MCQ 11 Mark
The smallest value of x satisfying the equation $\sqrt{3}(\cot\text{x}+\tan\text{x})=4$ is:
- A
$\frac{2\pi}{3}$
- B
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{6}$
- D
$\frac{\pi}{12}$
AnswerCorrect option: C. $\frac{\pi}{6}$
Given:
$\sqrt{3}(\cot\text{x}+\tan\text{x})=4$
$\Rightarrow\sqrt{3}\Big(\frac{\cos\text{x}}{\sin\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\Big)=4$
$\Rightarrow\sqrt{3}(\cos^2\text{x}+\sin^2\text{x})=4\sin\text{x}\cos\text{x}$
$\Rightarrow\sqrt{3}=2\sin2\text{x}$ $\big[\sin2\text{x}=2\sin\text{x}\cos\text{x}\big]$
$\Rightarrow\sin2\text{x}=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin2\text{x}=\sin\frac{\pi}{3}$
$\Rightarrow2\text{x}=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{3},\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2}+(-1)^{\text{n}}\frac{\pi}{6},\text{n}\in\text{Z}$
To obtain the smallest value of x, we will put n = 0n = 0 in the above equation.
Thus, we have:
$\text{x}=\frac{\pi}{6}$
Hence, the smallest value of x is $\frac{\pi}{6}.$
View full question & answer→MCQ 21 Mark
General solution of $\tan5\text{x}=\cot2\text{x}$ is:
- A
$\frac{\text{n}\pi}{7}+\frac{\pi}{2},\ \text{n}\in\text{Z}$
- B
$\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{3},\ \text{n}\in\text{Z}$
- ✓
$\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$
- D
$\text{x}=\frac{\text{n}\pi}{7}=\frac{\pi}{14},\ \text{n}\in\text{Z}$
AnswerCorrect option: C. $\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$
Given:
$\tan5\text{x}=\cot2\text{x}$
$\Rightarrow\tan5\text{x}=\tan\Big(\frac{\pi}{2}-2\text{x}\Big)$
$\Rightarrow5\text{x}=\text{n}\pi+\frac{\pi}{2}-2\text{x}$
$\Rightarrow7\text{x}=\text{n}\pi+\frac{\pi}{2}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$
View full question & answer→MCQ 31 Mark
The general value of x satisfying the equation $\sqrt{3}\sin\text{x}+\cos\text{x}=\sqrt{3}$ is given by:
- A
$\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}+\frac{\pi}{3},\ \text{n}\in\text{Z}$
- ✓
$\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}-\frac{\pi}{6},\text{n}\in\text{Z}$
- C
$\text{x}=\text{n}\pi\pm\frac{\pi}{6},\ \text{n}\in\text{Z}$
- D
$\text{x}=\text{n}\pi\pm\frac{\pi}{3},\ \text{n}\in\text{Z}$
AnswerCorrect option: B. $\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}-\frac{\pi}{6},\text{n}\in\text{Z}$
Given:
$\sqrt{3}\sin\text{x}+\cos\text{x}=\sqrt{3}\ .....(1)$
This equation is of the form $\text{a}\sin\theta+\text{b}\cos\theta=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt{3}$
Let:
$\text{a}=\text{r}\cos\alpha$ and $\text{b}=\text{r}\sin\alpha$
Now
$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^1}=2$ and $\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{1}{\sqrt{3}}$
$\Rightarrow\alpha=\frac{\pi}{6}$
On putting $\text{a}=\sqrt{3}=\text{r}\cos\alpha$ and $\text{b}=1=\text{r}\sin\alpha$ in equation (1), we get:
$\text{r}\cos\alpha\sin\text{x}+\text{r}\sin\alpha\cos\text{}x=\sqrt{3}$
$\Rightarrow\text{r}\sin\big(\text{x}+\alpha\big)=\sqrt{3}$
$\Rightarrow\text{2}\sin\big(\text{x}+\alpha\big)=\sqrt{3}$
$\Rightarrow\text{}\sin\big(\text{x}+\alpha\big)=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{}\sin\big(\text{x}+\alpha\big)=\sin\frac{\pi}{3}$
$\Rightarrow\text{}\sin\big(\text{x}+\frac{\pi}{6}\big)=\sin\frac{\pi}{3}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}-\frac{\pi}{6},\ \text{n}\in\text{Z}$
View full question & answer→MCQ 41 Mark
If $\cos\text{x}+\sqrt{3}\sin\text{x}=2,\text{then}\ \text{x}=$
- ✓
$\frac{\pi}{3}$
