JEE Advanced 2020 Concept Recapitulation Test 3 Paper 2

Vertex of f(x) will be at x=‌

a

2

Case - I: ‌‌ If a<0, then maximum value of f(x) is at x=0
∴f(0)=−5⇒a2+2a−3=0
⇒a=−3 or 1⇒a=−3
Case - II : ‌‌ If a>2, then maximum value of f(x) is at x=1 ∴f(1)=−5⇒a2+a−2=0
⇒a=−2 or 1which is not possible. Case −111:‌‌ If 0<a<2, then maximum value is at x=‌

a

2

⇒f(‌

a

2

)=−5⇒3a2+8a−12=0⇒a=‌

−4±2√13

3

⇒a=‌

−4+2√13

3

is possible