119620
Let \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\). Then,
1 \(a_0+a_2+\ldots+a_{18}=a_1+a_2+\ldots+a_{17}\)
2 \(a_0+a_2+\ldots+a_{18}\) is even
3 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is divisible by 9
4 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is dividable by 3 but not by 9
Explanation:
B Given, \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\) \(\text { Put, } x=-1 \text {, we get- }\) \((1-1+1)^9=a_0-a_1+a_2-\ldots+a_{18}\) \(1=a_0-a_1+a_2 \ldots .+a_{18}\) \(\text { And, put } x=1 \text {, we get - }\) \((1+1+1)^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9+1=2\left(a_0+a_2+\ldots+a_{18}\right)\) \(a_0+a_1+a_2+\ldots+a_{18}=\frac{3^9+1}{2}\) \(=\frac{19683+1}{2}\) \(=\frac{19684}{2}\) \(=9842 \text {, which is an even number. }\) On adding equation (i) and (ii), we get -
WB JEE-2017
Binomial Theorem and its Simple Application
119615
Let \(f(x)\) be a polynomial function such that \(f\) \((x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\). Then, the value of \(\lim _{x \rightarrow I} \frac{f(x)}{x-1}\) is equal to:
1 -15
2 -60
3 60
4 15
Explanation:
A Given, \(f(x)\) be a polynomial function such that. \(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^5+64\) Consider, \(f(x)=x^5+p x^4+q x^3+r x^2+s x+t\) \(f^{\prime}(x)=5 x^4+4 p x^3+3 q x^2+2 r x+s\) \(f^{\prime \prime}(x)=20 x^3+12 p x^2+6 q x+2 r\) Then, \(f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\) \(x^5+ (p+5) x^4+(q+4 p+20) x^3+(r+3 q+12 p) x^2\) \((s+2 r+6 q) x+(t+s+2 r)=x^2+64\) \(p+5=0\) \((q+4 p+20)=0\) \(r+3 q+12 p=0\) \(s+2 r+6 q=0\) \(t+s+2 r=64\) By solving, we get- \(p=-5, q=0, r=60, s=-120, t=64\) \(\therefore \quad f(x)=x^5-5 x^4+60 x^2-120 x+64\) \(\text { Then, } \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1} \text { is } \frac{0}{0} \text { from }\) Then, \(\quad \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1}\) is \(\frac{0}{0}\) from \(\therefore\) By L' Hospital rule- \(\lim _{x \rightarrow 1} \frac{5 x^4-20 x^3+120 x-120}{1}\) \(=5-20+120-120\) \(=-15\)
JEE Main-25.06.2022
Binomial Theorem and its Simple Application
119616
If \(x^2+p x+1\) is a factor of \(a x^3+b x+c\), then
1 \(a^2+c^2=a b\)
2 \(\mathrm{a}^2+\mathrm{c}^2=-\mathrm{ab}\)
3 \(a^2-c^2=a b\)
4 None of these
Explanation:
C Given \(x^2+p x+1\) is a factor of \(a x^3+b x+c\) Let, \(\alpha, \beta\) be the roots of \(\mathrm{x}^2+\mathrm{px}+1=0\) \(\alpha+\beta=-p\) \(\alpha \beta=1\) Now, \(\alpha, \beta, \mathrm{r}\) be the roots of \(\mathrm{ax}^3+\mathrm{bx}+\mathrm{c}=0\) Then, \(\alpha+\beta+r=0\) \(\alpha \beta+\beta r+r \alpha=\frac{b}{a}\) \(\alpha \beta r=-\frac{c}{a}\) From (i) and (iii) we get, \(\mathrm{r}=\mathrm{p}\) From (ii) and (v) we get, \(\mathrm{r}=\frac{-\mathrm{c}}{\mathrm{a}}\) And, from (ii) and (iv) we get- \(1+\beta p+p \alpha=\frac{b}{a}\) \(p(\alpha+\beta)=\frac{b}{a}-1\) \(p(-p)=\frac{b-a}{a}\) \(-p^2=\frac{b-a}{a} \Rightarrow-r^2=\frac{b-a}{a}\) \(-\left(\frac{-c}{a}\right)^2=\frac{b-a}{a}\) \(-c^2=\frac{b-a}{a}\) \(a^2-c^2=a b\)
AMU-2015]
Binomial Theorem and its Simple Application
119617
The minimum degree of a polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 i\) as two of its roots is
1 8
2 6
3 4
4 2
Explanation:
B Given, A polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 \mathrm{i}\) as two roots Then, other roots are- \(\sqrt{3}-\sqrt{27},-\sqrt{3}+\sqrt{27},-\sqrt{3}-\sqrt{27}\) and \(\sqrt{2}-5 \mathrm{i}\) So, the minimum degree of a polynomial equation having 6 distinct roots is 6 Hence, number of roots are 6.
AP EAMCET-22.04.2018
Binomial Theorem and its Simple Application
119618
If \(f(x)=2 x^4-13 x^2+a x+b\) is divisible by \(x^2-3 x+2\), then \((a, b)\) is equal to
1 \((-9,-2)\)
2 \((6,4)\)
3 \((9,2)\)
4 \((2,9)\)
Explanation:
C Given, \(f(x)=2 x^4-13 x^2+a x+b\) And is divisible by \(\mathrm{x}^2-3 \mathrm{x}+2=\mathrm{g}(\mathrm{x})\) \(x^2-2 x-x+2=g(x)\) \(x(x-2)-1(x-2)=g(x)\) \((x-1)(x-2)=g(x)\) \((x-1)=0 \text { or }(x-2)=g(x)\) \(x=1 \text { or } x=2\) Then, put \(\mathrm{x}=1\) or \(\mathrm{x}=2\) in equation (i), we get- \(f(1)=2 \times 1^4-13 \times 1^2+a \times 1+b\) \(f(1)=2-13+a+b\) \(f(1)=-11+a+b\) And, \(f(2)=2 \times 2^4-13 \times 2^2+\) \(f(2)=32-52+2 a+b\) \(f(2)=-20+2 a+b\) \(\because \quad \mathrm{f}(\mathrm{x})\) is divisible by \(\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}-2)\) \((x-1) \text { and }(x-2) \text { are factors of } f(x)\) \(f(1)=0 \text { and } f(2)=0\) Then, \(a+b=11\) \(2 a+b=20\) Subtract equation (i) from equation (ii) we get \(\mathrm{a}=9\) \(\mathrm{b}=2\)So, \(\quad(\mathrm{a}, \mathrm{b})=(9,2)\)
119620
Let \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\). Then,
1 \(a_0+a_2+\ldots+a_{18}=a_1+a_2+\ldots+a_{17}\)
2 \(a_0+a_2+\ldots+a_{18}\) is even
3 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is divisible by 9
4 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is dividable by 3 but not by 9
Explanation:
B Given, \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\) \(\text { Put, } x=-1 \text {, we get- }\) \((1-1+1)^9=a_0-a_1+a_2-\ldots+a_{18}\) \(1=a_0-a_1+a_2 \ldots .+a_{18}\) \(\text { And, put } x=1 \text {, we get - }\) \((1+1+1)^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9+1=2\left(a_0+a_2+\ldots+a_{18}\right)\) \(a_0+a_1+a_2+\ldots+a_{18}=\frac{3^9+1}{2}\) \(=\frac{19683+1}{2}\) \(=\frac{19684}{2}\) \(=9842 \text {, which is an even number. }\) On adding equation (i) and (ii), we get -
WB JEE-2017
Binomial Theorem and its Simple Application
119615
Let \(f(x)\) be a polynomial function such that \(f\) \((x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\). Then, the value of \(\lim _{x \rightarrow I} \frac{f(x)}{x-1}\) is equal to:
1 -15
2 -60
3 60
4 15
Explanation:
A Given, \(f(x)\) be a polynomial function such that. \(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^5+64\) Consider, \(f(x)=x^5+p x^4+q x^3+r x^2+s x+t\) \(f^{\prime}(x)=5 x^4+4 p x^3+3 q x^2+2 r x+s\) \(f^{\prime \prime}(x)=20 x^3+12 p x^2+6 q x+2 r\) Then, \(f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\) \(x^5+ (p+5) x^4+(q+4 p+20) x^3+(r+3 q+12 p) x^2\) \((s+2 r+6 q) x+(t+s+2 r)=x^2+64\) \(p+5=0\) \((q+4 p+20)=0\) \(r+3 q+12 p=0\) \(s+2 r+6 q=0\) \(t+s+2 r=64\) By solving, we get- \(p=-5, q=0, r=60, s=-120, t=64\) \(\therefore \quad f(x)=x^5-5 x^4+60 x^2-120 x+64\) \(\text { Then, } \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1} \text { is } \frac{0}{0} \text { from }\) Then, \(\quad \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1}\) is \(\frac{0}{0}\) from \(\therefore\) By L' Hospital rule- \(\lim _{x \rightarrow 1} \frac{5 x^4-20 x^3+120 x-120}{1}\) \(=5-20+120-120\) \(=-15\)
JEE Main-25.06.2022
Binomial Theorem and its Simple Application
119616
If \(x^2+p x+1\) is a factor of \(a x^3+b x+c\), then
1 \(a^2+c^2=a b\)
2 \(\mathrm{a}^2+\mathrm{c}^2=-\mathrm{ab}\)
3 \(a^2-c^2=a b\)
4 None of these
Explanation:
C Given \(x^2+p x+1\) is a factor of \(a x^3+b x+c\) Let, \(\alpha, \beta\) be the roots of \(\mathrm{x}^2+\mathrm{px}+1=0\) \(\alpha+\beta=-p\) \(\alpha \beta=1\) Now, \(\alpha, \beta, \mathrm{r}\) be the roots of \(\mathrm{ax}^3+\mathrm{bx}+\mathrm{c}=0\) Then, \(\alpha+\beta+r=0\) \(\alpha \beta+\beta r+r \alpha=\frac{b}{a}\) \(\alpha \beta r=-\frac{c}{a}\) From (i) and (iii) we get, \(\mathrm{r}=\mathrm{p}\) From (ii) and (v) we get, \(\mathrm{r}=\frac{-\mathrm{c}}{\mathrm{a}}\) And, from (ii) and (iv) we get- \(1+\beta p+p \alpha=\frac{b}{a}\) \(p(\alpha+\beta)=\frac{b}{a}-1\) \(p(-p)=\frac{b-a}{a}\) \(-p^2=\frac{b-a}{a} \Rightarrow-r^2=\frac{b-a}{a}\) \(-\left(\frac{-c}{a}\right)^2=\frac{b-a}{a}\) \(-c^2=\frac{b-a}{a}\) \(a^2-c^2=a b\)
AMU-2015]
Binomial Theorem and its Simple Application
119617
The minimum degree of a polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 i\) as two of its roots is
1 8
2 6
3 4
4 2
Explanation:
B Given, A polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 \mathrm{i}\) as two roots Then, other roots are- \(\sqrt{3}-\sqrt{27},-\sqrt{3}+\sqrt{27},-\sqrt{3}-\sqrt{27}\) and \(\sqrt{2}-5 \mathrm{i}\) So, the minimum degree of a polynomial equation having 6 distinct roots is 6 Hence, number of roots are 6.
