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 \(C\).

  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
0