# NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4

Go back to  'Application of Derivatives'

## Chapter 6 Ex.6.4 Question 1

Using differentials, find the approximate value of each of the following upto 3 places of decimals.

 i. $$\sqrt {25.3}$$ ii. $$\sqrt{49.5}$$ iii. $$\sqrt {0.6}$$ iv. $${\left( {0.009} \right)^{\frac{1}{3}}}$$ v. $${\left( {0.999} \right)^{\frac{1}{{10}}}}$$ vi.$${\left( {15} \right)^{\frac{1}{4}}}$$ vii. $${\left( {26} \right)^{\frac{1}{3}}}$$ viii. $${\left( {255} \right)^{\frac{1}{4}}}$$ ix. $${\left( {82} \right)^{\frac{1}{4}}}$$ x. $${\left( {401} \right)^{\frac{1}{2}}}$$ xi. $${\left( {0.0037} \right)^{\frac{1}{2}}}$$ xii. $${\left( {26.57} \right)^{\frac{1}{3}}}$$ xiii. $${\left( {81.5} \right)^{\frac{1}{4}}}$$ xiv. $${\left( {3.968} \right)^{\frac{3}{2}}}$$ xv. $${\left( {32.15} \right)^{\frac{1}{5}}}$$

### Solution

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

Consider $$y = \sqrt x$$.

Let $$x = 25\,\,{\rm{and}}\,\,\Delta x = 0.3$$

Then,

\begin{align}\Delta y &= \sqrt {x + \Delta x} - \sqrt x = \sqrt {25.3} - \sqrt {25} \\&= \sqrt {25.3} - 5\\ \Rightarrow \;\Delta y + 5 &= \sqrt {25.3} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{2\sqrt x }}\left( {0.3} \right) \qquad \left[ {{\text{as }}y = \sqrt x } \right]\\& = \frac{1}{{2\sqrt {25} }}\left( {0.3} \right) = 0.03\end{align}

Hence, the approximate value of $$\sqrt {25.3}$$  is  $$0.03 + 5 = 5.03$$ .

(ii) $$\sqrt{49.5}$$

Consider $$y = \sqrt x$$.

Let $$x = 49\,\,{\text{and}}\,\,\Delta x = 0.5$$

Then,

\begin{align}\Delta y& = \sqrt {x + \Delta x} - \sqrt x = \sqrt {49.5} - \sqrt {49} \\&= \sqrt {49.5} - 7\\ \Rightarrow \; 7 + \Delta y &= \sqrt {49.5} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{2\sqrt x }}\left( {0.5} \right) \qquad \left[ {{\text{as }}y = \sqrt x } \right]\\& = \frac{1}{{2\sqrt {49} }}\left( {0.5} \right) = \frac{1}{{14}}\left( {0.5} \right) = 0.035\end{align}

Hence, the approximate value of $$\sqrt {49.5}$$  is  $$7 + 0.035 = 7.035$$ .

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

Consider $$y = \sqrt x$$.

Let $$x = 1\,\,{\text{and}}\,\,\Delta x = - 0.4$$

Then,

\begin{align} \Delta y &= \sqrt {x + \Delta x} - \sqrt x \\&= \sqrt {0.6} - 1\\ \Rightarrow \; 1 + \Delta y &= \sqrt {0.6} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{2\sqrt x }}\left( { - 0.4} \right) \qquad \left[ {{\text{as }}y = \sqrt x } \right] \\ &= \frac{1}{2}\left( { - 0.4} \right) = - 0.2\end{align}

Hence, the approximate value of $$\sqrt {0.6}$$  is  $$1 + \left( { - 0.2} \right) = 1 - 0.2 = 0.8$$.

(iv) $${\left( {0.009} \right)^{\frac{1}{3}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{3}}}$$.

