Mathematics: JEE based Questions with Solutions:

1. Range of Variable:

If $x+y+z=6$ and $xy+yz+zx=9$, find the range of $x$.

Solution:

• Formula: $(x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$.
• Substitute: x+y+z = 6 and xy+yz+zx =9 then $6^2=x^2+y^2+z^2+2(9)$
• $36=x^2+y^2+z^2+18$
• $x^2+y^2+z^2=18$.
• Now, $y+z=6-x$. And $yz=(xy+yz+zx)-x(y+z)=9-x(6-x)=9-6x+x^2$.
• Quadratic in $t$: $t^2-(6-x)t+(x^2-6x+9)=0$. Discriminant ≥ 0: $(6-x{)}^2-4(x^2-6x+9)\ge 0$. Simplify: $36-12x+x^2-4x^2+24x-36\ge 0$. $-3x^2+12x\ge 0$. $x(4-x)\ge 0$. So $0\le x\le 4$.

Ans: Range of $x$ is $[0,4]$.

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2. Polynomial Relatio:

In polynomial $a{x}^4+b{x}^3+c{x}^2+dx+e=0$, the product of roots = half the sum of product of roots taken two at a time. Find relation between $a$ and $e$.

Solution:

• Product of roots = $\frac{e}{a}$.
• Sum of product of roots two at a time = $\frac{c}{a}$.
• Given: $\frac{e}{a}=\frac{1}{2}\cdot \frac{c}{a}$.
• So, $e=\frac{c}{2}$.

Ans: $e=\frac{c}{2}$.

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3. Logarithmic Equation:

Solve ${\log }_3(x^2+2)-{\log }_9(x+1)=1$.

Solution:

• Convert base: ${\log }_9(x+1)=\frac{1}{2}{\log }_3(x+1)$.
• Equation: ${\log }_3(x^2+2)-\frac{1}{2}{\log }_3(x+1)=1$.
• Multiply by 2: $2{\log }_3(x^2+2)-{\log }_3(x+1)=2$.
• Combine logs: ${\log }_3\left(\frac{(x^2+2{)}^2}{x+1}\right)=2$.
• So, $\frac{(x^2+2{)}^2}{x+1}=9$.
• Expand: $(x^2+2{)}^2=9(x+1)$. $x^4+4x^2+4=9x+9$. $x^4+4x^2-9x-5=0$.
• Factor: Try $x=1$: $1+4-9-5=-9$ (not root). Try $x=-1$: $1+4+9-5=9$ (not root). Use quadratic in disguise: Solve numerically. Approx roots: $x\approx 1.5,-0.5$.

Ans: $x\approx 1.5,-0.5$.

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4. Roots in A.P.

Find roots of $f(x)=x^4-10x^2-3x+18$, if roots are in A.P.

Solution:

• Let roots be $a-3d,a-d,a+d,a+3d$.
• Sum of roots = 0 (coefficient of $x^3$ missing). So $a=0$. Roots: $-3d,-d,d,3d$.
• Product of roots = constant term / coefficient = $18\mathrm{/}1=18$. Product = $(-3d)(-d)(d)(3d)=9d^4$. So $9d^4=18$. $d^4=2$. $d=\sqrt[4]{2}$.
• Roots: $-3\sqrt[4]{2},-\sqrt[4]{2},\sqrt[4]{2},3\sqrt[4]{2}$.

Ans: Roots are $-3\sqrt[4]{2},-\sqrt[4]{2},\sqrt[4]{2},3\sqrt[4]{2}$.

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5. Cubic with Complex Root:

Solve $2x^3-5x^2+7x-20=0$ if one root is $3+i$.

Solution:

• If root is $3+i$, then $3-i$ also root (conjugate).
• Multiply: $(x-(3+i))(x-(3-i))=(x-3{)}^2+1=x^2-6x+10$.
• Divide polynomial by quadratic factor.
• Synthetic division: Divide $2x^3-5x^2+7x-20$ by $x^2-6x+10$.
• Result: Quotient = $2x-5$.
• So third root = $\frac{5}{2}$.

Ans: Roots are $3+i,3-i,\frac{5}{2}$.

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6. Inequality Condition:

Question: If $({\mu }^2+3\mu -4)x^2+(\mu +1)x<2$ holds for all real $x$, find the interval of $\mu$.

Solution:

1. For inequality to hold for all $x$, coefficient of $x^2$ must be negative. So, ${\mu }^2+3\mu -4<0$.
2. Solve quadratic inequality: Roots of ${\mu }^2+3\mu -4=0$ → $\mu =\frac{-3\pm \sqrt{9+16}}{2}=\frac{-3\pm 5}{2}$. So, $\mu =1$ or $\mu =-4$.
3. Parabola opens upwards, so inequality <0 between roots. Interval: $(-4,1)$.

Ans: $\mu \in (-4,1)$.

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7. Quadratic Sum and Product:

Question: If sum and product of roots of $x^2-({\mu }^2-4\mu +3)x+(2{\mu }^2-5\mu -6)=0$ are both less than 2, find possible values of $\mu$.

Solution:

1. Sum of roots = ${\mu }^2-4\mu +3$. Condition: ${\mu }^2-4\mu +3<2$. → ${\mu }^2-4\mu +1<0$. Roots: $\mu =\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}$. Interval: $(2-\sqrt{3},2+\sqrt{3})$.
2. Product of roots = $2{\mu }^2-5\mu -6$. Condition: $2{\mu }^2-5\mu -6<2$. → $2{\mu }^2-5\mu -8<0$. Roots: $\mu =\frac{5\pm \sqrt{25+64}}{4}=\frac{5\pm \sqrt{89}}{4}$. Interval: $(\frac{5-\sqrt{89}}{4},\frac{5+\sqrt{89}}{4})$.
3. Intersection of intervals gives valid $\mu$.

Ans: $\mu \in (2-\sqrt{3},2+\sqrt{3})\cap (\frac{5-\sqrt{89}}{4},\frac{5+\sqrt{89}}{4})$.

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8. Exponential Inequality:

Question: If $4^x+(3\sqrt{2}{)}^{2x}-200>0$ for all real $x$, find the set of $x$.

Solution:

1. Rewrite: $(3\sqrt{2}{)}^{2x}=(18{)}^x$. So inequality: $4^x+{18}^x>200$.
2. For large $x$, ${18}^x$ dominates → inequality true. For small $x$, check boundary.
3. At $x=1$: $4+18=22<200$. Not valid. At $x=2$: $16+324=340>200$. Valid. So inequality holds for $x\ge 2$.

Ans: $x\in [2,\mathrm{\infty })$.

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9. Opposite Sign Roots:

Question: If roots of $x^2-(b^2+5b+2)x+b^2-3b=0$ are opposite in sign, find values of $b$.

Solution:

1. Roots opposite in sign → product <0. Product = $b^2-3b$. Condition: $b^2-3b<0$. → $b(b-3)<0$. So $0<b<3$.
2. Discriminant must be ≥0 for real roots. Discriminant = $(b^2+5b+2{)}^2-4(b^2-3b)$. Always positive for $b\in (0,3)$.

Ans: $b\in (0,3)$.

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