119619
Let \(f(x)=x^3+a x^2+b x+c\) be polynomial with integer coefficients. If the roots of \(f(x)\) are integer and are in Arithmetic Progression, then ' \(a\) ' cannot take the value
1 -642
2 1214
3 1323
4 1626
Explanation:
B We have, \(f(x)=x^3+a x^2+b x+c\) Let roots of \(\mathrm{f}(\mathrm{x})\) are \(\alpha, \beta, \gamma\) and since roots are in AP. \(\therefore \quad 2 \beta=\alpha+\gamma\) Now, sum of roots \(\alpha+\beta+\gamma=-\mathrm{a}\) or \(2 \beta+\beta=-\mathrm{a}\) or \(\beta=-\frac{a}{3}\) It is given that, roots are integers. \(\therefore \beta\) is an integer when a is multiple of 3 , Option (a), (c) and (d) are multiple of 3. \(a \neq 1214\) So, option (b) cannot multiple of 3.
AP EAMCET-19.08.2021
Binomial Theorem and its Simple Application
119621
Let \(p(x)\) be a quadratic polynomial with constant term 1 . Suppose \(p(x)\) when divided by \(x-1\) leaves remainder 2 and when divided by \(x\) +1 leaves reminder 4 . Then, the sum of the roots of \(p(x)=0\) is
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(\frac{1}{2}\)
Explanation:
D Given, \(\mathrm{p}(\mathrm{x})\) be a quadratic polynomial with constant term 1. Then, \(p(x)=a x^2+b x+c=a x^2+b x+1 \ldots\) (i) According to the question \(p(1)=2\) \(a+b+1=2\) \(a+b=1\) And, \(p(-1)=4\) \(a-b+1=4\) \(a-b=3\) On adding equation (ii) and equation (iii) we get \(2 a=4\) \(a=2\) \(b=-1\) \(\text { and } \quad b=-1\) On putting the values of a and be in equation (i), we get- \(\mathrm{p}(\mathrm{x})=2 \mathrm{x}^2-\mathrm{x}+1=0\) So, \(\text { Sum of the roots }=\frac{- \text { coefficient of } x}{\text { coefficient of } x^2}\) \(\frac{-(-1)}{2}=\frac{1}{2}\)
WB JEE-2013
Binomial Theorem and its Simple Application
119622
Let \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots .+a_n x^{2 n}\) be \(a\) polynomial in a real variable \(x\) with \(0\lt a_1\lt a_2\) \(\lt \ldots\lt \mathbf{a}_{\mathrm{n}}\). The function \(\mathrm{P}(\mathrm{x})\) has
1 neither a maximum nor a minimum
2 only one mammon
3 only one minimum
4 only one maximum and only one minimum
Explanation:
C Given that, \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots \ldots+a_n x^{2 n} \text { and } 0\lt a_1 c a_2\) \(\lt a_3 \ldots \ldots\lt a_n\) \(\text { On differentiating above equation, }\) \(P^{\prime}(x)=2 a_1 x+4 a_2 x^3+6 a_3 x^5+\ldots .+2 n a_n x^{2 n-1}\) \(P^{\prime}(x)=2 x\left(a_1+2 a_2 x^2+3 a_3 x^3+\ldots .+n a_n x^{2 n-2}\right.\) \(\text { There fore, }\) \(\quad P^{\prime}(x)=0\) \(\text { When, } \quad x=0 \text { and } p^{\prime \prime}(x)=2 a_1>0\) \(\therefore \quad x=0\) \(\text { Hence, only one minimum. }\) On differentiating above equation, There fore, When, Hence, only one minimum.
