1 Marks Question - MATHS STD 11 Science Questions - Vidyadip

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

Question 11 Mark

If $\tan\text{A}+\cot\text{A}=4,$ then write the value of $\tan^4\text{A}+\cot^4\text{A}.$

Answer

$(\tan\text{A}+\cot\text{A})^2=\tan^2\text{A}+\cot^2\text{A}+2\tan\text{A}\cot{A}$
$16=\tan^2\text{A}+\cot^2\text{A}+2$
$\tan^2\text{A}+\cot^2\text{A}=14$
$(\tan^2\text{A}+\cot\text{A})^4=\tan^4\text{A}+1\tan^3\text{A}\cot\text{A}6\tan^2\text{A}\cot^2\text{A}+4\tan\text{A}\cot^3\text{A}+\cot^4\text{A}$
$256=\tan^4\text{A}+4(\tan^2\text{A}+\cot^2\text{A})+\cot^4\text{A}+6$
$256=\tan^4\text{A}+4(14)+\cot^4\text{A}+6$
$\tan^4\text{A}+\cot^4\text{A}=194$

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Question 21 Mark

Write the maximum and minimum values of $\cos(\cos\text{x}).$

Answer

$-1\le\cos(\text{x})\le1$ For $\text{x }\in [0, 2\pi]$
$\cos(\text{x})$ is maximum at x = 0 and minimum at $\text{x}=\pi.$
$\cos(\cos(0))=\cos(1)=0.9984$
$\cos(\cos(\pi))=\cos(-1)=\cos(1)$
$\cos\Big(\cos\Big(\frac{\pi}{2}\Big)\Big)=\cos(0)=1$
The maximum value of $\cos(\cos(\text{x}))$ is 1.
The manimum value of
$\cos(\cos(\text{x}))$ is $\cos(1).$

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Question 31 Mark

Write the value of $2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1.$

Answer

$2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1.$
$=2((\sin^3\text{x})^2)+(\cos^2\text{x})^2)-3((\sin^2\text{x})^2+(\cos^2\text{x})^2)+1$
$=2(1)-3(1)+1$
$=2-3+1$
$=0$

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Question 41 Mark

If $\sin\text{x}=\text{cosec}\text{ x}=2,$ then write the value of $\sin^\text{n}\text{x}\text{+ cosec}^\text{n}\text{ x}.$

Answer

$\sin\text{x}+\text{cosec x}=2$
$\Rightarrow\sin\text{x}+\frac{1}{\sin\text{x}}=2$
$\Rightarrow\frac{\sin\text{x}+1}{\sin\text{x}}=2$
$\Rightarrow\sin^2\text{x}+1=2\sin\text{x}$
$\Rightarrow\sin^2\text{x}+1-2\sin\text{x}=0$
$\Rightarrow(\sin\text{x}-1)^2=0$
$\Rightarrow\sin\text{x}-1=0$
$\Rightarrow\sin\text{x}=1$
And, $\text{cosec x}=\frac{1}{\sin\text{x}}=1$
$\therefore\sin^\text{n}\text{x}+\text{cosec}^\text{n}\text{x}=1^\text{n}+1^\text{n}$
$=1+1=2$

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Question 51 Mark

If $\sin^2\text{x}+\sin^2\text{ x}=1,$ then write the value of $\cos^8\text{x}+2\cos^6\text{x}+\cos^4\text{ x}.$

Answer

$\sin\text{x}+\sin^2\text{x}=1$ $\sin\text{x}=1-\sin^2\text{x}=\cos^2\text{x}$$\cos^8+2\cos^6\text{x}+\sin^2\text{x}$
$=\sin^4\text{x}+2\sin^3\text{x}+\sin^2\text{x}$
$=(\sin\text{x}+\sin^2\text{x})^2$
$=1$

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Question 61 Mark

Write the maximum values of $\sin(\cos\text{x}).$

Answer

$\cos\text{x}$ is maximum at x = 0.
$\sin(\cos(0))=\sin(1)$
The maximum value of $\sin(\cos(\text{x}))$ is $\sin(1).$

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Question 71 Mark

If $\sin\text{x}=\sin^2\text{ x}=1,$ then write the value of $\cos^{12}\text{x}+3\cos^{10}\text{x}+3\cos^8\text{x}+\cos^6\text{x}.$

