MathematicsQ1–1 of 1 questions
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1
MathematicsEasyMCQGSEB · 2010
વક્ર $y = \cot x$,રેખાઓ $x = \frac{\pi}{4}$,$x = \frac{\pi}{2}$ અને $X$-અક્ષ દ્વારા આવૃત પ્રદેશનું ક્ષેત્રફળ . . . . . . ચોરસ એકમ છે.
A
$\log 2$
B
$\frac{3}{2} \log 2$
C
$\frac{1}{2} \log 2$
D
$2 \log 2$
Solution
(C) ક્ષેત્રફળ $A$ એ $x = \frac{\pi}{4}$ થી $x = \frac{\pi}{2}$ સુધીના વિધેય $y = \cot x$ ના સંકલન દ્વારા મળે છે.
$A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot x \, dx$
આપણે જાણીએ છીએ કે $\int \cot x \, dx = \log |\sin x| + C$.
સીમાઓ લાગુ પાડતા:
$A = [\log |\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$A = \log |\sin(\frac{\pi}{2})| - \log |\sin(\frac{\pi}{4})|$
$A = \log(1) - \log(\frac{1}{\sqrt{2}})$
કારણ કે $\log(1) = 0$ અને $\log(\frac{1}{\sqrt{2}}) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$:
$A = 0 - (-\frac{1}{2} \log 2) = \frac{1}{2} \log 2$.
આમ,સાચો વિકલ્પ $C$ છે.
$A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot x \, dx$
આપણે જાણીએ છીએ કે $\int \cot x \, dx = \log |\sin x| + C$.
સીમાઓ લાગુ પાડતા:
$A = [\log |\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$A = \log |\sin(\frac{\pi}{2})| - \log |\sin(\frac{\pi}{4})|$
$A = \log(1) - \log(\frac{1}{\sqrt{2}})$
કારણ કે $\log(1) = 0$ અને $\log(\frac{1}{\sqrt{2}}) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$:
$A = 0 - (-\frac{1}{2} \log 2) = \frac{1}{2} \log 2$.
આમ,સાચો વિકલ્પ $C$ છે.
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