Let $$x = 0.008\,\,{\text{and}}\,\,\Delta x = 0.001$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{3}}} - {\left( x \right)^{\frac{1}{3}}} = {\left( {0.009} \right)^{\frac{1}{3}}} - {\left( {0.008} \right)^{\frac{1}{3}}} \\&= {\left( {0.009} \right)^{\frac{1}{3}}} - 0.2\\ \Rightarrow \; 0.2 + \Delta y &= {\left( {0.009} \right)^{\frac{1}{3}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align} dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{3{{\left( x \right)}^{\frac{2}{3}}}}}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {x^{\frac{1}{3}}}} \right]\\ &= \frac{1}{{3 \times 0.04}}\left( {0.001} \right) = \frac{{0.001}}{{0.12}} = 0.008\end{align}

Hence, the approximate value of $${\left( {0.009} \right)^{\frac{1}{3}}}$$  is  $$0.2 + 0.008 = 0.208$$ .

(v) $${\left( {0.999} \right)^{\frac{1}{{10}}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{{10}}}}$$.

Let $$x = 1\,\,{\text{and}}\,\,\Delta x = - 0.001$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{{10}}}} - {\left( x \right)^{\frac{1}{{10}}}} \\&= {\left( {0.999} \right)^{\frac{1}{{10}}}} - 1\\ \Rightarrow \; 1 + \Delta y &= {\left( {0.999} \right)^{\frac{1}{{10}}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{10{{\left( x \right)}^{\frac{9}{{10}}}}}}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{{10}}}}} \right]\\ &= \frac{1}{{10}}\left( { - 0.001} \right) = - 0.0001\end{align}

Hence, the approximate value of $${\left( {0.999} \right)^{\frac{1}{{10}}}}$$  is  $$1 + \left( { - 0.0001} \right) = 0.9999$$ .

(vi) $${\left( {15} \right)^{\frac{1}{4}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{4}}}$$.

Let $$x = 16\,\,{\text{and}}\,\,\Delta x = - 1$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{4}}} - {\left( x \right)^{\frac{1}{4}}} = {\left( {15} \right)^{\frac{1}{4}}} - {\left( {16} \right)^{\frac{1}{4}}} \\&= {\left( {15} \right)^{\frac{1}{4}}} - 2\\ \Rightarrow \; 2 + \Delta y &= {\left( {15} \right)^{\frac{1}{4}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{4{{\left( x \right)}^{\frac{3}{4}}}}}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{4}}}} \right]\\ &= \frac{1}{{4{{\left( {16} \right)}^{\frac{3}{4}}}}}\left( { - 1} \right) = \frac{{ - 1}}{{4 \times 8}} = \frac{{ - 1}}{{32}} = - 0.03125\end{align}

Hence, the approximate value of $${\left( {15} \right)^{\frac{1}{4}}}$$  is  $$2 + \left( { - 0.03125} \right) = 1.96875$$ .

(vii) $${\left( {26} \right)^{\frac{1}{3}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{3}}}$$.

Let $$x = 27\,\,{\text{and}}\,\,\Delta x = - 1$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{3}}} - {\left( x \right)^{\frac{1}{3}}} = {\left( {26} \right)^{\frac{1}{3}}} - {\left( {27} \right)^{\frac{1}{3}}} \\&= {\left( {26} \right)^{\frac{1}{3}}} - 3\\ \Rightarrow \; 3 + \Delta y &= {\left( {26} \right)^{\frac{1}{3}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{3{{\left( x \right)}^{\frac{2}{3}}}}}\left( {\Delta x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{3}}}} \right]\\& = \frac{1}{{3{{\left( {27} \right)}^{\frac{2}{3}}}}}\left( { - 1} \right) = \frac{{ - 1}}{{27}} = - 0.0\overline {370} \end{align}

Hence, the approximate value of $${\left( {26} \right)^{\frac{1}{3}}}$$  is  $$3 + \left( { - 0.0370} \right) = 2.9629$$ .

(viii) $${\left( {255} \right)^{\frac{1}{4}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{4}}}$$.

Let $$x = 256\,\,{\text{and}}\,\,\Delta x = - 1$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{4}}} - {\left( x \right)^{\frac{1}{4}}} = {\left( {255} \right)^{\frac{1}{4}}} - {\left( {256} \right)^{\frac{1}{4}}} \\&= {\left( {255} \right)^{\frac{1}{4}}} - 4\\ \Rightarrow \; 4 + \Delta y &= {\left( {255} \right)^{\frac{1}{4}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align} dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{4{{\left( x \right)}^{\frac{3}{4}}}}}\left( {\Delta x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{4}}}} \right] \hfill \\& = \frac{1}{{4{{\left( {256} \right)}^{\frac{3}{4}}}}}\left( { - 1} \right) = \frac{{ - 1}}{{4 \times {4^3}}} = \frac{{ - 1}}{{32}} = - 0.0039 \hfill \\ \end{align}

Hence, the approximate value of $${\left( {255} \right)^{\frac{1}{4}}}$$  is  $$4 + \left( { - 0.0039} \right) = 3.9961$$ .