AP EAMCET-22.04.2018
Binomial Theorem and its Simple Application
119618
If \(f(x)=2 x^4-13 x^2+a x+b\) is divisible by \(x^2-3 x+2\), then \((a, b)\) is equal to
1 \((-9,-2)\)
2 \((6,4)\)
3 \((9,2)\)
4 \((2,9)\)
Explanation:
C Given, \(f(x)=2 x^4-13 x^2+a x+b\) And is divisible by \(\mathrm{x}^2-3 \mathrm{x}+2=\mathrm{g}(\mathrm{x})\) \(x^2-2 x-x+2=g(x)\) \(x(x-2)-1(x-2)=g(x)\) \((x-1)(x-2)=g(x)\) \((x-1)=0 \text { or }(x-2)=g(x)\) \(x=1 \text { or } x=2\) Then, put \(\mathrm{x}=1\) or \(\mathrm{x}=2\) in equation (i), we get- \(f(1)=2 \times 1^4-13 \times 1^2+a \times 1+b\) \(f(1)=2-13+a+b\) \(f(1)=-11+a+b\) And, \(f(2)=2 \times 2^4-13 \times 2^2+\) \(f(2)=32-52+2 a+b\) \(f(2)=-20+2 a+b\) \(\because \quad \mathrm{f}(\mathrm{x})\) is divisible by \(\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}-2)\) \((x-1) \text { and }(x-2) \text { are factors of } f(x)\) \(f(1)=0 \text { and } f(2)=0\) Then, \(a+b=11\) \(2 a+b=20\) Subtract equation (i) from equation (ii) we get \(\mathrm{a}=9\) \(\mathrm{b}=2\)So, \(\quad(\mathrm{a}, \mathrm{b})=(9,2)\)
119620
Let \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\). Then,
1 \(a_0+a_2+\ldots+a_{18}=a_1+a_2+\ldots+a_{17}\)
2 \(a_0+a_2+\ldots+a_{18}\) is even
3 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is divisible by 9
4 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is dividable by 3 but not by 9
Explanation:
B Given, \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\) \(\text { Put, } x=-1 \text {, we get- }\) \((1-1+1)^9=a_0-a_1+a_2-\ldots+a_{18}\) \(1=a_0-a_1+a_2 \ldots .+a_{18}\) \(\text { And, put } x=1 \text {, we get - }\) \((1+1+1)^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9+1=2\left(a_0+a_2+\ldots+a_{18}\right)\) \(a_0+a_1+a_2+\ldots+a_{18}=\frac{3^9+1}{2}\) \(=\frac{19683+1}{2}\) \(=\frac{19684}{2}\) \(=9842 \text {, which is an even number. }\) On adding equation (i) and (ii), we get -
WB JEE-2017
Binomial Theorem and its Simple Application
119615
Let \(f(x)\) be a polynomial function such that \(f\) \((x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\). Then, the value of \(\lim _{x \rightarrow I} \frac{f(x)}{x-1}\) is equal to:
1 -15
2 -60
3 60
4 15
Explanation:
A Given, \(f(x)\) be a polynomial function such that. \(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^5+64\) Consider, \(f(x)=x^5+p x^4+q x^3+r x^2+s x+t\) \(f^{\prime}(x)=5 x^4+4 p x^3+3 q x^2+2 r x+s\) \(f^{\prime \prime}(x)=20 x^3+12 p x^2+6 q x+2 r\) Then, \(f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\) \(x^5+ (p+5) x^4+(q+4 p+20) x^3+(r+3 q+12 p) x^2\) \((s+2 r+6 q) x+(t+s+2 r)=x^2+64\) \(p+5=0\) \((q+4 p+20)=0\) \(r+3 q+12 p=0\) \(s+2 r+6 q=0\) \(t+s+2 r=64\) By solving, we get- \(p=-5, q=0, r=60, s=-120, t=64\) \(\therefore \quad f(x)=x^5-5 x^4+60 x^2-120 x+64\) \(\text { Then, } \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1} \text { is } \frac{0}{0} \text { from }\) Then, \(\quad \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1}\) is \(\frac{0}{0}\) from \(\therefore\) By L' Hospital rule- \(\lim _{x \rightarrow 1} \frac{5 x^4-20 x^3+120 x-120}{1}\) \(=5-20+120-120\) \(=-15\)