- B
$\frac{2\pi}{3}$
- C
$\frac{4\pi}{3}$
- D
$\frac{5\pi}{3}$
AnswerCorrect option: A. $\frac{\pi}{3}$
Given: $\cos\text{x}+\sqrt{3}\sin\text{x}=2\ ......(1)$
This equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=2.$
Let:
$\text{a}=\text{r}\ \cos\alpha$ and $\text{b}=\sin\alpha$
Now,
$1=\text{r}\ \cos\alpha,\sqrt{3}=\text{r}\sin\alpha$
$\Rightarrow\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{1+3}=\sqrt{4}=2$
And,
$\tan\alpha=\frac{\text{b}}{\text{a}}$
$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$
$\Rightarrow\tan\alpha\sqrt{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
On putting $\text{a}=1=\text{r}\ \cos\alpha$ and $\text{b}=\sqrt{3}=r\ \sin\alpha$ in equation (1), we get:
$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$
$\Rightarrow2\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=1$
$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=\cos0$
$\Rightarrow\text{x}=\frac{\pi}{3}=2\text{n}\pi\pm0$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$
For $\text{n}=0,\text{x}=\frac{\pi}{3}.$
$\therefore\text{x}=\frac{\pi}{3}$
View full question & answer→MCQ 51 Mark
If $\cot\text{x}-\tan\text{x}=\sec\text{x},$ then, x is equal to:
- A
$2\text{n}\pi+\frac{3\pi}{2},\text{n}\in\text{Z}$
- ✓
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
- C
$\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{Z}$
- D
AnswerCorrect option: B. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
Given equation:
$\cot\text{x}-\tan\text{x}=\sec\text{x}$
$\Rightarrow\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}=\frac{1}{\cos\text{x}}$
$\Rightarrow\frac{\cos^2\text{x}-\sin^2}{\sin\text{x}\cosx}=\frac{1}{\cos\text{x}}$
$\Rightarrow\cos^2\text{x}\sin^2=\sin\text{x}$
$\Rightarrow(1-\sin^2\text{x})-\sin^2\text{x}=\sin\text{x}$
$\Rightarrow1-2\sin^2=\sin\text{x}$
$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$
$\Rightarrow2\sin^2\text{x}+2\sin\text{x}-\sin\text{x}-1=0$
$\Rightarrow2\sin\text{x}(\sin\text{x}+1)-1(\sin\text{x}+1)=0$
$\Rightarrow(\sin\text{x}+1)(2\sin\text{x}-1)=0$
$\Rightarrow\sin\text{x}+1=0$ or $2\sin\text{x}-1=0$
$\Rightarrow\sin\text{x}=-1$ or $\sin\text{x}=\frac{1}{2}$
Now,
$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=sin\frac{3\pi}{2}\Rightarrow\text{x}=\text{m}\pi+(-1)^\text{m}\frac{3\pi}{2},\ \text{m}\in\text{Z}$
And
$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
$\therefore\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
View full question & answer→MCQ 61 Mark
If a is any real number, the number of roots of $\cot\text{x}-\tan\text{x}=\text{a}$ in the first quadrant is (are):
AnswerGiven:
$\cot\text{x}-\tan\text{x}=\text{a}$
$\Rightarrow\frac{1}{\tan\text{x}}-\tan\text{x}=\text{a}$
$\Rightarrow1-\tan^2\text{x}=\text{a}\tan\text{x}$
$\Rightarrow\tan^2\text{x}+\text{a}\tan\text{x}-1=0$
It is a quadratic equation.
If $\tan\text{x}=\text{z},$ then the equation becomes
$\text{z}^2+\text{az}-1=0$
$\Rightarrow\text{z}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$
$\Rightarrow\tan\text{x}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$
$\Rightarrow\text{x}=\tan^{-1}\Big(\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}\Big)$
There are two roots of the given equation, but we need to find the number of roots in the first quadrant.
There is exactly one root of the equation, that is, $\text{x}=\tan^{-1}\Big(\frac{-\text{a}+\sqrt{\text{a}^2+4}}{2}\Big).$
View full question & answer→MCQ 71 Mark
The equation $3\cos\text{x}+4\sin\text{x}=6$ has .... solution.
