3 Marks Question - MATHS STD 11 Science Questions - Vidyadip

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22 questions · timed · auto-graded

Question 13 Marks

Solve the following equations:
$\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$

Answer

We have,
$\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$

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Question 23 Marks

Solve the following equations:
$(\sqrt{3}-1)\cos\text{x}+(\sqrt{3}+1)\sin\text{x}=2$

Answer

We have,
$(\sqrt{3}-1)\cos\text{x}+(\sqrt{3}+1)\sin\text{x}=2$
Divide on both side by $2\sqrt{2}$
$\frac{(\sqrt{3}-1)}{2\sqrt{2}}\cos\text{x}+\frac{(\sqrt{3}+1)}{2\sqrt{2}}\sin\text{x}=\frac{1}{\sqrt{2}}$
$\sin\Big(\text{x}+\tan^{-1}\Big(\frac{\sqrt{3}-1}{\sqrt{3}+1}\Big)\Big)=\sin\frac{\pi}{4}$
$\text{x}=2\text{n}\pi+\frac{\pi}{3}$ or $2\text{n}\pi-\frac{\pi}{6}\text{n}\in\text{z}$

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Question 33 Marks

Find the general solutions of the following equations:
$\sin2\text{x}+\cos\text{x}=0$

Answer

We have,
$\sin2\text{x}+\cos\text{x}=0$
$2\sin\text{x}\cos\text{x}+\cos\text{x}=0$
$\cos\text{x}\text{(2sinx+1)}=0$
$\cos\text{x}=0$ or $2\sin\text{x+1}=0$
$\text{x}=\text{(4m}-1)\frac{\pi}{2}$ or $\sin\text{x}=\frac{-1}{2}$
$\text{x}=\text{(4m}-1)\frac{\pi}{2}$ or $\text{x}=\text{(4n}-1)\frac{\pi}{6},\text{m,n}\in\text{z}$

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Question 43 Marks

Find the general solutions of the following equations:
$\tan\text{px}=\cot\text{qx}$

Answer

We have,
$\tan\text{px}=\cot\text{qx}$
$\Rightarrow\tan\text{px}=\tan\Big(\frac{\pi}{2}-\text{qx}\Big)$
$\Rightarrow\text{px}=\text{nx}\pm\Big(\frac{\pi}{2}-\text{qx}\Big),\text{n}\in\text{z}$
$\Rightarrow\text{(p+q)}\text{x}=\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$
$\Rightarrow\text{(p+q)}\text{x}=\text{(2n+1)}\frac{\pi}{2},\text{n}\in\text{z}$
$\Rightarrow\text{x}=\frac{\text{(2n+1)}}{\text{(p+q)}}\frac{\pi}{2},\text{n}\in\text{z}$

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Question 53 Marks

Solve the following equations:

$\cos\text{x}+\cos3\text{x}-\cos2\text{x}=0$

Answer

$\cos\text{x}+\cos3\text{x}-\cos2\text{x}=0$
$\Rightarrow2\cos2\text{x}.\cos\text{x}-\cos2\text{x}=0$
$\Rightarrow\cos2\theta(2\cos\theta-1)=1$
Either
$\cos2\theta=0\ \text{or}\ 2\cos\theta=1$
$\Rightarrow2\theta=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{ z}$or $\cos\theta=\frac{1}{2}=\cos\frac{\pi}{3}$
$\Rightarrow\theta=(2\text{n}+1)\frac{\pi}{4},\text{n }\in\ \text{z}$ or $\theta=2\text{m}\pi\pm\frac{\pi}{3},\text{m }\in\ \text{z}$

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Question 63 Marks

Solve the following equations:
$\sec\text{x}\cos5\text{x}+1=0,0<\text{x}<\frac{\pi}{2}$

Answer

$\sec\text{x}\cos5\text{x}+1=0$
$\frac{\cos5\text{x}+\cos\text{x}}{\cos\text{x}}=0\Rightarrow\cos\text{x}\neq0$
$2\cos3\text{x}\cos2\text{x}=0$
$\cos3\text{x}=0$ or $\cos2\text{x}=0$
$3\text{x}=\frac{\pi}{2}$ or $2\text{x}=\frac{\pi}{2}$
$\text{x}=\frac{\pi}{4},\frac{\pi}{6}$

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Question 73 Marks

Solve the following equations:
$\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$