Answer

$\sin\text{x}+\sin^2\text{x}=1$
$\sin\text{x}={1}-{\sin^2\text{x}}=\cos^2\text{x}$
$\cos^{12}\text{x}+3\cos^{10}\text{x}+3\cos^{8}\text{x}+\cos^6\text{x}$
$=\sin^6\text{x}+3\sin^5\text{x}+3\sin^4\text{x}\sin^3\text{x}$
$=(\sin\text{x}+\sin^2\text{x})^3$
$=1$

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Question 81 Mark

Write the value of $\sin10^\circ+\sin20^\circ+\sin30^\circ+...+\sin360^\circ.$

Answer

$\sin(360^\circ-\theta)=-\sin\theta$
$\sin10^\circ+\sin20^\circ+...+\sin360^\circ$
$=\sin(360^\circ=350^\circ)+\sin(360^\circ-340^\circ)+\ ...\ \\\ +\sin180^\circ+\sin190^\circ+\ ...\ +\sin360^\circ$
$=-\sin350^\circ-\sin340^\circ-\ ...\ =\sin190^\circ\\\ \ \ \ +\sin180^\circ+\sin190^\circ+\ ...\ +\sin360^\circ$
$=\sin180^\circ+\sin360^\circ$
$=0+0$
$=0$

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Question 91 Mark

Write the maximum and minimum values of $\sin(\sin\text{x}).$

Answer

$-1\le\sin(\text{x})\le1$ For $\text{x }\in [0, 2\pi]$
$\sin(\text{x})$ is maximum at $\text{x}=\frac{\pi}{2}$ and minimum at $\text{x}=\frac{3\pi}{2}.$
$\sin\Big(\sin\Big(\frac{\pi}{2}\Big)\Big)=\sin(1)=0.017$
$\sin(\sin(\pi))=\sin(1)=-\sin(1)=-0.017$
The maximum value of $\sin(\sin(\text{x}))$ is $\sin(1).$
The manimum value of $\sin(\sin(\text{x}))$ is $-\sin(1).$

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Question 101 Mark

If $\text{x}=\sin^{14}\text{x}+\cos^{20}\text{x},$ then write the smallest interval in which the value of x lie.

Answer

$\sin\text{x}$ lies between [-1, 1]
$\sin^{14}\text{x}$ (power is even) will lie between [0, 1]
Simillarly,
$\cos\text{x}$ lies between [-1, 1]
$\cos^{20}\text{x}$ (power is even) will lie between [0, 1]
$\therefore \sin^{14}\text{x}+\cos^{20}\text{x}$ Will have maximum value 1.
$(\because$ when $\sin\text{x}=0, \cos\text{x}=1$and vice a versa $)$
$\therefore\text{x}=\sin^{14}\text{x}+\cos^{20}\text{x}$ lies in the (0, 1) interval.

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Question 111 Mark

If $\cot(\alpha+\beta)=0,$ then write the value of $\sin(\alpha+2\beta).$

Answer

$\cot(\alpha+\beta)=0$
$\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}=0$
$\therefore\cos(\alpha+\beta)=0$
$\therefore\cos\alpha\cos\beta=\sin\alpha\sin\beta$
$\sin(\alpha+2\beta)$
$=\sin\alpha\cos2\beta+\cos\alpha\sin2\beta$
$=\sin\alpha(\cos^2\beta-\sin^2\beta)+\cos\alpha(2\sin\beta\cos\beta)$
$=\sin\alpha(\cos^2\beta-\sin^2\beta)+2\sin^2\beta(\sin\alpha\sin\beta)$
$=\sin\alpha(\cos^2\beta-\sin^2\beta+2\sin^2\beta)$
$=\sin\alpha(\cos^2\beta+\sin\beta)$
$=\sin\alpha$
$\cos(\alpha+\beta)=0$
$\Rightarrow\alpha+\beta=\frac\pi2$
$\Rightarrow\alpha=\frac\pi2-\beta$
$\therefore\sin(\alpha+2\beta)=\sin\alpha$
$\sin(\alpha+2\beta)=\sin\Big(\frac\pi2-\beta\Big)=\cos\beta$
$\therefore\sin(\alpha+2\beta)=\sin\alpha \text{ or }\cos\beta$

