JEE MAIN - Mathematics (2003 - No. 7)
Let $$f(a) = g(a) = k$$ and their nth derivatives
$${f^n}(a)$$, $${g^n}(a)$$ exist and are not equal for some n. Further if
$$\mathop {\lim }\limits_{x \to a} {{f(a)g(x) - f(a) - g(a)f(x) + f(a)} \over {g(x) - f(x)}} = 4$$
then the value of k is
$${f^n}(a)$$, $${g^n}(a)$$ exist and are not equal for some n. Further if
$$\mathop {\lim }\limits_{x \to a} {{f(a)g(x) - f(a) - g(a)f(x) + f(a)} \over {g(x) - f(x)}} = 4$$
then the value of k is
0
4
2
1
Explanation
$$\mathop {\lim }\limits_{x \to a} {{f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ (By $$L'$$ Hospital rule)
$$\mathop {\lim }\limits_{x \to a} {{k\,\,g'\left( x \right) - k\,\,f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}} = 4$$
$$\therefore$$ $$k=4.$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ (By $$L'$$ Hospital rule)
$$\mathop {\lim }\limits_{x \to a} {{k\,\,g'\left( x \right) - k\,\,f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}} = 4$$
$$\therefore$$ $$k=4.$$
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