AnswerGiven equation:
$3\cos\text{x}+4\sin\text{x}=6\ .....(1)$
Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where and $\text{c}=6.$
Let:
$\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$
Now,
$\tan\alpha=\frac{\text{b}}{\text{a}}=\frac{4}{3}$
$\Rightarrow\alpha=\tan^{-1}\Big(\frac{4}{3}\Big)$
Also,
$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{9+16}=\sqrt{25}=5$
On putting $\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$ in equation (i), we get:
$\Rightarrow\text{r}\cos(\theta-\alpha)=6$
$\Rightarrow\text{5}\cos(\theta-\alpha)=6$
$\Rightarrow\text{}\cos(\theta-\alpha)=\frac{6}{5}$
From here, we cannot find the value of $\theta.$
View full question & answer→MCQ 81 Mark
If $4\sin^2\text{x}=1$ then the values of x are:
- A
$2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
- B
$\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
- ✓
$\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
- D
$2\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
AnswerCorrect option: C. $\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
Given:
$\Rightarrow\sin^2\text{x}=1$
$\Rightarrow\sin^2\text{x}=\frac{1}{4}$
$\Rightarrow\sin\text{x}=\frac{1}{2}$ or $\sin\text{x}=-\frac{1}{2}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$ or $\sin\text{x}=\sin\Big(-\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$ or $\text{n}\pi+(-1)^\text{n}\Big(-\frac{\pi}{6}\Big),\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
View full question & answer→MCQ 91 Mark
If $\tan\text{px}-\tan\text{qx}=0,$ then the values of $\theta$ form a series in:
AnswerGiven:
$\tan\text{px}-\tan\text{qx}=0$
$\Rightarrow\tan\text{px}=\tan\text{qx}$
$\Rightarrow\frac{\sin\text{px}}{\cos\text{px}}=\frac{\sin\text{qx}}{\cos\text{qx}}$
$\Rightarrow\sin\text{px}\cos\text{qx}=\sin\text{qx}\cos\text{px}$
$\Rightarrow\frac{1}{2}\Big[\sin\Big(\frac{\text{p+q}}{2}\Big)\text{x}+\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}\Big]\\=\frac{1}{2}\Big[\sin\Big(\frac{\text{q+p}}{2}\Big)\text{x}+\sin\Big(\frac{\text{q}-\text{p}}{2}\text{x}\Big)\Big]$
Now,
$\Rightarrow\sin\text{A}\cos\text{B}=\frac{1}{2}\Big[\Big(\frac{\text{A+B}}{2}\Big)+\Big(\frac{\text{A}-\text{B}}{2}\Big)\sin\Big]$
$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$
$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=-\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$
$\Rightarrow2\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=0$
$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=0$
$\Rightarrow\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
Now, on putting the value of n, we get:
$\text{n}=1,\text{x}=\frac{2\pi}{(\text{p}-\text{q})}=\text{a}_1$
$\text{n}=2,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_2$
$\text{n}=3,\text{x}=\frac{6\pi}{(\text{p}-\text{q})}=\text{a}_3$
$\text{n}=4,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_4$
And so on.
Also,
$\text{d}=\text{a}_2-\text{a}_1=\frac{4\pi}{(\text{p}-\text{q})}-\frac{2\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$
$\text{d}=\text{a}_3-\text{a}_2=\frac{6\pi}{(\text{p}-\text{q})}-\frac{4\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$
$\text{d}=\text{a}_4-\text{a}_3=\frac{8\pi}{(\text{p}-\text{q})}-\frac{6\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$
And so on.
Thus, xx forms a series in AP.
View full question & answer→MCQ 101 Mark
The number of values of x in $[0,\ 2\pi]$ that satisfy the equation $\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$
Answer$\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$
$\Rightarrow(1-\cos^2\text{x})-\cos\text{x}=\frac{1}{4}$
$\Rightarrow4-4\cos^2\text{x}-4\cos\text{x}=1$
$\Rightarrow4\cos^2-4\cos\text{x}=1$
$\Rightarrow4\cos^2\text{x}+4\cos\text{x}-3=0$
$\Rightarrow4\cos^2\text{x}+6\cos\text{x}-2\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(2\cos\text{x}-1)=0$
$\Rightarrow(2\cos\text{x}+3)(2\cos\text{x}-1)=0$
$\Rightarrow2\cos\text{x}+3=0$ or $2\cos\text{x}-1=0$
$\Rightarrow\cos\text{x}=-\frac{3}{2}$ is not possible.
