Let f be a function defined by f x -2x2-log - x - x- 0- then

Let $g (x) = \int_{0}^{x} f (t) d t$ and $f (x)$ satisfies the equation $f (x + y) = f (x) + f (y) + 2 x y - 1$ for all $x, y ϵ R$ and $f^{'} (0) = 2$ then

Q. Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x)
(a) increases on [0, a]
(b) decreases on [0, a]
(c) increases on [−a, 0]
(d) decreases on [a, 2a]

Q. Prove that the function f given by f(x) = log cos x is strictly increasing on (−π/2, 0) and strictly decreasing on (0, π/2).

Q. Let

g (x)

be a polynomial of degree one and

f (x)

be defined by

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} g (x), & x \leq 0 {[\frac{1 + x}{2 + x}]}^{1 / x}, & x > 0 \end{matrix}

Let

f (x)

be a continuous function satisfying

f^{'} (1) = f (- 1) .

Then

f (- 2)

is equal to

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Let f be a function defined by f(x)=2x2−log|x|,x≠0, then

The correct option is C f increases on (−∞,−12]Given, f(x)=2x2−log|x|f′(x)=4x−1x=(2x+1)(2x−1)xFor f(x) to be increasing, f′(x)>0⇒(2x+1)(2x−1)x>0⇒xϵ(−12,0)∪(12,∞)For f(x) to be decreasing, f′(x)<00⇒(2x+1)(2x−1)x<0⇒xϵ(−∞,−12)∪(0,12)

Let f be a function defined by $f (x) = 2 x^{2} - log | x |, x \neq 0$ , then