118177
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{1}{a+b}\) is
1 \(\frac{4}{a+b}\)
2 \(\frac{1}{a+b}\)
3 0
4 -1
Explanation:
C Given, \(\alpha, \beta\) are the root of - \(\mathrm{x}^2-\mathrm{a}(\mathrm{x}-1)+\mathrm{b}=0\) Therefore, \(\alpha\) and \(\beta\) will satisfy the equation. \(\therefore \quad \alpha^2-a(\alpha-1)+b=0\) \(\text { or } \quad \alpha^2-a \alpha+a+b=0\) \(\Rightarrow \quad \alpha^2-a \alpha=-(a+b)\) \(\text { Similarity } \beta-a(\beta-1)+b=0\) \(\Rightarrow \quad \beta^2-a \beta+a+b=0\) \(\therefore \quad \beta^2-a \beta=-(a+b)\) \(\therefore \frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}=\frac{-1}{a+b}-\frac{1}{(a+b)}+\frac{2}{(a+b)}\) \(\quad=0\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118178
If \(a, b, c\) and \(x\) are non zero real numbers, then \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds
1 Only when a, b, c are in A.P.
2 Only when a, b, c are in G.P.
3 Only when a, b, c are in H.P.
4 Always
Explanation:
B Given, The equation \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds real \(\mathrm{x}\). \(\therefore\) \((\text { Discriminate })=[-2 b(a+c)]^2-4\left(a^2+b^2\right)\left(b^2+c^2\right) \geq 0\) \(\Rightarrow \quad b^2(a+c)^2=\left(a^2+b^2\right)\left(b^2+c^2\right)\) \(\Rightarrow \quad b^2\left(a^2+c^2+2 a c\right)=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\Rightarrow \quad b^2 a^2+b^2 c^2+b^2 \cdot 2 a c=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\mathrm{~b}^4+(\mathrm{ac})^2-2 \mathrm{ac} \cdot \mathrm{b}^2=0\) \(\left(b^2-a c\right)^2=0\) \(\mathrm{~b}^2=\mathrm{ac}\) \(\therefore \quad \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are is G.P. } \\ end{aligned}\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118179
If \(\alpha, \beta\) are the roots of \(x^2-p(x+1)-c=0\) then, \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2 \beta+c}\) is equal to :
1 3
2 2
3 1
4 0
Explanation:
C Given, \(\alpha, \beta\) are the roots of \(\mathrm{x}^2-\mathrm{p}(\mathrm{x}+1)-\mathrm{c}=0\) Let \(\mathrm{p}=-2\), then \(\mathrm{x}^2+2 \mathrm{x}+2-\mathrm{c}=0\) \(\therefore \quad \mathrm{a}^2+2 \mathrm{a}+1+1-\mathrm{c}=0\) and \(\quad \beta^2+2 \beta+1+1-c=0\) \((\because \alpha, \beta\) will satisfy the equation) \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2+c}\) \(\therefore \quad \frac{\mathrm{c}-1}{2 \mathrm{c}-2}+\frac{\mathrm{c}-1}{2 \mathrm{c}-2}=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}\) \(=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}=1\)
Assam CEE-2017
Complex Numbers and Quadratic Equation
118181
The number of integers \(n\) for which \(3 x^3-25 x+\) \(\mathbf{n}=\mathbf{0}\) has three real roots is
1 1
2 25
3 55
4 infinite
Explanation:
C We have, \(f(x)=3 x^3-25 x+n\) Using the fact that, between two roots of derivative, a function has at most 1 root \(f^{\prime}(x)=9 x^2-25\) \(\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=0\) for \(\mathrm{x}= \pm \frac{5}{3}\) \(\mathrm{f}\left[\frac{5}{3}\right]=3\left[\frac{5}{3}\right]^3-25 \times \frac{5}{3}+\mathrm{n}\) \(=\mathrm{n}-\frac{250}{9}\) \(\mathrm{f}\left[-\frac{5}{3}\right]=\mathrm{n}+\frac{250}{9}\) For the cubic to have 3 real roots, \(\mathrm{f}\left[-\frac{5}{3}\right]>0 \text { and } \mathrm{f}\left[\frac{5}{3}\right]\lt 0 \text { i.e. } \frac{-250}{9}\lt \mathrm{n}\lt \) There are 55 such \(n\) possible \(\text { i.e. } \mathrm{n} \in[-27,27]\)
