Area Ques 7
- $\int_{-1}^1 g^{\prime}(x) d x$ is equal to
(a) $2 g(-1)$
(b) 0
(c) $-2 g(1)$
(d) $2 g(1)$
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Answer:
Correct Answer: 7.(d)
Solution: (d) Let $I=\int_{-1}^1 g^{\prime}(x) d x=[g(x)]_{-1}^1=g(1)-g(-1)$
Since, $ y^3-3 y+x=0 $ $\quad$ ……..(i)
and $ y=g(x) $
$ \therefore \quad \{g(x)\}^3-3 g(x)+x=0 $
At $x=1, \quad\{g(1)\}^3-3 g(1)+1=0$ $\quad$ ……..(ii)
At $x=-1,\quad \{g(-1)\}^3-3 g(-1)-1=0$ $\quad$ ……..(iii)
On adding Eqs. (i) and (ii), we get
$ \begin{aligned} & \{g(1)\}^3+\{g(-1)\}^3-3\{g(1)+g(-1)\}=0 \\ & \Rightarrow \quad [g(1)+g(-1)]\left[\{g(1)\}^2+\{g(-1)\}^2-g(1) g(-1)-3\right]=0 \\ & \Rightarrow \quad g(1)+g(-1)=0 \\ & \Rightarrow \quad g(1)=-g(-1) \\ & \therefore \quad I=g(1)-g(-1) \\ & = \quad g(1)-\{-g(1)\}=2 g(1) \\ \end{aligned} $
BETA