(ix) $${\left( {82} \right)^{\frac{1}{4}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{4}}}$$.

Let $$x = 81\,\,{\text{and}}\,\,\Delta x = 1$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{4}}} - {\left( x \right)^{\frac{1}{4}}} = {\left( {82} \right)^{\frac{1}{4}}} - {\left( {81} \right)^{\frac{1}{4}}} \\&= {\left( {82} \right)^{\frac{1}{4}}} - 3\\ \Rightarrow \; \Delta y + 3 &= {\left( {82} \right)^{\frac{1}{4}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{4{{\left( x \right)}^{\frac{3}{4}}}}}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{4}}}} \right]\\ &= \frac{1}{{4{{\left( {81} \right)}^{\frac{3}{4}}}}}\left( 1 \right) = \frac{1}{{4 \times {3^3}}} = \frac{1}{{108}} = 0.009\end{align}

Hence, the approximate value of $${\left( {82} \right)^{\frac{1}{4}}}$$  is  $$3 + 0.009 = 3.009$$ .

(x) $${\left( {401} \right)^{\frac{1}{2}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{2}}}$$.

Let $$x = 400\,\,{\text{and}}\,\,\Delta x = 1$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{2}}} - {\left( x \right)^{\frac{1}{2}}} = {\left( {401} \right)^{\frac{1}{2}}} - {\left( {400} \right)^{\frac{1}{2}}} \\&= {\left( {401} \right)^{\frac{1}{2}}} - 20\\ \Rightarrow \; 20 + \Delta y &= {\left( {401} \right)^{\frac{1}{2}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{2\sqrt x }}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{2}}}} \right]\\& = \frac{1}{{2\left( {20} \right)}}\left( 1 \right) = \frac{1}{{40}} = 0.025\end{align}

Hence, the approximate value of $${\left( {401} \right)^{\frac{1}{2}}}$$  is  $$20 + 0.025 = 20.025$$ .

(xi) $${\left( {0.0037} \right)^{\frac{1}{2}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{2}}}$$.

Let $$x = 0.0036\,\,{\text{and}}\,\,\Delta x = 0.0001$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{2}}} - {\left( x \right)^{\frac{1}{2}}} = {\left( {0.0037} \right)^{\frac{1}{2}}} - {\left( {0.0036} \right)^{\frac{1}{2}}} \\&= {\left( {0.0037} \right)^{\frac{1}{2}}} - 0.06\\ \Rightarrow \; 0.06 + \Delta y &= {\left( {0.0037} \right)^{\frac{1}{2}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{2\sqrt x }}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{2}}}} \right]\\ &= \frac{1}{{2\left( {0.06} \right)}}\left( {0.0001} \right) = \frac{{0.0001}}{{0.12}} = 0.0008325\end{align}

Hence, the approximate value of $${\left( {0.0037} \right)^{\frac{1}{2}}}$$  is  $$0.06 + 0.00083 = 0.06083$$.

(xii) $${\left( {26.57} \right)^{\frac{1}{3}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{3}}}$$.

Let $$x = 27\,\,{\text{and}}\,\,\Delta x = - 0.43$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{3}}} - {\left( x \right)^{\frac{1}{3}}} = {\left( {26.57} \right)^{\frac{1}{3}}} - {\left( {27} \right)^{\frac{1}{3}}} \\&= {\left( {26.57} \right)^{\frac{1}{3}}} - 3\\ \Rightarrow \; 3 + \Delta y &= {\left( {26.57} \right)^{\frac{1}{3}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align} dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{3{{\left( x \right)}^{\frac{2}{3}}}}}\left( {\Delta x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{3}}}} \right] \hfill \\& = \frac{1 }{{3\left( 9 \right)}}\left( { - 0.43} \right) = \frac{{ - 0.43}}{{27}} = - 0.015 \end{align}

Hence, the approximate value of $${\left( {26.57} \right)^{\frac{1}{3}}}$$  is  $$3 + \left( { - 0.015} \right) = 2.984$$.

