- A$3abc + {a^3} + {b^3} + {c^3}$
- ✓$3abc - {a^3} - {b^3} - {c^3}$
- C$abc - {a^3} + {b^3} + {c^3}$
- D$abc + {a^3} - {b^3} - {c^3}$
$({R_1} \to {R_1} + {R_2} + {R_3})$
=$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}}{1}&{1}&{1}\\b&c&a\\c&a&b\end{array}\,} \right|$ =$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}} 1&1&1 \\ b&c&a \\ c&a&b \end{array}\,} \right|$ =$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}} 1&0&0 \\ b&{b - c}&{c - a} \\ c&{c - a}&{a - b} \end{array}\,} \right|$
= $3abc - {a^3} - {b^3} - {c^3}$, (After simplification).
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is:
$\frac{7}{64}$
$\frac{7}{128}$
$\frac{45}{1024}$
$\frac{7}{41}$
| List $I$ | List $II$ |
| $P.$ Let $y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1], x \neq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)}\left\{\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right\}$ equals | $1.$ $1$ |
| $Q.$ Let $A_1, A_2, \ldots \ldots, A_n(n>2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $\vec{a}_k$ be the position vector of the point $A_k, k=1,2, \ldots, n$. If $\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_k} \times \overrightarrow{a_{k+1}}\right)\right|=\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_k} \cdot \overrightarrow{a_{k+1}}\right)\right|$, then the minimum value of $n$ is | $2.$ $2$ |
| $R.$ If the normal from the point $P(h, 1)$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ is perpendicular to the line $x+y=8$, then the value of $h$ is | $3.$ $8$ |
| $S.$ Number of positive solutions satisfying the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right)$ is | $4.$ $9$ |
Codes: $ \quad P \quad Q \quad R \quad S $
$\frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right), y(0)=0 .$
Then, which of the following statements is/are $TRUE$?