Mathematics Test - 67

Given,

\(\frac{-1}{(|x|-2)}\geq 1\)

⇒\(\frac{-1}{(|x|-2)} \geq 0\)\)

⇒\({\{-1-(|x|-2)\}}{(|x|-2)} \geq 0\)

⇒\({\{1-|x|\}}{(|x|-2)} \geq 0\)

⇒\(\frac{-(|x|-1)}{(|x|-2)} \geq 0\)

Using number line rule:

1 ≤ |x| < 2

⇒ x ∈ (-2, -1) ∪ (1, 2)

Let, \(\overrightarrow{\mathrm{OA}}=-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{OD}}=-\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\hat{i}+\frac{1}{2} \hat{\jmath}+4 \hat{k}\right)-\left(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right)\)

\(=2 \hat{i}\)

\(|\overrightarrow{A B}|=\sqrt{2^{2}}=2\)

\(\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=\left(\hat{i}-\frac{1}{2} \hat{\jmath}+4 \hat{k}\right)-\left(-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\right)\)

\(=2 \hat{i}\)

\(|\overrightarrow{B C}|=\sqrt{2^{2}}=2\)

Area of rectangle, \(\mathrm{ABCD}=|\mathrm{AB}| \times|\mathrm{BC}|=2 \times 2\)

\(=4\) square units

Given,

\(\int_{2}^{3} \frac{\cos x-\sin x}{4} dx\)

\(=\frac{1}{4}\int_{2}^{3} (\cos x- \sin x)~dx\)

\(=\frac{1}{4}[\sin x -(-\cos x )]^{3}_{2}\)

\(=\frac{1}{4}[\sin x+\cos x]^{3}_{2}\)

Put the value of limit,

\(=\frac{1}{4}(\sin 3+\cos 3)-\frac{1}{4}(\sin 2+\cos 2)\)

\(=\frac{1}{4}(\sin 3+\cos 3-\sin 2-\cos 2)\)

Given,

\(2 \sin ^{2} \mathrm{~A}+3 \cos ^{2} \mathrm{~A}=2\)

\(\Rightarrow 2 \sin ^{2} A+3\left(1-\sin ^{2} A\right)=2\)\(\quad\quad(\because\cos ^{2} A=1-\sin ^{2} A)\)

\(\Rightarrow 2 \sin ^{2} A+ 3-3\sin ^{2} A=2\)

\(\Rightarrow -\sin ^{2} A=-1\)

\(\Rightarrow \sin ^{2} A=1\)

\(\Rightarrow \sin A=1\)

\(\Rightarrow \sin A=\sin 90^{\circ}\)

\(\Rightarrow A=90^{\circ}\)

Now, we will find

\((\tan A-\cot A)^{2}=\left(\tan 90^{\circ}-\cot 90^{\circ}\right)^{2}\)

\(\Rightarrow(\tan A-\cot A)^{2}=(\infty-0)^{2}\)

\(\Rightarrow(\tan A-\cot A)^{2}=\infty\)

Given:

\(\mathrm{a}_{1}=3\) and \(\mathrm{a}_{\mathrm{k}}=7 \mathrm{a}_{\mathrm{k}-1}\)

\(\Rightarrow \mathrm{a}_{2}=7 \times \mathrm{a}_{1}\)

\(=7 \times 3=21\)

\(=3.7^{2-1}\)

\(\Rightarrow \mathrm{a}_{3}=7 \times \mathrm{a}_{2}\)

\(=7^{2} \times \mathrm{a}_{1}=49 \times 3=147\)

\(=3.7^{3-1}\)

\(\Rightarrow \mathrm{a}_{4}=7 \times \mathrm{a}_{3}\)

\(=7^{2} \times \mathrm{a}_{2}\)

\(=7^{3} \times \mathrm{a}_{1}=343 \times 3\)

\(=1029=3.7^{4-1}\)

\(\Rightarrow a_{n}=7 a_{n-1}=3.7^{n-1}\)

Let \(a_{m}=7 a_{m-1}=3.7^{m-1}\) be true.

