Mathematics Test - 50

Every time the selected coupon must be from 1 to \(9 .\) The probability of such a selection at any time is \(\frac{9}{15}=\frac{3}{5}\) and this can happen 7 times i.e., \(\left(\frac{9}{15}\right)^{7}\) out of these \(\left(\frac{9}{15}\right)^{7},\left(\frac{8}{15}\right)^{7}\) cases do not contain the number 9 Thus, the required probability \(=\left(\frac{9}{15}\right)^{7}-\left(\frac{8}{15}\right)^{7}\)
Hence, the correct option is (D)

\(\mathrm{f}(1+1)=\mathrm{f}(1)+\mathrm{f}(1)-1-1=0\)
\(\mathrm{f}(2+1)=\mathrm{f}(2)+\mathrm{f}(1)-2-1=-2\)
\(\mathrm{f}(3+1)=-5\) and \(\mathrm{f}(4+1)=-9\)
\(\therefore\) Required sum \(=1+0-2-5-9=-15\)
Hence, the correct option is (B)

Let \(A_{i}=\) Events that Ace comes in \(i^{t h}\) draw
\(P\left(A_{1}\right)=\frac{1}{13} \quad P\left(\bar{A}_{1}\right)=\frac{12}{13}\)
\(P\left(\bar{A}_{1} \cap \bar{A}_{2}\right)=P\left(\bar{A}_{1}\right) P\left(\frac{\bar{A}_{2}}{\bar{A}_{1}}\right)\)
\(=\frac{12}{13} \times \frac{11}{12}\)
So the required probability that are comes in \(4^{\text {th }}\) draw
\(=P\left(\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3} \cap A_{4}\right)\)
\(=P\left(\bar{A}_{1}\right) P\left(\frac{\bar{A}_{2}}{\bar{A}_{1}}\right) P\left(\frac{\bar{A}_{3}}{\bar{A}_{1} \cap \bar{A}_{2}}\right) P\left(\frac{A_{4}}{\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}}\right)\)
\(=\frac{12}{13} \times \frac{11}{12} \times \frac{10}{11} \times \frac{1}{10}=\frac{1}{13}\)
Hence, the correct option is (A)

Let \(\overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}\)
\(\because|\vec{c}|=1 \Rightarrow x^{2}+y^{2}+z^{2}=1 \ldots(1)\)
\(\& \cos \frac{\pi}{3}=\frac{\vec{c} \cdot(i+j)}{|\vec{c}||(i+j)|}=\frac{x+y}{\sqrt{2}} \Rightarrow \frac{1}{2}=\frac{x+y}{\sqrt{2}} \Rightarrow x+y=\frac{1}{\sqrt{2}} \ldots(2)\)
\(\because\) We Have \(\frac{a^{n}+b^{n}}{2} \geq\left(\frac{a+b}{2}\right)^{\mathrm{n}} \Rightarrow \frac{x^{2}+y^{2}}{2} \geq\left(\frac{x+y}{2}\right)^{2}\)
from (1)\(\&(2)\)
\(\Rightarrow \frac{1-z^{2}}{2} \geq\left(\frac{1}{2 \sqrt{2}}\right)^{2}\)
\(\Rightarrow \quad z^{2} \leq \frac{3}{4}\)
\(\Rightarrow-\frac{\sqrt{3}}{2} \leq z \leq \frac{\sqrt{3}}{2} \quad \ldots .\)
\(\&(i \times j) . \vec{c}=(\hat{k} . \vec{c})=z\)
\(\therefore\) Min \(\&\) Max value of \(z\) are \(-\frac{\sqrt{3}}{2} \& \frac{\sqrt{3}}{2} \quad\) From (3)

(1) Scalar or Dot product of two vectors: If \(\vec{a}\) and \(\vec{b}\) are two non - zero vectors and \(\theta\) be the angle between them, then their scalar product (or dot product) is denoted by \(\vec{a} . \vec{b}\) and is defined as the scalar \(|\mathrm{a}||b| \cos \theta,\) where \(|\mathrm{a}|\) and \(|b|\) are modulit of \(\vec{a}\) and \(\vec{b}\) respectively and \(0 \leq \theta \leq \pi .\) Dot product of two vectors is a scalar quantity.

Some of the properties of the dot product are .
(i) \(\vec{u} . \vec{v}=\vec{v} . \vec{u}\) (Commutative property)
(ii) \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v} \cdot+\vec{u} . \vec{w}\) (Distributrive property)
(iii) \(\vec{u} \cdot \vec{v}=0\) when \(\vec{u}\) and \(\vec{v}\) are orthogonal.
\((\mathrm{iv})|\vec{v}|^{2}=\vec{v} \cdot \vec{v}\)
(v) \(a(\vec{u} \cdot \vec{v})=(a \vec{u}) \cdot \vec{v}\)
(vi) \(|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}\)
(vii) \(|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}\)
(viii) \(|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)
(ix) If \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} \& \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) then \(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\)
Hence, the correct option is (B)

\(\begin{aligned} \because S &=\sum_{r=1}^{16 n} \frac{8 r}{\left(4 r^{4}+1\right)} \\ &=\sum_{r=1}^{16} \frac{8 r}{\left(2 r^{2}-2 r+1\right)\left(2 r^{2}+2 r+1\right)} \\ \mathrm{S} &=2 \sum_{r=1}^{16}\left(\frac{1}{2 r^{2}-2 r+1}-\frac{1}{2 r^{2}+2 r+1}\right) \\ &=2\left\{\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{13}\right)+\ldots+\left(\frac{1}{481}-\frac{1}{545}\right)\right\} \\ &=2\left(1-\frac{1}{545}\right)=\frac{1088}{545} \end{aligned}\)
Hence, the correct option is (D)

