NCERT Solutions for Class 12 Maths Chapter 6: Application of Derivatives - Exercise 6.4

Exercise 6.4

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

(i)\[\mathbf{\sqrt {25.3}} \]

Ans: Let \[y = \sqrt x \]. Let \[x = 25\] and \[\Delta x = 0.3\].

Then,

$  \Delta y = \sqrt {x + \vartriangle x}  - \sqrt x  \\ $

$  \Delta y = \sqrt {25.3}  - \sqrt {25}  \\ $

$  \Delta y = \sqrt {25.3}  - 5 \\ $

$  \sqrt {25.3}  = \Delta y + 5 \\ $ 

Now, finding the value of \[\Delta y\].

As \[y = \sqrt x \]

On differentiating both side w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x $

$  \Delta y = \dfrac{1}{{2\sqrt x }}\Delta x $ 

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{2\sqrt {25} }}\left( {0.3} \right) \\ $

$  \Delta y = \dfrac{1}{{2 \times 5}} \times \left( {0.3} \right) \\ $

$  \Delta y = \dfrac{{0.3}}{{10}} \\ $

$  \Delta y = 0.03 \\ $

Therefore,

$ \sqrt {25.3}  = \Delta y + 5 \\ $

$  \sqrt {25.3}  = 5 + 0.03 \\ $

$   = 5.03 \\ $

So, the approximate value of \[\sqrt {25.3} \] is $5.03$.

(ii) \[\mathbf{\sqrt {49.5}} \]

Ans: Let \[y = \sqrt x \]. Let \[x = 49\] and \[\Delta x = 0.5\] .

Then,

$  \Delta y = \sqrt {x + \vartriangle x}  - \sqrt x  \\ $

$  \Delta y = \sqrt {49.5}  - \sqrt {49}  \\ $

 $ \Delta y = \sqrt {49.3}  - 7 \\ $

$  \sqrt {49.5}  = \Delta y + 7 \\ $

 Now, finding the value of \[\Delta y\].

As \[y = \sqrt x \], on differentiating both side w.r.t. \[x\] we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\] , it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{2\sqrt x }}\Delta x \\ $

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{2\sqrt {49} }}\left( {0.5} \right) \\ $

$  \Delta y = 0.03571 \\ $

 Therefore,

$  \sqrt {49.5}  = \Delta y + 7 \\ $

$  \sqrt {49.5}  = 7 + 0.03571 \\ $

$   = 7.03571 \\ $

So, the value of \[\sqrt {49.5} \] is \[7.03571\] .

(iii) \[\mathbf{\sqrt {0.6} }\]

Ans: Let \[y = \sqrt x \]. Let \[x = 1\] and \[\Delta x =  - 0.4\].

Therefore,

$  \Delta y = \sqrt {x + \vartriangle x}  - \sqrt x  \\ $

$  \Delta y = \sqrt {0.6}  - \sqrt 1  \\ $

$  \Delta y = \sqrt {0.6}  - 1 \\ $

$  \sqrt {0.6}  = \Delta y + 1 \\ $

Now, finding the value of \[\Delta y\].

As \[y = \sqrt x \], on differentiating both side w.r.t. \[x\] we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\] , it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{2\sqrt x }}\Delta x \\ $

 Now, substituting the value of \[x\] and \[\Delta x\].

$   \Delta y = \dfrac{1}{{2\sqrt 1 }}\left( { - 0.4} \right) \\ $

$  \Delta y =  - 0.2 \\ $

 Therefore,

$  \sqrt {0.6}  = \Delta y + 1 \\ $

$  \sqrt {0.6}  = 1 - 0.2 \\  $

 $  = 0.8 \\ $

So, the value of \[\sqrt {0.6} \] is \[0.8\].

(iv) \[\mathbf{{\left( {0.009} \right)^{\dfrac{1}{3}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{3}}}\]. Let \[x = 0.008\] and \[\Delta x = 0.001\].

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{3}}} - {\left( x \right)^{\dfrac{1}{3}}} $

$  \Delta y = {\left( {0.009} \right)^{\dfrac{1}{3}}} - {\left( {0.008} \right)^{\dfrac{1}{3}}} $

$  \Delta y = {\left( {0.009} \right)^{\dfrac{1}{3}}} - 0.2 $

$ {\left( {0.009} \right)^{\dfrac{1}{3}}} = \Delta y + 0.2 $

 Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{3}}}\], on differentiating both side w.r.t. \[x\] we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$ \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x $

$  \Delta y = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}\Delta x $ 

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{3{{\left( {0.008} \right)}^{\dfrac{2}{3}}}}}\left( {0.001} \right) \\ $

$  \Delta y = \dfrac{1}{{3 \times 0.04}} \times \left( {0.001} \right) \\ $

$  \Delta y = \dfrac{{0.001}}{{0.12}} \\ $

$  \Delta y = 0.008 \\ $

 Therefore,

$  {\left( {0.009} \right)^{\dfrac{1}{3}}} = \Delta y + 0.2 \\$ 

$  {\left( {0.009} \right)^{\dfrac{1}{3}}} = 0.2 + 0.008 \\ $

$   = 0.208 \\ $

So, the value of \[{\left( {0.009} \right)^{\dfrac{1}{3}}}\] is \[0.208\].

