Which of the following statements is/are False?
$S_1:\exists \,n\in N$, such that $n^2+n+2$ is divisible by 4.
$S_2:\exists \,x\in N$, such that $x-17 < 20$.
$S_3:\forall \,n\in N,\,x^2+3x-10 = 0$.
$S_4:\forall \,n\in N,\,n^2\geq 1$.

Question

If $ x = a(\cos \theta + \theta \sin \theta), \, y = a(\sin \theta - \theta \cos \theta) $, then $ \frac{d^2 y}{dx^2} $ equals

Answer 1

Step 1: Family of lines:
Any line through the intersection is $(x + 2y + 6) + k(2x - y - 2) = 0$.

Step 2: Impose the intercept:
Put $x = 0$: $2y + 6 + k(-y - 2) = 0$, so $y(2 - k) = 2k - 6$ and $y = \frac{2k - 6}{2 - k}$.
Set $y = 5$: $2k - 6 = 10 - 5k$, so $k = \frac{16}{7}$.
Substitute: $7x + 14y + 42 + 16(2x - y - 2) = 0$, which gives $39x - 2y + 10 = 0$. This is option (B).

Final Answer:
$39x - 2y + 10 = 0$. \[ \boxed{39x - 2y + 10 = 0} \]

Which of the following statements is/are False?
\(S_1:\exists \,n\in N\), such that \(n^2+n+2\) is divisible by 4.
\(S_2:\exists \,x\in N\), such that \(x-17 < 20\).
\(S_3:\forall \,n\in N,\,x^2+3x-10 = 0\).
\(S_4:\forall \,n\in N,\,n^2\geq 1\).

Show Hint

The Correct Option is C

Solution and Explanation

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Questions Asked in MHT CET exam

Which of the following statements is/are False?\(S_1:\exists \,n\in N\), such that \(n^2+n+2\) is divisible by 4.\(S_2:\exists \,x\in N\), such that \(x-17 < 20\).\(S_3:\forall \,n\in N,\,x^2+3x-10 = 0\).\(S_4:\forall \,n\in N,\,n^2\geq 1\).