(Each question 1 marks)

Question 11 Mark

Write 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

We have, $\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$ $\Rightarrow1-\cos^2\text{x}-\cos\text{x}=\frac{1}{4}$ $\Rightarrow\cos^2\text{x}+\cos\text{x}-\frac{3}{4}=0$ $\Rightarrow4\cos^2\text{x}+4\cos\text{x}-3=0$ $\Rightarrow4\cos^2\text{x}+6\cos\text{x}-2\cos\text{x}-3=0$ $\Rightarrow2\cos\text{x}[2\cos\text{x}+3]-[2\cos\text{x}+3]=0$ $\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+3)=0$ $\therefore\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=\frac{-3}{2}$ $\therefore$ only for 2 values in $[0,2\pi],$ $\cos\text{x}=\frac{1}2{},$ or the given equation is satisfies.

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Question 21 Mark

Write the number of solutions of the equation $\tan\text{x}+\sec\text{x}=2\cos\text{x}$ in the interval $[0, 2\pi].$

Answer

$\tan\text{x}+\sec\text{x}=2\cos\text{x}$ $1+\sin\text{x}=2\cos^2\text{x}$ $2(1-\sin^2\text{x})=1+\sin\text{x}$ $2\sin^2\text{x}+\sin\text{x}=1=0$ $(2\sin\text{x}-1)(\sin\text{x}+1)=0$ $\sin\text{x}=\frac{1}{2}$ or $-1$ $\text{x}=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}$ Thus, there are 3 solutions in the interval $[0, 2\pi]$

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Question 31 Mark

Write the values of x in $[0,\pi]$ for which $\sin2\text{x},\frac{1}{2}$ and $\cos2\text{x}$ are in A.P.

Answer

We want to find the value of x for which $\frac{1}{2}-\sin2\text{x}=\cos2\text{x}-\frac{1}{2}$ $\Rightarrow\cos2\text{x}+\sin2\text{x}=1$ $\Rightarrow1-2\sin^2\text{x}+2\sin\text{x}\cos\text{x}=1$ $\Rightarrow2\sin\text{x}(\cos\text{x}-\sin\text{x})=0$ $\Rightarrow\sin\text{x}=0$ or $\cos\text{x}=\sin\text{x}$ $\Rightarrow\text{x}=0,\frac{\pi}{4},\pi$

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Question 41 Mark

If $2\sin^2\text{x}=3\cos\text{x},$ where $0\leq\text{x}\leq2\pi,$ then find the value of x.

Answer

The given equation is $2\sin^2\text{x}=3\cos\text{x}.$ Now, $2\sin^2\text{x}=3\cos\text{x}$ $\Rightarrow2(1-\cos^2\text{x})=3\cos\text{x}$ $\Rightarrow2\cos^2\text{x}+3\cos\text{x}-2=0$ $\Rightarrow(2\cos\text{x}-1)(\cos\text{x}+2)=0$ $\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-2$ But, $\cos\text{x}=-2$ is not possible. $(-1\leq\cos\text{x}\leq1)$ $\therefore\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$ $(\cos\text{x}=\cos\alpha\Rightarrow\text{x}=2\text{n}\pi\pm\alpha,\text{n}\in\text{Z})$ putting n = 0 and n = 1, we get $\text{x}=\frac{\pi}{3},\frac{5\pi}{3}$ $(0\leq\text{x}\leq2\pi)$

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Question 51 Mark

Write the number of points of intersection of the curves $2\text{y}=1$ and $\text{y}=\cos\text{x},0\leq\text{x}\leq2\pi.$

Answer

We have, $2\text{y}=1$ and $\text{y}=\cos\text{x}$ These curves intersect where $\cos\text{x}=\frac{1}{2}$ $\therefore$ These curves intersect at 2 points.