(xiii) $${\left( {81.5} \right)^{\frac{1}{4}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{4}}}$$.

Let $$x = 81\,\,{\text{and}}\,\,\Delta x = 0.5$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{4}}} - {\left( x \right)^{\frac{1}{4}}} = {\left( {81.5} \right)^{\frac{1}{4}}} - {\left( {81} \right)^{\frac{1}{4}}} \\&= {\left( {81.5} \right)^{\frac{1}{4}}} - 3\\ \Rightarrow \; 3 + \Delta &= y{\left( {81.5} \right)^{\frac{1}{4}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{4{{\left( x \right)}^{\frac{3}{4}}}}}\left( {\Delta x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{4}}}} \right] \\&= \frac{1}{{4{{\left( 3 \right)}^3}}}\left( {0.5} \right) = \frac{{0.5}}{{108}} = 0.0046\end{align}

Hence, the approximate value of $${\left( {81.5} \right)^{\frac{1}{4}}}$$  is  $$3 + 0.0046 = 3.0046$$ .

(xiv) $${\left( {3.968} \right)^{\frac{3}{2}}}$$

Consider $$y = {\left( x \right)^{\frac{3}{2}}}$$.

Let $$x = 4\,\,{\text{and}}\,\,\Delta x = - 0.032$$

Then,

\begin{align} \Delta y &= {\left( {x + \Delta x} \right)^{\frac{3}{2}}} - {\left( x \right)^{\frac{3}{2}}} = {\left( {3.968} \right)^{\frac{3}{2}}} - {\left( 4 \right)^{\frac{3}{2}}} \\&= {\left( {3.968} \right)^{\frac{3}{2}}} - 8\\ \Rightarrow \;8 + \Delta y &= {\left( {3.968} \right)^{\frac{3}{2}}}\end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{3}{2}{\left( x \right)^{\frac{1}{2}}}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{3}{2}}}} \right]\\& = \frac{3}{2}\left( 2 \right)\left( { - 0.032} \right) = - 0.096\end{align}

Hence, the approximate value of $${\left( {3.968} \right)^{\frac{3}{2}}}$$  is  $$8 + \left( { - 0.096} \right) = 7.904$$ .

(xv) $${\left( {32.15} \right)^{\frac{1}{5}}}$$

Consider $$y = {\left( x \right)^{\frac{1}{4}}}$$.

Let $$x = 32{\text{ }}\,{\text{and}}\,\,\Delta x = 0.15$$

Then,

\begin{align} \Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{4}}} - {\left( x \right)^{\frac{1}{4}}} = {\left( {32.15} \right)^{\frac{1}{5}}} - {\left( {32} \right)^{\frac{1}{5}}} \\&= {\left( {32.15} \right)^{\frac{1}{5}}} - 2\\ \Rightarrow \; 2 + \Delta y &= {\left( {32.15} \right)^{\frac{1}{5}}} \end{align}

Now, dy is approximately equal to $$\Delta y$$ and is given by,

\begin{align}dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x = \frac{1}{{5{{\left( x \right)}^{\frac{4}{5}}}}}\left( {\Delta x} \right) \qquad \left[ {{\text{as }}y = {{\left( x \right)}^{\frac{1}{5}}}} \right]\\ &= \frac{1}{{5{{\left( 2 \right)}^4}}}\left( {0.15} \right) = \frac{{0.15}}{{80}} = 0.00187\end{align}

Hence, the approximate value of $${\left( {32.15} \right)^{\frac{1}{5}}}$$  is  $$2 + 0.00187 = 2.00187$$.

## Chapter 6 Ex.6.4 Question 2

Find the approximate value of $$f\left( {2.01} \right)$$ , where $$f\left( x \right) = 4{x^2} + 5x + 2$$ .

