1 Marks Question - Maths STD 11 Science Questions - Vidyadip

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark

If $\tan(\text{A+B})=\text{p}$ and $\tan\text{(A - B)}=\text{q},$ then write the value of $\tan2\text{B}.$

Answer

We have,
$\tan(\text{A+B})=\text{p}$ and $\tan(\text{A - B})=\text{q}$
Now, $\tan2\text{B}=\tan\big[(\text{A+B})-(\text{A}-\text{B})\big]$
$=\frac{\tan(\text{A+B})-\tan\text{(A - B)}}{1+\tan(\text{A+B})\times\tan(\text{A - B})}$
$=\frac{\text{p - q}}{1+\text{pq}}$
$\therefore\ \tan2\text{B}=\frac{\text{p - q}}{1+\text{pq}}$

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Question 21 Mark

Write the maximum and minimum values of $3\cos\text{x}+4\sin\text{x}+5.$

Answer

Let $\text{f(x)}=3\cos\text{x}+4\sin\text{x}+5$
We know that,
$-\sqrt{3^2+4^2}\le3\cos\text{x}+4\sin\text{x}\le\sqrt{3^2+4^2}$
$\Rightarrow-\sqrt{9+16}\le3\cos\text{x}+4\sin\text{x}\le\sqrt{9+16}$
$\Rightarrow-\sqrt{25}\le3\cos\text{x}+4\sin\text{x}\le\sqrt{25}$
$\Rightarrow-5\le3\cos\text{x}+4\sin\text{x}\le5$
$\Rightarrow-5+5\le3\cos\text{x}+4\sin\text{x}+5\le5+5$
$\Rightarrow0\le3\cos\text{x}+4\sin\text{x}+5\le10$
Hence, minimum and maximum value of $3\cos\text{x}+4\sin\text{x}+5$ are 0 and 10 respectively.

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Question 31 Mark

Evaluate the following:
$\cos47^\circ\cos13^\circ-\sin47^\circ\sin13^\circ$

Answer

$=\cos(47^\circ+13^\circ)$ $[\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}]$
$=\cos60^\circ$
$=\frac{1}{2}$

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Question 41 Mark

Evaluate the following:
$\cos80^\circ\cos20^\circ+\sin80^\circ\sin20^\circ$

Answer

$=\cos(80^\circ-20^\circ)$$[\cos(\text{A-B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}]$
$=\cos60^\circ$
$=\frac{1}{2}$

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Question 51 Mark

If $\text{x}\cos\theta=\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{z}\Big(\theta+\frac{4\pi}{3}\Big),$ then write the value of $\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}.$

Answer

Let, $\text{x}\cos\theta=\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{z}\cos\Big(\theta+\frac{4\pi}{3}\Big)=\text{k}$
$\Rightarrow\text{x}\cos\theta=\text{k},\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{k}$ and, $\text{z}\cos\Big(\theta+\frac{4\pi}{3}\Big)=\text{k}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\cos\theta}{\text{k}},\frac{1}{\text{y}}=\frac{\cos\Big(\frac{2\pi}{3}+\theta\Big)}{\text{k}}$ and, $\frac{1}{\text{z}}=\frac{\cos\Big(\frac{4\pi}{3}+\theta\Big)}{\text{k}}$
Now, $\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=\frac{\cos\theta}{\text{k}}+\frac{\cos\big(\frac{2\pi}{3}+\theta\Big)}{\text{k}}+\frac{\cos\Big(\frac{4\pi}{3}+\theta\Big)}{\text{k}}$
$=\frac{1}{\text{k}}\Big[\cos\theta+\cos\Big\{\frac{\pi}{2}+\Big(\frac{\pi}{6}+\theta\Big)\Big\}\cos\Big\{\pi+\Big(\frac{\pi}{3}+\theta\Big)\Big\}\Big]$
$=\frac{1}{\text{k}}\Big[\cos\theta-\sin\Big(\frac{\pi}{6}+\theta\Big)-\cos\Big(\frac{\pi}{3}+\theta\Big)\Big]$
$=\frac{1}{\text{k}}\Big[\cos\theta-\Big(\sin\frac{\pi}{6}\cos\theta+\cos\frac{\pi}{6}\sin\theta\Big)-(\cos\frac\pi3\times\cos\theta-\sin\frac{\pi}{3}\times\sin\theta\Big)\Big]$
$=\frac{1}{\text{k}}\Big[\cos\theta-\Big(\frac12\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\Big)-\Big(\frac12\cos\theta-\frac{\sqrt{3}}{2}\sin\theta\Big)\Big]$
$=\frac{1}{\text{k}}\Big[\cos\theta-\frac12\cos\theta-\frac{\sqrt{3}}{2}\sin\theta-\frac12\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\Big]$
$=0$

