JEE Novice - (19)

Algebra Level 4

− m ≤ x − 20 m \Large -m\leq x - 20\sqrt{m}

Find the smallest value of x x which satisfies the above inequality for all non-negative ' m m '.

This question is a part of JEE Novices .


The answer is 100.

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Nihar Mahajan by Nihar Mahajan , 20, India

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1 solution

We require that m − 20 m + x = ( m ) 2 − 20 m + x ≥ 0. m - 20\sqrt{m} + x = (\sqrt{m})^{2} - 20\sqrt{m} + x \ge 0.

This is a quadratic in m , \sqrt{m}, so the minimum possible value of x x that allows this inequality to be satisfied will be such that the discriminant is 0 , 0, , (i.e., when the parabola is tangent to the x x -axis). This means that

( − 20 ) 2 − 4 x = 0 ⟹ x = 100 . (-20)^{2} - 4x = 0 \Longrightarrow x = \boxed{100}.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Another way would be to complete the square.

∀ m ∈ R 0 + , ( m ) 2 − 20 m + x ≥ 0 ⟺ ( m − 10 ) 2 + x ≥ 100 \forall~m\in\Bbb{R_0^+}~,~(\sqrt m)^2-20\sqrt m+x\geq 0\iff (\sqrt m-10)^2+x\geq 100

By the trivial inequality , we have ( m − 10 ) 2 ≥ 0 ∀ m ∈ R 0 + (\sqrt m-10)^2\geq 0~\forall~m\in\Bbb{R_0^+} . Hence, we have,

x ≥ 100 − ( m − 10 ) 2 ≥ 100 − 0 = 100 ⟹ x ≥ 100 x\geq 100-(\sqrt m-10)^2\geq 100-0=100\implies x\geq 100

with equality (for all the inequalities so far) iff ( m − 10 ) = 0 ⟹ m = 100 (\sqrt m-10)=0\implies m=100 .

Prasun Biswas - 6 years ago

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∵ ( m − 10 ) 2 ≥ 0 ⇒ 100 − ( m − 10 ) 2 ≤ 100 \because \left(\sqrt{m} - 10\right)^2 \geq 0 \Rightarrow 100 - \left(\sqrt{m} - 10\right)^2 \leq 100 and not ≥ 100 \geq 100

Ankit Kumar Jain - 4 years, 2 months ago

what if M=1, shouldn't then the answer will be x>20 or x=20. why 100 when we can have x=20!?

Harshit Sharma - 5 years, 11 months ago

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But we need to find the smallest value of x x such that the given inequality holds true for all non-negative m , m, and not just m = 1. m = 1. For example, the inequality is not true for x = 20 , m = 4. x = 20, m = 4.

Brian Charlesworth - 5 years, 11 months ago

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