### Solution

Let $$x = 2{\text{ }}\,{\text{and}}\,\,\Delta x = 0.01$$

Then,

\begin{align}f\left( {2.01} \right) &= f\left( {x + \Delta x} \right) \\&= 4{\left( {x + \Delta x} \right)^2} + 5\left( {x + \Delta x} \right) + 2\\\Delta y &= f\left( {x + \Delta x} \right) - f\left( x \right)\\\therefore f\left( {x + \Delta x} \right) &= f\left( x \right) + \Delta y\\ &\approx f\left( x \right) + f'\left( x \right) \cdot \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{as }}dx = \Delta x} \right)\\\\ \Rightarrow \; f\left( {2.01} \right) &\approx \left( {4{x^2} + 5x + 2} \right) + \left( {8x - 5} \right)\Delta x\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left[ {4{{\left( 2 \right)}^2} + 5\left( 2 \right) + 2} \right] + \left[ {8\left( 2 \right) + 5} \right]\left( {0.01} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}x = 2,\Delta x = 0.01} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {16 + 10 + 2} \right) + \left( {16 + 5} \right)\left( {0.01} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 28 + 21\left( {0.01} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 28 + 0.21\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 28.21\end{align}

Hence, the approximate value of $$f\left( {2.01} \right) = 28.21$$ .

## Chapter 6 Ex.6.4 Question 3

Find the approximate value of $$f\left( {5.001} \right)$$ , where $$f\left( x \right) = {x^3} - 7{x^2} + 15$$ .

### Solution

Let $$x = 5{\text{ }}\,{\text{and}}\,\,\Delta x = 0.001$$

Then,

\begin{align}f\left( {5.001} \right) &= f\left( {x + \Delta x} \right) \\&= {\left( {x + \Delta x} \right)^3} - 7{\left( {x + \Delta x} \right)^2} + 15\\\Delta y &= f\left( {x + \Delta x} \right) - f\left( x \right)\\\therefore f\left( {x + \Delta x} \right) &= f\left( x \right) + \Delta y\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\approx f\left( x \right) + f'\left( x \right) \cdot \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{as }}dx = \Delta x} \right)\\\\ \Rightarrow \; f\left( {5.001} \right) &\approx \left( {{x^3} - 7{x^2} + 15} \right) + \left( {3{x^2} - 14x} \right)\Delta x\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left[ {{{\left( 5 \right)}^3} - 7{{\left( 5 \right)}^2} + 15} \right] + \left[ {3{{\left( 5 \right)}^2} - 14\left( 5 \right)} \right]\left( {0.001} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}x = 5,\Delta x = 0.001} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {125 - 175 + 15} \right) + \left( {75 - 70} \right)\left( {0.001} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left( { - 35} \right) + \left( 5 \right)\left( {0.001} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= - 35 + 0.005\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= - 34.995\end{align}

Hence, the approximate value of $$f\left( {5.001} \right) = - 34.995$$.

## Chapter 6 Ex.6.4 Question 4

Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by $$1\%$$ .

### Solution

The volume of a cube (V) of side x is given by $$V = {x^3}$$ .

\begin{align} \therefore dV &= \left( {\frac{{dV}}{{dx}}} \right)\Delta x \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\, &= \left( {3{x^2}} \right)\Delta x \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\, &= \left( {3{x^2}} \right)\left( {0.01x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}1\% {\text{ of }}x{\text{ is }}0.01x} \right] \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\, &= 0.03{x^3} \hfill \\ \end{align}

Hence, the approximate change in the volume V of the cube $$0.03{x^3}{m^3}$$.

## Chapter 6 Ex.6.4 Question 5

Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by $$1\%$$ .

### Solution

The surface area of a cube (S) of side x is given by $$S = 6{x^2}$$ .

\begin{align}\therefore dS &= \left( {\frac{{dS}}{{dx}}} \right)\Delta x\\\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {12x} \right)\Delta x\\\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {12x} \right)\left( {0.01x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}1\% {\text{ of }}x{\text{ is }}0.01x} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,& = 0.12{x^2}\end{align}

Hence, the approximate change in the volume V of the cube is $$0.12{x^2}{m^2}$$.

## Chapter 6 Ex.6.4 Question 6

If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.

### Solution

Let r be the radius of the sphere and $$\Delta r$$ be the error in measuring the radius.