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Question 61 Mark

Evaluate the following:
$\sin36^\circ\cos9^\circ+\cos36^\circ\sin9^\circ$

Answer

$=\sin(36^\circ+9^\circ)$ $[\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}]$
$=\sin45^\circ$
$=\frac{1}{\sqrt{2}}$

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Question 71 Mark

If $12\sin\text{x}-9\sin^2\text{x}$ attains its maximum value at $\text{x}=\alpha,$ then write the value of $\sin\alpha.$

Answer

$12\sin\text{x}-9\sin^2\text{x}$ has maximum value as 4, so $12\sin\alpha-9\sin^2\alpha=4$ $9\sin^2\alpha-612\sin\alpha+4=0$ $9\sin^2\alpha-6\sin\alpha-6\sin\alpha+4=0$ $3\sin\alpha(3\sin\alpha-2)-2(3\sin\alpha-2)=0$ $(3\sin\alpha-2)(3\sin\alpha-2)=0$$3\sin\alpha-2=0$
$\sin\alpha=\frac23$

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Question 81 Mark

If $\sin\alpha-\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ then write the value of $\cos(\alpha+\beta).$

Answer

We have,
$\sin\alpha-\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b}$
Now, $\text{a}^2+\text{b}^2=(\sin\alpha-\sin\beta)^2+(\cos\alpha+\cos\beta)^2$
$=\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta$
$=(\sin^2\alpha+\cos^2\alpha)+(\sin^2\beta+\cos^2\beta)+2[\cos\alpha\cos\beta-\sin\alpha\sin\beta]$
$=1+1+2\cos(\alpha+\beta)$
$=2+2\cos(\alpha+\beta)$
$\Rightarrow\text{a}^2+\text{b}^2=2+2\cos(\alpha+\beta)$
$\Rightarrow\text{a}^2+\text{b}^2-2=2\cos(\alpha+\beta)$
$\Rightarrow2\cos(\alpha+\beta)=\text{a}^2+\text{b}^2-2$
$\Rightarrow\cos(\alpha+\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$

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Question 91 Mark

If $\alpha+\beta-\gamma=\pi,$ and $\sin^2\alpha+\sin^2\beta-\sin^2\gamma=\lambda\sin\alpha\sin\beta\cos\gamma,$ then write the value of $\lambda.$

Answer

We have:
$\alpha+\beta-\gamma=\pi$
Now, $\sin^2\alpha+\sin^2\beta-\sin^2\gamma$
$=\frac{1-\cos2\alpha}{2}+\frac{1\cos2\beta}{2}-\frac{(1-\cos2\gamma)}{2}$
$\Big(\frac12+\frac12+\frac12\Big)-\Big(\frac{\cos2\alpha}{2}+\frac{\cos2\beta}{2}-\frac{\cos2\gamma}{2}\Big)$
$=\frac12-\frac{1}{2}\Big[2\cos\frac{2\alpha+2\beta}{2}\cos\frac{2\alpha+2\beta}{2}-\cos2\gamma\Big]$
$=\frac12-\frac12[2\cos(\alpha+\beta)\cos(\alpha-\beta)-\cos2\gamma]$
$=\frac12-\frac12[2\cos(\pi+\gamma)\cos(\alpha-\beta)-\cos2\gamma]$
$=\frac12-\frac12[2\cos\gamma\cos(\alpha-\beta)-\cos2\gamma]$
$=\frac12-\frac12\big[2\cos\gamma\cos(\alpha-\beta)-\big(2\cos^2\gamma-1\big]$
$=\frac12+\frac12\big[2\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma-1\big]$
$=\frac12-\frac12\big[2\{\cos(\alpha+\beta)\cos(\alpha-\beta)-\cos2\gamma\}\big]$
$=\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma$
$=\cos\gamma[\cos(\alpha-\beta)+\cos\gamma]$
$=\cos\Big[\cos(\alpha-\beta)+\cos\{-(\pi-(\alpha+\beta))\}\Big]$
$=\cos\gamma[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$
$=\cos\gamma[2\sin\alpha\sin\beta]$
$=2\cos\gamma\sin\alpha\sin\beta$
$=2\sin\alpha\sin\beta\sin\gamma$
$\therefore\sin^2\alpha+\sin^2\beta-\sin^2\gamma=2\sin\alpha\sin\beta\cos\gamma$
$\therefore\lambda=2$