$\therefore\cos\text{x}=\frac{1}{2}$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$
Now for n = 0 and 1, the values of x are $\frac{\pi}{3},\ \frac{5\pi}{3}$ and $\frac{7\pi}{3},$ but $\frac{7\pi}{3}$ is not in $[0,\ 2\pi]$
Hence, there are two solutions in $[0,\ 2\pi].$
View full question & answer→MCQ 111 Mark
The number of values of x in the interval $[0,\ 5\pi]$ satisfying the equation $3\sin^2\text{x}-7\sin\text{x}+2=0$ is:
AnswerGiven:
$3\sin^2\text{x}-7\sin\text{x}+2=0$
$\Rightarrow3\sin^2\text{x}-6\sin\text{x}-\sin\text{x}+2=0$
$\Rightarrow3\sin\text{x}(\sin\text{x}-2)-1(\sin\text{x}-2)=0$
$\Rightarrow(3\sin\text{x}-1)(\sin\text{x}-2)=0$
$\Rightarrow3\sin\text{x}-1=0$ or $\sin\text{x}-2=0$
Now, $\sin\text{x}=2$ is not possible, as the value of sin xsin x lies between -1 and 1.
$\Rightarrow\sin\text{x}=\frac{1}{4}$
Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval $[0,\ \pi].$
Hence, it is positive six times in the interval $[0,\ 5\pi]$ viz $[0,\ \pi],$ $2\pi,\ 3\pi$ and $[4\pi,\ 5\pi].$
View full question & answer→MCQ 121 Mark
The general solution of the equation $7\cos^2\text{x}+3\sin^2\text{x}=4$ is:
- A
$\text{x}=2\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
- B
$\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$
- ✓
$\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$
- D
AnswerCorrect option: C. $\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$
Given:
$7\cos^2\text{x}+3\sin^2\text{x}=4$
$\Rightarrow7\cos^2\text{x}+3\Big(1-\cos^2\text{x}\Big)=4$
$\Rightarrow7\cos^2\text{x}+3-3\cos^2\text{x}=4$
$\Rightarrow4\cos^2\text{x}+3=4$
$\Rightarrow4\Big(1-\cos^2\text{x}\Big)=3$
$\Rightarrow4\sin^2\text{x}=3$
$\Rightarrow\sin^2\text{x}=\frac{3}{4}$
$\Rightarrow\sin\text{x}=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{3}$
$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
View full question & answer→MCQ 131 Mark
In $(0,\ \pi)$ the number of solutions of the equation$\tan\text{x}+\tan2\text{x}+\tan3\text{x}=\tan\text{x}\tan2\text{x}\tan3\text{x}$ is:
AnswerGiven equation:
$\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$
$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$
$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}\Big(1-\tan\text{x}\tan2\text{x}\Big)$
$\Rightarrow\frac{\tan\text{x}+\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}=-\tan3\text{x}$
$\Rightarrow\tan(\text{x}+2\text{x})=-\tan3\text{x}$
$\Rightarrow\tan3\text{x}=-\tan3\text{x}$
$\Rightarrow2\tan3\text{x}=0$
$\Rightarrow\tan3\text{x}=0$
$\Rightarrow3\text{x}=\text{n}\pi$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}$
Now,
$\text{x}=\frac{\pi}{3},\ \text{n}=1$
$\text{x}=\frac{2\pi}{3},\ \text{n}=2$
$\text{x}=\frac{3\pi}{3}=180^\circ,$ which is not possible, as it is not in the interval $(0,\ 2\pi).$
Hence, the number of solutions of the given equation is 2.