Jamia Millia Islamia-2009
Binomial Theorem and its Simple Application
119623
Suppose \(f(x)\) is a polynomial of degree four, having critical points at \(-1,0,1\). If \(T=\{x \in\) \(R\} f(x)=f(0)\), then the sum of squares of all the elements of \(T\) is
1 2
2 4
3 8
4 6
Explanation:
B We know that, For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(f^{\prime}(x)=(x+1) x(x-1)\) \(f^{\prime}(x)=\left(x^2-1\right) x\) \(f^{\prime}(x)=x^3-x\) On integrating both the side, we get - \(f(x)=\int\left(x^3-x\right) d x\) \(=\frac{x^4}{4}-\frac{x^2}{2}+c\) \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\frac{\mathrm{x}^4}{4}-\frac{\mathrm{x}^2}{2}+\mathrm{c}=\mathrm{c}\) \(\frac{x^4}{4}-\frac{x^2}{2}=0\) \(\frac{x^4-2 x^2}{4}=0\) \(x^4-2 x^2=0\) \(x^2\left(x^2-2\right)=0\) \(x=0,-\sqrt{2}, \sqrt{2}\) The sum and square all the elements of 7 , \(x=(0)^2+(-\sqrt{2})^2+(\sqrt{2})^2\) \(x=0+2+2=4\)
119619
Let \(f(x)=x^3+a x^2+b x+c\) be polynomial with integer coefficients. If the roots of \(f(x)\) are integer and are in Arithmetic Progression, then ' \(a\) ' cannot take the value
1 -642
2 1214
3 1323
4 1626
Explanation:
B We have, \(f(x)=x^3+a x^2+b x+c\) Let roots of \(\mathrm{f}(\mathrm{x})\) are \(\alpha, \beta, \gamma\) and since roots are in AP. \(\therefore \quad 2 \beta=\alpha+\gamma\) Now, sum of roots \(\alpha+\beta+\gamma=-\mathrm{a}\) or \(2 \beta+\beta=-\mathrm{a}\) or \(\beta=-\frac{a}{3}\) It is given that, roots are integers. \(\therefore \beta\) is an integer when a is multiple of 3 , Option (a), (c) and (d) are multiple of 3. \(a \neq 1214\) So, option (b) cannot multiple of 3.
AP EAMCET-19.08.2021
Binomial Theorem and its Simple Application
119621
Let \(p(x)\) be a quadratic polynomial with constant term 1 . Suppose \(p(x)\) when divided by \(x-1\) leaves remainder 2 and when divided by \(x\) +1 leaves reminder 4 . Then, the sum of the roots of \(p(x)=0\) is
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(\frac{1}{2}\)
Explanation:
D Given, \(\mathrm{p}(\mathrm{x})\) be a quadratic polynomial with constant term 1. Then, \(p(x)=a x^2+b x+c=a x^2+b x+1 \ldots\) (i) According to the question \(p(1)=2\) \(a+b+1=2\) \(a+b=1\) And, \(p(-1)=4\) \(a-b+1=4\) \(a-b=3\) On adding equation (ii) and equation (iii) we get \(2 a=4\) \(a=2\) \(b=-1\) \(\text { and } \quad b=-1\) On putting the values of a and be in equation (i), we get- \(\mathrm{p}(\mathrm{x})=2 \mathrm{x}^2-\mathrm{x}+1=0\) So, \(\text { Sum of the roots }=\frac{- \text { coefficient of } x}{\text { coefficient of } x^2}\) \(\frac{-(-1)}{2}=\frac{1}{2}\)
WB JEE-2013
Binomial Theorem and its Simple Application
119622
Let \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots .+a_n x^{2 n}\) be \(a\) polynomial in a real variable \(x\) with \(0\lt a_1\lt a_2\) \(\lt \ldots\lt \mathbf{a}_{\mathrm{n}}\). The function \(\mathrm{P}(\mathrm{x})\) has
1 neither a maximum nor a minimum
2 only one mammon
3 only one minimum
4 only one maximum and only one minimum
Explanation:
C Given that, \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots \ldots+a_n x^{2 n} \text { and } 0\lt a_1 c a_2\) \(\lt a_3 \ldots \ldots\lt a_n\) \(\text { On differentiating above equation, }\) \(P^{\prime}(x)=2 a_1 x+4 a_2 x^3+6 a_3 x^5+\ldots .+2 n a_n x^{2 n-1}\) \(P^{\prime}(x)=2 x\left(a_1+2 a_2 x^2+3 a_3 x^3+\ldots .+n a_n x^{2 n-2}\right.\) \(\text { There fore, }\) \(\quad P^{\prime}(x)=0\) \(\text { When, } \quad x=0 \text { and } p^{\prime \prime}(x)=2 a_1>0\) \(\therefore \quad x=0\) \(\text { Hence, only one minimum. }\) On differentiating above equation, There fore, When, Hence, only one minimum.