KVPY SB/SX-2014
Complex Numbers and Quadratic Equation
118182
Let \(r(x)\) be the remainder when the polynomial \(x^{135}+x^{125}-x^{115}+x^5+1\) is divided by \(x^3-x\). Then
1 \(r(x)\) is the zero polynomial
2 \(r(x)\) is a nonzero constant
3 degree of \(r(x)\) is one
4 degree of \(r(x)\) is two
Explanation:
C According to given summation, we can write, \(\mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+\mathrm{Bx}^2+\mathrm{cx}+\mathrm{d}\) Putting \(\mathrm{x}=0 \quad \Rightarrow \mathrm{d}=1\) Putting \(\mathrm{x}=1 \quad \Rightarrow \mathrm{B}+\mathrm{C}=2\) putting \(\mathrm{x}=-1 \Rightarrow \mathrm{B}-\mathrm{C}=-2\) Using equation (i) and (ii), \(\mathrm{B}=0, \mathrm{C}=2\) \(\therefore \quad \mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+2 \mathrm{x}+1\)Hence, \(r(x)=2 x+1\)
118177
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{1}{a+b}\) is
1 \(\frac{4}{a+b}\)
2 \(\frac{1}{a+b}\)
3 0
4 -1
Explanation:
C Given, \(\alpha, \beta\) are the root of - \(\mathrm{x}^2-\mathrm{a}(\mathrm{x}-1)+\mathrm{b}=0\) Therefore, \(\alpha\) and \(\beta\) will satisfy the equation. \(\therefore \quad \alpha^2-a(\alpha-1)+b=0\) \(\text { or } \quad \alpha^2-a \alpha+a+b=0\) \(\Rightarrow \quad \alpha^2-a \alpha=-(a+b)\) \(\text { Similarity } \beta-a(\beta-1)+b=0\) \(\Rightarrow \quad \beta^2-a \beta+a+b=0\) \(\therefore \quad \beta^2-a \beta=-(a+b)\) \(\therefore \frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}=\frac{-1}{a+b}-\frac{1}{(a+b)}+\frac{2}{(a+b)}\) \(\quad=0\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118178
If \(a, b, c\) and \(x\) are non zero real numbers, then \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds
1 Only when a, b, c are in A.P.
2 Only when a, b, c are in G.P.
3 Only when a, b, c are in H.P.
4 Always
Explanation:
B Given, The equation \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds real \(\mathrm{x}\). \(\therefore\) \((\text { Discriminate })=[-2 b(a+c)]^2-4\left(a^2+b^2\right)\left(b^2+c^2\right) \geq 0\) \(\Rightarrow \quad b^2(a+c)^2=\left(a^2+b^2\right)\left(b^2+c^2\right)\) \(\Rightarrow \quad b^2\left(a^2+c^2+2 a c\right)=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\Rightarrow \quad b^2 a^2+b^2 c^2+b^2 \cdot 2 a c=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\mathrm{~b}^4+(\mathrm{ac})^2-2 \mathrm{ac} \cdot \mathrm{b}^2=0\) \(\left(b^2-a c\right)^2=0\) \(\mathrm{~b}^2=\mathrm{ac}\) \(\therefore \quad \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are is G.P. } \\ end{aligned}\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118179
If \(\alpha, \beta\) are the roots of \(x^2-p(x+1)-c=0\) then, \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2 \beta+c}\) is equal to :
1 3
2 2
3 1
4 0
Explanation:
C Given, \(\alpha, \beta\) are the roots of \(\mathrm{x}^2-\mathrm{p}(\mathrm{x}+1)-\mathrm{c}=0\) Let \(\mathrm{p}=-2\), then \(\mathrm{x}^2+2 \mathrm{x}+2-\mathrm{c}=0\) \(\therefore \quad \mathrm{a}^2+2 \mathrm{a}+1+1-\mathrm{c}=0\) and \(\quad \beta^2+2 \beta+1+1-c=0\) \((\because \alpha, \beta\) will satisfy the equation) \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2+c}\) \(\therefore \quad \frac{\mathrm{c}-1}{2 \mathrm{c}-2}+\frac{\mathrm{c}-1}{2 \mathrm{c}-2}=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}\) \(=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}=1\)
Assam CEE-2017
Complex Numbers and Quadratic Equation
118181
The number of integers \(n\) for which \(3 x^3-25 x+\) \(\mathbf{n}=\mathbf{0}\) has three real roots is
1 1
2 25
3 55
4 infinite
Explanation:
C We have, \(f(x)=3 x^3-25 x+n\) Using the fact that, between two roots of derivative, a function has at most 1 root \(f^{\prime}(x)=9 x^2-25\) \(\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=0\) for \(\mathrm{x}= \pm \frac{5}{3}\) \(\mathrm{f}\left[\frac{5}{3}\right]=3\left[\frac{5}{3}\right]^3-25 \times \frac{5}{3}+\mathrm{n}\) \(=\mathrm{n}-\frac{250}{9}\) \(\mathrm{f}\left[-\frac{5}{3}\right]=\mathrm{n}+\frac{250}{9}\) For the cubic to have 3 real roots, \(\mathrm{f}\left[-\frac{5}{3}\right]>0 \text { and } \mathrm{f}\left[\frac{5}{3}\right]\lt 0 \text { i.e. } \frac{-250}{9}\lt \mathrm{n}\lt \) There are 55 such \(n\) possible \(\text { i.e. } \mathrm{n} \in[-27,27]\)
KVPY SB/SX-2014
Complex Numbers and Quadratic Equation
118182
Let \(r(x)\) be the remainder when the polynomial \(x^{135}+x^{125}-x^{115}+x^5+1\) is divided by \(x^3-x\). Then
1 \(r(x)\) is the zero polynomial
2 \(r(x)\) is a nonzero constant
3 degree of \(r(x)\) is one
4 degree of \(r(x)\) is two
Explanation:
C According to given summation, we can write, \(\mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+\mathrm{Bx}^2+\mathrm{cx}+\mathrm{d}\) Putting \(\mathrm{x}=0 \quad \Rightarrow \mathrm{d}=1\) Putting \(\mathrm{x}=1 \quad \Rightarrow \mathrm{B}+\mathrm{C}=2\) putting \(\mathrm{x}=-1 \Rightarrow \mathrm{B}-\mathrm{C}=-2\) Using equation (i) and (ii), \(\mathrm{B}=0, \mathrm{C}=2\) \(\therefore \quad \mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+2 \mathrm{x}+1\)Hence, \(r(x)=2 x+1\)
118177
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{1}{a+b}\) is
1 \(\frac{4}{a+b}\)
2 \(\frac{1}{a+b}\)
3 0
4 -1
Explanation:
C Given, \(\alpha, \beta\) are the root of - \(\mathrm{x}^2-\mathrm{a}(\mathrm{x}-1)+\mathrm{b}=0\) Therefore, \(\alpha\) and \(\beta\) will satisfy the equation. \(\therefore \quad \alpha^2-a(\alpha-1)+b=0\) \(\text { or } \quad \alpha^2-a \alpha+a+b=0\) \(\Rightarrow \quad \alpha^2-a \alpha=-(a+b)\) \(\text { Similarity } \beta-a(\beta-1)+b=0\) \(\Rightarrow \quad \beta^2-a \beta+a+b=0\) \(\therefore \quad \beta^2-a \beta=-(a+b)\) \(\therefore \frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}=\frac{-1}{a+b}-\frac{1}{(a+b)}+\frac{2}{(a+b)}\) \(\quad=0\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118178
If \(a, b, c\) and \(x\) are non zero real numbers, then \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds
1 Only when a, b, c are in A.P.
2 Only when a, b, c are in G.P.
3 Only when a, b, c are in H.P.
4 Always
Explanation:
B Given, The equation \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds real \(\mathrm{x}\). \(\therefore\) \((\text { Discriminate })=[-2 b(a+c)]^2-4\left(a^2+b^2\right)\left(b^2+c^2\right) \geq 0\) \(\Rightarrow \quad b^2(a+c)^2=\left(a^2+b^2\right)\left(b^2+c^2\right)\) \(\Rightarrow \quad b^2\left(a^2+c^2+2 a c\right)=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\Rightarrow \quad b^2 a^2+b^2 c^2+b^2 \cdot 2 a c=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\mathrm{~b}^4+(\mathrm{ac})^2-2 \mathrm{ac} \cdot \mathrm{b}^2=0\) \(\left(b^2-a c\right)^2=0\) \(\mathrm{~b}^2=\mathrm{ac}\) \(\therefore \quad \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are is G.P. } \\ end{aligned}\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118179
If \(\alpha, \beta\) are the roots of \(x^2-p(x+1)-c=0\) then, \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2 \beta+c}\) is equal to :
1 3
2 2
3 1
4 0
Explanation:
C Given, \(\alpha, \beta\) are the roots of \(\mathrm{x}^2-\mathrm{p}(\mathrm{x}+1)-\mathrm{c}=0\) Let \(\mathrm{p}=-2\), then \(\mathrm{x}^2+2 \mathrm{x}+2-\mathrm{c}=0\) \(\therefore \quad \mathrm{a}^2+2 \mathrm{a}+1+1-\mathrm{c}=0\) and \(\quad \beta^2+2 \beta+1+1-c=0\) \((\because \alpha, \beta\) will satisfy the equation) \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2+c}\) \(\therefore \quad \frac{\mathrm{c}-1}{2 \mathrm{c}-2}+\frac{\mathrm{c}-1}{2 \mathrm{c}-2}=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}\) \(=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}=1\)
Assam CEE-2017
Complex Numbers and Quadratic Equation
118181
The number of integers \(n\) for which \(3 x^3-25 x+\) \(\mathbf{n}=\mathbf{0}\) has three real roots is
1 1
2 25
3 55
4 infinite
Explanation:
C We have, \(f(x)=3 x^3-25 x+n\) Using the fact that, between two roots of derivative, a function has at most 1 root \(f^{\prime}(x)=9 x^2-25\) \(\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=0\) for \(\mathrm{x}= \pm \frac{5}{3}\) \(\mathrm{f}\left[\frac{5}{3}\right]=3\left[\frac{5}{3}\right]^3-25 \times \frac{5}{3}+\mathrm{n}\) \(=\mathrm{n}-\frac{250}{9}\) \(\mathrm{f}\left[-\frac{5}{3}\right]=\mathrm{n}+\frac{250}{9}\) For the cubic to have 3 real roots, \(\mathrm{f}\left[-\frac{5}{3}\right]>0 \text { and } \mathrm{f}\left[\frac{5}{3}\right]\lt 0 \text { i.e. } \frac{-250}{9}\lt \mathrm{n}\lt \) There are 55 such \(n\) possible \(\text { i.e. } \mathrm{n} \in[-27,27]\)
KVPY SB/SX-2014
Complex Numbers and Quadratic Equation
118182
Let \(r(x)\) be the remainder when the polynomial \(x^{135}+x^{125}-x^{115}+x^5+1\) is divided by \(x^3-x\). Then
1 \(r(x)\) is the zero polynomial
2 \(r(x)\) is a nonzero constant
3 degree of \(r(x)\) is one
4 degree of \(r(x)\) is two
Explanation:
