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Algebra Level 5

Find the range of value of λ \lambda for which the expression 2 x 2 5 x + 3 4 x λ \dfrac{2x^2-5x+3}{4x-\lambda} can take all real values for x R { λ 4 } x \in \mathbb R - \left\{ \frac{\lambda}{4} \right\} .

( 4 , 6 ) (4,6) [ 4 , 6 ) [4,6) [ 4 , 6 ] [4,6] ( 4 , 6 ] (4,6] None of these

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1 solution

Tanishq Varshney
Nov 14, 2015

A standard approach will be to find the derivative of the above expression. If a function is strictly increasing or strictly decreasing along the real line then the function takes all real values.

In this case the derivative comes out to be 8 x 2 4 λ x + 5 λ 12 ( 4 x λ ) 2 \large{\frac{8x^2-4\lambda x+5\lambda -12}{(4x-\lambda)^2}}

We notice that the numerator part is an concave up parabola and the denominator is always greater than zero in the provided domain.

Thus the derivative must be greater than zero.

8 x 2 4 λ x + 5 λ 12 > 0 \large{8x^2-4\lambda x+5\lambda -12>0}

This implies that it's discriminant should be less than zero.

16 λ 2 32 ( 5 λ 12 ) < 0 \large{16 \lambda^2-32(5\lambda -12)<0}

( λ 4 ) ( λ 6 ) < 0 \large{(\lambda-4)(\lambda-6)<0}

λ ( 4 , 6 ) \large{\boxed{\lambda \in (4,6)}}

But if lambda = 4 or 6 then the given function becomes linear so it is capable of taking all values? .

please reply @Sandeep Bhardwaj sir

Prakhar Bindal - 5 years, 3 months ago

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The linear function will not have 4 and 6 in its domain , so the corresponding values will be missing.

Deepanshu Vijay - 5 years, 2 months ago

@Deepanshu why does linear function not have all the values in its domain?

ABC def - 4 years, 3 months ago

Lambda couldn't be 4 or 6. Why ??

Vishal Yadav - 4 years, 2 months ago

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