Jamia Millia Islamia-2009
Binomial Theorem and its Simple Application
119623
Suppose \(f(x)\) is a polynomial of degree four, having critical points at \(-1,0,1\). If \(T=\{x \in\) \(R\} f(x)=f(0)\), then the sum of squares of all the elements of \(T\) is
1 2
2 4
3 8
4 6
Explanation:
B We know that, For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(f^{\prime}(x)=(x+1) x(x-1)\) \(f^{\prime}(x)=\left(x^2-1\right) x\) \(f^{\prime}(x)=x^3-x\) On integrating both the side, we get - \(f(x)=\int\left(x^3-x\right) d x\) \(=\frac{x^4}{4}-\frac{x^2}{2}+c\) \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\frac{\mathrm{x}^4}{4}-\frac{\mathrm{x}^2}{2}+\mathrm{c}=\mathrm{c}\) \(\frac{x^4}{4}-\frac{x^2}{2}=0\) \(\frac{x^4-2 x^2}{4}=0\) \(x^4-2 x^2=0\) \(x^2\left(x^2-2\right)=0\) \(x=0,-\sqrt{2}, \sqrt{2}\) The sum and square all the elements of 7 , \(x=(0)^2+(-\sqrt{2})^2+(\sqrt{2})^2\) \(x=0+2+2=4\)
119619
Let \(f(x)=x^3+a x^2+b x+c\) be polynomial with integer coefficients. If the roots of \(f(x)\) are integer and are in Arithmetic Progression, then ' \(a\) ' cannot take the value
1 -642
2 1214
3 1323
4 1626
Explanation:
B We have, \(f(x)=x^3+a x^2+b x+c\) Let roots of \(\mathrm{f}(\mathrm{x})\) are \(\alpha, \beta, \gamma\) and since roots are in AP. \(\therefore \quad 2 \beta=\alpha+\gamma\) Now, sum of roots \(\alpha+\beta+\gamma=-\mathrm{a}\) or \(2 \beta+\beta=-\mathrm{a}\) or \(\beta=-\frac{a}{3}\) It is given that, roots are integers. \(\therefore \beta\) is an integer when a is multiple of 3 , Option (a), (c) and (d) are multiple of 3. \(a \neq 1214\) So, option (b) cannot multiple of 3.