View full question & answer→MCQ 141 Mark
If $\text{e}^{\sin\text{x}}-\text{e}^{-\sin\text{x}}-4=0,$ then $\text{x}=$
AnswerGiven equation: $\text{e}^{\sin\text{x}}-\text{e}^{-\sin\text{x}}-4=0,$
Let:
$\text{e}^{\sin\text{x}}=\text{y}$
Now,
$\text{y}-\text{y}^{-1}-4=0$
$\Rightarrow\text{y}^2-4\text{y}-1=0$
$\therefore\text{y}=\frac{4\pm\sqrt{16+4}}{2}$
$\Rightarrow\text{y}=\frac{4\pm\sqrt{20}}{2}$
$\Rightarrow\text{y}=\frac{4\pm2\sqrt{5}}{2}=2\pm\sqrt{5}$
And
$\text{y}=\text{e}^{\sin\text{x}}$
$\Rightarrow\text{y}^{\sin\text{x}}={2\pm\sqrt{5}}$
Taking log on both sides, we get:
$\sin\text{x}=\log_\text{e}\big(2\pm\sqrt{5}\big)$
$\Rightarrow\sin\text{x}=\log_{e}\big(2+\sqrt{5}\big)$ or $\sin\text{x}=\log_\text{e}\big(2-\sqrt{5}\big)$
$\Rightarrow\sin\text{x}=\log_{e}\big(4.24\big)$ or $\sin\text{x}=\log_\text{e}\big(-0.24\big)$
$\log(4.24)>1$ and $\sin\text{x}$ cannot be greater than 1.
In the other case, the log of negative term occurs, which is not defined.
View full question & answer→MCQ 151 Mark
A solution of the equation $\cos^2\text{x}+\sin\text{x}+1=0,$ lies in the interval:
- A
$\Big(\frac{-\pi}{4},\ \frac{\pi}{4}\Big)$
- B
$\Big(\frac{\pi}{4},\ \frac{3\pi}{4}\Big)$
- C
$\Big(\frac{3\pi}{4},\ \frac{5\pi}{4}\Big)$
- ✓
$\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
AnswerCorrect option: D. $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
Given:
$\cos^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow(1-\sin^2\text{x})+\sin\text{x}+1=0$
$\Rightarrow1-\sin^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow\sin^2\text{x}-\sin\text{x}-2=0$
$\Rightarrow\sin^2\text{x}-2\sin\text{x}+\sin\text{x}-2=0$
$\Rightarrow\sin\text{x}(\sin\text{x}-2)+1(\sin\text{x}-2)=0$
$\Rightarrow(\sin\text{x}-2)(\sin\text{x}+1)=0$
$\Rightarrow\sin\text{x}-2=0$ or $\sin\text{x}+1=0$
$\Rightarrow\sin\text{x}=2$ or $\sin\text{x}=-1$
$\sin\text{x}=2$ is not possible.
$\Rightarrow\sin\text{x}=-1$
$\therefore\sin\text{x}=\sin\frac{3\pi}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{3\pi}{2},\ \text{n}\in\text{Z}$
The values of x lies in the third and fourth quadrants.
Hence, x lies in $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$
View full question & answer→MCQ 161 Mark
The smallest positive angle which satisfies the equation $2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$ is:
- ✓
$\frac{5\pi}{6}$
- B
$\frac{2\pi}{3}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: A. $\frac{5\pi}{6}$
Given:
$2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2(1-\cos^2\text{x})+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2-2\cos^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2\cos^2\text{x}-\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos^2\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}\Big(\cos\text{x}-\sqrt{3}\Big)+\sqrt{3}\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(2\cos\text{x}+\sqrt{3}\Big)\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\therefore\cos\text{x}+\sqrt{3}=0$ or, $\cos\text{x}=\sqrt{3}$ is not possible.
$\Rightarrow\cos\text{x}=\cos\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\ \text{n}\in\text{Z}$
For n = 0, the value of x is $\pm\frac{5\pi}{6}.$
Hence, the smallest positive angle is $\frac{5\pi}{6}.$
View full question & answer→MCQ 171 Mark
The solution of the equation $\cos^2\text{x}+\sin\text{x}+1=0$ lies in the interval:
- A
$\Big(\frac{-\pi}{4},\ \frac{\pi}{4}\Big)$
- B
$\Big(\frac{\pi}{4},\ \frac{3\pi}{4}\Big)$
- C
$\Big(\frac{3\pi}{4},\ \frac{5\pi}{4}\Big)$
- ✓
$\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
AnswerCorrect option: D. $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
Given equation:
$\cos^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow(1-\sin^2\text{x})+\sin\text{xx}+1=0$
$\Rightarrow2-\sin^2\text{x}+\sin\text{x}=0$
$\Rightarrow\sin^2\text{x}=\sin\text{x}-2=0$
$\Rightarrow\sin^2\text{x}-2\sin\text{x}+\sin\text{x}-2=0$
$\Rightarrow\sin\text{x}(\sin\text{x}-2)+1(\sin\text{x}-2)=0$
$\Rightarrow(\sin\text{x}-2)(\sin\text{x}+1)=0$
$\Rightarrow\sin\text{x}-2=0$ or $\sin\text{x}+1=0$
$\Rightarrow\sin\text{x}=2$ or $\sin\text{x}=-1$
Now, $\sin\text{x}=2$ is not possible.