Answer

$\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$
$\Rightarrow\cos2\text{x}+2\cos2\text{x}.\cos\text{x}=0$ $[\because\cos\text{x}+\cos3\text{x}=2\cos2\text{x}.\cos\text{x}]$
$\Rightarrow\cos2\text{x}(1+2\cos\text{x})=0$
Either
$\cos2\text{x}=0$ or $1+2\cos\text{x}=0$
$\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=-\frac{1}{2}$
$\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=+\cos\Big(\pi-\frac{\pi}{3}\Big)$
or $\cos\text{x}=\cos2\frac{\pi}{3}$
or $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{z}$
Thus,
$\text{x}=(2\text{n}+1)\frac{\pi}{4}$ or $\Big(2\text{n}\pi\pm\frac{2\pi}{3}\Big),\text{n}\in\text{z}$

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Question 83 Marks

Solve the following equations:
$\sin3\text{x}-\sin\text{x}=4\cos^{2}\text{x}-2$

Answer

$\sin3\text{x}-\sin\text{x}=4\cos^{2}\text{x}-2$
$\Rightarrow2\cos2\text{x},\sin\text{x}=2(2\cos^{2}\text{x}-1)$
$\Rightarrow2\cos2\text{x},\sin\text{x}=2\cos2\text{x}$ $[\because\cos2\text{x}=2\cos^{2}\text{x}-1]$
$\Rightarrow2\cos2\text{x}(\sin\text{x}-1)=0$
Either
$\cos2\text{x}=0$ or $\sin\text{x}-1=0$
$\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{z}$or $\sin\text{x}=1=\sin\frac{\pi}{2}$
$\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\text{x}=\text{m}\pi+(-1)^\text{m}\frac{\pi}{2},\text{m}\in\text{z}$
Thus,
$\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\text{m}\pi+(-1)^\text{m}\frac{\pi}{2},\text{m}\in\text{z}$

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Question 93 Marks

Find the general solutions of the following equations:
$\tan\text{mx}+\cot\text{nx}=0$

Answer

We have,
$\tan\text{mx}+\cot\text{nx}=0$
$\sin\text{mx}\sin\text{nx}+\cos\text{mx}\cos\text{nx}=0$
$\cos\text{(m-n)}\text{x}=0$
$\text{(m-n)}\text{x}=\Big(\frac{2\text{k}+1}{2}\Big)\pi$
$\text{x}=\Big(\frac{2\text{k}+1}{2\text{(m-n)}}\Big)\pi,\text{k}\in\text{z}$

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Question 103 Marks

Solve the following equations:

$\sin\theta+\sin5\theta=\sin3\theta$

Answer

$\sin\text{x}+\sin5\text{x}=\sin3\text{x}$
$\Rightarrow2\sin3\text{x},\cos2\text{x}-\sin3\text{x}=0$
$\Big[\because\sin\text{C}+\sin\text{D}=2\sin\frac{\text{C}+\text{D}}{2}.\cos\frac{\text{C}-\text{D}}{2}\Big] $
Either
$\Rightarrow\sin3\text{x}=0$ or $2\cos2\text{x}-1=0 $
$\Rightarrow3\text{x}=\text{n}\pi,\text{n }\in\ \text{z}$ or $\cos2\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3},\text{n }\in\ \text{z}$ or $2\text{x}=2\text{m}\pi\pm\frac{\pi}{3},\text{m }\in\ \text{z}$
or $\text{x}=\text{m}\pi\pm\frac{\pi}{6}$
Thus,
$\text{x}=\frac{\text{n}\pi}{3}$ or $\text{m}\pi\pm\frac{\pi}{6},\text{n, m }\in\ \text{z}$

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Question 113 Marks

Find the general solutions of the following equations:
$\tan3\text{x}=\cot\text{x}$

Answer

We have,
$\tan3\text{x}=\cot\text{x}$
$\Rightarrow\tan3\text{x}=\tan\Big(\frac{\pi}{2}-\text{x}\Big)\Big[\because\tan\Big(\frac{\pi}{2}-\text{x}\Big)=\cot\text{x}\Big]$
$\Rightarrow3\text{x}=\text{n}\pi+\frac{\pi}{2}-\text{x},\text{n}\in\text{z}$
$\Rightarrow4\text{x}=\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{4}+\frac{\pi}{8},\text{n}\in\text{z}$

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Question 123 Marks

Solve the following equations:
$5\cos^{2}\text{x}+7\sin^{2}\text{x}-6=0$

Answer

$2\sin^{2}\text{x}+5-6=0$
$\sin^{2}\text{x}=\frac{1}{2}$
$\sin\text{x}=\pm\frac{1}{\sqrt{2}}$
$\text{x}=\text{n}\pi\pm\frac{\pi}{4}$

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Question 133 Marks

Find the general solutions of the following equations:
$\sin\text{x}=\tan\text{x}$

