Question 9:
In order that the given function #f# be differentiable, we must have
#f'(1-)=f'(1+)........................(ast)#.
Recall that, #f'(1-)=lim_(x to 1-) (f(x)-f(1))/(x-1)#.
Now, as # x to 1-, x > 1 :. f(x)=3x." Also, "f(1)=3#.
#:. f'(1-)=lim_(x to 1-)(3x-3)/(x-1)#,
#=lim_(xto 1-){3(x-1)}/(x-1)#,
#=lim_(x to 1-)3#.
#rArr f'(1-)=3........................(ast^1)#.
Next, #f'(1+)=lim_(x to 1+)(Bx^2+Cx-3)/(x-1)#.
#(ast) and (ast^1) rArr lim_(x to 1+)(Bx^2+Cx-3)/(x-1)=3#.
In order that the above finite limit exist, we must have,
#(x-1)|(Bx^2+Cx-3) rArr B+C=3............(1)#.
Replacing #3" by "(B+C)# in the limit, we get,
#f'(1+)=lim_(x to 1+)(Bx^2+Cx-B-C)/(x-1)#,
#=lim_(x to 1+){B(x^2-1)+C(x-1)}/(x-1)#,
#=lim_(x to 1+){B(x+1)+C}#.
# rArr f'(1+)=2B+C#.
Hence, by #(ast) and (ast^1) rArr 2B+C=3...(2)#.
Solving #(1) and (2), B=0, C=3#.
So, the Right Option is #c)@{B=0,C=3}#.
Question 10:
#f(-3)=-4#, means the point of contact of the tgt. line is
#P(-3,-4)#.
#f'(-3)=6#, means the slope of the tgt. at the point #P# is #6#.
Utilising the Slope-Point Form of line, the eqn. of tgt. is,
#y-(-4)=6(x-(-3)) rArr y=6x+14#.
#:." The Right Option is "b)@y=6x+14#.
Enjoy Maths., and Spread the Joy!
