If 0 ≤ x ≤ π and x lies in the IInd quadrant such that sin x

Question: (i) If $0 \leq x \leq \pi$ and $x$ lies in the llnd quadrant such that $\sin x=\frac{1}{4}$. Find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\tan \frac{x}{2}$ (ii) If $\cos x=\frac{4}{5}$ and $x$ is acute, find $\tan 2 x$ (iii) If $\sin x=\frac{4}{5}$ and $0x\frac{\pi}{2}$, find the value of $\sin 4 x$. Solution: (i) $\sin x=\frac{1}{4}$ $\therefore \sin x=\sqrt{1-\cos ^{2} x}$ $\Rightarrow\left(\frac{1}{4}\right)^{2}=1-\cos ^{2} x$ $\Rightarrow \frac{1}{16}-1=-\cos ^{2} x...

Find the angle between the following pairs of lines:

Question: Find the angle between the following pairs of lines: (i) $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ (ii) $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$ Solution: i. Let $\vec{b}_{1}$ and $\vec{b}_{2}$ be the vectors parallel to the pair of lines, $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$, respectively. $\therefore \vec{b}_{1}=2 \hat{i}+5 \hat{j}-...

Find the values of

Question: Find the values of $n$ and $\bar{X}$ in each of the following cases: (i). $\sum_{i=1}^{n}\left(x_{i}-12\right)=-10$ and $\sum_{i=1}^{n}\left(x_{i}-3\right)=62$ (ii). $\sum_{i=1}^{n}\left(x_{i}-10\right)=30$ and $\sum_{i=1}^{n}\left(x_{i}-6\right)=150$ Solution: (i). Given $\sum_{i=1}^{n}\left(x_{i}-12\right)=-10$ $\Rightarrow\left(x_{1}-12\right)+\left(x_{2}-12\right)+\ldots \ldots \ldots+\left(x_{n}-12\right)=-10$ $\Rightarrow\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+\cdots+x_{n}\right)-(12...

If 0 ≤ x ≤ π and x lies in the IInd quadrant such that sin x

Question: (i) If $0 \leq x \leq \pi$ and $x$ lies in the llnd quadrant such that $\sin x=\frac{1}{4}$. Find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\tan \frac{x}{2}$ (ii) If $\cos x=\frac{4}{5}$ and $x$ is acute, find $\tan 2 x$ (iii) If $\sin x=\frac{4}{5}$ and $0x\frac{\pi}{2}$, find the value of $\sin 4 x$. Solution: (i) $\sin x=\frac{1}{4}$ $\therefore \sin x=\sqrt{1-\cos ^{2} x}$ $\Rightarrow\left(\frac{1}{4}\right)^{2}=1-\cos ^{2} x$ $\Rightarrow \frac{1}{16}-1=-\cos ^{2} x...

Find the angle between the following pairs of lines:

Question: Find the angle between the following pairs of lines: (i) $\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})$ and $\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$ (ii) $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$ Solution: (i)Let Q be the angle between the given lines. The angle between the given pairs of lines is given by, $\cos Q=\left|\frac{...

If sin x=

Question: If $\sin x=\frac{\sqrt{5}}{3}$ and $x$ lies in Ilnd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\tan \frac{x}{2}$ Solution: Given: $\sin x=\frac{\sqrt{5}}{3}$ Using the identity $\cos x=\sqrt{1-\sin ^{2} x}$, we get $\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\left(\frac{\sqrt{5}}{3}\right)^{2}}=\pm \frac{2}{3}$ Sincexlies in the 2nd quadrant, cosxis negative. $\therefore \cos x=-\frac{2}{3}$ Now, $\cos x=1-2 \sin ^{2} \frac{x}{2}$ $\Rightarrow-\frac{2}{3}=1-2 \sin ...

If sec A=178, verify that 3−4 sin2 A4 cos2 A−3=3−tan2 A1−3 tan2 A.

Question: If sec $A=\frac{17}{8}$, verify that $\frac{34 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}$ Solution: Given: $\sec A=\frac{17}{8}$.....(1) To verify: $\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}$...... (2) Now we know that $\sec A=\frac{1}{\cos A}$ Therefore $\cos A=\frac{1}{\sec A}$ Now, by substituting the value of $\sec A$ from equation (1) We get, $\cos A=\frac{1}{\frac{17}{8}}$ $=\frac{8}{17}$ Therefore, $\cos A=\frac{8}{17...

