লঘুমান গুরুমান বিষয়ক
x এর মান কত হলে f(x)=sinx−cosx−xf (x)=sinx-cosx-xf(x)=sinx−cosx−x এর চরম মান পাওয়া যাবে?
π2\frac{\pi}{2} 2π
π6\frac{\pi}{6} 6π
π3\frac{\pi}{3} 3π
π4\frac{\pi}{4} 4π
f′(x)=cosx+sinx−1=0 f^\prime\left(x\right)=cosx+sinx-1=0\ f′(x)=cosx+sinx−1=0
⇒12cosx+12 sinx=12\Rightarrow\frac{1}{\sqrt2}cosx+\frac{1}{\sqrt2}\ sinx=\frac{1}{\sqrt2}⇒21cosx+21 sinx=21
⇒cos(x−π4)=cosπ4∴ x=2nπ±π4+π4\Rightarrow\cos{\left(x-\frac{\pi}{4}\right)}=\cos{\frac{\pi}{4}} \therefore\ x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{4}⇒cos(x−4π)=cos4π∴ x=2nπ±4π+4π
n=0n=0n=0 হলে, x=0,π2x=0,\frac{\pi}{2}x=0,2π
দৃশ্যকল্প-I: y(x+1)(x+2)−x+4 y(x+1)(x+2)-x+4 y(x+1)(x+2)−x+4
দৃশ্যকল্প-II: g(x)=3x3−6x2−5x+1 \mathrm{g}(\mathrm{x})=3 \mathrm{x}^{3}-6 \mathrm{x}^{2}-5 \mathrm{x}+1 g(x)=3x3−6x2−5x+1
Let f(x)={x3/5x≤1−(x−2)3x>1f\left( x \right) =\left\{ \begin{matrix} { x }^{ { 3 }/{ 5 } }\quad \quad \quad x\le 1 \\ -{ \left( x-2 \right) }^{ 3 }\quad x>1 \end{matrix} \right. f(x)={x3/5x≤1−(x−2)3x>1
then the number of critical points on the graph of the function is
If for all x,yx, yx,y the function f is defined by; f(x)+f(y)+f(x)⋅f(y)=1f(x)+f(y)+f(x)\cdot f(y)=1f(x)+f(y)+f(x)⋅f(y)=1 and f(x)>0f(x) > 0f(x)>0.When f(x)f(x)f(x) is differentiable f′(x)=f'(x)= f′(x)=,
x এর কোন মানের জন্য f(x)=xlnx f{\left ( x \right )} = \frac{x}{\ln{x}} f(x)=lnxx সর্বনিম্ন হবে-
