ফাংশনের মান নির্ণয়
If f(x)=coshx+sinhxf(x)=\cosh x+\sinh x f(x)=coshx+sinhx then f(x1+x2+.......+xn)=f(x_{1}+x_{2}+.......+x_{n})=f(x1+x2+.......+xn)=
111
f(x1+x2+.......)f(x_{1}+x_{2}+.......)f(x1+x2+.......)
000
f(x1).f(x2).......f(x3)f(x_{1}).f(x_{2}).......f(x_{3})f(x1).f(x2).......f(x3)
coshx,sinhx \cosh x, \sinh x coshx,sinhx এই গুলো Hyperbolic function.
f(x)=coshx+sinhx f(x)=\cosh x+\sinh x f(x)=coshx+sinhx
⇒f(x)=ex+e−x2+ex−e−x2 \Rightarrow f(x)=\frac{e^{x}+e^{-x}}{2}+\frac{e^{x}-e^{-x}}{2} ⇒f(x)=2ex+e−x+2ex−e−x
⇒f(x)=ex+e−x+ex−e−x2 \Rightarrow f(x)=\frac{e^{x}+e^{-x}+e^{x}-e^{-x}}{2} ⇒f(x)=2ex+e−x+ex−e−x
⇒f(x)=2ex2⇒f(x)=ex \begin{array}{l}\Rightarrow f(x)=\frac{2 e^{x}}{2} \\ \Rightarrow f(x)=e^{x}\end{array} ⇒f(x)=22ex⇒f(x)=ex
⇒f(x1+x2+x3⋯⋅)=ex1+x2+x3⋯⇒f(x1+x2+x3⋯ )=ex1⋅ex2⋅ex3⋯⇒f(x1+x2+x3⋯ )=f(x1)⋅f(x2)⋅f(x3)⋯ \begin{array}{l}\Rightarrow f\left(x_{1}+x_{2}+x_{3} \cdots \cdot\right)=e^{x_{1}+x_{2}+x_{3} \cdots} \\ \Rightarrow f\left(x_{1}+x_{2}+x_{3} \cdots\right)=e^{x_{1}} \cdot e^{x_{2}} \cdot e^{x_{3}} \cdots \\ \Rightarrow f\left(x_{1}+x_{2}+x_{3} \cdots\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right) \cdot f\left(x_{3}\right) \cdots\end{array} ⇒f(x1+x2+x3⋯⋅)=ex1+x2+x3⋯⇒f(x1+x2+x3⋯)=ex1⋅ex2⋅ex3⋯⇒f(x1+x2+x3⋯)=f(x1)⋅f(x2)⋅f(x3)⋯
Ai এর মাধ্যমে
১০ লক্ষ+ প্রশ্ন ডাটাবেজ
প্র্যাকটিস এর মাধ্যমে নিজেকে তৈরি করে ফেলো
উত্তর দিবে তোমার বই থেকে ও তোমার মত করে।
সারা দেশের শিক্ষার্থীদের মধ্যে নিজের অবস্থান যাচাই
f(x)=sinx f(x)=\sin x f(x)=sinx
f(π10)+f(π2−π10)f(π2−3π20) \frac{f\left(\frac{\pi}{10}\right)+f\left(\frac{\pi}{2}-\frac{\pi}{10}\right)}{f\left(\frac{\pi}{2}-\frac{3 \pi}{20}\right)} f(2π−203π)f(10π)+f(2π−10π) এর মান কত?
The solution set of x2+5x+6=0x^{2}+5x+6=0 x2+5x+6=0 is ........
If (2x+3)(3x−4)(x−1)(4x+5)=f(x)−59(x−1)+3118(4x+5)\dfrac {(2x+3)(3x-4)}{(x-1)(4x+5)}=f(x)-\dfrac {5}{9(x-1)}+\dfrac {31}{18(4x+5)}(x−1)(4x+5)(2x+3)(3x−4)=f(x)−9(x−1)5+18(4x+5)31, then f(x)=f(x)=f(x)=
Let A,B,CA,B,CA,B,C finite sets. Suppose then n(A)=10,n(B)=15,n(C)=20,n(A∩B)=8n(A)=10, n(B)=15, n(C)=20, n(A\cap B)=8n(A)=10,n(B)=15,n(C)=20,n(A∩B)=8 and n(B∩C)=9n(B\cap C)=9n(B∩C)=9. Then the possible value of n(A∪B∪C)n(A\cup B\cup C)n(A∪B∪C) is