Let f : R rightarrow R$f : R \rightarrow R$ be a twice differentiable function such that f''(x) > 0$f''(x) > 0$ for all x in R$x \in R$ and f'(a-1) = 0$f'(a-1) = 0$, where a$a$ is a real number. Let g(x) = f(tan^2 x - 2tan x + a)$g(x) = f(\tan^2 x - 2\tan x + a)$, 0 < x < fracpi2$0 < x < \frac{\pi}{2}$. Consider the following two statements: (I) g$g$ is increasing in left(0, fracpi4right)$\left(0, \frac{\pi}{4}\right)$ (II) g$g$ is decreasing in left(fracpi4, fracpi2right)$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Then,

### Related Formula g'(x) = f'(h(x)) cdot h'(x)$$g'(x) = f'(h(x)) \cdot h'(x)$$ For monotonicity, g'(x) > 0 implies$g'(x) > 0 \implies$ increasing, and g'(x) < 0 implies$g'(x) < 0 \implies$ decreasing. ### Core Logic Given g(x) = f((tan x - 1)^2 + a - 1)$g(x) = f((\tan x - 1)^2 + a - 1)$. Differentiating w.r.t. x$x$: g'(x) = f'((tan x - 1)^2 + a - 1) cdot 2(tan x - 1)sec^2 x$$g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2(\tan x - 1)\sec^2 x$$ ### Step 1: Analyze the Sign of the Derivative We know f''(x) > 0 implies f'(x)$f''(x) > 0 \implies f'(x)$ is strictly increasing. Since f'(a-1) = 0$f'(a-1) = 0$, for any input X > a-1$X > a-1$, f'(X) > 0$f'(X) > 0$. Here, the input to f'$f'$ is (tan x - 1)^2 + a - 1$(\tan x - 1)^2 + a - 1$. Since (tan x - 1)^2 geq 0$(\tan x - 1)^2 \geq 0$, it is strictly positive for x neq fracpi4$x \neq \frac{\pi}{4}$ in the given interval. Thus, (tan x - 1)^2 + a - 1 geq a - 1 implies f'((tan x - 1)^2 + a - 1) > 0$(\tan x - 1)^2 + a - 1 \geq a - 1 \implies f'((\tan x - 1)^2 + a - 1) > 0$ for all valid x$x$. ### Step 2: Determine Intervals of Monotonicity Now, the sign of g'(x)$g'(x)$ depends solely on the term (tan x - 1)$(\tan x - 1)$ because 2sec^2 x > 0$2\sec^2 x > 0$. Case 1: x in left(0, fracpi4right)$x \in \left(0, \frac{\pi}{4}\right)$ Here tan x < 1 implies tan x - 1 < 0$\tan x < 1 \implies \tan x - 1 < 0$. So, g'(x) < 0 implies g(x)$g'(x) < 0 \implies g(x)$ is decreasing. Case 2: x in left(fracpi4, fracpi2right)$x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Here tan x > 1 implies tan x - 1 > 0$\tan x > 1 \implies \tan x - 1 > 0$. So, g'(x) > 0 implies g(x)$g'(x) > 0 \implies g(x)$ is increasing. Therefore, neither statement (I) nor (II) is true. ### Pattern Recognition When a function wraps a quadratic, f((u-k)^2 + c)$f((u-k)^2 + c)$, the critical points match the inner function's extrema. Evaluate the sign directly from the inner derivative u'$u'$ and (u-k)$(u-k)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives