Mathematics Test - 14

Given that:

\(\underset{{{x \rightarrow 0}}}{\lim}\frac{\tan ^{2} 3 x}{x^{2}}\)

Dividing by \(9\) in both numerator and denominator, we get:

\(=\underset{{{x \rightarrow 0}}}{\lim}\frac{\tan ^{2} 3 x}{x^{2}} \times \frac{9}{9}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{9 \tan ^{2} 3 x}{(3 x)^{2}}\)

\(=9 \times \underset{{{x \rightarrow 0}}}{\lim} \frac{\tan 3 x}{3 x} \times \underset{{{x \rightarrow 0}}}{\lim} \frac{\tan 3 x}{3 x}\)

We know that:

\(\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\tan {x}}{{x}}=1\)

Therefore,

\(=9 \times 1 \times 1\)

\(=9\)

Therefore, the value of \(\underset{{{x \rightarrow 0}}}{\lim} \frac{\tan ^{2} 3 x}{x^{2}}\) is\(9\).

\(\omega^{6}+\omega^{7}+\omega^{5}\)

\(=\omega^{5}\left(\omega+\omega^{2}+1\right)\)

\(=\omega^{5} \times\left(1+\omega+\omega^{2}\right)\)

\(=\omega^{5} \times 0\)

\(=0\)

Let two lines \(\ell_{1}, \ell_{2} \in \mathrm{L}\)

Now \(\left(\ell_{1}, \ell_{2}\right) \in R\)

Only when \(\ell_{1} \perp \ell_{2}\)

1. For reflexivity:

Let \(\ell_{1} \in \mathrm{L}\)

But \(\left(\ell_{1}, \ell_{1}\right) \notin R\)

Since, no line is perpendicular to itself.

Therefore, \(\mathrm{R}\) is not reflexive.

2. For symmetric:

Let, \(\left(\ell_{1}, \ell_{2}\right) \in \mathrm{L}\)

Now, if \(\ell_{1} \perp \ell_{2}\), then this implies that \(\ell_{2}\) is also perpendicular to \(\ell_{1}\).

So, \(\left(\ell_{1}, \ell_{2}\right) \in L\)

Therefore, \(\mathrm{R}\) is symmetric.

3. For Transitive:

Let \(\left(\ell_{1}, \ell_{2}\right) \in L\) and \(\left(\ell_{2}, \ell_{3}\right) \in L\)

\(\ell_{1} \perp \ell_{2}\) and \(\ell_{2} \perp \ell_{3}\)

\(\ell_{1}\) is not perpendicular to \(\ell_{3}\).

\(\left(\ell_{1}, \ell_{3}\right) \notin \mathrm{R}\)

So, R is not transitive.

\(\mathrm{S}=\{1,2,3, \ldots \ldots . .2022\} \\\)

\( \operatorname{HCF}(\mathrm{n}, 2022)=1\)

\(\Rightarrow \mathrm{n}\) and 2022 have no common factor

Total elements \(=2022\)

\(2022=2 \times 3 \times 337\)

M : numbers divisible by 2 .

\(\{2,4,6, \ldots \ldots ., 2022\} n(M)=1011\)

\(\mathrm{N}\) : numbers divisible by 3 .

\(\{3,6,9, \ldots \ldots . .2022\} n(N)=674\)

\(\mathrm{L}:\) numbers divisible by 6 .

\( \{6,12,18, \ldots \ldots, 2022\} n(L)=337 \\\)

\( n(M \cup N)=n(M)+n(N)-n(L) \\\)

\( =1011+674-337 \\\)

\( =1348\)

\(0=\) Number divisible by 337 but not in M UN

\(\{337,1685\}\)

Number divisible by 2,3 or 337

\( =1348+2=1350 \\\)

\( \text { Required probability }=\frac{2022-1350}{2022} \\\)

\( =\frac{672}{2022} \\\)

\( =\frac{112}{337}\)

Given, \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\)

\(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=\mathrm{k}\) and \(\mathrm{f}(0)=\mathrm{k}\)

\(\therefore \lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow 0} \frac{\log (1+\operatorname{ax})-\log (1-\mathrm{bx})}{\mathrm{x}}\left(\frac{0}{0}\right.\) form \()\)

\(=\lim _{x \rightarrow 0}\left(\frac{1}{1+\operatorname{ax}} a+\frac{b}{1-b x}\right)\)

\(=\mathrm{a}+\mathrm{b}\)

\(\therefore \quad \mathrm{a}+\mathrm{b}=\mathrm{f}(0)=\mathrm{k}\)

Here,

\(x^2=y\) and line \(y=1\) cut the parabola

\(\therefore x^2=1\)

\(\Rightarrow x=1\) and -1

Area \(=\int_{-1}^1 y d x\)

Here, the area is symmetric about the \(y\)-axis, we can find the area on one side and then multiply it by 2, we will get the area,

Area \(_1=\int_0^1 y d x\)

Area \(_1=\int_0^1 x^2 d x\)

\(=\left[\frac{ x ^3}{3}\right]_0^1=\frac{1}{3}\)

This area is between \(y=x^2\) and the positive \(x\)-axis.

To get the area of the shaded region, we have to subtract this area from the area of square i.e. \((1 \times 1)-\frac{1}{3}=\frac{2}{3}\)

Total Area \(=2 \times \frac{2}{3}=\frac{4}{3}\) square units

\(\frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-2 \mathrm{ax}+48>0\)

\((\because \mathrm{y}\) is an increasing function)

\(\therefore\) Discriminant, \(\mathrm{D} \leq 0\)

\(\Rightarrow 4 \mathrm{a}^{2}-4 \times 3 \times 48<0\)

\(\Rightarrow \mathrm{a}^{2}-144<0\)

\(\Rightarrow \mathrm{a} \epsilon(-12,12)\)

The number of variables in dual is equal to the number of constraints in the primal and the number of variables in primal is equal to the number of constraints in the dual.

Therefore, the primal and dual doesn't have equal number of variables.

Given,

\(\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

As we know \(\sin ^{-1}(-x)=-\sin ^{-1} x\)

So, \(\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)

Let \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\theta\)

\(\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\)

\(\therefore \theta=-45^{\circ}\)

We know that:

L-Hospital Rule as:

\(\underset{{{x \rightarrow c}}}{\lim} \frac{f(x)}{g(x)}=\underset{{{x \rightarrow c}}}{\lim} \frac{f^{\prime}(x)}{g(x)}\)

Given that:

\(\lim _{x \rightarrow 0} \frac{\cos 4 x-1}{1-\cos x}\)

Let,

\(L=\underset{{{x \rightarrow 0}}}{\lim}\frac{\cos 4 x-1}{1-\cos x}\)....(1)

On putting the limits in above, we get:

\(L=\frac{0}{0}\)

Applying L-Hospital Rule on equation (1), we get:

\(L=\underset{{{x \rightarrow 0}}}{\lim}\frac{\cos 4 x-1}{1-\cos x} =\underset{{{x \rightarrow 0}}}{\lim}\frac{\frac{d}{dx}(\cos 4 x-1)}{\frac{d}{dx}(1-\cos x)}\)

\({L}=\underset{{{{x} \rightarrow 0}}}{\lim} \frac{-4 \sin 4 {x}}{\sin {x}}\)...(2)

\(L=\frac{0}{0}\)

Again applying L-Hospital Rule on equation (2), we get:

\({L}=\underset{{{{x} \rightarrow 0}}}{\lim}\frac{-16 \cos 4 {x}}{\cos {x}}\)

\(L=\frac{-16}{1}\)

\(L=-16\)