Q.4. Let the set of all values of such that the equation , , has at least one solution, be the inte

Q: What is the correct answer for JEE Main Mathematics Q4?

The correct answer is A. Sol. Put $z = x + iy$, so $\bar{z} = x - iy$. The equation is $(x+iy)(x-iy+2+i)+k(2+3i)=0$. $(x+iy)((x+2)+i(1-y))+k(2+3i)=0$. Real part: $x(x+2) - y(1-y) + 2k = 0 \Rightarrow x^2+2x-y+y^2+2k=0$. Imaginary part: $y(x+2) + x(1-y) + 3k = 0 \Rightarrow xy+2y+x-xy+3k=0 \Rightarrow x+2y+3k=0$. From the imaginary part, $x = -2y-3k$. Substitute $x$ into the real part: $(-2y-3k)^2+2(-2y-3k)-y+y^2+2k=0$. $4

Q: Q.4. Let the set of all values of such that the equation , , has at least one solution, be the interval [, ]. Then is equal to:

The correct answer is A. Sol. Put $z = x + iy$, so $\bar{z} = x - iy$. The equation is $(x+iy)(x-iy+2+i)+k(2+3i)=0$. $(x+iy)((x+2)+i(1-y))+k(2+3i)=0$. Real part: $x(x+2) - y(1-y) + 2k = 0 \Rightarrow x^2+2x-y+y^2+2k=0$. Imaginary part: $y(x+2) + x(1-y) + 3k = 0 \Rightarrow xy+2y+x-xy+3k=0 \Rightarrow x+2y+3k=0$. From the imaginary part, $x = -2y-3k$. Substitute $x$ into the real part: $(-2y-3k)^2+2(-2y-3k)-y+y^2+2k=0$. $4

Q: What are the options for this JEE Main Mathematics question?

A. -10 · B. -8 · C. 10/13 · D. $8\sqrt{13}$

Question

Q.4. Let the set of all values of  such that the equation , , has at least one solution, be the interval [, ]. Then  is equal to:

FirstInTest · Accepted Answer

Sol. Put $z = x + iy$, so $\bar{z} = x - iy$. The equation is $(x+iy)(x-iy+2+i)+k(2+3i)=0$. $(x+iy)((x+2)+i(1-y))+k(2+3i)=0$. Real part: $x(x+2) - y(1-y) + 2k = 0 \Rightarrow x^2+2x-y+y^2+2k=0$. Imaginary part: $y(x+2) + x(1-y) + 3k = 0 \Rightarrow xy+2y+x-xy+3k=0 \Rightarrow x+2y+3k=0$. From the imaginary part, $x = -2y-3k$. Substitute $x$ into the real part: $(-2y-3k)^2+2(-2y-3k)-y+y^2+2k=0$. $4y^2+12ky+9k^2-4y-6k-y+y^2+2k=0$. $5y^2 + (12k-5)y + (9k^2-4k) = 0$. For this quadratic equation in $y$ to have at least one real solution, the discriminant $D$ must be greater than or equal to zero. $D = (12k-5)^2 - 4(5)(9k^2-4k) \ge 0$. $144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$. $-36k^2 - 40k + 25 \ge 0$. $36k^2 + 40k - 25 \le 0$. To find the roots of $36k^2 + 40k - 25 = 0$: $k = \frac{-40 \pm \sqrt{40^2 - 4(36)(-25)}}{2(36)} = \frac{-40 \pm \sqrt{1600 + 3600}}{72} = \frac{-40 \pm \sqrt{5200}}{72} = \frac{-40 \pm 20\sqrt{13}}{72} = \frac{-10 \pm 5\sqrt{13}}{18}$. So, $k \in \left[\frac{-10 - 5\sqrt{13}}{18}, \frac{-10 + 5\sqrt{13}}{18}\right]$. This means $\alpha = \frac{-10 - 5\sqrt{13}}{18}$ and $\beta = \frac{-10 + 5\sqrt{13}}{18}$. Then $\alpha + \beta = \frac{-10 - 5\sqrt{13} - 10 + 5\sqrt{13}}{18} = \frac{-20}{18} = -\frac{10}{9}$. Finally, $9(\alpha + \beta) = 9\left(-\frac{10}{9}\right) = -10$.

Answer

-10

Answer

-8

Answer

10/13

Answer

$8\sqrt{13}$

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