AP EAMCET-19.08.2021
Binomial Theorem and its Simple Application
119621
Let \(p(x)\) be a quadratic polynomial with constant term 1 . Suppose \(p(x)\) when divided by \(x-1\) leaves remainder 2 and when divided by \(x\) +1 leaves reminder 4 . Then, the sum of the roots of \(p(x)=0\) is
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(\frac{1}{2}\)
Explanation:
D Given, \(\mathrm{p}(\mathrm{x})\) be a quadratic polynomial with constant term 1. Then, \(p(x)=a x^2+b x+c=a x^2+b x+1 \ldots\) (i) According to the question \(p(1)=2\) \(a+b+1=2\) \(a+b=1\) And, \(p(-1)=4\) \(a-b+1=4\) \(a-b=3\) On adding equation (ii) and equation (iii) we get \(2 a=4\) \(a=2\) \(b=-1\) \(\text { and } \quad b=-1\) On putting the values of a and be in equation (i), we get- \(\mathrm{p}(\mathrm{x})=2 \mathrm{x}^2-\mathrm{x}+1=0\) So, \(\text { Sum of the roots }=\frac{- \text { coefficient of } x}{\text { coefficient of } x^2}\) \(\frac{-(-1)}{2}=\frac{1}{2}\)
WB JEE-2013
Binomial Theorem and its Simple Application
119622
Let \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots .+a_n x^{2 n}\) be \(a\) polynomial in a real variable \(x\) with \(0\lt a_1\lt a_2\) \(\lt \ldots\lt \mathbf{a}_{\mathrm{n}}\). The function \(\mathrm{P}(\mathrm{x})\) has
1 neither a maximum nor a minimum
2 only one mammon
3 only one minimum
4 only one maximum and only one minimum
Explanation:
C Given that, \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots \ldots+a_n x^{2 n} \text { and } 0\lt a_1 c a_2\) \(\lt a_3 \ldots \ldots\lt a_n\) \(\text { On differentiating above equation, }\) \(P^{\prime}(x)=2 a_1 x+4 a_2 x^3+6 a_3 x^5+\ldots .+2 n a_n x^{2 n-1}\) \(P^{\prime}(x)=2 x\left(a_1+2 a_2 x^2+3 a_3 x^3+\ldots .+n a_n x^{2 n-2}\right.\) \(\text { There fore, }\) \(\quad P^{\prime}(x)=0\) \(\text { When, } \quad x=0 \text { and } p^{\prime \prime}(x)=2 a_1>0\) \(\therefore \quad x=0\) \(\text { Hence, only one minimum. }\) On differentiating above equation, There fore, When, Hence, only one minimum.
Jamia Millia Islamia-2009
Binomial Theorem and its Simple Application
119623
Suppose \(f(x)\) is a polynomial of degree four, having critical points at \(-1,0,1\). If \(T=\{x \in\) \(R\} f(x)=f(0)\), then the sum of squares of all the elements of \(T\) is
1 2
2 4
3 8
4 6
Explanation:
B We know that, For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(f^{\prime}(x)=(x+1) x(x-1)\) \(f^{\prime}(x)=\left(x^2-1\right) x\) \(f^{\prime}(x)=x^3-x\) On integrating both the side, we get - \(f(x)=\int\left(x^3-x\right) d x\) \(=\frac{x^4}{4}-\frac{x^2}{2}+c\) \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\frac{\mathrm{x}^4}{4}-\frac{\mathrm{x}^2}{2}+\mathrm{c}=\mathrm{c}\) \(\frac{x^4}{4}-\frac{x^2}{2}=0\) \(\frac{x^4-2 x^2}{4}=0\) \(x^4-2 x^2=0\) \(x^2\left(x^2-2\right)=0\) \(x=0,-\sqrt{2}, \sqrt{2}\) The sum and square all the elements of 7 , \(x=(0)^2+(-\sqrt{2})^2+(\sqrt{2})^2\) \(x=0+2+2=4\)
119619
Let \(f(x)=x^3+a x^2+b x+c\) be polynomial with integer coefficients. If the roots of \(f(x)\) are integer and are in Arithmetic Progression, then ' \(a\) ' cannot take the value
1 -642
2 1214
3 1323
4 1626
Explanation:
B We have, \(f(x)=x^3+a x^2+b x+c\) Let roots of \(\mathrm{f}(\mathrm{x})\) are \(\alpha, \beta, \gamma\) and since roots are in AP. \(\therefore \quad 2 \beta=\alpha+\gamma\) Now, sum of roots \(\alpha+\beta+\gamma=-\mathrm{a}\) or \(2 \beta+\beta=-\mathrm{a}\) or \(\beta=-\frac{a}{3}\) It is given that, roots are integers. \(\therefore \beta\) is an integer when a is multiple of 3 , Option (a), (c) and (d) are multiple of 3. \(a \neq 1214\) So, option (b) cannot multiple of 3.
