Mathematics Test - 17

The scalar product (or dot product) of two non-zero vector and denoted by \(\vec{a} \cdot \vec{b}\) is defined as,

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=a b \cos \theta \quad \ldots\) (1)

Case: If both the vectors are perpendicular then, \(\theta=90^{\circ}\).

Form equation (1) \(\vec{a} \cdot \vec{b}=a b \cdot \cos 90^{\circ}\)

\(\vec{a} \cdot \vec{b}=a b \times 0 \Rightarrow \vec{a} \cdot \vec{b}=0 \quad \ldots(2)\)

Given, \(\vec{a}=2 \hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}-4 \hat{j}+\lambda \hat{k}\)

Are mutually perpendicular then by equation (2),

\(\vec{a} \cdot \vec{b}=0\)

\(\Rightarrow(2 \hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-4 \hat{j}+\lambda \hat{k})=0 \)

\(\Rightarrow(2 \times 1)(-4 \times 1)+(1 \times \lambda)=0 \)

\(\Rightarrow 2-4+\lambda=0 \)

\(\lambda=2\)

Given,

\(\sin \theta+\cos \theta=p \), \(\sec \theta+\operatorname{cosec} \theta=q \)

\(p^{2}=(\sin \theta+\cos \theta)^{2} \)= \(p^{2}=\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta \)

\(p^{2}=1+2 \sin \theta \cos \theta \)

\(q\left(p^{2}-1\right)=(\sec \theta+\operatorname{cosec} \theta)(1+2 \sin \theta \cos \theta-1) \)

\(\Rightarrow p\left(p^{2}-1\right)=(\sec \theta+\operatorname{cosec} \theta)(2 \sin \theta \cos \theta) \) \(=\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)(2 \sin \theta \cos \theta) \)

\(=2 \sin \theta+2 \cos \theta \) \(=2(\sin \theta+\cos \theta) \)

\(=2 p \)

Given,

\(f(t)=\sqrt{t}(a+b t)\)

As we know,

\(\frac{d}{d x}[f(x) g(x)]=f(x) \frac{d}{d x}[g(x)]+g(x) \frac{d}{d x}[f(x)] \)

\(\frac{d}{d x}\left(x^n\right)=n x^{n-1}\)

Differentiating with respect to \(t\), we get

\(f^{\prime}(t)=\sqrt{t} \frac{d}{d t}(a+b t)+(a+b t) \frac{d}{d t}(\sqrt{t})\)

\(=\sqrt{t} \cdot b+(a+b t) \cdot \frac{1}{2} t^{\frac{-1 }{ 2}} \)

\(=b \sqrt{t}+\frac{a+b t}{2 \sqrt{t}}\)

\(=\frac{a+3 b t}{2 \sqrt{t}}\)

Given:

\(\cos ^{4} \theta-\sin ^{4} \theta=\frac{1}{3}\)

\(a^{2}-b^{2}=(a-b)(a+b)\)

\(\Rightarrow\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{1}{3}\)

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

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

and \(\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\)

\(\frac{1}{3}=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\)

By cross multiplication,

\(1+\tan ^{2} \theta=3\left(1-\tan ^{2} \theta\right)\)

\(\Rightarrow 1+\tan ^{2} \theta=3-3 \tan ^{2} \theta\)

\(\Rightarrow 4 \tan ^{2} \theta=3-1\)

\(\Rightarrow 4 \tan ^{2} \theta=2\)

\(\therefore \tan ^{2} \theta=\frac{1}{2}\)

Given,

\((-1+i \sqrt{3})^{48}\)

Consider \(z=(x+i y)=(-1+i \sqrt{3})\)

\(\therefore x=1 \text { and } y=\sqrt{3}\)

The polar form of the complex number is given by,

\(z=r(\cos \theta+i \sin \theta)\)

Where \(r=\sqrt{x^{2}+y^{2}}\) and \(\theta=\tan ^{-1}\left(\frac{y}{x}\right)\)

So,

\(r=\sqrt{1^{2}+(\sqrt{3})^{2}}=2\)

\(\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi}{3}\)

\(\Rightarrow z=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\)

\(\Rightarrow z^{48}=\left[2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\right]^{48}\)

\(\Rightarrow z^{48}=2^{48}\left(\cos \left(48 \times \frac{\pi}{3}\right)+i \sin \left(48 \times \frac{\pi}{3}\right)\right.\)

\(\Rightarrow z^{48}=2^{48}[\cos (16 \pi)+i \sin (16 \pi)]\)

\(\Rightarrow z^{48}=2^{48}[1+i(0)]\)

\(\Rightarrow z^{48}=2^{48}\)

So, the value of \((-1+i \sqrt{3})^{48}=2^{48}\)

Given,

\(\frac{d}{d x}\left(\frac{x^2+x-2}{x^3+6}\right)\)

As we know,

\(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x) \frac{d}{d x}[f(x)]-f(x) \frac{d}{d x}[g(x)]}{[g(x)]^2} \)

\(\frac{d}{d x}\left(x^n\right)=n x^{n-1}\)

Differentiating with respect to \(x\), we get

\(=\frac{\left(x^3+6\right) \frac{d}{d x}\left(x^2+x-2\right)-\left(x^2+x-2\right) \frac{d}{d x}\left(x^3+6\right)}{\left(x^3+6\right)^2}\)

