NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.4


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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