Given $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$       such that $$x = 0$$  is the only real root of $$P'\left( x \right) = 0.$$   If $$P\left( { - 1} \right) < {\text{ }}P\left( 1 \right),$$   then in the interval [1, -1]:

Solution :
$$\eqalign{ & {\text{We}}\,{\text{have}}\,P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d \cr & \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx + c \cr & {\text{But }}P'\left( 0 \right) = 0 \Rightarrow c = 0 \cr & \therefore \,P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d \cr & {\text{As}}\,{\text{given}}\,{\text{that}}\,P\left( { - 1} \right) < P\left( a \right) \cr & \Rightarrow \,1 - a + b + d\, < \,1 + a + b + d\, \Rightarrow a > 0 \cr & {\text{Now }}P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2}{\text{ }} + {\text{ }}3ax + 2b} \right) \cr & {\text{As }}P'\left( x \right) = 0,{\text{ there is only one solution }}x = 0, \cr & {\text{therefore }}4{x^2} + 3ax + 2b = 0{\text{ should not have any real roots i}}{\text{.e }}D < 0 \cr & \Rightarrow \,9{a^2} - 32b < 0 \Rightarrow b > \frac{{9{a^2}}}{{32}} > 0 \cr & {\text{Hence}}\,a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0\,\forall x > 0 \cr & \therefore \,P\left( x \right)\,{\text{is}}\,{\text{an}}\,{\text{increasing}}\,{\text{function}}\,{\text{on}}\,\left( {0,1} \right) \cr & \therefore \,P\left( 0 \right) < P\left( a \right) \cr & {\text{Similarly we can prove }}P\left( x \right)\,{\text{is decreasing on}}\,\left( { - 1,0} \right) \cr & \therefore \,P\left( { - 1} \right) > P\left( 0 \right) \cr & {\text{So we can conclude that}} \cr & {\text{Max}}\,P\left( x \right) = P\left( 1 \right)\,{\text{and}}\,{\text{Min}}\,P\left( x \right) = P\left( 0 \right) \cr & \Rightarrow P\left( { - 1} \right)\,{\text{is}}\,{\text{not minimum but }}P\left( 1 \right){\text{ is the maximum of }}P. \cr} $$