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Question 61 Mark

Write the number of points of intersection of the curves $2\text{y}=-1$ and $\text{y = cosec x}.$

Answer

Given: $2\text{y}=-1$ and $\text{y = cosec x}.$ Now, $2\text{y}=-1\Rightarrow\text{y}=-\frac{1}{2}$ Also, $\text{cosec x = y}$ $\Rightarrow\text{cosec x}=-\frac{1}{2}$ $\Rightarrow\frac{1}{\sin\text{x}}=-\frac{1}{2}$ $\Rightarrow\sin\text{x}=-2$ The value of sinesine function lies between -1 and 1. Therefore, the two curves will not intersect at any point. Hence, the number of points of intersection of the curves is 0.

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Question 71 Mark

Write the number of solutions of the equation $4\sin\text{x}-3\cos\text{x}=7.$

Answer

$4\sin\text{x}-3\cos\text{x}=7$ Divide both sides by $\sqrt{3^2+4^2}=5$ $\frac{4}{5}\sin\text{x}-\frac{3}{5}\cos\text{x}=\frac{7}{5}$ $\sin\Big(\text{x}-\tan^{-1}\Big(\frac{3}{4}\Big)\Big)=1.4$ we know that range of sine function is $[-1,1]$ $\Rightarrow\sin\Big(\text{x}-\tan^{-1}\Big(\frac{3}{4}\Big)\Big)\neq1.4$ so number of solution is zero

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Question 81 Mark

If $\sec\text{x}\cos5\text{x}+1=0,$ where $0<\text{x}\leq\frac{\pi}{2},$ find the value of x.

Answer

The given equation is $\sec\text{x}\cos5\text{x}+1=0.$ Now, $\sec\text{x}\cos5\text{x}+1=0$ $\Rightarrow\frac{\cos5\text{x}}{\cos\text{x}}+1=0$ $\Rightarrow\cos5\text{x}+\cos\text{x}=0$ $\Rightarrow2\cos3\text{x}\cos2\text{x}=0$ $\Rightarrow\cos3\text{x}=0$ or $\cos2\text{x}=0$ $\Rightarrow3\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$ or $2\text{x}=(2\text{m}+1)\frac{\pi}{2},\text{m}\in\text{Z}$ $\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{6}$ or $\text{x}=(2\text{m}+1)\frac{\pi}{4}$ Putting n = 0 and n = 1, we get $\text{x}=\frac{\pi}{6},\frac{\pi}{2}$ $\Big(0<\text{x}\leq\frac{\pi}{2}\Big)$ Also, putting m = 0, e get $\text{x}=\frac{\pi}{4}$ $\Big(0<\text{x}\leq\frac{\pi}{2}\Big)$ Hence, the values of x are $\frac{\pi}{6},\frac{\pi}{4}$ and $\frac{\pi}{2}.$

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Question 91 Mark

Write the general solutions of $\tan^22\text{x}=1.$

Answer

$\Rightarrow\tan^22\text{x}=1$ $\Rightarrow\tan^22\text{x}-1=0$ $\Rightarrow(\tan2\text{x}-1)(\tan2\text{x}+1)=0$ $\Rightarrow\tan2\text{x}=1$ or $\tan2\text{x}=-1$ $\Rightarrow\tan2\text{x}=\tan\frac{\pi}{4}$ or $\tan2\text{x}=-\tan\Big(\frac{\pi}{4}\Big)=\tan\Big(\pi-\frac{\pi}{4}\Big)$ $=\tan\Big(\frac{3\pi}{4}\Big)$ $\therefore2\text{x}=\text{x}\pi+\frac{\pi}{4}$ or $2\text{x}=\text{x}\pi+\frac{3\pi}{4},\text{x}\in\mathcal{Z}$ $\Rightarrow\text{x}=\frac{\text{x}\pi}{2}+\frac{\pi}{8}$ or $\text{x}=\frac{\text{x}\pi}{2}+\frac{3\pi}{8}$ $\Rightarrow\text{x}=\frac{\text{x}\pi}{2}+\frac{\pi}{8},\text{x}\in\text{z}$

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Question 101 Mark

If $3\tan(\text{x}-15^{\circ})=\tan(\text{x}+15^{\circ}),0<\text{x}<90^{\circ},$ find x.

