Mathematics Test - 19

\( A=\left\{m \in R: x^2-(m+1) x+m+4=0\right. \text { has real roots } \\\)

\( D \geq 0 \\\)

\( \Rightarrow(m+1)^2-4(m+4) \geq 0 \\\)

\( \Rightarrow m^2-2 m-15 \geq 0\)

A={(−∞,−3]∪[5,∞)}
B=[−3,5)⇒A−B=(−∞,−3)∪[5,∞)

Given equation \(a x^{2}+b x+c=0\) and \(a^{\prime} x^{2}+b^{\prime} x+c=0\)\(\alpha\) is the common root thus,

\(a a^{2}+b a+c=0 \quad\quad\ldots \ldots(1)\)

and \(a^{\prime} a^{2}+b^{\prime} \alpha+c=0 \quad\quad\ldots .(2)\)

Solving both equation (1) and (2) by cross multiple method:

\(\frac{\alpha^{2}}{\left(b c^{\prime}-b^{\prime} c\right)}=\frac{\alpha}{\left(c a^{\prime}-a c^{\prime}\right)}=\frac{1}{\left(a b^{\prime}-a^{\prime} b\right)}\)
\(\quad(I) \quad\quad\quad(I I) \quad\quad\quad(I I I)\)

From (I) and (II) we get, \(\alpha=\frac{b c^{\prime}-b^{\prime} c}{c a^{\prime}-a^{\prime} c^{\prime}} \quad\quad\ldots .(A)\)

From (II) and (III) we get, \(\alpha=\frac{c a^{\prime}-a c^{\prime}}{a b^{\prime}-a^{\prime} b} \quad\quad\ldots(B)\)

From equation (A) and (B) \(\frac{b c^{\prime}-b^{\prime} c}{c a^{\prime}-a c^{\prime}}=\frac{c a^{\prime}-a c^{\prime}}{a b^{\prime}-a^{\prime} b}\)

\(\left({ca}^{\prime}-{ac}^{\prime}\right)^{2}=\left({bc}^{\prime}-{b}^{\prime} {c}\right)\left({ab}^{\prime}-{a}^{\prime} {b}\right)\)

Given:

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|\)

To Find: Angle between \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\)

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|\)

\(\Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta|\hat{\mathrm{n}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta\)

\(\Rightarrow \sin \theta=\cos \theta\) \(\quad(\because|\hat{n}|=1)\)

\(\Rightarrow \tan \theta=1\)

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

Rate of change of '\(x\)' is given by \(\frac{d x}{d t}\)

Given that, \(y=2 x-x^{2}\) and \(\frac{d x}{d t}=3\) units\(/\)sec

Then, the slope of the curve, \(\frac{d y}{d x}=2-2 x=m\)

\(\frac{ dm }{ dt }=0 - 2 \times \frac{ dx }{ dt }\)

\(=-2(3)=-6\) units per second

Thus, the slope of the curve is decreasing at the rate of 6 units per second when \(x\) is increasing at the rate of 3 units per second.

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

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

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

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

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

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

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

\( \Rightarrow \lambda=\frac{1}{2} \therefore \overrightarrow{B_1}=\frac{\alpha}{2} \\\)

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

Cross product with \(\vec{B}_1\) in equation (1)

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

\( \Rightarrow \vec{\beta} \times \vec{B}_1=\vec{B}_1 \times \vec{B}_2 \Rightarrow \vec{\beta}_1 \times \vec{\beta}_2=\frac{\vec{\beta} \times \vec{\alpha}}{2} \\\)

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

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

\( x^2 \sin \theta-x(\sin \theta \cdot \cos \theta+1)+\cos \theta=0 . \\\)

\( x^2 \sin \theta-x \sin \theta \cdot \cos \theta-x+\cos \theta=0 \\\)

\( x \sin \theta(x-\cos \theta)-1(x-\cos \theta)=0 \\\)

\( (x-\cos \theta)(x \sin \theta-1)=0 \\\)

\( \therefore x=\cos \theta, \operatorname{cosec} \theta, \theta \in\left(0,45^{\circ}\right) \\\)

\( \therefore \alpha=\cos \theta, \beta=\operatorname{cosec} \theta \\\)

