JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 1)
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4.
Then $$\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$$ equals :
Then $$\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$$ equals :
4 $$-$$ 2a
2a + 4
a + 4
2a $$-$$ 4
Explanation
$$L = \mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$$ [$${0 \over 0}$$ form]
Using L' Hospital rule we get
$$L = \mathop {\lim }\limits_{x \to a} {{f(a) - af'(x)} \over 1}$$
$$f(a) - af'(a) = 4 - 2a$$
Using L' Hospital rule we get
$$L = \mathop {\lim }\limits_{x \to a} {{f(a) - af'(x)} \over 1}$$
$$f(a) - af'(a) = 4 - 2a$$
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