Answer

We have,

$\sin\text{x}=\tan\text{x}$

$\Rightarrow\sin\text{x}=\frac{\sin\text{x}}{\cos\text{x}}$

$\Rightarrow\sin\text{x}=\frac{\sin\text{x}}{\cos\text{x}}=0$

$\Rightarrow\sin\text{x}(\cos\text{x}-1)=0$

$\Rightarrow$ Either $\sin\text{x}=0$ or $\cos\text{x}-1=0$

Now,

$\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $\cos\text{x}=1$

$\Rightarrow\cos\text{x}=\cos0^\circ$

$\text{x}=2\text{m}\pi,\text{m}\in\text{z}$

Thus,

$\text{x}=\text{n}\pi\text{n}\in\text{z}$ or $\text{x}=2\text{m}\pi,\text{m}\in\text{z}$

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Question 143 Marks

Solve the following equation:
$2\sin^{2}\text{x}+\sqrt{3}\cos\text{x}+1=0$

Answer

We have,
$2\sin^{2}\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2(1-\cos^{2}\text{x})+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2\cos^{2}\text{x}-\sqrt{3}\cos\text{x}-3=0$
factorise it,we get,
$\Rightarrow2\cos^{2}\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}(\cos\text{x}-\sqrt{3})+\sqrt{3}(\cos\text{x}-\sqrt{3})=0$
$\Rightarrow(2\cos\text{x}+\sqrt{3})(\cos\text{x}-\sqrt{3})=0$
$\Rightarrow\text{Either}$
$\cos\text{x}=-\frac{\sqrt{3}}{2}$ or $\cos\text{x}=\sqrt{3}$ [This is not possible as$-1<\cos\text{x}<1$]
$\Rightarrow\cos\text{x}=\cos\Big(\pi-\frac{\pi}{6}\Big)$
$\Rightarrow\cos\text{x}=\cos\frac{5\pi}{6}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\text{n}\in\text{z}$

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Question 153 Marks

Find the general solutions of the following equations:
$\tan2\text{x}\tan\text{x}=1$

Answer

We have,
$\tan2\text{x}\tan\text{x}=1$
$\Rightarrow\tan2\text{x}=\frac{1}{\tan\text{x}}$
$\Rightarrow\tan2\text{x}=\cot\text{x}$
$\Rightarrow\tan2\text{x}=\tan\Big(\frac{\pi}{2}-\text{x}\Big)$
$\Rightarrow2\text{x}=\text{n}\pi+\frac{\pi}{2}-\text{x},\text{n}\in\text{z}$
$\Rightarrow3\text{x}=\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}+\frac{\pi}{6},\text{n}\in\text{z}$

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Question 163 Marks

Find the general solutions of the following equations:
$\tan\text{x}+\cot2\text{x}=0$

Answer

We have,
$\tan\text{x}+\cot2\text{x}=0$
$\tan\text{x}=-\cot2\text{x}$
$\Rightarrow\cot2\text{x}=-\tan\text{x}$
$\Rightarrow\tan2\text{x}=-\cot\text{x}$
$\Rightarrow\tan2\text{x}=-\tan\Big(\frac{\pi}{2}-\text{x}\Big)$
$\Rightarrow\tan2\text{x}=\tan\Big(\text{x}-\frac{\pi}{2}\Big)$
$\Rightarrow2\text{x}=\text{n}\pi+\Big(\text{x}-\frac{\pi}{2}\Big),\text{n}\in\text{z}$
$\text{x}=\text{n}\pi-\frac{\pi}{2},\text{n}\in\text{z}$

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Question 173 Marks

Find the general solutions of the following equations:
$\tan^{2}\text{x}+(1-\sqrt{3})\tan\text{x}-\sqrt{3}=0$

Answer

We have,
$\tan^{2}\text{x}+(1-\sqrt{3})\tan\text{x}-\sqrt{3}=0$
$\Rightarrow\tan^{2}\text{x}+\tan\text{x}-\sqrt{3}\tan\text{x}-\sqrt{3}=0$
$\Rightarrow\tan\text{x}(\tan\text{x}+1)-\sqrt{3}(\tan\text{x}+1)=0$
$\Rightarrow(\tan\text{x}-\sqrt{3})(\tan\text{x}+1)=0$
$\Rightarrow\text{Either}$
$\tan\text{x}=\sqrt{3}$ or $\tan\text{x}=-1$
$\Rightarrow\tan\text{x}=\tan\frac{\pi}{3}$ or $\tan\text{x} =-\tan\frac{\pi}{4}$
$\Rightarrow\text{x}=\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{z}$ or $\text{x}=\text{m}\pi-\frac{\pi}{4},\text{m}\in\text{z}$
$\therefore\text{x}=\text{x}\pi+\frac{\pi}{3}$ or $\text{m}\pi-\frac{\pi}{4},\text{n},\text{m}\in\text{z}$

