Loading [MathJax]/jax/output/HTML-CSS/jax.js


1)

 Let f.g:[-1,2]→  R,  be continuous  functions which are twice differentiable  on the interval (-1,2) . Let the values  of f and g at the points -1,0 and 2 be as given in the following table .

1132021436_m2.JPG

 In each of the intervals (-1,0) and (0,2), the function (f-3g)" never  vanishes. Then the correct statement(s) is/are


A) f'(x)-3g'(x)=0 has exactly three solutions in (1,0)(0,2)

B) f'(x)-3g'(x)=0 has exactly one solution in (-1,0)

C) f'(x)-3g'(x)=0 has exactly one solution (0,2)

D) f'(x)-3g'(x)=0 has exactly two solution in (-1,0) and exactly two solutions in (0,2)

Answer:

Option B,C

Explanation:

Let F(x)=f(x)-3g(x)

       F(-1)=3, F(0)=3 and F(2)=3

 So, F'(x) will vanish atleast twice in   (1,0)(0,2)

     F"(x)>0  or <0,xϵ(1,0)(0,2)

 Hence, f'(x) -3g'(x)=0 has exactly  one solution in (-1,0) and one solution in (0,2)