\(y=\frac{a^{2}-a x}{1+a^{4}} \quad \cdots\)
\(y=\frac{x^{2}+2 a x+3 a^{2}}{1+a^{4}}\)
point of intersection of (i) and (ii)
\(\frac{a^{2}-a x}{1+a^{4}}=\frac{x^{2}+2 a x+3 a^{2}}{1+a^{4}}\)
\((x+a)(x+2 a)=0\)
\(x=-a,-2 a\)
Req area
\(f(a)=\int_{-2 a}^{-a}\left[\left(\frac{a^{2}-a x}{1+a^{4}}\right)-\left(\frac{x^{2}+2 a x+3 a^{2}}{1+a^{4}}\right)\right] d x\)
\(f(a)=\frac{a^{3}}{6\left(1+a^{4}\right)}\)
\(\mathrm{f}(\mathrm{a})\) is max is when
\(f^{\prime}(a)=0\)
\(\Rightarrow 3+3 a^{4}-4 a^{4}=0\)
\(\Rightarrow \mathrm{a}^{4}=3 \Rightarrow \mathrm{a}=3^{1 / 4}\)
Hence, the correct option is (A)

Given \(:\left(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{\sqrt{a}}}\right)^{21}\)
\(t_{r+1}={ }^{21} C_{r}\left(\frac{a}{b}\right)^{\frac{21-r}{3}} \frac{b^{r / 3}}{a^{r / 6}}={ }^{21} C_{r} \quad a^{\frac{\Delta-3 r}{6}} \quad b^{\frac{2 r-21}{3}}\)
\(\therefore 42-3 r=4 r-42\) i.e. \(\quad r=12\)
\(\therefore 13^{\text {th }}\) term contains same power of a \(\&\) b.
rth term of binomial expansion The \(r^{\text {th }}\) term of the expansion of \((a+b)^{n}\) is :
\(\left(\begin{array}{c}\mathbf{n} \\ \mathbf{r}-1\end{array}\right) \mathbf{a}^{\mathrm{n}-(\mathbf{r}-1)} \mathbf{b}^{\mathbf{r}-\mathbf{1}}\)
\(\left(\begin{array}{c}\mathbf{n} \\ \mathbf{r}-\mathbf{1}\end{array}\right),\) means \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\)
Hence, the correct option is (D)

Let \(\Delta=\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ \cos (\mathrm{p}-\mathrm{d}) \mathrm{x} & \cos \mathrm{px} & \cos (\mathrm{p}+\mathrm{d}) \mathrm{x} \\ \sin (\mathrm{p}-\mathrm{d}) \mathrm{x} & \sin \mathrm{px} & \sin (\mathrm{p}+\mathrm{d}) \mathrm{x}\end{array}\right|\)
Applying \(C_{1} \rightarrow C_{1}+C_{3}\)
\(\Rightarrow \Delta=\left|\begin{array}{ccc}1+\mathrm{a}^{2} & \mathrm{a} & \mathrm{a}^{2} \\ \cos (\mathrm{p}-\mathrm{d}) \mathrm{x}+\cos (\mathrm{p}+\mathrm{d}) \mathrm{x} & \cos \mathrm{px} & \cos (\mathrm{p}+\mathrm{d}) \mathrm{x} \\ \sin (\mathrm{p}-\mathrm{d}) \mathrm{x}+\sin (\mathrm{p}+\mathrm{d}) \mathrm{x} & \sin \mathrm{px} & \sin (\mathrm{p}+\mathrm{d}) \mathrm{x}\end{array}\right|\)
\(\Rightarrow \Delta=\left|\begin{array}{ccc}1+\mathrm{a}^{2} & \mathrm{a} & \mathrm{a}^{2} \\ 2 \cos \mathrm{px} \cos \mathrm{dx} & \cos \mathrm{px} & \cos (\mathrm{p}+\mathrm{d}) \mathrm{x} \\ 2 \sin \mathrm{px} \cos \mathrm{dx} & \sin \mathrm{px} & \sin (\mathrm{p}+\mathrm{d}) \mathrm{x}\end{array}\right|\)
Applying \(C_{1} \rightarrow C_{1}-(2 \cos d x) C_{2}\)
\(\Rightarrow \Delta=\left|\begin{array}{ccc}1+\mathrm{a}^{2}-2 \mathrm{a} \cos \mathrm{dx} & \mathrm{a} & \mathrm{a}^{2} \\ 0 & \cos \mathrm{px} & \cos (\mathrm{p}+\mathrm{d}) \mathrm{x} \\ 0 & \sin \mathrm{px} & \sin (\mathrm{p}+\mathrm{d}) \mathrm{x}\end{array}\right|\)
\(\Rightarrow \Delta=\left(1+a^{2}-2 a \cos d x\right)[\sin (p+d) x \cos p x-\sin p x \cos (p+d) x]\)
\(\Rightarrow \Delta=\left(1+a^{2}-2 a \cos d x\right) \sin d x\)
Which is independent of \(\mathrm{p}\)
Hence, the correct option is (B)