JEE Main-25.06.2022
Binomial Theorem and its Simple Application
119616
If \(x^2+p x+1\) is a factor of \(a x^3+b x+c\), then
1 \(a^2+c^2=a b\)
2 \(\mathrm{a}^2+\mathrm{c}^2=-\mathrm{ab}\)
3 \(a^2-c^2=a b\)
4 None of these
Explanation:
C Given \(x^2+p x+1\) is a factor of \(a x^3+b x+c\) Let, \(\alpha, \beta\) be the roots of \(\mathrm{x}^2+\mathrm{px}+1=0\) \(\alpha+\beta=-p\) \(\alpha \beta=1\) Now, \(\alpha, \beta, \mathrm{r}\) be the roots of \(\mathrm{ax}^3+\mathrm{bx}+\mathrm{c}=0\) Then, \(\alpha+\beta+r=0\) \(\alpha \beta+\beta r+r \alpha=\frac{b}{a}\) \(\alpha \beta r=-\frac{c}{a}\) From (i) and (iii) we get, \(\mathrm{r}=\mathrm{p}\) From (ii) and (v) we get, \(\mathrm{r}=\frac{-\mathrm{c}}{\mathrm{a}}\) And, from (ii) and (iv) we get- \(1+\beta p+p \alpha=\frac{b}{a}\) \(p(\alpha+\beta)=\frac{b}{a}-1\) \(p(-p)=\frac{b-a}{a}\) \(-p^2=\frac{b-a}{a} \Rightarrow-r^2=\frac{b-a}{a}\) \(-\left(\frac{-c}{a}\right)^2=\frac{b-a}{a}\) \(-c^2=\frac{b-a}{a}\) \(a^2-c^2=a b\)
AMU-2015]
Binomial Theorem and its Simple Application
119617
The minimum degree of a polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 i\) as two of its roots is
1 8
2 6
3 4
4 2
Explanation:
B Given, A polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 \mathrm{i}\) as two roots Then, other roots are- \(\sqrt{3}-\sqrt{27},-\sqrt{3}+\sqrt{27},-\sqrt{3}-\sqrt{27}\) and \(\sqrt{2}-5 \mathrm{i}\) So, the minimum degree of a polynomial equation having 6 distinct roots is 6 Hence, number of roots are 6.
AP EAMCET-22.04.2018
Binomial Theorem and its Simple Application
119618
If \(f(x)=2 x^4-13 x^2+a x+b\) is divisible by \(x^2-3 x+2\), then \((a, b)\) is equal to
1 \((-9,-2)\)
2 \((6,4)\)
3 \((9,2)\)
4 \((2,9)\)
Explanation:
C Given, \(f(x)=2 x^4-13 x^2+a x+b\) And is divisible by \(\mathrm{x}^2-3 \mathrm{x}+2=\mathrm{g}(\mathrm{x})\) \(x^2-2 x-x+2=g(x)\) \(x(x-2)-1(x-2)=g(x)\) \((x-1)(x-2)=g(x)\) \((x-1)=0 \text { or }(x-2)=g(x)\) \(x=1 \text { or } x=2\) Then, put \(\mathrm{x}=1\) or \(\mathrm{x}=2\) in equation (i), we get- \(f(1)=2 \times 1^4-13 \times 1^2+a \times 1+b\) \(f(1)=2-13+a+b\) \(f(1)=-11+a+b\) And, \(f(2)=2 \times 2^4-13 \times 2^2+\) \(f(2)=32-52+2 a+b\) \(f(2)=-20+2 a+b\) \(\because \quad \mathrm{f}(\mathrm{x})\) is divisible by \(\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}-2)\) \((x-1) \text { and }(x-2) \text { are factors of } f(x)\) \(f(1)=0 \text { and } f(2)=0\) Then, \(a+b=11\) \(2 a+b=20\) Subtract equation (i) from equation (ii) we get \(\mathrm{a}=9\) \(\mathrm{b}=2\)So, \(\quad(\mathrm{a}, \mathrm{b})=(9,2)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Binomial Theorem and its Simple Application
119620
Let \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\). Then,
1 \(a_0+a_2+\ldots+a_{18}=a_1+a_2+\ldots+a_{17}\)
2 \(a_0+a_2+\ldots+a_{18}\) is even
3 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is divisible by 9
4 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is dividable by 3 but not by 9
Explanation:
B Given, \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\) \(\text { Put, } x=-1 \text {, we get- }\) \((1-1+1)^9=a_0-a_1+a_2-\ldots+a_{18}\) \(1=a_0-a_1+a_2 \ldots .+a_{18}\) \(\text { And, put } x=1 \text {, we get - }\) \((1+1+1)^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9+1=2\left(a_0+a_2+\ldots+a_{18}\right)\) \(a_0+a_1+a_2+\ldots+a_{18}=\frac{3^9+1}{2}\) \(=\frac{19683+1}{2}\) \(=\frac{19684}{2}\) \(=9842 \text {, which is an even number. }\) On adding equation (i) and (ii), we get -
WB JEE-2017
Binomial Theorem and its Simple Application
119615
Let \(f(x)\) be a polynomial function such that \(f\) \((x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\). Then, the value of \(\lim _{x \rightarrow I} \frac{f(x)}{x-1}\) is equal to:
1 -15
2 -60
3 60
4 15
Explanation:
A Given, \(f(x)\) be a polynomial function such that. \(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^5+64\) Consider, \(f(x)=x^5+p x^4+q x^3+r x^2+s x+t\) \(f^{\prime}(x)=5 x^4+4 p x^3+3 q x^2+2 r x+s\) \(f^{\prime \prime}(x)=20 x^3+12 p x^2+6 q x+2 r\) Then, \(f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\) \(x^5+ (p+5) x^4+(q+4 p+20) x^3+(r+3 q+12 p) x^2\) \((s+2 r+6 q) x+(t+s+2 r)=x^2+64\) \(p+5=0\) \((q+4 p+20)=0\) \(r+3 q+12 p=0\) \(s+2 r+6 q=0\) \(t+s+2 r=64\) By solving, we get- \(p=-5, q=0, r=60, s=-120, t=64\) \(\therefore \quad f(x)=x^5-5 x^4+60 x^2-120 x+64\) \(\text { Then, } \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1} \text { is } \frac{0}{0} \text { from }\) Then, \(\quad \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1}\) is \(\frac{0}{0}\) from \(\therefore\) By L' Hospital rule- \(\lim _{x \rightarrow 1} \frac{5 x^4-20 x^3+120 x-120}{1}\) \(=5-20+120-120\) \(=-15\)
JEE Main-25.06.2022
Binomial Theorem and its Simple Application
119616
If \(x^2+p x+1\) is a factor of \(a x^3+b x+c\), then
1 \(a^2+c^2=a b\)
2 \(\mathrm{a}^2+\mathrm{c}^2=-\mathrm{ab}\)
3 \(a^2-c^2=a b\)
4 None of these
Explanation:
C Given \(x^2+p x+1\) is a factor of \(a x^3+b x+c\) Let, \(\alpha, \beta\) be the roots of \(\mathrm{x}^2+\mathrm{px}+1=0\) \(\alpha+\beta=-p\) \(\alpha \beta=1\) Now, \(\alpha, \beta, \mathrm{r}\) be the roots of \(\mathrm{ax}^3+\mathrm{bx}+\mathrm{c}=0\) Then, \(\alpha+\beta+r=0\) \(\alpha \beta+\beta r+r \alpha=\frac{b}{a}\) \(\alpha \beta r=-\frac{c}{a}\) From (i) and (iii) we get, \(\mathrm{r}=\mathrm{p}\) From (ii) and (v) we get, \(\mathrm{r}=\frac{-\mathrm{c}}{\mathrm{a}}\) And, from (ii) and (iv) we get- \(1+\beta p+p \alpha=\frac{b}{a}\) \(p(\alpha+\beta)=\frac{b}{a}-1\) \(p(-p)=\frac{b-a}{a}\) \(-p^2=\frac{b-a}{a} \Rightarrow-r^2=\frac{b-a}{a}\) \(-\left(\frac{-c}{a}\right)^2=\frac{b-a}{a}\) \(-c^2=\frac{b-a}{a}\) \(a^2-c^2=a b\)
AMU-2015]
Binomial Theorem and its Simple Application
119617
The minimum degree of a polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 i\) as two of its roots is
1 8
2 6
3 4
4 2
Explanation:
B Given, A polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 \mathrm{i}\) as two roots Then, other roots are- \(\sqrt{3}-\sqrt{27},-\sqrt{3}+\sqrt{27},-\sqrt{3}-\sqrt{27}\) and \(\sqrt{2}-5 \mathrm{i}\) So, the minimum degree of a polynomial equation having 6 distinct roots is 6 Hence, number of roots are 6.
