NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

Part (i): $\sqrt{25.3}$

Given:
We need to approximate $\sqrt{25.3}$.
Let the function be $y = f(x) = \sqrt{x}$.
We choose $x=25$ (a perfect square close to 25.3) and $\Delta x = 0.3$.

Formula Used:
$f(x + \Delta x) \approx f(x) + \Delta y$, where $\Delta y \approx f'(x)\Delta x$.

Step 1: Differentiate $y = \sqrt{x}$ with respect to $x$.
$f'(x) = \frac{dy}{dx} = \frac{1}{2\sqrt{x}}$

Step 2: Find the approximate change $\Delta y$.
$\Delta y \approx f'(x)\Delta x = \left(\frac{1}{2\sqrt{x}}\right) \Delta x$
Substituting $x=25$ and $\Delta x=0.3$:
$\Delta y \approx \left(\frac{1}{2\sqrt{25}}\right) (0.3) = \frac{1}{10}(0.3) = 0.03$

Step 3: Calculate the approximate value.
$\sqrt{25.3} = f(x + \Delta x) \approx f(x) + \Delta y$
$\sqrt{25.3} \approx \sqrt{25} + 0.03 = 5 + 0.03 = 5.03$

Final Answer (i): $\sqrt{25.3} \approx 5.030$

Part (ii): $\sqrt{49.5}$

Given:
Let $y = f(x) = \sqrt{x}$.
We choose $x=49$ and $\Delta x = 0.5$.

Step 1: The derivative is $f'(x) = \frac{1}{2\sqrt{x}}$.

Step 2: Find the approximate change $\Delta y$.
$\Delta y \approx \left(\frac{1}{2\sqrt{x}}\right) \Delta x$
Substituting $x=49$ and $\Delta x=0.5$:
$\Delta y \approx \left(\frac{1}{2\sqrt{49}}\right) (0.5) = \frac{1}{14}(0.5) \approx 0.0357$

Step 3: Calculate the approximate value.
$\sqrt{49.5} \approx f(49) + \Delta y = \sqrt{49} + 0.0357 = 7 + 0.0357 = 7.0357$

Final Answer (ii): $\sqrt{49.5} \approx 7.036$ (rounded to 3 decimal places)

Part (iii): $\sqrt{0.6}$

Given:
Let $y = f(x) = \sqrt{x}$.
We choose $x=0.64$ (closest perfect square) and $\Delta x = 0.6 - 0.64 = -0.04$.

Step 1: The derivative is $f'(x) = \frac{1}{2\sqrt{x}}$.

Step 2: Find the approximate change $\Delta y$.
$\Delta y \approx \left(\frac{1}{2\sqrt{0.64}}\right) (-0.04) = \frac{1}{2(0.8)}(-0.04) = \frac{-0.04}{1.6} = -0.025$

Step 3: Calculate the approximate value.
$\sqrt{0.6} \approx \sqrt{0.64} + \Delta y = 0.8 - 0.025 = 0.775$

Final Answer (iii): $\sqrt{0.6} \approx 0.775$

Given:
Function $f(x) = 4x^2 + 5x + 2$.
Let $x = 2$ and $\Delta x = 0.01$.

Formula Used:
$f(x + \Delta x) \approx f(x) + f'(x)\Delta x$

Step 1: Find the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(4x^2 + 5x + 2) = 8x + 5$

Step 2: Calculate $f(2)$ and $f'(2)$.
$f(2) = 4(2)^2 + 5(2) + 2 = 16 + 10 + 2 = 28$
$f'(2) = 8(2) + 5 = 16 + 5 = 21$

Step 3: Apply the approximation formula.
$f(2.01) \approx f(2) + f'(2)\Delta x$
$f(2.01) \approx 28 + (21)(0.01)$
$f(2.01) \approx 28 + 0.21 = 28.21$

Final Answer: $f(2.01) \approx 28.21$

Given:
Function $f(x) = x^3 - 7x^2 + 15$.
Let $x = 5$ and $\Delta x = 0.001$.

Step 1: Find the derivative $f'(x)$.
$f'(x) = 3x^2 - 14x$

Step 2: Calculate $f(5)$ and $f'(5)$.
$f(5) = (5)^3 - 7(5)^2 + 15 = 125 - 175 + 15 = -35$
$f'(5) = 3(5)^2 - 14(5) = 75 - 70 = 5$

Step 3: Apply the approximation formula.
$f(5.001) \approx f(5) + f'(5)\Delta x$
$f(5.001) \approx -35 + (5)(0.001)$
$f(5.001) \approx -35 + 0.005 = -34.995$

Final Answer: $f(5.001) \approx -34.995$

Given:
Side of the cube = $x$ metres.
Increase in side = 1%.
Change in side $\Delta x = 1\% \text{ of } x = 0.01x$.