Answer

Given: $3\tan(\text{x}-15^{\circ})=\tan(\text{x}+15^{\circ})$ $\Rightarrow\frac{\tan(\text{x}+15^{\circ})}{\tan(\text{x}-15^{\circ})}=3$ Applying componendo and dividendo, we have $\frac{\tan(\text{x}+15^{\circ})+\tan(\text{x}-15^{\circ})}{\tan(\text{x}+15)-\tan(\text{x}-15^{\circ})}=\frac{3+1}{3-1}$ $\Rightarrow\frac{\frac{\sin(\text{x}+15^{\circ})}{\cos(\text{x}+15^{\circ})}+\frac{\sin(\text{x}-15^{\circ})}{\cos(\text{x}-15^{\circ})}}{\frac{\sin(\text{x}+15^{\circ})}{\cos(\text{x}+15^{\circ})}-\frac{\sin(\text{x}-15^{\circ})}{\cos(\text{x}-15^{\circ})}}=\frac{4}{2}$ $\Rightarrow\frac{\sin(\text{x}+15^{\circ})\cos(\text{x}-15^{\circ})+\cos(\text{x}+15^{\circ})\sin(\text{x}-15^{\circ})}{\sin(\text{x}+15^{\circ})\cos(\text{x}-15^{\circ})-\cos(\text{x}+15^{\circ})\sin(\text{x}-15^{\circ})}=2$ $\Rightarrow\frac{\sin(\text{x}+15^{\circ}+\text{x}-15^{\circ})}{\sin(\text{x}+15^{\circ}-\text{x}+15^{\circ})}=2$ $\Rightarrow\frac{\sin2\text{x}}{\sin30^{\circ}}=2$ $\Rightarrow\sin2\text{x}=2\times\frac{1}{2}=1$ $\Big(\sin30^{\circ}=\frac{1}{2}\Big)$ $\Rightarrow\sin2\text{x}=\sin90^{\circ}$ $\Rightarrow2\text{x}=90^{\circ}$ $(0<\text{x}<90^{\circ})$ $\Rightarrow\text{x}=45^{\circ}$

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Question 111 Mark

Write the solution set of the equation $(2\cos\text{x}+1)(4\cos\text{x}+5)=0$ in the interval $[0,2\pi].$

Answer

We have, $(2\cos\text{x}+1)(4\cos\text{x}+5)=0$ $\Rightarrow\cos\theta=\frac{-1}{2}$ or $\cos\theta=\frac{-5}{4}$ Since $\cos\theta\neq\frac{-5}{4}$ for any value of x the only solutions of the given equation in $[0,2\pi]$ are $\text{x}=\frac{2\pi}{3},\frac{4\pi}{3}$

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Question 121 Mark

If $\cos\text{x = k}$ has exactly one solution in $[0,2\pi],$, then write the values(s) of k.

Answer

We know that $\cos\text{x = k}$ Has only one solution in $[0,2\pi]$ This can happen onl at $\text{x}=\pi$ $\therefore\text{k}=\cos\pi=-1$

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Question 131 Mark

Write the set of values of a for which the equation $\sqrt{3}\sin\text{x}-\cos\text{x}=\text{a}$ has no solution.

Answer

We have $\sqrt{3}\sin\text{x}-\cos\text{x = a}$ The LHS of this equation is of the form $\text{a}\sin\text{x + b}\cos\text{x}$ Where $\text{a}=\sqrt{3},\text{b}=-1$ $\therefore\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{3+1}=2$ So that $\text{a = r}\cos\text{a}$ $\text{b = r}\sin\text{a}$ $\therefore\text{LHS}=\text{r}\cos\text{a}\sin\text{x + r}\sin\text{a}\cos\text{x}$ $=\text{r}[\sin(\text{x + a})]$ $=2\sin(\text{x + a})$ So, $\sqrt{3}\sin\text{x}-\cos\text{x = a}$ $\Rightarrow2\sin(\text{x + a})=\text{a}$ $\Rightarrow\sin(\text{x + a})=\frac{\text{a}}{2}$ This equation has a solution only if $-2\leq\text{a}\leq2.$ $\therefore$ The equation has no solution if $\text{a}\in(-\infty,-2)\cup(2,\infty).$

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