C According to given summation, we can write, \(\mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+\mathrm{Bx}^2+\mathrm{cx}+\mathrm{d}\) Putting \(\mathrm{x}=0 \quad \Rightarrow \mathrm{d}=1\) Putting \(\mathrm{x}=1 \quad \Rightarrow \mathrm{B}+\mathrm{C}=2\) putting \(\mathrm{x}=-1 \Rightarrow \mathrm{B}-\mathrm{C}=-2\) Using equation (i) and (ii), \(\mathrm{B}=0, \mathrm{C}=2\) \(\therefore \quad \mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+2 \mathrm{x}+1\)Hence, \(r(x)=2 x+1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118177
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{1}{a+b}\) is
1 \(\frac{4}{a+b}\)
2 \(\frac{1}{a+b}\)
3 0
4 -1
Explanation:
C Given, \(\alpha, \beta\) are the root of - \(\mathrm{x}^2-\mathrm{a}(\mathrm{x}-1)+\mathrm{b}=0\) Therefore, \(\alpha\) and \(\beta\) will satisfy the equation. \(\therefore \quad \alpha^2-a(\alpha-1)+b=0\) \(\text { or } \quad \alpha^2-a \alpha+a+b=0\) \(\Rightarrow \quad \alpha^2-a \alpha=-(a+b)\) \(\text { Similarity } \beta-a(\beta-1)+b=0\) \(\Rightarrow \quad \beta^2-a \beta+a+b=0\) \(\therefore \quad \beta^2-a \beta=-(a+b)\) \(\therefore \frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}=\frac{-1}{a+b}-\frac{1}{(a+b)}+\frac{2}{(a+b)}\) \(\quad=0\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118178
If \(a, b, c\) and \(x\) are non zero real numbers, then \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds
1 Only when a, b, c are in A.P.
2 Only when a, b, c are in G.P.
3 Only when a, b, c are in H.P.
4 Always
Explanation:
B Given, The equation \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds real \(\mathrm{x}\). \(\therefore\) \((\text { Discriminate })=[-2 b(a+c)]^2-4\left(a^2+b^2\right)\left(b^2+c^2\right) \geq 0\) \(\Rightarrow \quad b^2(a+c)^2=\left(a^2+b^2\right)\left(b^2+c^2\right)\) \(\Rightarrow \quad b^2\left(a^2+c^2+2 a c\right)=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\Rightarrow \quad b^2 a^2+b^2 c^2+b^2 \cdot 2 a c=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\mathrm{~b}^4+(\mathrm{ac})^2-2 \mathrm{ac} \cdot \mathrm{b}^2=0\) \(\left(b^2-a c\right)^2=0\) \(\mathrm{~b}^2=\mathrm{ac}\) \(\therefore \quad \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are is G.P. } \\ end{aligned}\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118179
If \(\alpha, \beta\) are the roots of \(x^2-p(x+1)-c=0\) then, \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2 \beta+c}\) is equal to :
1 3
2 2
3 1
4 0
Explanation:
C Given, \(\alpha, \beta\) are the roots of \(\mathrm{x}^2-\mathrm{p}(\mathrm{x}+1)-\mathrm{c}=0\) Let \(\mathrm{p}=-2\), then \(\mathrm{x}^2+2 \mathrm{x}+2-\mathrm{c}=0\) \(\therefore \quad \mathrm{a}^2+2 \mathrm{a}+1+1-\mathrm{c}=0\) and \(\quad \beta^2+2 \beta+1+1-c=0\) \((\because \alpha, \beta\) will satisfy the equation) \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2+c}\) \(\therefore \quad \frac{\mathrm{c}-1}{2 \mathrm{c}-2}+\frac{\mathrm{c}-1}{2 \mathrm{c}-2}=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}\) \(=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}=1\)
