Q1
Using differentials, the approximate value of $\sqrt{25.2}$ is closest to:
📝 SolutionLet $f(x) = \sqrt{x}$, $x = 25$, $dx = 0.2$. $f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{10}$. Approximation: $\sqrt{25.2} \approx 5 + \frac{1}{10}(0.2) = 5.02$.
Q2
Assertion (A): The function $f(x) = x^3$ is increasing for all real $x$.
Reason (R): $f'(x) = 3x^2 \ge 0$ for all real $x$.
Choose the correct option:
Reason (R): $f'(x) = 3x^2 \ge 0$ for all real $x$.
Choose the correct option:
📝 Solution$f'(x) = 3x^2 \ge 0$ for all $x$, with equality only at $x=0$, so $f(x) = x^3$ is (non-strictly) increasing everywhere. R gives the exact reason, correctly explaining A.
Q3
The maximum value of $f(x) = -x^2 + 4x + 1$ occurs at:
📝 Solution$f'(x) = -2x + 4 = 0 \implies x = 2$. Since $f''(x) = -2 < 0$, this is a maximum, occurring at $x = 2$.
Q4
Assertion (A): $f(x) = \sin x$ has a local maximum at $x = \frac{\pi}{2}$.
Reason (R): $f'\left(\frac{\pi}{2}\right) = 0$ and $f''\left(\frac{\pi}{2}\right) < 0$.
Choose the correct option:
Reason (R): $f'\left(\frac{\pi}{2}\right) = 0$ and $f''\left(\frac{\pi}{2}\right) < 0$.
Choose the correct option:
📝 Solution$f'(x) = \cos x = 0$ at $x = \frac{\pi}{2}$, and $f''(x) = -\sin x = -1 < 0$ there, confirming a local maximum. R states exactly this second derivative test, explaining A.
Q5
The point on the curve $y = x^2$ where the tangent is parallel to the x-axis is:
📝 SolutionTangent parallel to the x-axis means slope $= 0$: $\frac{dy}{dx} = 2x = 0 \implies x = 0$. At $x=0$, $y=0$, giving the point $(0,0)$.
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Class 12 · Maths
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