Then,

$$r = 7{\text{ and }}\Delta r = 0.02$$

Now, the volume V of the sphere is given by,

\begin{align}V &= \frac{4}{3}\pi {r^3}\\\therefore \frac{{dV}}{{dr}} &= 4\pi {r^2}\\\therefore dV &= \left( {\frac{{dV}}{{dr}}} \right)\Delta r\\\,\,\,\,\,\,\,\,\,\,\,\,& = \left( {4\pi {r^2}} \right)\Delta r\\\,\,\,\,\,\,\,\,\,\,\,\, &= 4\pi {\left( 7 \right)^2}\left( {0.02} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, &= 3.92\pi {m^3}\end{align}

Hence, the approximate error in calculating its volume is $$3.92\pi {m^3}$$.

## Chapter 6 Ex.6.4 Question 7

If the radius of a sphere is measured as 9 m with an error of 0.03m, then find the approximate error in calculating its surface area.

### Solution

Let r be the radius of the sphere and $$\Delta r$$ be the error in measuring the radius.

Then,

$$r = 9{\text{ and }}\Delta r = 0.03$$

Now, the volume V of the sphere is given by,

\begin{align}S &= 4\pi {r^2}\\\therefore \frac{{dS}}{{dr}} &= 8\pi r\\\therefore dS &= \left( {\frac{{dS}}{{dr}}} \right)\Delta r\\\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {8\pi r} \right)\Delta r\\\,\,\,\,\,\,\,\,\,\,\,\, &= 8\pi \left( 9 \right)\left( {0.03} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, &= 2.16\pi {m^2}\end{align}

Hence, the approximate error in calculating its surface area is $$2.16\pi {m^2}$$

## Chapter 6 Ex.6.4 Question 8

If$$f\left( x \right) = 3{x^2} + 15x + 5$$ , then the approximate value of $$f\left( {3.02} \right)$$ is

(A) $$47.66$$                (B) $$57.66$$              (C) $$67.66$$              (D) $$77.66$$

### Solution

Let $$x = 3{\text{ }}\,{\text{and}}\,\,\Delta x = 0.02$$

Then,

\begin{align}f\left( {3.02} \right) &= f\left( {x + \Delta x} \right) \\&= 3{\left( {x + \Delta x} \right)^2} + 15\left( {x + \Delta x} \right) + 5\\\Delta y &= f\left( {x + \Delta x} \right) - f\left( x \right)\\\therefore f\left( {x + \Delta x} \right) &= f\left( x \right) + \Delta y\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\approx f\left( x \right) + f'\left( x \right) \cdot \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{as }}dx = \Delta x} \right)\\\\ \Rightarrow \; f\left( {3.02} \right) &\approx \left( {3{x^2} + 15x + 5} \right) + \left( {6x + 15} \right)\Delta x\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left[ {3{{\left( 3 \right)}^2} + 15\left( 3 \right) + 5} \right] + \left[ {6\left( 3 \right) + 15} \right]\left( {0.02} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as }}x = 3,\Delta x = 0.02} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {27 + 45 + 5} \right) + \left( {18 + 15} \right)\left( {0.02} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 77 + \left( {33} \right)\left( {0.02} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 77 + 0.66\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&= 77.66\end{align}

Hence, the approximate value of $$f\left( {3.02} \right) = 77.66$$.

The correct answer is $$D$$.

## Chapter 6 Ex.6.4 Question 9

The approximate change in the volume of a cube of side x metres caused by increasing the side by $$3\%$$ is

(A) $$0.06{x^3}{m^3}$$                (B) $$0.6{x^3}{m^3}$$               (C) $$0.09{x^3}{m^3}$$              (D) $$0.9{x^3}{m^3}$$

### Solution

The volume of a cube (V) of side x is given by $$V = {x^3}$$ .

\begin{align} \therefore dV &= \left( {\frac{{dV}}{{dx}}} \right)\Delta x \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\, &= \left( {3{x^2}} \right)\Delta x \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\, &= \left( {3{x^2}} \right)\left( {0.03x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{as 3}}\% {\text{ of }}x{\text{ is }}0.03x} \right] \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\, &= 0.09{x^3} \hfill \\ \end{align}

Hence, the approximate change in the volume V of the cube $$0.09{x^3}{m^3}$$.

The correct answer is $$C$$.

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