Consider the following four statements
Statement:1 Let $f (x)$ and $g (x)$ be differentiable functions on R, such that $f (a) \geq g (a), a \in R$ and $f' (x) > g' (x)$ for all $x$ , then $f (x) > g (x)$ for all $x > a$ ,
Statement:2 If $f' (x) = g' (x)$ for all $x$ in an interval [a, b], that there is a constant $K$ such that $f (x) = g (x) + K$ for all $x$ in [a, b]. (where f(x) and g(x) are differentiable functions)
Statement:3 Let $f (x)$ and $g (x)$ be continuous functions such that $f (x) = g (x)$ for all rational $x$ . then $f (x) = g (x)$ for all .
Statement:4 Let $f (x)$ be a continuous function defined for every real $x \in R$ . For any real numbers $' a'$ and $' b'$ that satisfy $a < b, f (x)$ always satisfies $f (a) > f (b)$ , there exists a real number x for which $f (x) = f (x + 1)$
Number of correct statements, is

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answer is C.

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Detailed Solution

Statement:1 The function $h (x) = f (x) - g (x)$ is differentiable, $h (a) \geq 0$ , and $h^{'} (x) > 0$ for all $x$ . By the latter condition, $h (x)$ is increasing, and, therefore, since $h (a) \geq 0, h (x) > 0$ for all $x > a$ . Thus, $f (x) > g (x)$ for all $x > a$ .
Statement:2 Let $h (x) = f (x) - g (x)$ . Then $h^{'} (x) = 0$ for all $x$ in [a, b]. There exists a constant $K$ such that $h (x) = K$ for all $x$ in [a, b]. Hence, $f (x) = g (x) + K$ for all $x$ in [a, b].Statement:3 Consider any real number $c$ . Since $f (x)$ is continuous at $c, \lim_{x \to c} f (x) = f (c)$ . But, since there are rational numbers arbitrarily close to $c, f (x) = 0$ for values of arbitrarily close to , and, therefore, $\lim_{x \to c} f (x) = 0$ . Hence, $f (c) = 0$ .