Formula Used:
Approximate change in Volume $\Delta V \approx \frac{dV}{dx} \Delta x$.
Volume of a cube $V = x^3$.

Step 1: Differentiate $V$ with respect to $x$.
$\frac{dV}{dx} = 3x^2$

Step 2: Calculate the approximate change in volume $\Delta V$.
$\Delta V \approx \left(\frac{dV}{dx}\right) \Delta x$
$\Delta V \approx (3x^2)(0.01x)$

Final Answer: The approximate change in volume is $\Delta V \approx 0.03x^3 \text{ m}^3$.

Given:
Side of the cube = $x$ metres.
Decrease in side = 1%.
Change in side $\Delta x = -1\% \text{ of } x = -0.01x$.

Formula Used:
Approximate change in Surface Area $\Delta S \approx \frac{dS}{dx} \Delta x$.
Surface area of a cube $S = 6x^2$.

Step 1: Differentiate $S$ with respect to $x$.
$\frac{dS}{dx} = 12x$

Step 2: Calculate the approximate change in surface area $\Delta S$.
$\Delta S \approx \left(\frac{dS}{dx}\right) \Delta x$
$\Delta S \approx (12x)(-0.01x)$

Final Answer: The approximate change in surface area is $\Delta S \approx -0.12x^2 \text{ m}^2$. (The negative sign indicates a decrease).

Given:
Radius of sphere $r = 7 \text{ m}$.
Error in radius $\Delta r = 0.02 \text{ m}$.

Formula Used:
Approximate error in Volume $\Delta V \approx \frac{dV}{dr} \Delta r$.
Volume of a sphere $V = \frac{4}{3}\pi r^3$.

Step 1: Differentiate $V$ with respect to $r$.
$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$

Step 2: Calculate the approximate error in volume $\Delta V$.
$\Delta V \approx (4\pi r^2) \Delta r$
$\Delta V \approx 4\pi (7)^2 (0.02)$
$\Delta V \approx 4\pi (49) (0.02) = 196\pi(0.02) = 3.92\pi$

Final Answer: The approximate error in calculating the volume is $3.92\pi \text{ m}^3$.

Given:
Radius of sphere $r = 9 \text{ m}$.
Error in radius $\Delta r = 0.03 \text{ m}$.

Formula Used:
Approximate error in Surface Area $\Delta S \approx \frac{dS}{dr} \Delta r$.
Surface area of a sphere $S = 4\pi r^2$.

Step 1: Differentiate $S$ with respect to $r$.
$\frac{dS}{dr} = \frac{d}{dr}(4\pi r^2) = 8\pi r$

Step 2: Calculate the approximate error in surface area $\Delta S$.
$\Delta S \approx (8\pi r) \Delta r$
$\Delta S \approx 8\pi (9) (0.03)$
$\Delta S \approx 72\pi(0.03) = 2.16\pi$

Final Answer: The approximate error in calculating the surface area is $2.16\pi \text{ m}^2$.

Given: $f(x) = 3x^2 + 15x + 5$.
Let $x=3$ and $\Delta x = 0.02$.

Step 1: Find the derivative $f'(x)$.
$f'(x) = 6x + 15$

Step 2: Calculate $f(3)$ and $f'(3)$.
$f(3) = 3(3)^2 + 15(3) + 5 = 27 + 45 + 5 = 77$
$f'(3) = 6(3) + 15 = 18 + 15 = 33$

Step 3: Apply the approximation formula.
$f(3.02) \approx f(3) + f'(3)\Delta x$
$f(3.02) \approx 77 + (33)(0.02) = 77 + 0.66 = 77.66$

Final Answer: The correct option is (D) 77.66.

Given: Side of cube = $x$, Volume $V=x^3$.
Increase in side = 3%, so $\Delta x = 0.03x$.

Step 1: Find the derivative of the volume.
$\frac{dV}{dx} = 3x^2$

Step 2: Calculate the approximate change in volume, $\Delta V$.
$\Delta V \approx \frac{dV}{dx} \Delta x$
$\Delta V \approx (3x^2)(0.03x) = 0.09x^3$

Final Answer: The correct option is (C) $0.09x^3 \text{ m}^3$.