JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 13)
For every pair of continuous function f, g : [0, 1] $$\to$$ 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]
Explanation
Suppose f(x) is maximum at c1 and g(x) is maximum at c2. When f(x) is maximum g(x) may or may not be maximum.
Therefore, in the function h(x) = f(x) $$-$$ g(x), we get
$$h({c_1}) = f({c_1}) - g({c_1}) \ge 0$$ and $$h({c_2}) = f({c_2}) - g({c_2}) \ge 0$$.
Therefore, h(x) = 0 for some c $$\in$$ [0, 1].
Therefore, $$h(c) = 0 \Rightarrow f(c) - g(c) = 0$$.
Therefore, $$f(c) = g(c)$$.
Option (a) $$ \Rightarrow {f^2}(c) - {g^2}(c) + 3[f(c) - g(c)] = 0$$ which is true from Eq. (i).
Option (d) $$ \Rightarrow {f^2}(c) - {g^2}(c) = 0$$ which is true from Eq. (i)
Now, if we take
f(x) = 1 and g(x) = 1, $$\forall$$x $$\in$$[0, 1]
Option (b) and (c) does not hold. Hence, option (a) and (d) are correct.
Therefore, in the function h(x) = f(x) $$-$$ g(x), we get
$$h({c_1}) = f({c_1}) - g({c_1}) \ge 0$$ and $$h({c_2}) = f({c_2}) - g({c_2}) \ge 0$$.
Therefore, h(x) = 0 for some c $$\in$$ [0, 1].
Therefore, $$h(c) = 0 \Rightarrow f(c) - g(c) = 0$$.
Therefore, $$f(c) = g(c)$$.
Option (a) $$ \Rightarrow {f^2}(c) - {g^2}(c) + 3[f(c) - g(c)] = 0$$ which is true from Eq. (i).
Option (d) $$ \Rightarrow {f^2}(c) - {g^2}(c) = 0$$ which is true from Eq. (i)
Now, if we take
f(x) = 1 and g(x) = 1, $$\forall$$x $$\in$$[0, 1]
Option (b) and (c) does not hold. Hence, option (a) and (d) are correct.
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