AP EAMCET-22.04.2018
Binomial Theorem and its Simple Application
119618
If \(f(x)=2 x^4-13 x^2+a x+b\) is divisible by \(x^2-3 x+2\), then \((a, b)\) is equal to
1 \((-9,-2)\)
2 \((6,4)\)
3 \((9,2)\)
4 \((2,9)\)
Explanation:
C Given, \(f(x)=2 x^4-13 x^2+a x+b\) And is divisible by \(\mathrm{x}^2-3 \mathrm{x}+2=\mathrm{g}(\mathrm{x})\) \(x^2-2 x-x+2=g(x)\) \(x(x-2)-1(x-2)=g(x)\) \((x-1)(x-2)=g(x)\) \((x-1)=0 \text { or }(x-2)=g(x)\) \(x=1 \text { or } x=2\) Then, put \(\mathrm{x}=1\) or \(\mathrm{x}=2\) in equation (i), we get- \(f(1)=2 \times 1^4-13 \times 1^2+a \times 1+b\) \(f(1)=2-13+a+b\) \(f(1)=-11+a+b\) And, \(f(2)=2 \times 2^4-13 \times 2^2+\) \(f(2)=32-52+2 a+b\) \(f(2)=-20+2 a+b\) \(\because \quad \mathrm{f}(\mathrm{x})\) is divisible by \(\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}-2)\) \((x-1) \text { and }(x-2) \text { are factors of } f(x)\) \(f(1)=0 \text { and } f(2)=0\) Then, \(a+b=11\) \(2 a+b=20\) Subtract equation (i) from equation (ii) we get \(\mathrm{a}=9\) \(\mathrm{b}=2\)So, \(\quad(\mathrm{a}, \mathrm{b})=(9,2)\)
119620
Let \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\). Then,
1 \(a_0+a_2+\ldots+a_{18}=a_1+a_2+\ldots+a_{17}\)
2 \(a_0+a_2+\ldots+a_{18}\) is even
3 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is divisible by 9
4 \(\mathrm{a}_0+\mathrm{a}_2+\ldots+\mathrm{a}_{18}\) is dividable by 3 but not by 9
Explanation:
B Given, \(\left(1+x+x^2\right)^9=a_0+a_1 x+a_2 x^2+\ldots+a_{18} x^{18}\) \(\text { Put, } x=-1 \text {, we get- }\) \((1-1+1)^9=a_0-a_1+a_2-\ldots+a_{18}\) \(1=a_0-a_1+a_2 \ldots .+a_{18}\) \(\text { And, put } x=1 \text {, we get - }\) \((1+1+1)^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9=a_0+a_1+a_2+\ldots+a_{18}\) \(3^9+1=2\left(a_0+a_2+\ldots+a_{18}\right)\) \(a_0+a_1+a_2+\ldots+a_{18}=\frac{3^9+1}{2}\) \(=\frac{19683+1}{2}\) \(=\frac{19684}{2}\) \(=9842 \text {, which is an even number. }\) On adding equation (i) and (ii), we get -
WB JEE-2017
Binomial Theorem and its Simple Application
119615
Let \(f(x)\) be a polynomial function such that \(f\) \((x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\). Then, the value of \(\lim _{x \rightarrow I} \frac{f(x)}{x-1}\) is equal to:
1 -15
2 -60
3 60
4 15
Explanation:
A Given, \(f(x)\) be a polynomial function such that. \(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^5+64\) Consider, \(f(x)=x^5+p x^4+q x^3+r x^2+s x+t\) \(f^{\prime}(x)=5 x^4+4 p x^3+3 q x^2+2 r x+s\) \(f^{\prime \prime}(x)=20 x^3+12 p x^2+6 q x+2 r\) Then, \(f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^5+64\) \(x^5+ (p+5) x^4+(q+4 p+20) x^3+(r+3 q+12 p) x^2\) \((s+2 r+6 q) x+(t+s+2 r)=x^2+64\) \(p+5=0\) \((q+4 p+20)=0\) \(r+3 q+12 p=0\) \(s+2 r+6 q=0\) \(t+s+2 r=64\) By solving, we get- \(p=-5, q=0, r=60, s=-120, t=64\) \(\therefore \quad f(x)=x^5-5 x^4+60 x^2-120 x+64\) \(\text { Then, } \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1} \text { is } \frac{0}{0} \text { from }\) Then, \(\quad \lim _{x \rightarrow 1} \frac{x^5-5 x^4+60 x^2-120 x+64}{x-1}\) is \(\frac{0}{0}\) from \(\therefore\) By L' Hospital rule- \(\lim _{x \rightarrow 1} \frac{5 x^4-20 x^3+120 x-120}{1}\) \(=5-20+120-120\) \(=-15\)
