JEE Advanced 2014 · Paper 1 · Mathematics Q42

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MathematicsMultiple correct+3Question 42 of 60

For every pair of continuous functions $f,g:[0,1]\to\mathbb R$ such that
$\max\{f(x):x\in[0,1]\}=\max\{g(x):x\in[0,1]\},$
the correct statement(s) is(are):

(f(c))^2+3f(c)=(g(c))^2+3g(c)

for some

c\in[0,1]

(f(c))^2+f(c)=(g(c))^2+3g(c)

for some

c\in[0,1]

(f(c))^2+3f(c)=(g(c))^2+g(c)

for some

c\in[0,1]

(f(c))^2=(g(c))^2

for some

c\in[0,1]

Answer and worked solution

Official answer: A, D

A common point. Let $h=f-g$ , and let the common maximum $m$ be attained by $f$ at $x_1$ and by $g$ at $x_2$ . Then
$h(x_1)=m-g(x_1)\ge0,\qquad h(x_2)=f(x_2)-m\le0.$
By the intermediate value theorem there is $c$ with $h(c)=0$ , i.e. $f(c)=g(c)$ .

(A) At that $c$ , $f(c)^2+3f(c)=g(c)^2+3g(c)$ . True.
(D) Likewise $f(c)^2=g(c)^2$ . True.
(B), (C) Take $f=g\equiv1$ : (B) would need $2=4$ and (C) $4=2$ at some point. Not always true.

(A), (D)

Topics:Continuity

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JEE Advanced 2014 · Paper 1, Mathematics question 42