Assam CEE-2017
Complex Numbers and Quadratic Equation
118181
The number of integers \(n\) for which \(3 x^3-25 x+\) \(\mathbf{n}=\mathbf{0}\) has three real roots is
1 1
2 25
3 55
4 infinite
Explanation:
C We have, \(f(x)=3 x^3-25 x+n\) Using the fact that, between two roots of derivative, a function has at most 1 root \(f^{\prime}(x)=9 x^2-25\) \(\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=0\) for \(\mathrm{x}= \pm \frac{5}{3}\) \(\mathrm{f}\left[\frac{5}{3}\right]=3\left[\frac{5}{3}\right]^3-25 \times \frac{5}{3}+\mathrm{n}\) \(=\mathrm{n}-\frac{250}{9}\) \(\mathrm{f}\left[-\frac{5}{3}\right]=\mathrm{n}+\frac{250}{9}\) For the cubic to have 3 real roots, \(\mathrm{f}\left[-\frac{5}{3}\right]>0 \text { and } \mathrm{f}\left[\frac{5}{3}\right]\lt 0 \text { i.e. } \frac{-250}{9}\lt \mathrm{n}\lt \) There are 55 such \(n\) possible \(\text { i.e. } \mathrm{n} \in[-27,27]\)
KVPY SB/SX-2014
Complex Numbers and Quadratic Equation
118182
Let \(r(x)\) be the remainder when the polynomial \(x^{135}+x^{125}-x^{115}+x^5+1\) is divided by \(x^3-x\). Then
1 \(r(x)\) is the zero polynomial
2 \(r(x)\) is a nonzero constant
3 degree of \(r(x)\) is one
4 degree of \(r(x)\) is two
Explanation:
C According to given summation, we can write, \(\mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+\mathrm{Bx}^2+\mathrm{cx}+\mathrm{d}\) Putting \(\mathrm{x}=0 \quad \Rightarrow \mathrm{d}=1\) Putting \(\mathrm{x}=1 \quad \Rightarrow \mathrm{B}+\mathrm{C}=2\) putting \(\mathrm{x}=-1 \Rightarrow \mathrm{B}-\mathrm{C}=-2\) Using equation (i) and (ii), \(\mathrm{B}=0, \mathrm{C}=2\) \(\therefore \quad \mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+2 \mathrm{x}+1\)Hence, \(r(x)=2 x+1\)
118177
If \(\alpha\) and \(\beta\) are the roots of \(x^2-a(x-1)+b=0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{1}{a+b}\) is
1 \(\frac{4}{a+b}\)
2 \(\frac{1}{a+b}\)
3 0
4 -1
Explanation:
C Given, \(\alpha, \beta\) are the root of - \(\mathrm{x}^2-\mathrm{a}(\mathrm{x}-1)+\mathrm{b}=0\) Therefore, \(\alpha\) and \(\beta\) will satisfy the equation. \(\therefore \quad \alpha^2-a(\alpha-1)+b=0\) \(\text { or } \quad \alpha^2-a \alpha+a+b=0\) \(\Rightarrow \quad \alpha^2-a \alpha=-(a+b)\) \(\text { Similarity } \beta-a(\beta-1)+b=0\) \(\Rightarrow \quad \beta^2-a \beta+a+b=0\) \(\therefore \quad \beta^2-a \beta=-(a+b)\) \(\therefore \frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}=\frac{-1}{a+b}-\frac{1}{(a+b)}+\frac{2}{(a+b)}\) \(\quad=0\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118178
If \(a, b, c\) and \(x\) are non zero real numbers, then \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds
