JEE MAIN - Mathematics (2002 - No. 5)
Let $$f(2) = 4$$ and $$f'(x) = 4.$$
Then $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$ is given by
Then $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$ is given by
$$2$$
$$- 2$$
$$- 4$$
$$3$$
Explanation
Given,
$$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$
= $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( 2 \right) + 2f\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$
= $$\mathop {\lim }\limits_{x \to 2} {{f\left( 2 \right)\left( {x - 2} \right) - 2\left\{ {f\left( x \right) - f\left( 2 \right)} \right\}} \over {x - 2}}$$
= $$f\left( 2 \right) - 2\mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}$$
$$\left[ {As\,f'\left( x \right) = \mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}} \right]$$
= $$f\left( 2 \right) - 2f'\left( x \right)$$
= 4 - 2 $$ \times $$ 4
= - 4
$$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$
= $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( 2 \right) + 2f\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$
= $$\mathop {\lim }\limits_{x \to 2} {{f\left( 2 \right)\left( {x - 2} \right) - 2\left\{ {f\left( x \right) - f\left( 2 \right)} \right\}} \over {x - 2}}$$
= $$f\left( 2 \right) - 2\mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}$$
$$\left[ {As\,f'\left( x \right) = \mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}} \right]$$
= $$f\left( 2 \right) - 2f'\left( x \right)$$
= 4 - 2 $$ \times $$ 4
= - 4
Comments (0)