If sin x=

Question: If $\sin x=\frac{\sqrt{5}}{3}$ and $x$ lies in Ilnd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\tan \frac{x}{2}$ Solution: Given: $\sin x=\frac{\sqrt{5}}{3}$ Using the identity $\cos x=\sqrt{1-\sin ^{2} x}$, we get $\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\left(\frac{\sqrt{5}}{3}\right)^{2}}=\pm \frac{2}{3}$ Sincexlies in the 2nd quadrant, cosxis negative. $\therefore \cos x=-\frac{2}{3}$ Now, $\cos x=1-2 \sin ^{2} \frac{x}{2}$ $\Rightarrow-\frac{2}{3}=1-2 \sin ...

The mean of 200 items was 50. Later on, it was on discovered that

Question: The mean of 200 items was 50. Later on, it was on discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean. Solution: The mean of 200 items = 50 Then the sum of 200 items =200 50= 10,000 Correct values = 192 and 88. Incorrect values = 92 and 8. correct sum = 10000 92 8 + 192 + 88 = 10180 correct mean = 10180/200 = 101.8/2 = 50.9 ....

Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Question: Find the vector and the Cartesian equations of the line that passes through the points (3, 2, 5), (3, 2, 6). Solution: Let the line passing through the points, P (3, 2, 5) and Q (3, 2, 6), be PQ. Since PQ passes through P (3, 2, 5), its position vector is given by, $\vec{a}=3 \hat{i}-2 \hat{j}-5 \hat{k}$ The direction ratios of PQ are given by, $(3-3)=0,(-2+2)=0,(6+5)=11$ $\vec{b}=0 . \hat{i}-0 . \hat{j}+11 \hat{k}=11 \hat{k}$ The equation of $\mathrm{PQ}$ in vector form is given by, $...

The mean of 5 numbers is 18.

Question: The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number. Solution: The mean of 5 numbers is 18 Then, the sum of 5 numbers =5 18= 90 If one number is excluded Then, the mean of 4 numbers = 16 sum of 4 numbers =4 16= 64 Excluded number = 90 64 = 26....

If cos x

Question: (i) If $\cos x=-\frac{3}{5}$ and $x$ lies in the Illrd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}, \sin 2 x$. (ii) If $\cos x=-\frac{3}{5}$ and $x$ lies in IInd quadrant, find the values of $\sin 2 x$ and $\sin \frac{x}{2}$ Solution: (i) $\cos x=-\frac{3}{5}$ Using the identity $\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta$, we get $\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}$ $\Rightarrow-\frac{3}{5}=2 \cos ^{2} \frac{x}{2}-1$ $\Rightarrow 1-\frac{3}{5}...

Question: (i) If $\cos x=-\frac{3}{5}$ and $x$ lies in the Illrd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}, \sin 2 x$. (ii) If $\cos x=-\frac{3}{5}$ and $x$ lies in IInd quadrant, find the values of $\sin 2 x$ and $\sin \frac{x}{2}$ Solution: (i) $\cos x=-\frac{3}{5}$ Using the identity $\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta$, we get $\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}$ $\Rightarrow-\frac{3}{5}=2 \cos ^{2} \frac{x}{2}-1$ $\Rightarrow 1-\frac{3}{5}...

The mean weight of 8 numbers is 15.

Question: The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean? Solution: We have, The mean weight of 8 numbers is 15 Then, the sum of 8 numbers =8 15= 120 If each number is multiplied by 2 Then, new mean =120 2= 240 new mean = 240/8 = 30....

The mean weight per student in a group of 7 students is 55 kg.

Question: The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student. Solution: The mean weight per student in a group of 7 students = 55 kg Weight of 6 students (in kg) =52, 54, 55, 53, 56 and 54 Let the weight of seventh student = x kg $\therefore$ Mean Weight $=\frac{\text { Sum of weights }}{\text { Total no. of students }}$ $\Rightarrow 55=\frac{52+54+55+53+56+54+x}{7}$ ⇒ 38...