1 Only when a, b, c are in A.P.
2 Only when a, b, c are in G.P.
3 Only when a, b, c are in H.P.
4 Always
Explanation:
B Given, The equation \(\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0\) holds real \(\mathrm{x}\). \(\therefore\) \((\text { Discriminate })=[-2 b(a+c)]^2-4\left(a^2+b^2\right)\left(b^2+c^2\right) \geq 0\) \(\Rightarrow \quad b^2(a+c)^2=\left(a^2+b^2\right)\left(b^2+c^2\right)\) \(\Rightarrow \quad b^2\left(a^2+c^2+2 a c\right)=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\Rightarrow \quad b^2 a^2+b^2 c^2+b^2 \cdot 2 a c=a^2 b^2+a^2 c^2+b^2 c^2+b^4\) \(\mathrm{~b}^4+(\mathrm{ac})^2-2 \mathrm{ac} \cdot \mathrm{b}^2=0\) \(\left(b^2-a c\right)^2=0\) \(\mathrm{~b}^2=\mathrm{ac}\) \(\therefore \quad \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are is G.P. } \\ end{aligned}\)
Assam CEE-2021
Complex Numbers and Quadratic Equation
118179
If \(\alpha, \beta\) are the roots of \(x^2-p(x+1)-c=0\) then, \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2 \beta+c}\) is equal to :
1 3
2 2
3 1
4 0
Explanation:
C Given, \(\alpha, \beta\) are the roots of \(\mathrm{x}^2-\mathrm{p}(\mathrm{x}+1)-\mathrm{c}=0\) Let \(\mathrm{p}=-2\), then \(\mathrm{x}^2+2 \mathrm{x}+2-\mathrm{c}=0\) \(\therefore \quad \mathrm{a}^2+2 \mathrm{a}+1+1-\mathrm{c}=0\) and \(\quad \beta^2+2 \beta+1+1-c=0\) \((\because \alpha, \beta\) will satisfy the equation) \(\frac{\alpha^2+2 \alpha+1}{\alpha^2+2 \alpha+c}+\frac{\beta^2+2 \beta+1}{\beta^2+2+c}\) \(\therefore \quad \frac{\mathrm{c}-1}{2 \mathrm{c}-2}+\frac{\mathrm{c}-1}{2 \mathrm{c}-2}=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}\) \(=\frac{2(\mathrm{c}-1)}{(2 \mathrm{c}-2)}=1\)
Assam CEE-2017
Complex Numbers and Quadratic Equation
118181
The number of integers \(n\) for which \(3 x^3-25 x+\) \(\mathbf{n}=\mathbf{0}\) has three real roots is
1 1
2 25
3 55
4 infinite
Explanation:
C We have, \(f(x)=3 x^3-25 x+n\) Using the fact that, between two roots of derivative, a function has at most 1 root \(f^{\prime}(x)=9 x^2-25\) \(\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=0\) for \(\mathrm{x}= \pm \frac{5}{3}\) \(\mathrm{f}\left[\frac{5}{3}\right]=3\left[\frac{5}{3}\right]^3-25 \times \frac{5}{3}+\mathrm{n}\) \(=\mathrm{n}-\frac{250}{9}\) \(\mathrm{f}\left[-\frac{5}{3}\right]=\mathrm{n}+\frac{250}{9}\) For the cubic to have 3 real roots, \(\mathrm{f}\left[-\frac{5}{3}\right]>0 \text { and } \mathrm{f}\left[\frac{5}{3}\right]\lt 0 \text { i.e. } \frac{-250}{9}\lt \mathrm{n}\lt \) There are 55 such \(n\) possible \(\text { i.e. } \mathrm{n} \in[-27,27]\)
KVPY SB/SX-2014
Complex Numbers and Quadratic Equation
118182
Let \(r(x)\) be the remainder when the polynomial \(x^{135}+x^{125}-x^{115}+x^5+1\) is divided by \(x^3-x\). Then
1 \(r(x)\) is the zero polynomial
2 \(r(x)\) is a nonzero constant
3 degree of \(r(x)\) is one
4 degree of \(r(x)\) is two
Explanation:
C According to given summation, we can write, \(\mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+\mathrm{Bx}^2+\mathrm{cx}+\mathrm{d}\) Putting \(\mathrm{x}=0 \quad \Rightarrow \mathrm{d}=1\) Putting \(\mathrm{x}=1 \quad \Rightarrow \mathrm{B}+\mathrm{C}=2\) putting \(\mathrm{x}=-1 \Rightarrow \mathrm{B}-\mathrm{C}=-2\) Using equation (i) and (ii), \(\mathrm{B}=0, \mathrm{C}=2\) \(\therefore \quad \mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^5+1\) \(=\mathrm{A}\left(\mathrm{x}^3-\mathrm{x}\right)+2 \mathrm{x}+1\)Hence, \(r(x)=2 x+1\)