1}{6}}\)

\(=\frac{10}{5}\)

\(=2\)

Let 'A' be the event of smokers and non-vegetarian.

Let ' B ' be the event of smokers and vegetarian.

Let 'C ' be the event of non-smokers and vegetarian.

Let ' E' be the event of chest disorders.

According to question,

\(P(A)=\frac{160}{400}=\frac{2}{5}, P(B)=\frac{100}{400}=\frac{1}{4} \\\)

\( P(C)=\frac{140}{400}=\frac{7}{20} P\left(\frac{E}{A}\right)=35 \%=\frac{35}{100} \\\)

\( P\left(\frac{E}{B}\right)=\frac{20}{100}, P\left(\frac{E}{C}\right)=\frac{10}{100}\)

Thus,

\(P\left(\frac{A}{E}\right)=\frac{P(A) \cdot P\left(\frac{E}{A}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) P\left(\frac{E}{B}\right)+P(C) P\left(\frac{E}{C}\right)} \\\)

\(=\frac{\frac{2}{5} \cdot \frac{35}{100}}{\frac{2}{5} \cdot \frac{35}{100}+\frac{1}{4} \cdot \frac{20}{100}+\frac{7}{20} \cdot \frac{10}{100}} \\\)

\( =\frac{14}{100} \\\)

\( \frac{14}{100}+\frac{5}{100}+\frac{\frac{7}{100}}{14} \\\)

\( =\frac{14}{14+5+\frac{7}{2}}=\frac{28}{45}\)

\(\overrightarrow{\mathrm{PQ}}=(2 \lambda+1,3 \lambda+1,4 \lambda+1) \\\)

\( \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{n}}=0 \Rightarrow(2 \lambda+1) \cdot(2)+(3 \lambda+1)(1)+(4 \lambda+1)(-3)=0 \\\)

\( \Rightarrow-5 \lambda=0 \\\)

\( \Rightarrow \lambda=0 \\\)

\( Q=(1,2,3) \\\)

\( \frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu \\\)

\( \text { distance of line from }(1,-9,2) \\\)

\( \left(P^{\prime} Q\right) \cdot(1,1,1)=0 \\\)

\( \Rightarrow[\mu-1, \mu+10, \mu] \cdot[1,1,1]=0 \\\)

\( \Rightarrow \mu-1+\mu+10+\mu=0 \\\)

\(\mu=-3 \\\)

\( Q^{\prime}=(-3,-2,1) \\\)

\( P^{\prime} Q^{\prime}=\sqrt{16+49+9}=\sqrt{74}\)

Given:

\(\mathrm{P}(\mathrm{n})=3^{(2 \mathrm{n}+2)}-8 \mathrm{n}-9=8 \mathrm{~m}\)

Where \(\mathrm{a} \in \mathrm{N}\), i.e. a is a natural number.

For \(\mathrm{n}=1\)

\(\text { LHS }=3^{(2 \times 1+2)}-8 \times 1-9 \)

\(=3^{4}-17 \)

\(=81-17 \)

\(=64=8 \times 8\)

Therefore, \(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=1\)

Suppose, \(\mathrm{P}(\mathrm{k})\) is true.

\(3^{(2 k+2)}-8 k-9=8 a\).....(1) [where \(\mathrm{a} \in \mathrm{N}\)]

Now, we will prove that \(\mathrm{P}(\mathrm{k}+1)\) is true.

\(\text { LHS }=3^{2(\mathrm{k}+1)+2}-8(\mathrm{k}+1)-9 \)

\(=9\left[3^{(2 \mathrm{k}+2)}\right]-8 \mathrm{k}-17\)

From equation (1), we get

\(=9[8 \mathrm{a}+9+8 \mathrm{k}]-8 \mathrm{k}-17 \)

\(=72 \mathrm{a}+64 \mathrm{k}+64 \)

\(=8(9 \mathrm{a}+8 \mathrm{k}+8) \)

\(=8 \mathrm{~b}\)

Where \(\mathrm{b}=9 \mathrm{a}+8 \mathrm{k}+8\) and \(\mathrm{b}\) is natural number.

Therefore, \(\mathrm{P}(\mathrm{k}+1)\) is true whenever \(\mathrm{P}(\mathrm{k})\) is true.

Therefore, \(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}\), where \(\mathrm{n}\) is a natural number.

So, the given equation is divisible by 8.