And,
$\Rightarrow\sin\text{x}=\sin\frac{3\pi}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{3\pi}{2}$
For $\text{n}=0,\ \text{x}=\frac{3\pi}{2},$ for $\text{n}=1,\ \text{x}=\frac{7\pi}{2}$ and so on.
Hence, $\frac{3\pi}{2}$ lies in the interval $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$
View full question & answer→MCQ 181 Mark
If $\cos\text{x}=-\frac{1}{2}$ and $0<\text{x}<2\pi,$ then the solutions are:
- A
$\text{x}=\frac{\pi}{3},\ \frac{4\pi}{3}$
- B
$\text{x}=\frac{2\pi}{3},\ \frac{4\pi}{3}$
- C
$\text{x}=\frac{2\pi}{3},\ \frac{7\pi}{3}$
- ✓
$\theta=\frac{2\pi}{3},\ \frac{5\pi}{3}$
AnswerCorrect option: D. $\theta=\frac{2\pi}{3},\ \frac{5\pi}{3}$
Given equation:
$\cos\text{x}=-\frac{1}{2}$
$\Rightarrow\cos\text{x}=\cos\frac{2\pi}{3}$
$\Rightarrow\text{x}=\frac{2\pi}{3}$
or
$\cos\text{x}=\cos\frac{4\pi}{3}$
$\Rightarrow\text{x}=\frac{4\pi}{3}$
so, both $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ line in $0<\text{}\text{x}<2\pi.$
View full question & answer→MCQ 191 Mark
A value of x satisfying $\cos\text{x}+\sqrt{3}\sin\text{x}=2$ is:
- A
$\frac{5\pi}{3}$
- B
$\frac{4\pi}{3}$
- C
$\frac{2\pi}{3}$
- ✓
$\frac{\pi}{3}$
AnswerCorrect option: D. $\frac{\pi}{3}$
Given equation:
$\cot\text{x}+\sqrt{3}\sin\text{x}=2\ .....(1)$
Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=3$
Let:
$\text{a}=\text{r}\cos\alpha$ and $\text{b}=\text{r}\sin\alpha$
$1=\text{r}\cos\alpha$ and $\sqrt{3}=\text{r}\sin\alpha$
$\Rightarrow\text{r}\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^2}=2$ and
$\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}\Rightarrow\tan\alpha\tan\frac{\pi}{3}\Rightarrow\alpha=\frac{\pi}{3}$
On putting $\text{a}=1=\text{r}\cos\alpha$ and $\text{b}=\sqrt{3}=\text{r}\sin\alpha$ in equation (1) we get:
$\text{r}\cos\alpha\cos\text{x}+\text{r}\sin\alpha\sin\text{x}=2$
$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$
$\Rightarrow\text{r}\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\text{2}\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=1$
$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=\cos0$
$\Rightarrow\text{x}-\frac{\pi}{3}=0$
$\Rightarrow\text{x}=\frac{\pi}{3}$
View full question & answer→MCQ 201 Mark
If $\sqrt{3}\cos\text{x}+\sin\text{x}=\sqrt{2},$ then general value of $\theta$ is:
- A
$\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4},\ \text{n}\in\text{Z}$
- B
$(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
- C
$\text{n}\pi+\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
- ✓
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
AnswerCorrect option: D. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
Given equation:
$\sqrt{3}\cos\text{x}+\sin\text{x}\sqrt{2}\ .....(1)$
Thus, is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt2$
Let:
$\text{a}=\text{r}\sin\alpha$ and $\text{b}=\text{r}\cos\alpha$
Now,
$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\big(\sqrt{3}\big)^2+1^2}=2$
