If graph of the function `y=f(x)` is continuous and passes through point (3, 1) then `lim_(xrarr3) (log_(e)(3f(x)-2))/(2(1-f(x)))` is equal

If graph of the function y=f(x) is continuous and passes through point (3,1) then lim_(x rarr3)(log_(e)(3f(x)-2))/(2(1-f(x))) is equal (3)/(2) b.(1)/(2)c.-(3)/(2)d

If graph of the function y=f(x) is continuous and passes through point (3,1) then (lim)_(xvec3)((log)_e(3f(x)-2))/(2(1-f(x))) is equal 3/2 b. 1/2 c. -3/2 d. -1/2

If the graph of the function y = f(x) passes through the points (1, 2) and (3, 5), then int_(1)^(3)f'(x) dx=

Statement-1: If the graph of the function y=f(x) has a unique tangent at the point (a, 0) through which lim_(x->a) log_e (1+6(f(x)))/(3f(x))=2 Statement-2: Since the graph passes through (a, 0). Therefore f(a) = 0. When f(a) = 0 given limit iszero by zero form so that it can be evaluate by using L-Hospital's rule.

The graph of function y=f(x) has a unique tangent at (e^a,0) , through which the graph passes, then lim_(xrarre^a)(log(1+7f(x))-sin(f(x)))/(3f(x)) is equal to

y=f(x) is a continuous function such that its graph passes through (a,0). Then lim_(x rarr a)(log_(e)(1+3f(x)))/(2f(x))is:1(b)0(c)(3)/(2)(d)(2)/(3)

The graph of the function y=f(x) has a unique tangent at the point (a,0) through which the graph passes then,lim_(x rarr a)((ln(1+6f(x)))/(3f(x))

The value of lim_(xrarr3) ((x^(3)+27)log_(e)(x-2))/(x^(2)-9) is

The value of lim_(xrarr3) ((x^(3)+27)log_(e)(x-2))/(x^(2)-9) is

The value of lim_(xrarr3) ((x^(3)+27)log_(e)(x-2))/(x^(2)-9) is