118092
A polynomial function \(f(x)\) satisfies the condition for all \(x \in R, x \neq 0\). If \(f(3)=-26\) then \(f(4)\) is
1 -35
2 -63
3 65
4 None of these
Explanation:
B We know that, if a polynomial function \(f(\mathrm{x})\) satisfies the condition \(f(\mathrm{x}) f\left(\frac{1}{\mathrm{x}}\right)=f(\mathrm{x})+f\left(\frac{1}{\mathrm{x}}\right)\) Then it will be in the form of \(f(\mathrm{x})=1 \pm \mathrm{x}^{\mathrm{n}}\) \(\because\) Given \(f(3)=-26\) \(\therefore f(3)=1 \pm 3^{\mathrm{n}}\) \(-26=1-3^{\mathrm{n}}\) \(=-27=-3^{\mathrm{n}}\) \(3^{\mathrm{n}}=27\) \(3^{\mathrm{n}}=3^3\) \(\mathrm{n}=3\) Now, \(f(4)=\) ? \(f(4)=1-4^3\) \(=1-64\) \(f(4)=-63\)
BCECE-2018
Complex Numbers and Quadratic Equation
118093
The number of solutions of the equation \(x^3+x^2+4 x+2 \sin x=0\) in \([0,2 \pi]\), is
1 0
2 1
3 2
4 4
Explanation:
B \( \mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}+2 \sin \mathrm{x}=0\) \(-\left(\frac{\mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}}{2}\right)=\sin \mathrm{x}\) \(\text { Plotting the graph of both } \sin \mathrm{x} \text { and cubic equation }\) Plotting the graph of both \(\sin \mathrm{x}\) and cubic equation They meets at only one point that is \(x=0\) \(\therefore\) Only one solution is possible.
BCECE-2018
Complex Numbers and Quadratic Equation
118094
If the roots of the given equation \((\cos p-1) x^2+(\cos p) x+\sin p=0\) are real, then
118095
If \(1,2,3\) and 4 are the roots of the equations \(\mathbf{x}^4+\mathbf{a} \mathbf{x}^3+\mathbf{b x ^ { 2 }}+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(\mathbf{a}+\mathbf{2 b}+\mathbf{c}\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given equation is \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}=0\) and \(\because\) Roots are \(-1,2,3\) and 4 \(\therefore(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\) \(\mathrm{d}\) \(\left(\mathrm{x}^2-3 \mathrm{x}+2\right)\left(\mathrm{x}^2-7 \mathrm{x}+12\right)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-7 \mathrm{x}^3+12 \mathrm{x}^2-3 \mathrm{x}^3+21 \mathrm{x}^2-36 \mathrm{x}+2 \mathrm{x}^2-14 \mathrm{x}+24=\) \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-10 \mathrm{x}^3+35 \mathrm{x}^2-50 \mathrm{x}+24=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) Comparing coefficient \(\mathrm{a}=-10, \mathrm{~b}=35, \mathrm{c}=-50, \mathrm{~d}=24\) \(\text { Now, the value of } \mathrm{a}+2 \mathrm{~b}+\mathrm{c} \text { is }\) \(\mathrm{a}+2 \mathrm{~b}+\mathrm{c}\) \(=-10+2 \times 35-50\) \(=-60+70=10 right.\)
118092
A polynomial function \(f(x)\) satisfies the condition for all \(x \in R, x \neq 0\). If \(f(3)=-26\) then \(f(4)\) is
1 -35
2 -63
3 65
4 None of these
Explanation:
B We know that, if a polynomial function \(f(\mathrm{x})\) satisfies the condition \(f(\mathrm{x}) f\left(\frac{1}{\mathrm{x}}\right)=f(\mathrm{x})+f\left(\frac{1}{\mathrm{x}}\right)\) Then it will be in the form of \(f(\mathrm{x})=1 \pm \mathrm{x}^{\mathrm{n}}\) \(\because\) Given \(f(3)=-26\) \(\therefore f(3)=1 \pm 3^{\mathrm{n}}\) \(-26=1-3^{\mathrm{n}}\) \(=-27=-3^{\mathrm{n}}\) \(3^{\mathrm{n}}=27\) \(3^{\mathrm{n}}=3^3\) \(\mathrm{n}=3\) Now, \(f(4)=\) ? \(f(4)=1-4^3\) \(=1-64\) \(f(4)=-63\)
BCECE-2018
Complex Numbers and Quadratic Equation
118093
The number of solutions of the equation \(x^3+x^2+4 x+2 \sin x=0\) in \([0,2 \pi]\), is
1 0
2 1
3 2
4 4
Explanation:
B \( \mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}+2 \sin \mathrm{x}=0\) \(-\left(\frac{\mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}}{2}\right)=\sin \mathrm{x}\) \(\text { Plotting the graph of both } \sin \mathrm{x} \text { and cubic equation }\) Plotting the graph of both \(\sin \mathrm{x}\) and cubic equation They meets at only one point that is \(x=0\) \(\therefore\) Only one solution is possible.