\(=\frac{\left(x^3+6\right)(2 x+1)-\left(x^2+x-2\right)\left(3 x^2\right)}{\left(x^3+6\right)^2} \)

\(=\frac{\left(2 x^4+x^3+12 x+6\right)-\left(3 x^4+3 x^3-6 x^2\right)}{\left(x^3+6\right)^2} \)

\(=\frac{-x^4-2 x^3+6 x^2+12 x+6}{\left(x^3+6\right)^2}\)

Given:

\(x+y \leq 3, y \leq 6\) and \(x \geq 0, y \geq 0\)

Converting the given in equations into equations

\(x+y=3\).......(1)

\(y=6 \ldots \text {...(2) }\)

Reglon represented by \(x+y \leq 3\) :

The line \(x+y=3\) meets the coordinate axes are \(\mathrm{A}(3,0)\) and \(\mathrm{B}(0,3)\) respectively,

\(x+y=3\)

\(\begin{array}{|l|l|l|} \hline \mathrm{x} & 3 & 0 \\ \hline \mathrm{y} & 0 & 3 \\ \hline \end{array}\)

\(\mathrm{A}(3,0) ; \mathrm{B}(0,3)\)

Join points \(A\) and \(B\) to obtain the line.

Clearly, \((0,0)\) satisfies the in equation \(x+y \leq 3\).

So, the region containing the origin represent the solution set of the in equation.

Reglon represented by \(y \leq 6\) :

The line \(y=6\) is parallel to \(x\)-axis and its every point will satisfy the in equation in first, quadrant, region containing the origin represents the solution set of this in equation.

Reglon represented by \(x \geq 0\) and \(y \geq 0\) :

Since every point in first quadrant satisfy the in equations, so the first quadrant is the solution set of these in equations.

The shaded region is the common region of in equations. This is feasible region of solution which is bounded and is in first quadrant.

Given,

\(z=\frac{(2-i)(1+2 i)}{(3+i)(2-3 i)}\)

\(\Rightarrow z=\frac{2+4 i-i-2 i^{2}}{6-9 i+2 i-3 i^{2}}\)

\(\Rightarrow z=\frac{2+4 i-i-2 i^{2}}{6-9 i+2 i-3 i^{2}}\)

\(\Rightarrow z=\frac{2+4 i-i+2}{6-9 i+2 i+3} \quad\left[\because i^{2}=-1\right]\)

\(\Rightarrow z=\frac{4+3 i}{9-7 i}\)

Multiplying by \((9+7 i)\) in numerator and denominator, we get

\(z=\frac{4+3 i}{9-7 i} \times \frac{9+7 i}{9+7 i}\)

\(\Rightarrow z=\frac{36+28 i+27 i+21 i^{2}}{81-49 i^{2}}\)

\(\Rightarrow z=\frac{36+28 i+27 i-21}{81+49} \quad\left[\because i^{2}=-1\right]\)

\(\Rightarrow z=\frac{15+55 i}{130}\)

\(\Rightarrow z=\frac{15}{130}+i \frac{55}{130}\)

As we know,

Conjugate of \(z=\bar{z}=x-i y\)

\(\therefore \bar{z}=\frac{15}{130}-i \frac{55}{130}\)

Event A is choosing a yellow pencil first, and Event \(\mathrm{B}\) is choosing a yellow pencil second.

Initially, there are 12 pencils, 7 of which are yellow.

Probability the first pencil is yellow \(=\mathrm{P}(\mathrm{A})=\frac{7}{12}\)

If a yellow pencil is chosen, there will be 11 pencils left, 6 of which are yellow.

Probability the second pencil is yellow \(=\mathrm{P}(\mathrm{B})=\frac{6}{11}\)

Two pencils are chosen at random from the box without replacement.

So events are independent of each other.

Probability they are both yellow:

\(=P(A \cap B)=P(A) \times P(B)\)

\(=\frac{7}{12} \times \frac{6}{11}=\frac{7}{22}\)

The given question is equivalent to finding the area bounded by the lines f(x, y) = x + y - 1 = 0, g(x, y) = 2x + 3y - 6 = 0, the x-axis (y = 0) and the y-axis (x = 0)

f(x, y) meets the x-axis at x + 0 - 1 = 0 ⇒ x = 1 ⇒ (1, 0)

f(x, y) meets the y-axis at 0 + y - 1 = 0 ⇒ y = 1 ⇒ (0, 1)

g(x, y) meets the x-axis at 2x + 0 - 6 = 0 ⇒ x = 3 ⇒ (3, 0)

g(x, y) meets the y-axis at 0 + 3y - 6 = 0 ⇒ y = 2 ⇒ (0, 2)

The resulting figure made by the four lines is as follows:

The required (shaded) area is: (Area of Δ made by g(x, y), the x-axis and the y-axis) - (Area of Δ made by f(x, y), the x-axis and the y-axis)

\(=\frac{1}{2} \times 2 \times 3-\frac{1}{2} \times 1 \times 1\)

\(=3-\frac{1}{2}\)

\(=2 \frac{1}{2}\) sq.units