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Question 183 Marks

Solve the following equations:
$\tan3\text{x}+\tan\text{x}=2\tan2\text{x}$

Answer

$\tan3\text{x}+\tan\text{x}=2\tan2\text{x}$
$\Rightarrow\tan3\text{x}-\tan2\text{x}=\tan2\text{x}-\tan\text{x}$
$\Rightarrow\tan3\text{x}-\tan2\text{x}=\tan2\text{x}-\tan\text{x}$
$\Rightarrow2\sin^{2}\text{x}\sin2\text{x}=0$
$\Rightarrow\text{Either}$
$\sin\text{x}=0$ or $\sin2\text{x}=0$
$\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $2\text{x}=\text{m}\pi,\text{m}\in\text{z}$
$\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $\text{x}=\text{m}\frac{\pi}{2},\text{m}\in\text{z}$

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Question 193 Marks

Solve the following equation:
$4\sin^{2}\text{x}-8\cos\text{x}+1=0$

Answer

We have,

$4\sin^{2}\text{x}-8\cos\text{x}+1=0$

$\Rightarrow4(1-\cos^{2}\text{x})-8\cos\text{x}+1=0$

$\Rightarrow4\cos^{2}\text{x}+8\cos\text{x}-5=0$

factorise it,we get,

$\Rightarrow4\cos^{2}\text{x}+10\cos\text{x}-2\cos\text{x}-5=0$

$\Rightarrow2\cos\text{x}(2\cos\text{x}+5)-1(2\cos\text{x}+5)=0$

$\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+5)=0$

Either $2\cos\text{x}-1=0$ or $2\cos\text{x}+5=0$

$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-\frac{5}{2}$

$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$

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Question 203 Marks

Find the general solutions of the following equations:
$\sin2\text{x}=\cos3\text{x}$

Answer

We have,
$\sin2\text{x}=\cos3\text{x}$
$\Rightarrow\cos3\text{x}=\cos\Big(\frac{\pi}{2}-2\text{x}\Big)\Big[\because\cos\Big(\frac{\pi}{2}-\text{x}\Big)=\sin\text{x}\Big]$
$\Rightarrow3\text{x}=2\text{n}\pi\pm\Big(\frac{\pi}{2}-2\text{x}\Big)\text{n}\in\text{z}$
$\Rightarrow\text{ Either}$
$5\text{x}=2\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{z}$ or$\text{x}=2\text{n}\pi-\frac{\pi}{2},\text{n}\in\text{z}$
$\Rightarrow5\text{x}(4\text{n}+1)\frac{\pi}{2},\text{n}\in\text{z}$ or $\Rightarrow\text{x}(4\text{n}-1)\frac{\pi}{2}$
$\Rightarrow\text{x}(4\text{n}-1)\frac{\pi}{10},\text{n}\in\text{z}$or$\Rightarrow\text{x}(4\text{n}-1)\frac{\pi}{2}\text{n}\in\text{z}$

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Question 213 Marks

Find the general solutions of the following equations:
$\sin9\text{x}=\sin\text{x}$

Answer

$\sin9\text{x}=\sin\text{x}$
$\sin9\text{x}-\sin\text{x}=0$
Apply $\sin\text{A}-\sin\text{B}$ Formula
$\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A-B}}{2}\Big)$
$\sin9\text{x}-\sin\text{x}2\cos5\text{x}\sin4\text{x}=0$
$\cos5\text{x}\sin4\text{x}=0$
$\Rightarrow\cos5\text{x}=0$ or $\sin4\text{x}=0$
$5\text{x}=\frac{(2\text{n}+1)\pi}{2}$ (or) $4\text{x}=\text{n}\pi$
$\text{x}=\Big\{\frac{(2\text{n}+1)\pi}{10}\Big\}$ {or}$\text{x}=\Big\{\frac{\text{n}\pi}{4}\Big\}$where$\text{n}\in\text{z}$

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Question 223 Marks

Solve the following equations:
$\text{cosec }\text{x}=1+\cot\text{x}$

Answer

We have,
$\text{cosec }\text{x}=1+\cot\text{x}$
$\Rightarrow\frac{1}{\sin\text{x}}=1+\frac{\cos\text{x}}{\sin\text{x}}$
$\Rightarrow1=\sin\text{x}+\cos\text{x}$
Divide both side by $\sqrt{2}$, We get,
$\Rightarrow\frac{1}{\sqrt2}\sin\text{x}+\frac{1}{\sqrt2}\cos\text{x}=\frac{1}{\sqrt2}$
$\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\frac{1}{\sqrt2}$
$\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$
$\Rightarrow\text{x}=\frac{\pi}{4}=2\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{z}$
$\therefore\text{x}\Big(2\text{n}\pi+\frac{\pi}{2}\Big)$ Or $2\text{n }\pi,\text{n}\in\text{z}$

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3 Marks Question - MATHS STD 11 Science Questions - Vidyadip