BCECE-2018
Complex Numbers and Quadratic Equation
118094
If the roots of the given equation \((\cos p-1) x^2+(\cos p) x+\sin p=0\) are real, then
118095
If \(1,2,3\) and 4 are the roots of the equations \(\mathbf{x}^4+\mathbf{a} \mathbf{x}^3+\mathbf{b x ^ { 2 }}+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(\mathbf{a}+\mathbf{2 b}+\mathbf{c}\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given equation is \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}=0\) and \(\because\) Roots are \(-1,2,3\) and 4 \(\therefore(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\) \(\mathrm{d}\) \(\left(\mathrm{x}^2-3 \mathrm{x}+2\right)\left(\mathrm{x}^2-7 \mathrm{x}+12\right)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-7 \mathrm{x}^3+12 \mathrm{x}^2-3 \mathrm{x}^3+21 \mathrm{x}^2-36 \mathrm{x}+2 \mathrm{x}^2-14 \mathrm{x}+24=\) \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-10 \mathrm{x}^3+35 \mathrm{x}^2-50 \mathrm{x}+24=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) Comparing coefficient \(\mathrm{a}=-10, \mathrm{~b}=35, \mathrm{c}=-50, \mathrm{~d}=24\) \(\text { Now, the value of } \mathrm{a}+2 \mathrm{~b}+\mathrm{c} \text { is }\) \(\mathrm{a}+2 \mathrm{~b}+\mathrm{c}\) \(=-10+2 \times 35-50\) \(=-60+70=10 right.\)
118092
A polynomial function \(f(x)\) satisfies the condition for all \(x \in R, x \neq 0\). If \(f(3)=-26\) then \(f(4)\) is
1 -35
2 -63
3 65
4 None of these
Explanation:
B We know that, if a polynomial function \(f(\mathrm{x})\) satisfies the condition \(f(\mathrm{x}) f\left(\frac{1}{\mathrm{x}}\right)=f(\mathrm{x})+f\left(\frac{1}{\mathrm{x}}\right)\) Then it will be in the form of \(f(\mathrm{x})=1 \pm \mathrm{x}^{\mathrm{n}}\) \(\because\) Given \(f(3)=-26\) \(\therefore f(3)=1 \pm 3^{\mathrm{n}}\) \(-26=1-3^{\mathrm{n}}\) \(=-27=-3^{\mathrm{n}}\) \(3^{\mathrm{n}}=27\) \(3^{\mathrm{n}}=3^3\) \(\mathrm{n}=3\) Now, \(f(4)=\) ? \(f(4)=1-4^3\) \(=1-64\) \(f(4)=-63\)
BCECE-2018
Complex Numbers and Quadratic Equation
118093
The number of solutions of the equation \(x^3+x^2+4 x+2 \sin x=0\) in \([0,2 \pi]\), is
1 0
2 1
3 2
4 4
Explanation:
B \( \mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}+2 \sin \mathrm{x}=0\) \(-\left(\frac{\mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}}{2}\right)=\sin \mathrm{x}\) \(\text { Plotting the graph of both } \sin \mathrm{x} \text { and cubic equation }\) Plotting the graph of both \(\sin \mathrm{x}\) and cubic equation They meets at only one point that is \(x=0\) \(\therefore\) Only one solution is possible.
