l_(1)((f_(1)-f_(0))/(2f_(1)-f_(0)-f_(2)))times h

Mode can be calculated by = l+((f_1-f_0)/(2f_1-f_0-f_2))xxh here f_1 represents ......

The maximum energies of photoelectrons emitted by a metal are E_(1) and E_(2) when the incident radiation has frequencies f_(1) and f_(2) respectively. Show that Planck's constant h and the work function W_(0) of the metal are h = (E_(1) - E_(2))/(f_(1) - f_(2)), W_(0) = (E_(1) f_(2) - E_(2) f_(1))/(f_(1) - f_(2))

The formula of mode is....a) L+[(f_1-f_0)/(2f_1-f_0-f_2)] xxh b) L-[(f_1-f_0)/(2f_1-f_0-f_2)] xxh c) L+[(f_1-f_0)/(2f_1-f_0+f_2)] xxh d) L+[(f_1-f_0)/(2f_2-f_0-f_2)] xxh

For x in R , x ne0, 1, let f_(0)(x)=(1)/(1-x) and f_(n+1)(x)=f_(0)(f_(n)(x)),n=0,1,2….. Then the value of f_(100)(3)+f_(1)((2)/(3))+f_(2)((3)/(2)) is equal to

For x in R , x ne0, 1, let f_(0)(x)=(1)/(1-x) and f_(n+1)(x)=f_(0)(f_(n)(x)),n=0,1,2….. Then the value of f_(100)(3)+f_(1)((2)/(3))+f_(2)((3)/(2)) is equal to

Let f(x)=a x^2+b x+cw h e ra in R^+a n db^2-4a c<0. Area bounded by y=f(x), x-axis and the lines x=0,x=1 is equal to 1/6(3f(1)+f(-1)+2f(0)) b. 1/(12)(5f(1)+f(-1)+8f(0)) c. 1/6(3f(1)-f(-1)+2f(0)) d. 1/(12)(5f(1)-f(-1)+8f(0))

Show that if f_(1) and f_(2) are one-one maps from R to R, then the product f_(1)xx f_(2):R rarr R defined by (f_(1)xx f_(2))(x)=f_(1)(x)f_(2)(x) need not be one-one.

f'(3)+f'(2)=0 Find the lim_(x rarr0)((1+f(3+x)-f(3))/(1+f(2-x)-f(2)))^((1)/(x))

If f'(2)=6 and f'(1)=4 ,then lim_(x rarr0)(f(x^(2)+2x+2)-f(2))/(f(1+x-x^(2))-f(1)) is equal to ?