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Question 121 Mark

If $3\sin\text{x}+5\cos\text{x}=5,$ then write the value of $5\sin\text{x}-3\cos\text{x}.$

Answer

$3\sin\text{x}+5\cos\text{x}=5$ (Given)
Squaring both the side:
$9\sin^2\text{x}+25\cos^2\text{x}+30\sin\text{x}\cos\text{x}=25$
$30\sin\text{x}\cos\text{x}=25-9\sin^2\text{x}-25\cos^2\text{x}\cdots(1)$
We have to find the value of $5\sin\text{x}-3\cos\text{x}.$
$(5\sin\text{x}-3\cos\text{x})^2=25\sin^2\text{x}+9\cos^2\text{x}-30\sin\text{x}\cos\text{x}$
$(5\sin\text{x}-3\cos\text{x})^2=25\sin^2\text{x}+6\cos^2\text{x}-(25-9\sin^2\text{x}-25\cos^2\text{x})$ [From (1)]
$(5\sin\text{x}-3\cos\text{x})^2=34\sin^2\text{x}+34\cos^2\text{x}-25$
$(5\sin\text{x}-3\cos\text{x})^2=34-25$ $(\because\sin^2\text{x}+\cos^2\text{x}=1)$
$(5\sin\text{x}-3\cos\text{x})^2=9$
$5\sin\text{x}-3\cos\text{x}=\pm3$

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Question 131 Mark

Write the value of $\cos1^\circ+\cos2^\circ+\cos3^\circ+\ ...\ +\cos180^\circ.$

Answer

$\cos(180^\circ-\theta)=-\cos\theta$
$\cos1^\circ+\cos2^\circ+\ ...\ +\cos180^\circ$
$=\cos(180^\circ-179^\circ)+\cos(180^\circ-178^\circ)+\ ...\ +\cos90^\circ+\cos91^\circ+\ ...\ +\cos180^\circ$
$=-\cos179^\circ-\cos178^\circ-\ ...\ -\cos91^\circ+\cos90^\circ+\cos91^\circ+\ ...\ +\cos180^\circ$
$=\cos90^\circ+\cos180^\circ$
$=0-1$
$=-1$

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Question 141 Mark

Write the value of $2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1.$

Answer

$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)+1$
$=2((\sin^3\theta)^2)+(\cos^2\theta)^2)-3((\sin^2\theta)^2+(\cos^2\theta)^2)+1$
$=2(1)-3(1)+1$
$=2-3+1$
$=0$

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Question 151 Mark

Write the least value of $\cos^2\text{x}+\sec^2\text{x}.$

Answer

$-1\le\cos\text{x}\le1$
$\Rightarrow1\le\cos^2\text{x}\le1$
$-1\le\frac{1}{\cos\text{x}}\le1$
$\Rightarrow1\le\frac{1}{\cos^2\text{x}}\le1$
$\Rightarrow1\le\sec^2\text{x}\le1$
$2\le\cos^2\text{x}+\sec^2\text{x}$.
So the least value of $\cos^2\text{x}+\sec^2\text{x}$ is 2.

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Question 161 Mark

If $\sin\theta_1+\sin\theta_2+\sin\theta_3=3,$ then write the valuue of $\cos\theta_1+\cos\theta_2+\cos\theta_3.$

Answer

The maximum value for $\sin(\text{x})$ is 1 for all x.
So,
$\Rightarrow\sin\theta_1+\sin\theta_2+\sin\theta_3=3,$
$\Rightarrow\sin\theta_1=\sin\theta_2=\sin\theta_3=1$
$\Rightarrow\theta_1=\theta_2=\theta_3=90^\circ$
$\therefore\cos\theta_1=\cos\theta_2=\cos\theta_3=0$
$\therefore\cos\theta_1+\cos\theta_2+\cos\theta_3=0$

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Question 171 Mark

If $\sin\text{x}=\cos^2\text{x},$ then write the value of $\cos^2\text{x}(1+\cos^2\text{x}).$

Answer

$\cos^2\text{x}(1=\cos^2\text{x})=\cos^2\text{x}+(\cos^2\text{x})^2$
$=\cos^2\text{x}+\sin^2\text{x}$ $(\because\sin\text{x}=\cos^2\text{x})$
$=1$

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