BCECE-2018
Complex Numbers and Quadratic Equation
118094
If the roots of the given equation \((\cos p-1) x^2+(\cos p) x+\sin p=0\) are real, then
118095
If \(1,2,3\) and 4 are the roots of the equations \(\mathbf{x}^4+\mathbf{a} \mathbf{x}^3+\mathbf{b x ^ { 2 }}+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(\mathbf{a}+\mathbf{2 b}+\mathbf{c}\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given equation is \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}=0\) and \(\because\) Roots are \(-1,2,3\) and 4 \(\therefore(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\) \(\mathrm{d}\) \(\left(\mathrm{x}^2-3 \mathrm{x}+2\right)\left(\mathrm{x}^2-7 \mathrm{x}+12\right)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-7 \mathrm{x}^3+12 \mathrm{x}^2-3 \mathrm{x}^3+21 \mathrm{x}^2-36 \mathrm{x}+2 \mathrm{x}^2-14 \mathrm{x}+24=\) \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-10 \mathrm{x}^3+35 \mathrm{x}^2-50 \mathrm{x}+24=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) Comparing coefficient \(\mathrm{a}=-10, \mathrm{~b}=35, \mathrm{c}=-50, \mathrm{~d}=24\) \(\text { Now, the value of } \mathrm{a}+2 \mathrm{~b}+\mathrm{c} \text { is }\) \(\mathrm{a}+2 \mathrm{~b}+\mathrm{c}\) \(=-10+2 \times 35-50\) \(=-60+70=10 right.\)
118092
A polynomial function \(f(x)\) satisfies the condition for all \(x \in R, x \neq 0\). If \(f(3)=-26\) then \(f(4)\) is
1 -35
2 -63
3 65
4 None of these
Explanation:
B We know that, if a polynomial function \(f(\mathrm{x})\) satisfies the condition \(f(\mathrm{x}) f\left(\frac{1}{\mathrm{x}}\right)=f(\mathrm{x})+f\left(\frac{1}{\mathrm{x}}\right)\) Then it will be in the form of \(f(\mathrm{x})=1 \pm \mathrm{x}^{\mathrm{n}}\) \(\because\) Given \(f(3)=-26\) \(\therefore f(3)=1 \pm 3^{\mathrm{n}}\) \(-26=1-3^{\mathrm{n}}\) \(=-27=-3^{\mathrm{n}}\) \(3^{\mathrm{n}}=27\) \(3^{\mathrm{n}}=3^3\) \(\mathrm{n}=3\) Now, \(f(4)=\) ? \(f(4)=1-4^3\) \(=1-64\) \(f(4)=-63\)
BCECE-2018
Complex Numbers and Quadratic Equation
118093
The number of solutions of the equation \(x^3+x^2+4 x+2 \sin x=0\) in \([0,2 \pi]\), is
1 0
2 1
3 2
4 4
Explanation:
B \( \mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}+2 \sin \mathrm{x}=0\) \(-\left(\frac{\mathrm{x}^3+\mathrm{x}^2+4 \mathrm{x}}{2}\right)=\sin \mathrm{x}\) \(\text { Plotting the graph of both } \sin \mathrm{x} \text { and cubic equation }\) Plotting the graph of both \(\sin \mathrm{x}\) and cubic equation They meets at only one point that is \(x=0\) \(\therefore\) Only one solution is possible.
BCECE-2018
Complex Numbers and Quadratic Equation
118094
If the roots of the given equation \((\cos p-1) x^2+(\cos p) x+\sin p=0\) are real, then
118095
If \(1,2,3\) and 4 are the roots of the equations \(\mathbf{x}^4+\mathbf{a} \mathbf{x}^3+\mathbf{b x ^ { 2 }}+\mathbf{c x}+\mathbf{d}=\mathbf{0}\), then \(\mathbf{a}+\mathbf{2 b}+\mathbf{c}\) is equal to
1 -25
2 0
3 10
4 24
Explanation:
C Given equation is \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}=0\) and \(\because\) Roots are \(-1,2,3\) and 4 \(\therefore(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\) \(\mathrm{d}\) \(\left(\mathrm{x}^2-3 \mathrm{x}+2\right)\left(\mathrm{x}^2-7 \mathrm{x}+12\right)=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-7 \mathrm{x}^3+12 \mathrm{x}^2-3 \mathrm{x}^3+21 \mathrm{x}^2-36 \mathrm{x}+2 \mathrm{x}^2-14 \mathrm{x}+24=\) \(\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) \(\mathrm{x}^4-10 \mathrm{x}^3+35 \mathrm{x}^2-50 \mathrm{x}+24=\mathrm{x}^4+\mathrm{ax}^3+\mathrm{bx}^2+\mathrm{cx}+\mathrm{d}\) Comparing coefficient \(\mathrm{a}=-10, \mathrm{~b}=35, \mathrm{c}=-50, \mathrm{~d}=24\) \(\text { Now, the value of } \mathrm{a}+2 \mathrm{~b}+\mathrm{c} \text { is }\) \(\mathrm{a}+2 \mathrm{~b}+\mathrm{c}\) \(=-10+2 \times 35-50\) \(=-60+70=10 right.\)
