Find the exact values of \(a\) and \(b\) if \(f\) is continuous and differentiable at \(x = 1\).

(i)     Use Rolle’s theorem, applied to \(f\), to prove that \(2{x^4} - 4{x^3} - 5{x^2} + 4x + 1 = 0\) has a root in the interval \(\left] { - 1,1} \right[\).

(ii)     Hence prove that \(2{x^4} - 4{x^3} - 5{x^2} + 4x + 1 = 0\) has at least two roots in the interval \(\left] { - 1,1} \right[\).

\(\mathop {{\text{lim}}}\limits_{x \to {1^ - }} {{\text{e}}^{ - {x^2}}}\left( { - {x^3} + 2{x^2} + x} \right) = \mathop {{\text{lim}}}\limits_{x \to {1^ + }}  (ax + b)\)   \(( = a + b)\)     M1

\(2{{\text{e}}^{ - 1}} = a + b\)     A1

differentiability: attempt to differentiate both expressions     M1

\(f'(x) =  - 2x{{\text{e}}^{ - {x^2}}}\left( { - {x^3} + 2{x^2} + x} \right) + {{\text{e}}^{ - {x^2}}}\left( { - 3{x^2} + 4x + 1} \right)\)   \((x < 1)\)     A1

(or \(f'(x) = {{\text{e}}^{ - {x^2}}}\left( {2{x^4} - 4{x^3} - 5{x^2} + 4x + 1} \right)\))

\(f'(x) = a\)   \((x > 1)\)     A1

substitute \(x = 1\) in both expressions and equate

\( - 2{{\text{e}}^{ - 1}} = a\)     A1

substitute value of \(a\) and find \(b = 4{{\text{e}}^{ - 1}}\)     M1A1

[8 marks]

(i)     \(f'(x) = {{\text{e}}^{ - {x^2}}}\left( {2{x^4} - 4{x^3} - 5{x^2} + 4x + 1} \right)\)   (for \(x \leqslant 1\))     M1

\(f(1) = f( - 1)\)     M1

Rolle’s theorem statement     (A1)

by Rolle’s Theorem, \(f'(x)\) has a zero in \(\left] { - 1,1} \right[\)     R1

hence quartic equation has a root in \(\left] { - 1,1} \right[\)     AG

(ii)     let \(g(x) = 2{x^4} - 4{x^3} - 5{x^2} + 4x + 1\).

\(g( - 1) = g(1) < 0\) and \(g(0) > 0\)     M1

as \(g\) is a polynomial function it is continuous in \(\left[ { - 1,0} \right]\) and \(\left[ {0,{\text{ 1}}} \right]\).     R1

(or \(g\) is a polynomial function continuous in any interval of real numbers)

then the graph of \(g\) must cross the x-axis at least once in \(\left] { - 1,0} \right[\)     R1

and at least once in \(\left] {0,1} \right[\).

[7 marks]