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Question 101 Mark

If $\text{a}=\text{b}\cos\frac{2\pi}{3}=\text{c}\cos\frac{4\pi}{3},$ then write the value of ab + bc + ca.

Answer

Let, $\text{a}=\text{b}\cos\frac{2\pi}{3}=\text{c}\cos\frac{4\pi}{3}=\text{k}$
$\Rightarrow\text{a}=\text{k},\text{b}=\frac{\text{k}}{\cos\frac{2\pi}{3}}$ and $\text{c}=\frac{\text{k}}{\cos\frac{4\pi}{3}}$
Now, ab + bc + ca $=\text{k}\times\frac{\text{k}}{\cos\frac{2\pi}{3}}+\frac{\text{k}}{\cos\frac{2\pi}{3}}\times\frac{\text{k}}{\cos\frac{4\pi}{3}}+\frac{\text{k}}{\cos\frac{4\pi}{3}}\times\text{k}$
$=\text{k}^2\sec\frac{2\pi}{3}+\text{k}^2\sec\frac{2\pi}{3}\sec\frac{4\pi}{3}+\text{k}^2\sec\frac{4\pi}{3}$
$=\text{k}^2\sec\Big(\frac{\pi}{2}+\frac\pi6\Big)+\text{k}^2\sec\Big(\frac{\pi}{2}+\frac\pi6\Big)\sec\Big(\pi+\frac{\pi}{3}\Big)+\text{k}^2\sec\Big(\pi+\frac{\pi}{3}\Big)$
$=-\text{k}^2\text{cosec}\frac{\pi}{6}+\text{k}^2\text{cosec}\frac{\pi}{6}\sec\frac{\pi}{3}-\text{k}^2\sec\frac{\pi}{3}$
$=-\text{k}^2\times2+\text{k}^2\times2\times2-\text{k}^2\times2$
$=-4\text{k}^2+4\text{k}^2=0$
$\therefore\text{ ab + bc + ca}=0$

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Question 111 Mark

Evaluate the following:
$\sin78^\circ\cos18^\circ-\cos78^\circ\sin18^\circ$

Answer

$=\sin(78^\circ-18^\circ)$ $[\sin(\text{A-B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}]$
$=\sin60^\circ$
$=\frac{\sqrt{3}}{2}$

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Question 121 Mark

If $\tan\alpha=\frac{1}{1+2^{-\text{x}}}$ and $\tan\beta=\frac{1}{1+2^{\text{x+1}}},$ then write the value of $\alpha+\beta$ lying in the interval $\Big(0,\frac\pi2\Big).$

Answer

We have,
$\tan\alpha=\frac{1}{1+2^{-\text{x}}}$ and $\tan\beta=\frac{1}{1+2^{\text{x+1}}}$
Now, $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\times\tan\beta}$
$=\frac{\frac{1}{1+2^{-\text{x}}}+\frac{1}{1+2^{\text{x+1}}}}{1-\frac{1}{1+2^{-\text{x}}}\times\frac{1}{1+2^{\text{x+1}}}}$
$=\frac{\frac{1+2^{\text{x+1}}+1+2^{-\text{x}}}{(1+2^{-\text{x}})(1+2^{\text{x+1}})}}{1-\frac{1}{(1+2^{-\text{x}})(1+2^{\text{x+1}})}}$
$=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{1+2^{\text{x}+1}+2^{-\text{x}}+2}}{1-\frac{1}{1+2^{\text{x}+1}+2^{-\text{x}}+2}}$
$=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{3+2^{\text{x}+1}+2^{-\text{x}}}}{1-\frac{1}{3+2^{\text{x}+1}+2^{-\text{x}}}}$
$=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{3+2^{\text{x}+1}+2^{-\text{x}}}}{\frac{3+2^{\text{x}+1}+2^{-\text{x}}-1}{3+2^{\text{x}+1}+2^{-\text{x}}}}$
$=\frac{2+2^{\text{x+1}}+2^{-\text{x}}}{2+2^{\text{x+1}}+2^{-\text{x}}}$
$=1$
$\Rightarrow\tan(\alpha+\beta)=1=\tan\Big(\frac{\pi}{4}\Big)$
$\Rightarrow\tan(\alpha+\beta)=\tan\Big(\frac\pi4\Big)$
$\Rightarrow\alpha+\beta=\frac\pi4$