JEE Main-25.06.2022
Binomial Theorem and its Simple Application
119616
If \(x^2+p x+1\) is a factor of \(a x^3+b x+c\), then
1 \(a^2+c^2=a b\)
2 \(\mathrm{a}^2+\mathrm{c}^2=-\mathrm{ab}\)
3 \(a^2-c^2=a b\)
4 None of these
Explanation:
C Given \(x^2+p x+1\) is a factor of \(a x^3+b x+c\) Let, \(\alpha, \beta\) be the roots of \(\mathrm{x}^2+\mathrm{px}+1=0\) \(\alpha+\beta=-p\) \(\alpha \beta=1\) Now, \(\alpha, \beta, \mathrm{r}\) be the roots of \(\mathrm{ax}^3+\mathrm{bx}+\mathrm{c}=0\) Then, \(\alpha+\beta+r=0\) \(\alpha \beta+\beta r+r \alpha=\frac{b}{a}\) \(\alpha \beta r=-\frac{c}{a}\) From (i) and (iii) we get, \(\mathrm{r}=\mathrm{p}\) From (ii) and (v) we get, \(\mathrm{r}=\frac{-\mathrm{c}}{\mathrm{a}}\) And, from (ii) and (iv) we get- \(1+\beta p+p \alpha=\frac{b}{a}\) \(p(\alpha+\beta)=\frac{b}{a}-1\) \(p(-p)=\frac{b-a}{a}\) \(-p^2=\frac{b-a}{a} \Rightarrow-r^2=\frac{b-a}{a}\) \(-\left(\frac{-c}{a}\right)^2=\frac{b-a}{a}\) \(-c^2=\frac{b-a}{a}\) \(a^2-c^2=a b\)
AMU-2015]
Binomial Theorem and its Simple Application
119617
The minimum degree of a polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 i\) as two of its roots is
1 8
2 6
3 4
4 2
Explanation:
B Given, A polynomial equation with rational coefficients having \(\sqrt{3}+\sqrt{27}, \sqrt{2}+5 \mathrm{i}\) as two roots Then, other roots are- \(\sqrt{3}-\sqrt{27},-\sqrt{3}+\sqrt{27},-\sqrt{3}-\sqrt{27}\) and \(\sqrt{2}-5 \mathrm{i}\) So, the minimum degree of a polynomial equation having 6 distinct roots is 6 Hence, number of roots are 6.
AP EAMCET-22.04.2018
Binomial Theorem and its Simple Application
119618
If \(f(x)=2 x^4-13 x^2+a x+b\) is divisible by \(x^2-3 x+2\), then \((a, b)\) is equal to
1 \((-9,-2)\)
2 \((6,4)\)
3 \((9,2)\)
4 \((2,9)\)
Explanation:
C Given, \(f(x)=2 x^4-13 x^2+a x+b\) And is divisible by \(\mathrm{x}^2-3 \mathrm{x}+2=\mathrm{g}(\mathrm{x})\) \(x^2-2 x-x+2=g(x)\) \(x(x-2)-1(x-2)=g(x)\) \((x-1)(x-2)=g(x)\) \((x-1)=0 \text { or }(x-2)=g(x)\) \(x=1 \text { or } x=2\) Then, put \(\mathrm{x}=1\) or \(\mathrm{x}=2\) in equation (i), we get- \(f(1)=2 \times 1^4-13 \times 1^2+a \times 1+b\) \(f(1)=2-13+a+b\) \(f(1)=-11+a+b\) And, \(f(2)=2 \times 2^4-13 \times 2^2+\) \(f(2)=32-52+2 a+b\) \(f(2)=-20+2 a+b\) \(\because \quad \mathrm{f}(\mathrm{x})\) is divisible by \(\mathrm{g}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}-2)\) \((x-1) \text { and }(x-2) \text { are factors of } f(x)\) \(f(1)=0 \text { and } f(2)=0\) Then, \(a+b=11\) \(2 a+b=20\) Subtract equation (i) from equation (ii) we get \(\mathrm{a}=9\) \(\mathrm{b}=2\)So, \(\quad(\mathrm{a}, \mathrm{b})=(9,2)\)