The number of ways in which 5 boys and 3 girls can be seated in a row so that no two girls sit together is divisible by 144. If the number of ways is $N$, find the value of $N/144$.
Let $f(x) = \int_0^x \fract1+t^2 dt$. If $f(2) = 0.5 \ln k$, then the value of $k$ is: (A) 5 (B) 4 (C) 3 (D) 2 tata mcgraw hill mathematics for iit jee
If the system of equations $x + ay = 0$, $az + y = 0$, and $ax + z = 0$ has infinitely many solutions, then $a$ can be: (A) $1$ (B) $-1$ (C) $0$ (D) $2$ The number of ways in which 5 boys
Hint: Vector triple product is always coplanar. Box product $[\veca \vecb \vecc] = \veca \cdot (\vecb \times \vecc)$. Angle $\vecb, \vecc = \pi/3$. Let $\vecb=\hatj, \vecc=\frac12\hatj+\frac\sqrt32\hatk$. $\vecb\times\vecc = \frac\sqrt32\hati$. $\veca \cdot (\vecb\times\vecc) = \frac\sqrt32 (\veca \cdot \hati)$. $\veca \cdot \vecb = a_y = 0$. $\veca \cdot \vecc = \frac12a_y + \frac\sqrt32a_k = \frac12$. If $a_y=0 \implies \frac\sqrt32a_k = 1/2 \implies a_k = 1/\sqrt3$. Since $\veca$ is unit vector, $a_x^2 + a_y^2 + a_z^2 = 1 \implies a_x^2 + 1/3 = 1 \implies a_x = \pm \sqrt2/3$. Box product = $\frac\sqrt32 a_x = \frac\sqrt32 (\pm \sqrt\frac23) = \pm \frac1\sqrt2$. Wait, check Option B. It says $\sqrt3/2$. Let me re-read $\veca \cdot \vecc = 1/2$. Check option C. Formula identity: $[\veca \vecb \vecc]^2 = \dots$ Lagrange Identity. Check D: $|\veca+\vecb+\vecc|^2 = 1+1+1+0+1+1 = 5 \implies \sqrt5$. Incorrect. Answer: (A, C) . (Verify B calculation again). If $f(2) = 0