CAT Exam Model Paper 1 with solutions for free online practice

Question : 94 of 100

Marks: +1, -0

If a is a non-zero real number and b is a positive real number and

2 b a ^ { 4 } + b a ^ { 3 } + a ^ { 2 } + b a + 2 b = k a ^ { 2 }

, then the least possible value of k is

Solution:

(a)

2 b a ^ { 4 } + b a ^ { 3 } + a ^ { 2 } + b a + 2 b = k a ^ { 2 }

⇒2 b a ^ { 2 } + b a + 1 + { b }/ { a } + { 2 b }/ { a ^ { 2 } } = k

⇒ 2 b ( a ^ { 2 } + { 1 } /{ a ^ { 2 } } ) + b ( a + { 1 }/ { a } ) + 1 = k

Let

\;\;\;\;\;\;a + { 1 } /{ a } = x

⇒

\;\a ^ { 2 } + { 1 } /{ a ^ { 2 } }+2 = x ^ { 2 }

⇒

\;\;\;\;\;\;a ^ { 2 } + { 1 } /{ a ^ { 2 } } = ( x ^ { 2 } - 2 )

⇒

2 b ( x ^ { 2 } - 2 ) + b x + 1 = k

⇒

2 b x ^ { 2 } + b x + ( 1 - 4 b ) = k

⇒

2 b [ x ^ { 2 } + { x }/ { 2 } ] + ( 1 - 4 b ) = k

⇒

2 b [ x + { 1 }/ { 4 } ] ^ { 2 } - { b }/ { 8 } + ( 1 - 4 b ) = k

⇒

2 b [ x + { 1 }/ { 4 } ] ^ { 2 } + 1 - { 33 b } /{ 8 } = k

k will be minimum when

2 b [ x + { 1 } /{ 4 } ] ^ { 2 }

is minimum.

{ \Now } , ( a + { 1 }/ { a } ) ^ { 2 } ≥4

⇒

x = ( a + { 1 }/ { a } ) = ( ∞ , - 2 ) ∪ ( 2 , ∞)

k will be minimum at x = − 2 or x = 2
As

( - 2 + { 1 } /{ 4 } ) ^ { 2 } < ( 2 + { 1 } /{ 4 } ) ^ { 2 }

So, k will be minimum at x = − 2.
∴

a + { 1 } /{ a } = - 2

∴a = − 1

k _ { \min } = { 2 b - b + 1 - b + 2 b } /{ 1 } = 2 b + 1

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