Search
You can find the results of your search below.
Fulltext results:
- Question 7 and 8, Exercise 6.3 @math-11-nbf:sol:unit06
- has exactly $2$ women. ** Solution. ** (i) If there are exactly $2$ women then there will be $3$ man in committee. Total possible ways $={ }^{4} C_{2} \tim... women. ** Solution. ** At least $2$ women means there could be more than $2$ women as well.\\ So we wil... having $2,3$ and four women in committee.\\ So if there are $2$ women we already had calculated $120$ pos
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- sqrt{1-\sin^2 2\theta}$$ Since $2\theta$ in QII, therefore $\cos 2\theta$ is negative. \begin{align*}\c... \pm\sqrt{1-\sin\theta}$$ As $\theta$ lies in QI, therefore $\cos\theta >0$, thus \begin{align*} \cos\the... lies in QIII, we know that \(\sin 2\theta < 0\). Therefore: \begin{align*} \sin 2\theta & = -\sqrt{1 - \... lies in QIII, we know that \(\cos 2\theta < 0\). Therefore: \begin{align*} \cos 2\theta & = -\sqrt{1 - \
- Question 23 and 24, Exercise 4.3 @math-11-nbf:sol:unit04
- rchers are in the last row? How many marchers are there altogether? ** Solution. ** From the statement,... $62$ marchers in last row and $950$ marchers are there altogether. GOOD m( =====Question 24===== How m... any poles will be in a pile of telephone poles if there are 50 in the first layer, 49 in the second and so on, until there are 6 in the last layer? ** Solution. ** From t
- Question 5 and 6, Exercise 6.3 @math-11-nbf:sol:unit06
- ny ways can $11$ players be chosen out of $16$ if there is no restriction. ** Solution. ** $11$ players... in ${ }^{16} C_{11}$ ways.\\ i.e. $4368$ ways are there to choose $11$ players.\\ =====Question 5(ii)===... ? ** Solution. ** Case I: If one man is chosen there will be two women in committee.\\ $1$ man may be ... \times{ }^{3} C_{2}=5 \times 3=15$\\ Case II: If there are $2$ men and one woman in committee. Total po
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- } \\ 0 & 0 & 1 \end{array} \right] \end{align*} There are $3$ non-zero rows.\\ The rank of the matrix i... nd{array} \right] \quad R3 - 12R2 \\ \end{align*} There are $2$ non-zero rows.\\ The rank of the matrix i... rray} \right] \quad \frac{3}{124}R3 \end{align*} There are $3$ non-zero rows.\\ The rank of the matrix i... \right] \quad R1 - 3R2, \, R3 - 3R2 \end{align*} There are $2$ non-zero rows.\\ The rank of the matrix i
- Exercise 6.3 (Solutions) @math-11-nbf:sol:unit06
- ways can 11 players be chosen out of 16 if\\ (i) there is no restriction. (ii) a particular player is al... -3-p7|Solution: Question 7 & 8]] **Question 8.** There are 10 points on a circle. Find the number of $(\... 2]] **Question 12.** For the post of 6 officers, there are 100 applicants, 2 posts are reserved for serving candidates and remaining for others.\\ Thereare 20 serving candidates among the applicants. In
- Question 6, Exercise 9.1 @math-11-nbf:sol:unit09
- lution. ** Since period of the $\sec$ is $2\pi$, therefore \begin{align*} 6 \sec(2 x-3) & = 6 \sec(2 x-3... lution. ** Since period of the $\cos$ is $2\pi$, therefore \begin{align*} \cos (5 x+4) & = 6 \cos(5x+4+2... ** Since the period of \( \sin \) is \( 2\pi \), therefore: \begin{align*} 7 \sin(3x + 3) &= 7 \sin(3x +... ** Since the period of \( \sin \) is \( 2\pi \), therefore: \begin{align*} 5 \sin(2x + 3) &= 5 \sin(2x +
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- {(\sqrt{13})^4}\\ &= \frac{12}{169}. \end{align} Therefore, the real part is \(\dfrac{5}{169}\) and the ... c{-7 - 6}{49 + 4} = \frac{-13}{53}. \end{align} Therefore, the real part is \(\dfrac{19}{53}\) and the ... right]}{1681}\\ =& \frac{-720}{1681} \end{align} Therefore, the real part of \(\left(\dfrac{5 - 4i}{5 +
- Question 3, Exercise 1.3 @math-11-nbf:sol:unit01
- } \\ &= \dfrac{1 \pm \sqrt{271}i}{4} \end{align} Therefore, the solution set is: $\left\{\dfrac{1 \pm \s... \dfrac{{6 \pm 8i}}{2} \\ &= 3 \pm 4i \end{align} Therefore, the solution set is: $\{3 \pm 4i\}$ ====Que... } \\ &= \dfrac{{9 \pm \sqrt{37}}}{2} \end{align} Therefore, the solution set $=\left\{ \dfrac{9 \pm \sqr
- Question 4, Exercise 2.1 @math-11-nbf:sol:unit02
- 9 & 5 & 0 \end{array}\right] $$ Since $D^t=-D$, therefore $D$ is skew-symmetric. =====Question 4(v)===... \ 9 & 0 & 0 \end{array}\right] $$ Since $E^t=E$, therefore $E$ is symmetric. =====Question 4(vi)===== F
- Question 1, Exercise 2.3 @math-11-nbf:sol:unit02
- (-2i) (1) \\ &= 2i - 6 - 2i \\ &= -6\end{align*} Therefore, the determinant of the matrix \(\left[\begin... \))} \\ &= 22 + 8i - 6 \\ &= 16 + 8i \end{align*} Therefore, the determinant of the matrix \(\left[\begin
- Question 1, Exercise 2.6 @math-11-nbf:sol:unit02
- \end{align*} With the different values of $x_3$, there are infinite solutions. Hence solution is; \beg... {5}x_3 = 0 \\ &x_1 = -\frac{7}{5}x_3 \end{align*} Therefore, the solution set is: \begin{align*} &\left[
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- 1} \\ \frac{70}{11} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: ... 7}{3} \\ \frac{11}{12} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is:
- Question 11 and 12, Exercise 4.2 @math-11-nbf:sol:unit04
- f the pattern is consistent, how many plants will there be in the eighth row? ** Solution. ** Given seq... lign*} Hence $a_8=7$, that is, $7$ plants will be there in the 8th row. GOOD ====Go to ==== <text align
- Question 11 and 12, Exercise 6.3 @math-11-nbf:sol:unit06
- ===Question 12===== For the post of $6$ officers, there are $100$ appliciants,\\ $2$ posts are reserved f... serving candidiates and remaining for others. \\ There are $20$ serving candidates among the appliciants