Volume of the parallelepiped

\(V =[\hat{i}+a \hat{j}+\hat{k} \hat{j}+a \hat{k} \quad a \hat{i}+\hat{k}]\)

\(=(\hat{i}+a \hat{j}+\hat{k}) \cdot\{(\hat{j}+a \hat{k}) \times(a \hat{i}+\hat{k})\}\)

\(=(\hat{i}+a \hat{j}+\hat{k}) \cdot\left\{\hat{i}+a^{2} \hat{j}-a \hat{k}\right\}\)

\(=1+a^{3}-a\)

Now,

\(\frac{d V}{d a}=3 a^{2}-1 \Rightarrow \frac{d^{2} V}{d a^{2}}=6 a\)

\(\frac{d V}{d a}=0 \Rightarrow 3 a^{2}-1=0 \Rightarrow a=\pm \frac{1}{\sqrt{3}}\)

At \(a=\frac{1}{\sqrt{3}}\),

\(\frac{d^{2} V}{d a^{2}}=\frac{6}{\sqrt{3}}>0\)

Therefore, \(V\) is minimum at \(a=\frac{1}{\sqrt{3}}\)

Given:

\(y=3 t^{2}-4 t-3\)

Differentiating above with respect to t, we get:

\(\frac{\mathrm{dy}}{\mathrm{dt}}=6 \mathrm{t}-4\)

Similarly,

\(x=8 t+5\)

Differentiating above with respect to t, we get:

\(\frac{\mathrm{dx}}{\mathrm{dt}}=8\)

\(\frac{\mathrm{dx}}{\mathrm{dt}}=8\)

As we know:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}\)

Then,

\(\frac{d y}{d x}=\frac{6 t-4}{8}\)

At t = 6, we get:

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(6 \times 6)-4}{8}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{32}{8}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}=4\)

Given: \(y=e^{2 x}\)

\(d y / d x=2 e^{2 x}\)

Hence, dy/dx at \((0,1)\) is \(=2 \mathrm{e}^{0}=2(1)=2\)

Therefore, the equation of a tangent is \(y-1=2(x-0)\)

\(\Rightarrow 2 x-y+1=0\), if it meet the \(x\)-axis, then \(y=0\)

Hence, \(x=-1 / 2\)

Hence, the point is \((-1 / 2,0)\).

Hence, the correct option is (C).

\(\vec{\beta}=\vec{\beta}_1-\vec{\beta}_2 \ldots \ldots(1)\)

Since, \(\overrightarrow{\beta_2}\) is perpendicular to \(\vec{\alpha}.\)

\(\therefore \overrightarrow{\beta_2} \cdot \vec{\alpha}=0\)

Since, \(\overrightarrow{\beta_1}\) is parallel to \(\vec{a}.\)

then \(\overrightarrow{\beta_1}=\lambda \vec{a}\) (say)

\(\vec{a} \cdot \vec{\beta}=\vec{a} \cdot \overrightarrow{\beta_1}-\alpha \cdot \overrightarrow{\beta_2} \\\)

\(\Rightarrow 5=\lambda \alpha^2 \Rightarrow 5=\lambda \times 10(\because|\vec{\alpha}|=\sqrt{10}) \\\)

\(\Rightarrow \lambda=\frac{1}{2} \therefore \overrightarrow{\mathrm{B}_1}=\frac{\alpha}{2}\)

\(\therefore \vec{\beta}_1=\frac{\vec{\alpha}}{2}\)

\(\text { Cross product with } \overrightarrow{\mathrm{B}}_1 \text { in equation (1) }\)

\(\Rightarrow \vec{\beta} \times \vec{B}_1=-\vec{B}_2 \times \vec{B}_1\)

\(\Rightarrow \vec{\beta} \times \vec{B}_1=\vec{B}_1 \times \vec{B}_2 \Rightarrow \vec{\beta}_1 \times \vec{\beta}_2=\frac{\vec{\beta} \times \vec{a}}{2}\)
\(\Rightarrow \vec{B}_1 \times \vec{B}_2=\frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\\hat{2} & -1 & 3 \\3 & 1 & 0\end{array}\right|\)


\(=\frac{1}{2}[-3 \hat{i}-\hat{j}(-9)+\hat{k}(5)]=\frac{1}{2}[-3 \hat{i}+9 \hat{j}+5 \hat{k}]\)