JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 20)
Let a function g : [ 0, 4 ] $$\to$$ R be defined as
$$g(x) = \left\{ {\matrix{ {\mathop {\max }\limits_{0 \le t \le x} \{ {t^3} - 6{t^2} + 9t - 3),} & {0 \le x \le 3} \cr {4 - x,} & {3 < x \le 4} \cr } } \right.$$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.
$$g(x) = \left\{ {\matrix{ {\mathop {\max }\limits_{0 \le t \le x} \{ {t^3} - 6{t^2} + 9t - 3),} & {0 \le x \le 3} \cr {4 - x,} & {3 < x \le 4} \cr } } \right.$$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.
Answer
1
Explanation
$$f(x) = {x^3} - 6{x^2} + 9x - 3$$
$$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$$
$$f(1) = 1$$, $$f(3) = 3$$
$$g(x) = \left[ {\matrix{ {f(9x)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$$
g(x) is continuous
$$g'(x) = \left[ {\matrix{ {3(x - 1)(x - 3)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$$
g(x) is non-differentiable at x = 3
$$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$$
$$f(1) = 1$$, $$f(3) = 3$$
$$g(x) = \left[ {\matrix{ {f(9x)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$$
g(x) is continuous
$$g'(x) = \left[ {\matrix{ {3(x - 1)(x - 3)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$$
g(x) is non-differentiable at x = 3
Comments (0)