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

Question: Find the vector and the Cartesian equations of the lines that pass through the origin and (5, 2, 3). Solution: The required line passes through the origin. Therefore, its position vector is given by, $\vec{a}=\overrightarrow{0}$ $\ldots(1)$ The direction ratios of the line through origin and (5, 2, 3) are (5 0) = 5, (2 0) = 2, (3 0) = 3 The line is parallel to the vector given by the equation, $\vec{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}$ The equation of the line in vector form through a poi...

The mean of five numbers is 27. If one number is excluded, their mean is 25.

Question: The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number. Solution: The mean of five numbers is 27 The sum of five numbers =5 27= 135 If one number is excluded, the new mean is 25 Sum of 4 numbers =4 25= 100 Excluded number = 135 100 = 35...

The traffic police recorded the speed (in km/hr) of 10 motorists

Question: The traffic police recorded the speed (in km/hr) of 10 motorists as 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 . Later on, an error in recording instrument was found. Find the correct average speed of the motorists if the instrument is recorded 5 km/hr less in each case. Solution: The speed of 10 motorists are 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 . Later on it was discovered that the instrument recorded 5 km/hr less than in each case correct values are = 52 , 58 , 54 , ...

The Cartesian equation of a line is

Question: The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$ Write its vector form. Solution: The Cartesian equation of the line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ ...(1) The given line passes through the point $(5,-4,6)$. The position vector of this point is $\vec{a}=5 \hat{i}-4 \hat{j}+6 \hat{k}$ Also, the direction ratios of the given line are 3, 7, and 2. This means that the line is in the direction of vector, $\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}$...

The mean of marks scored by 100 students was found to be 40. Later on,

Question: The mean of marks scored by 100 students was found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean. Solution: Mean marks of 100 students = 40 Sum of marks of 100 students =100 40 = 4000 Correct value = 53 Incorrect value = 83 Correct sum = 4000 83 + 53 = 3970 correct mean = 3970/100 = 39.7...

Explain, by taking a suitable example, how the arithmetic mean alters

Question: Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) Subtracting a constant k from each term, (iii) multiplying each term by a constant k and (iv) dividing each term by non-zero constant k. Solution: Let say numbers are 3, 4, 5 $\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $=\frac{3+4+5}{3}=4$ (i).Adding constant term k = 2 in each term. New numbers are = 5, 6, 7 $\therefore$ Mean $=\frac{\te...

Prove that:

Question: Prove that: $\cot \frac{\pi}{8}=\sqrt{2}+1$ Solution: $\frac{\pi}{8}=\left(22 \frac{1}{2}\right)^{\circ}$ Let $A=\left(22 \frac{1}{2}\right)^{\circ}$ Using the identity $\cot 2 A=\frac{\cot ^{2} A-1}{2 \cot A}$, we get $\cot 45^{\circ}=\frac{\cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}-1}{2 \cot \left(22 \frac{1}{2}\right)^{\circ}}$ $\Rightarrow 1=\frac{\cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}-1}{2 \cot \left(22 \frac{1}{2}\right)^{\circ}} \quad\left(\because \cot 45^{\circ}=1\righ...

If sin θ=34, prove that cosec2 θ−cot2 θsec2 θ−1−−−−−−−−−−√=7√3.

Question: If $\sin \theta=\frac{3}{4}$, prove that $\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3}$. Solution: Given: $\sin \theta=\frac{3}{4} \ldots \ldots$(1) To prove: $\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3}$.....(2) By definition, $\sin \mathrm{A}=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Hypotenuse }}$.....(3) By Comparing (1) and (3) We get, Perpendic...

Find the Cartesian equation of the line which passes through the point

Question: Find the Cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ Solution: It is given that the line passes through the point $(-2,4,-5)$ and is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ The direction ratios of the line, $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$, are 3,5, and 6 . The required line is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ Therefore, its direc...

tan

Question: $\tan 82 \frac{1^{\circ}}{2}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$ Solution: Here, $\tan (82.5)^{\circ}=\tan (90-7.5)^{\circ}$ $=\cot (7.5)^{\circ}$ $=\frac{1}{\tan (7.5)^{\circ}}$ We know, $\tan \left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}$ On putting $x=15^{\circ}$, we get $\tan \left(\frac{15}{2}\right)^{\circ}=\frac{\sin 15^{\circ}}{1+\cos 15^{\circ}}$ $=\frac{\sin (45-30)^{\circ}}{1+\cos (45-30)^{\circ}}$ $=\frac{\sin 45^{\circ} \cos 30^{\circ}-\...