119624
Let \(x=0\) be a polynomial of second degree. If \(f(\mathrm{x})\) and \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{AP}\), then \(\mathrm{x}=0\), and \(f(|\mathrm{x}|)\) \(M\) are in
1 AGP
2 \(\mathrm{AP}\)
3 GP
4 HP
Explanation:
B Suppose that, \(f(x)=p x^2+q(x)+r\) On differentiating with respect to \(\mathrm{x}\), we get - \(f^{\prime}(x)=2 p x+q\) Now, \(\mathrm{f}(-1)=\mathrm{f}(1)\) \(\therefore \quad \mathrm{p}(-1)^2+\mathrm{q}(-1)+\mathrm{r}=\mathrm{p}(1)^2+\mathrm{q}(1)+\mathrm{r}\) \(\mathrm{p}-\mathrm{q}+\mathrm{r}=\mathrm{p}=\mathrm{q}+\mathrm{r}\) \(q=0\) Therefore, \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{px}\) \(f(a)=2 p a, f^{\prime}(b)=2 p b, f^{\prime}(c)=2 p c\) We know that, a, b, c are in AP. It means \(2 b=a+c\). \(4 \mathrm{pb}=2 \mathrm{pa}+2 \mathrm{pc}\) \(2\mathrm{f}^{\prime}(\mathrm{b})=\mathrm{f}^{\prime}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{c})\) Hence, \(f^{\prime}(a), f^{\prime}(b), f^{\prime}(c)\) are in AP.
Jamia Millia Islamia-2015
Binomial Theorem and its Simple Application
119614
Let \(a, b\) be the two distinct roots of a polynomial \(f(x)\). Then there exists at least one root lying between \(a\) and \(b\) of the polynomial
1 \(f(x)\)
2 \(f^{\prime}(x)\)
3 f''(x)
4 none of these
Explanation:
B Given, a, b are the two distinct roots of a polynomial \(\mathrm{f}(\mathrm{x})\) Now, Rolle's theorem states that if function \(f(x)\) be continuous on \([\mathrm{a}, \mathrm{b}]\) differentiable on \((\mathrm{a}, \mathrm{b})\) and \(f(a)=f(b)\) then there exists some \(C\) between \(a\) and \(b\) such that \(\mathrm{f}^{\prime}(\mathrm{c})=0\)
119624
Let \(x=0\) be a polynomial of second degree. If \(f(\mathrm{x})\) and \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{AP}\), then \(\mathrm{x}=0\), and \(f(|\mathrm{x}|)\) \(M\) are in
1 AGP
2 \(\mathrm{AP}\)
3 GP
4 HP
Explanation:
B Suppose that, \(f(x)=p x^2+q(x)+r\) On differentiating with respect to \(\mathrm{x}\), we get - \(f^{\prime}(x)=2 p x+q\) Now, \(\mathrm{f}(-1)=\mathrm{f}(1)\) \(\therefore \quad \mathrm{p}(-1)^2+\mathrm{q}(-1)+\mathrm{r}=\mathrm{p}(1)^2+\mathrm{q}(1)+\mathrm{r}\) \(\mathrm{p}-\mathrm{q}+\mathrm{r}=\mathrm{p}=\mathrm{q}+\mathrm{r}\) \(q=0\) Therefore, \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{px}\) \(f(a)=2 p a, f^{\prime}(b)=2 p b, f^{\prime}(c)=2 p c\) We know that, a, b, c are in AP. It means \(2 b=a+c\). \(4 \mathrm{pb}=2 \mathrm{pa}+2 \mathrm{pc}\) \(2\mathrm{f}^{\prime}(\mathrm{b})=\mathrm{f}^{\prime}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{c})\) Hence, \(f^{\prime}(a), f^{\prime}(b), f^{\prime}(c)\) are in AP.
Jamia Millia Islamia-2015
Binomial Theorem and its Simple Application
119614
Let \(a, b\) be the two distinct roots of a polynomial \(f(x)\). Then there exists at least one root lying between \(a\) and \(b\) of the polynomial
1 \(f(x)\)
2 \(f^{\prime}(x)\)
3 f''(x)
4 none of these
Explanation:
B Given, a, b are the two distinct roots of a polynomial \(\mathrm{f}(\mathrm{x})\) Now, Rolle's theorem states that if function \(f(x)\) be continuous on \([\mathrm{a}, \mathrm{b}]\) differentiable on \((\mathrm{a}, \mathrm{b})\) and \(f(a)=f(b)\) then there exists some \(C\) between \(a\) and \(b\) such that \(\mathrm{f}^{\prime}(\mathrm{c})=0\)