\( \sum_{n=0}^{\infty} \alpha^n=1+\cos \theta+\cos ^2 \theta+\ldots \infty=\frac{1}{1-\cos \theta} \\\)

\( \sum_{n=0}^{\infty} \frac{(-1)^n}{\beta^n}=1-\frac{1}{\operatorname{cosec}^2}+\frac{1}{\operatorname{cosec}^2 \theta}-\frac{1}{\operatorname{cosec}^3 \theta}+\ldots \infty \\\)

\( =1-\sin \theta+\sin ^2 \theta-\sin ^3 \theta+\ldots \infty \\\)

\( =\frac{1}{1+\sin \theta} \\\)

\( \therefore \sum_{n=0}^{\infty}\left(\alpha^n+\frac{(-1)^n}{\beta^n}\right)=\sum_{n=0}^{\infty} \alpha^n+\sum_{n=0}^{\infty} \frac{(-1)^n}{\beta^n} \\\)

\( =\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta}\)

\(\text { As } \angle \mathrm{PRQ}=\frac{\pi}{2} \\\)

\( \left(\frac{\frac{2}{\mathrm{t}}}{3-\frac{1}{\mathrm{t}^2}}\right) \cdot\left(\frac{-2 \mathrm{t}}{3-\mathrm{t}^2}\right)=-1 \\\)

\( \Rightarrow \mathrm{t}= \pm 1 \\\)

\(\therefore \mathrm{P} \equiv(1,2) \& Q(1,-2) \\\)

\(\therefore \text { for ellipse } \frac{1}{\mathrm{a}^2}+\frac{4}{\mathrm{~b}^2}=1 \text { and } \mathrm{ae}=1 \\\)

\( \Rightarrow \frac{1}{\mathrm{a}^2}+\frac{4}{\mathrm{a}^2\left(1-\mathrm{e}^2\right)}=1 \\\)

\(\Rightarrow 1+\frac{4}{\left(1-\mathrm{e}^2\right)}=\frac{1}{\mathrm{e}^2} \\\)

\( \Rightarrow\left(5-\mathrm{e}^2\right) \mathrm{e}^2=1-\mathrm{e}^2 \\\)

\( \Rightarrow \mathrm{e}^4-6 \mathrm{e}^2+1=0 \\\)

\( \Rightarrow \mathrm{e}^2=\frac{1}{3-2 \sqrt{2}} \Rightarrow \frac{1}{\mathrm{e}^2}=3+2 \sqrt{2}\)

Given,

\(\frac{d^{2} y}{d x^{2}}+3\left(\frac{d y}{d x}\right)^{2}=x^{2} \log \left(\frac{d^{2} y}{d x^{2}}\right)\)

For the given differential equation the highest order derivative is \(2\).

The given differential equation is not a polynomial equation because it involved a logarithmic term in its derivatives so, its degree is not defined.

Given,

n^th term of the sequence is \(a_{n}=\frac{n^{2}}{2^{n}}\)

To find the 8th term for the same, substitute n = 8

We obtain

\(a_{8}=\frac{8^{2}}{2^{8}}\)

\(=\frac{64}{256}\)

As we know that \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\)

\(I=\int_{0}^{a} \frac{f(a-x)}{f(x)+f(a-x)} d x\).......(1)

So, we have lower limit \(a=0\), upper limit \(b=a\)

\(\Rightarrow I=\int_{0}^{a} \frac{f[(0+a)-(a-x)]}{f[(a+0)-x]+f[(a+0)-(a-x)]} d x \)

\(\Rightarrow I=\int_{0}^{a} \frac{f(x)}{f(a-x)+f(x)} d x \ldots \ldots . .(2)\)

Now adding equation (1) and equation (2)

\(\Rightarrow 2 I=\int_{0}^{a} \frac{f(a-x)}{f(x)+f(a-x)} d x+\int_{0}^{a} \frac{f(x)}{f(a-x)+f(x)} d x\)

\(\Rightarrow 2 I=\int_{0}^{a} 1 d x\)

\(\Rightarrow 2 I=a\)

\(\therefore I=\frac{a}{2}\)