And
$\tan\alpha=\frac{\sqrt{3}}{1}$
$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$
$\Rightarrow\tan\alpha=\tan\frac{\pi}{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
Putting $\text{a}=\sqrt{3}=\text{r}\sin\alpha$ and $\text{b}=1=\text{r}\cos\alpha$ in equation (i), we get:
$\text{r}\cos\text{x}\sin\alpha+\text{r}\sin\text{x}\cos\alpha=\sqrt{2}$
$\Rightarrow\text{r}\sin(\text{x}+\alpha)=\sqrt{2}$
$\Rightarrow2\sin(\text{x}+\alpha)=\sqrt{2}$
$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\frac{1}{\sqrt{2}}$
$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\cos\frac{\pi}{4}$
$\Rightarrow\text{x}\frac{\pi}{3}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \ \text{n}\in\text{Z}$
View full question & answer→MCQ 211 Mark
The number of solution in $\big[0,\ \frac{\pi}{2}\big]$ of the equation $3\text{x}\tan5\text{x}=\sin7\text{x}$ is:
AnswerGiven:
$\Rightarrow\cos(5\text{x}-2\text{x})\tan5\text{x}=\sin(5\text{x}+2\text{x})$
$\Rightarrow\tan5\text{x}=\frac{\sin(5\text{x}+2\text{x})}{\cos(5\text{x}-2\text{x})}$
$\Rightarrow\tan5\text{x}=\frac{\sin5\text{x}\cos2\text{x}+\cos5\text{x}\sin2\text{x}}{\cos5\text{x}\cos2\text{x}+\sin5\text{x}\cos2\text{x}}$
$\Rightarrow\frac{\sin5\text{x}}{\cos5\text{x}}=\frac{\sin5\text{x}\cos2\text{x}+\cos5\text{x}\sin2\text{x}}{\cos5\text{x}\cos2\text{x}+\sin5\text{x}\cos2\text{x}}$
$\Rightarrow\sin5\text{x}\cos5\text{x}+\sin^25\text{x}\sin2\text{x}\\=\sin5\text{x}\cos5\text{x}\cos2\text{x}+\cos^25\text{x}\sin2\text{x}$
$\Rightarrow\sin^25\text{x}\sin2\text{x}=\cos^25\text{x}\sin2\text{x}$
$\Rightarrow\Big(\sin^25\text{x}\cos^25\text{x}\Big)\sin2\text{x}=0$
$\Rightarrow\Big(\sin5\text{x}-\cos5\text{x}\Big)\Big(\sin5\text{x}+\cos5\text{x}\Big)\sin2\text{x}=0$
$\Rightarrow\sin5\text{x}-\cos5\text{x}=0,\ \sin5\text{x}+\cos5\text{x}=0$ or $\sin2\text{x}=0$
$\Rightarrow\frac{\sin5\text{x}}{\cos5\text{x}}=1,\ \frac{\sin5\text{x}}{\cos5\text{x}}=-1$ or $\sin2\text{x}=0$
Now,
$\Rightarrow\tan5\text{x}=\tan\frac{\pi}{4}$
$\Rightarrow5\text{x}=\text{n}\pi+\frac{\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{5}+\frac{\pi}{20},\text{n}\in\text{Z}$
F or n = 0, 1 and 2, the values of x are $\frac{\pi}{20},\ \frac{\pi}{4}$ and $\frac{9\pi}{20},$ respectively.
Or,
$\tan5\text{x}=1$
$\Rightarrow\tan5\text{x}=\tan\frac{3\pi}{4}$
$\Rightarrow5\pi=\text{n}\pi+\frac{3\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrowx\text{x}=\frac{\text{n}\pi}{5}+\frac{3\pi}{20},\ \text{n}\in\text{Z}$
For n = 0 and 1, the values of x are $\frac{3\pi}{20}$ and $\frac{7\pi}{20},$ respectively.
And,
$\sin2\text{x}=\sin0$
$\Rightarrow\sin2\text{x}=\sin0$
$\Rightarrow2\text{x}=\text{n}\pi,\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2},\ \text{n}\in\text{Z}$
For n = 0, the value of x is 0.
Also, for the odd multiple of $\frac{\pi}{2,}$ tanx is not defined.
Hence, there are six solutions.
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