JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 8)
If $$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|,$$ then $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$
does not exist.
exists and is equal to 2.
existsand is equal to 0.
exists and is equal to $$-$$ 2.
Explanation
Given,
$$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|$$
= cosx(x2 - 2x2) - x(2 sinx - 2x tanx) + (2x sinx - x2 tanx)
= x2 (tanx - cosx)
$$ \therefore $$ $${f^{'}}(x)$$ = 2x (tanx - cosx) + x2(sec2x + sinx)
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$
= $$\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$$
= $$\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$$
= 2 (0-1) + 0
= -2
$$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|$$
= cosx(x2 - 2x2) - x(2 sinx - 2x tanx) + (2x sinx - x2 tanx)
= x2 (tanx - cosx)
$$ \therefore $$ $${f^{'}}(x)$$ = 2x (tanx - cosx) + x2(sec2x + sinx)
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$
= $$\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$$
= $$\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$$
= 2 (0-1) + 0
= -2
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