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Question 131 Mark

Write the maximum values of $12\sin\text{x}-9\sin^2\text{x}.$

Answer

$12\sin\text{x}-9\sin^2\text{x}=3\sin\text{x}(4-3\sin\text{x})$
We know that $0\le\sin\text{x}\le1$
$\therefore$ Maximum value of $12\sin\text{x}-9\sin^2\text{x}$
$=12\Big(\frac23\Big)-9\Big(23\Big)^2$
$=\frac{24}{3}-\frac{36}{9}$
$=\frac{36}{9}$
$=4$
$\because$ for other than $\sin\text{x}=\frac{4}{3}$ it is less than 4 Maximum value of $12\sin\text{x}-9\sin^2\text{x}$
$=4$

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Question 141 Mark

If A + B = C, then write the value of $\tan\text{A}\tan\text{B}\tan\text{C}.$

Answer

We have, A + B = C
$\Rightarrow\tan(\text{A+B})=\tan\text{C}$
$\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=\tan\text{C}$
$\Rightarrow\tan\text{A}+\tan\text{B}=\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow\tan\text{A}\tan\text{B}\tan\text{C}=\tan\text{C}-\tan\text{A}-\tan\text{B}$

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Question 151 Mark

Write the interval in which the value of $5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3$ lie.

Answer

If $\text{f(x)}=5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3,$ then
$\text{f(x)}=5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3$
$=5\cos+3\Big(\cos\theta\cos\frac\pi3-\sin\text{x}\sin\frac\pi3\Big)+3$
$=5\cos\text{x}+\frac32\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3$
$=\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3\cdots(\text{i})$
Now,
$-\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{3\sqrt{3}}{2}\Big)^3}\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}\le\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{3\sqrt{3}}{2}\Big)^2}$
$\Rightarrow-7\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}\le7,$ for all x
$\Rightarrow-7+3\le\frac{13}{2}\sin\text{x}+3\le7+3$
$\Rightarrow-4\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3\le10$
$\Rightarrow-4\le5\cos\text{x}-3\cos\Big(\text{x}+\frac\pi3\Big)+3\le10$ [Using 1]
Hence, the required in terval is [-4, 10]

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Question 161 Mark

If $\frac{\cos(\text{x - y})}{\cos(\text{x+y})}=\frac{\text{m}}{\text{n}},$ then write the value of $\tan\text{x}\tan\text{y}.$

Answer

We have, $\frac{\cos(\text{x - y})}{\cos(\text{x+y})}=\frac{\text{m}}{\text{n}}$
$\Rightarrow\frac{\cos\text{x}\cos\text{y}+\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}-\sin\text{x}\sin\text{y}}=\frac{\text{m}}{\text{n}}$
$\Rightarrow\frac{\frac{\cos\text{x}\cos\text{y}+}{\cos\text{x}\cos\text{y}}+\frac{\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}}}{\frac{\cos\text{x}\cos\text{y}}{\cos\text{x}\cos\text{y}}-\frac{\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}}}=\frac{\text{m}}{\text{n}}$ $[$Dividing and multiplying numerator and denominator by $\cos\text{x}\cos\text{y}]$
$\Rightarrow\frac{1+\tan\text{x}\tan\text{y}}{1-\tan\text{x}\tan\text{y}}=\frac{\text{m}}{\text{n}}$
$\Rightarrow\text{n+n}\tan\text{x}\tan\text{y}=\text{m - m}\tan\text{x}\tan\text{y}$
$\Rightarrow\tan\text{x}\tan\text{y}(\text{n+m})=\text{m - n}$
$\Rightarrow\tan\text{x}\tan\text{y}=\frac{\text{m - n}}{\text{m+n}}$

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