Let gx - -0x ft dt- where f is such that 12- fx - 1 for t- -0- 1- and 0- ft -12 for t- -1- 2- Then- g2 satisfies the inequality-

Let $g (x) = \int_{0}^{x} f (t) d t$ where $\frac{1}{2} \leq f (t) \leq 1, t ϵ [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t ϵ (1, 2)$ , then [IIT Screening 2000]

Q. A function

f : R \to R

satisfies the following conditions:
1)

f (- x) = f (x)

f (x + 2) = f (x)

g (x) = \int_{0}^{x} f (t) d t

and

g (1) = α

Then, the value of

g (x + 2) - g (2)

Let $g (x) = \int_{0}^{x} f (t) d t$ where $\frac{1}{2} \leq f (t) \leq 1, t ϵ [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t ϵ (1, 2)$ , then [IIT Screening 2000]

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Let g(x)=∫x0f(t)dt, where f is such that 12≤f(x)≤1 for tϵ[0,1] and 0≤f(t)≤12 for tϵ[1,2]. Then, g(2) satisfies the inequality.

The correct option is C 12≤g(2)≤32If g(x)=∫x0f(t)dt then, g(2)=∫20f(t)dt=∫10f(t)dt+∫21f(t)dtgiven, if t∈[0,1]then0.5≤f(t)≤10.5≤l1≤1…eqn(1)and if t∈[1,2]then0≤f(t)≤0.50≤l2≤0.5…eqn(2)adding eqn(1) and eqn(2), we get0.5≤l1+l2≤1.50.5≤g(2)≤1.5

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (x) \leq 1$ for $t ϵ [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t ϵ [1, 2]$ . Then, $g (2)$ satisfies the inequality.