\(\Rightarrow a_{m+1}=7 a_{m+1-1}\)

\(=7 a_{m}=7.3 .7^{m-1}\)

\(=3.7^{(m+1)-1}\)

\(\Rightarrow a_{m+1}\) is true when \(a_{m}\) is true.

\(\therefore\) By Mathematical Induction \(a_{n}=3.7^{n-1}\) Is true for all natural numbers \(n .\)

Suppose angular bisector of \(A\) meets \(B C\) at \(D\left(x_z, y_z z\right)\)

Using angular bisector theorem,

\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}} \\\)

\(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\sqrt{(4-2)^2+(7-3)^2+(8-4)^2}}{\sqrt{(4-2)^2+(7-5)^2+(8-7)^2}} \\\)

\( =\frac{\sqrt{2^2+4^2+4^2}}{\sqrt{2^2+2^2+1^2}}=\frac{6}{3}=2\)

\( \text { So, } \mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv\left(\frac{(2)(2)+(1)(2)}{2+1}, \frac{(2)(5)+(1)(3)}{2+1}\right. \\\)

\( \left.\frac{(2)(7)+(1)(4)}{2+1}\right) \\\)

\( \mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv\left(\frac{6}{3}, \frac{13}{3}, \frac{18}{3}\right)\)

Therefore, position vector of point \(P=\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})\)

Given here:

\(\lim _{x \rightarrow 0} \frac{\sqrt[p]{1+x}-1}{x}\)

Apply L-Hospital rule,

\(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Therefore,

\(\lim _{x \rightarrow 0} \frac{\sqrt[p]{1+x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left((1+x)^{\frac{1}{p}}-1\right)}{\left(\frac{d x}{d x}\right)}\)

\(=\lim _{x \rightarrow 0} \frac{1}{p} \times(1+x)^{\frac{1}{p}-1}\)

\(=\frac{1}{p}(1+0)^{\frac{1}{p}-1}\)

\(=\frac{1}{p}\)

We have,

\(e^{x}=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

By the expansion of e^x, we get,

\(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\ldots+\frac{x^{n}}{n !}+\ldots\)

\(=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

Equating the coefficient of xⁿ on both sides, we get,

\(B_{n}-B_{n-1}=\frac{1}{n !}\)

Given:

Equation of the line which makes an angle of \(30^{\circ}\) with the positive direction of the x-axis and cuts off an intercept of 4 units with the negative direction of the \(y\) - axis.

Equation of a line whose slope is \(\mathrm{m}\) and which makes an intercept \(\mathrm{c}\) on the Y-axis is given by: 

\(y=m x+c\)

Here, \(c=-4\) and \(m=\tan 30=\frac{1}{\sqrt{3}}\)

So, the equation of the required line is: 

\(y=\frac{1}{\sqrt{3}} \times x-4\)

\( \sqrt{3} y=x-4 \sqrt{3}\)

So, the equation of the required line is \(x-\sqrt{3} y-4 \sqrt{3}=0\).

\(\begin{aligned} & \because \vec{a}+2 \vec{b}+2 \vec{c}=\overrightarrow{0} \text { [Given] } \\ & \Rightarrow \vec{a}+2 \vec{c}=-2 \vec{b} \Rightarrow(\vec{a}+2 \vec{c}) \cdot(\vec{a}+2 \vec{c})=(-2 \vec{b})(-2 \vec{b}) \\ & \Rightarrow \vec{a} \cdot \vec{a}+4 \vec{c} \cdot \vec{c}+4 \vec{a} \cdot \vec{c}=4 \vec{b} \cdot \vec{b} \Rightarrow 1+4+4 \vec{a} \cdot \vec{c}=4 \\ & \Rightarrow \vec{a} \cdot \vec{c}=\frac{-1}{4} \\ & \because|\vec{a} \cdot \vec{c}|^2+|\vec{a} \times \vec{c}|^2=1(\vec{a} \text { is unit vector) } \\ & \Rightarrow \frac{1}{16}+|\vec{a} \times \vec{c}|^2=1 \\ & \Rightarrow|\vec{a} \times \vec{c}|^2=\frac{15}{16} \Rightarrow|\vec{a} \times \vec{c}|=\frac{\sqrt{15}}{4}\end{aligned}\)