AP EAMCET-19.08.2021
Binomial Theorem and its Simple Application
119621
Let \(p(x)\) be a quadratic polynomial with constant term 1 . Suppose \(p(x)\) when divided by \(x-1\) leaves remainder 2 and when divided by \(x\) +1 leaves reminder 4 . Then, the sum of the roots of \(p(x)=0\) is
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(\frac{1}{2}\)
Explanation:
D Given, \(\mathrm{p}(\mathrm{x})\) be a quadratic polynomial with constant term 1. Then, \(p(x)=a x^2+b x+c=a x^2+b x+1 \ldots\) (i) According to the question \(p(1)=2\) \(a+b+1=2\) \(a+b=1\) And, \(p(-1)=4\) \(a-b+1=4\) \(a-b=3\) On adding equation (ii) and equation (iii) we get \(2 a=4\) \(a=2\) \(b=-1\) \(\text { and } \quad b=-1\) On putting the values of a and be in equation (i), we get- \(\mathrm{p}(\mathrm{x})=2 \mathrm{x}^2-\mathrm{x}+1=0\) So, \(\text { Sum of the roots }=\frac{- \text { coefficient of } x}{\text { coefficient of } x^2}\) \(\frac{-(-1)}{2}=\frac{1}{2}\)
WB JEE-2013
Binomial Theorem and its Simple Application
119622
Let \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots .+a_n x^{2 n}\) be \(a\) polynomial in a real variable \(x\) with \(0\lt a_1\lt a_2\) \(\lt \ldots\lt \mathbf{a}_{\mathrm{n}}\). The function \(\mathrm{P}(\mathrm{x})\) has
1 neither a maximum nor a minimum
2 only one mammon
3 only one minimum
4 only one maximum and only one minimum
Explanation:
C Given that, \(P(x)=a_0+a_1 x^2+a_2 x^4+\ldots \ldots+a_n x^{2 n} \text { and } 0\lt a_1 c a_2\) \(\lt a_3 \ldots \ldots\lt a_n\) \(\text { On differentiating above equation, }\) \(P^{\prime}(x)=2 a_1 x+4 a_2 x^3+6 a_3 x^5+\ldots .+2 n a_n x^{2 n-1}\) \(P^{\prime}(x)=2 x\left(a_1+2 a_2 x^2+3 a_3 x^3+\ldots .+n a_n x^{2 n-2}\right.\) \(\text { There fore, }\) \(\quad P^{\prime}(x)=0\) \(\text { When, } \quad x=0 \text { and } p^{\prime \prime}(x)=2 a_1>0\) \(\therefore \quad x=0\) \(\text { Hence, only one minimum. }\) On differentiating above equation, There fore, When, Hence, only one minimum.
Jamia Millia Islamia-2009
Binomial Theorem and its Simple Application
119623
Suppose \(f(x)\) is a polynomial of degree four, having critical points at \(-1,0,1\). If \(T=\{x \in\) \(R\} f(x)=f(0)\), then the sum of squares of all the elements of \(T\) is
1 2
2 4
3 8
4 6
Explanation:
B We know that, For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(f^{\prime}(x)=(x+1) x(x-1)\) \(f^{\prime}(x)=\left(x^2-1\right) x\) \(f^{\prime}(x)=x^3-x\) On integrating both the side, we get - \(f(x)=\int\left(x^3-x\right) d x\) \(=\frac{x^4}{4}-\frac{x^2}{2}+c\) \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\frac{\mathrm{x}^4}{4}-\frac{\mathrm{x}^2}{2}+\mathrm{c}=\mathrm{c}\) \(\frac{x^4}{4}-\frac{x^2}{2}=0\) \(\frac{x^4-2 x^2}{4}=0\) \(x^4-2 x^2=0\) \(x^2\left(x^2-2\right)=0\) \(x=0,-\sqrt{2}, \sqrt{2}\) The sum and square all the elements of 7 , \(x=(0)^2+(-\sqrt{2})^2+(\sqrt{2})^2\) \(x=0+2+2=4\)