(v) \[\mathbf{{\left( {0.999} \right)^{\dfrac{1}{{10}}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{{10}}}}\]. Let \[x = 1\] and \[\Delta x =  - 0.001\] .

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{{10}}}} - {\left( x \right)^{\dfrac{1}{{10}}}} $

$  \Delta y = {\left( {0.999} \right)^{\dfrac{1}{{10}}}} - {\left( 1 \right)^{\dfrac{1}{{10}}}} $

$  \Delta y = {\left( {0.999} \right)^{\dfrac{1}{{10}}}} - 1$

$  {\left( {0.999} \right)^{\dfrac{1}{{10}}}} = \Delta y + 1 $

 Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{{10}}}}\], on differentiating both side w.r.t. \[x\] we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{10{{\left( x \right)}^{\dfrac{9}{{10}}}}}}\]

As, \[dy\] and \[dx\]are approximately equal to \[\Delta y\]and \[\Delta x\] , it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x $

$  \Delta y = \dfrac{1}{{10{{\left( x \right)}^{\dfrac{9}{{10}}}}}}\Delta x $

Now, substituting the value of \[x\]and\[\Delta x\].

$  \Delta y = \dfrac{1}{{10{{\left( 1 \right)}^{\dfrac{9}{{10}}}}}}\left( { - 0.001} \right) \\ $

$  \Delta y = \dfrac{1}{{10 \times 1}} \times \left( { - 0.001} \right) \\ $

$  \Delta y =  - \dfrac{{0.001}}{{10}} \\ $

$   =  - 0.0001 \\ $

 Therefore,

$  {\left( {0.999} \right)^{\dfrac{1}{{10}}}} = \Delta y + 1 \\ $

$  {\left( {0.999} \right)^{\dfrac{1}{{10}}}} = 1 + \left( { - 0.0001} \right) \\ $

$   = 0.9999 \\ $

So, the value of \[{\left( {0.999} \right)^{\dfrac{1}{{10}}}}\]is \[0.9999\].

(vi) \[\mathbf{{\left( {15} \right)^{\dfrac{1}{4}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{4}}}\]. Let \[x = 16\] and \[\Delta x =  - 1\] .

Therefore,

$ \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{4}}} - {\left( x \right)^{\dfrac{1}{4}}} \\$

$  \Delta y = {\left( {15} \right)^{\dfrac{1}{4}}} - {\left( {16} \right)^{\dfrac{1}{4}}} \\ $

$  \Delta y = {\left( {15} \right)^{\dfrac{1}{4}}} - 2\\ $

$  {\left( {15} \right)^{\dfrac{1}{4}}} = \Delta y + 2 \\ $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{4}}}\], on differentiating both side w.r.t. \[x\] .

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$ \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x  \\ $

$  \Delta y = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\Delta x  \\ $

Now, substituting the value of \[x\]and\[\Delta x\].

$  \Delta y = \dfrac{1}{{4{{\left( {16} \right)}^{\dfrac{3}{4}}}}}\left( { - 1} \right) \\  $

$  \Delta y = \dfrac{1}{{4 \times 8}} \times \left( { - 1} \right) \\ $

$  \Delta y =  - \dfrac{1}{{32}} \\ $

$  \Delta y =  - 0.003125 \\ $

Therefore,

$  {\left( {15} \right)^{\dfrac{1}{4}}} = \Delta y + 2 \\ $

$  {\left( {15} \right)^{\dfrac{1}{4}}} = 2 + \left( { - 0.03125} \right) \\ $

$   = 1.96875 \\  $

So, the value of \[{\left( {15} \right)^{\dfrac{1}{4}}}\] is \[1.96875\].

(vii) \[{\mathbf{\left( {26} \right)^{\dfrac{1}{3}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{3}}}\]. Let \[x = 27\] and \[\Delta x =  - 1\] .

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{3}}} - {\left( x \right)^{\dfrac{1}{3}}} \\ $

$  \Delta y = {\left( {26} \right)^{\dfrac{1}{3}}} - {\left( {27} \right)^{\dfrac{1}{3}}} \\ $

$  \Delta y = {\left( {26} \right)^{\dfrac{1}{3}}} - 3 \\ $

$  {\left( {26} \right)^{\dfrac{1}{3}}} = \Delta y + 3 \\ $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{3}}}\], on differentiating both side w.r.t. \[x\] .

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\] , it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}\Delta x \\ $

Now, substituting the value of \[x\]and\[\Delta x\].

$  \Delta y = \dfrac{1}{{3{{\left( {27} \right)}^{\dfrac{2}{3}}}}}\left( { - 1} \right) \\ $

$  \Delta y = \dfrac{1}{{3{{\left( {27} \right)}^{\dfrac{2}{3}}}}}\left( { - 1} \right) \\  $

$  \Delta y = \dfrac{1}{{3 \times 9}} \times \left( { - 1} \right) \\ $

$  \Delta y =  - \dfrac{1}{{27}} \\ $

$  \Delta y =  - 0.0370 \\ $

Therefore,

$  {\left( {26} \right)^{\dfrac{1}{3}}} = \Delta y + 3 \\ $

$  {\left( {26} \right)^{\dfrac{1}{3}}} = 3 + \left( { - 0.0370} \right) \\ $

$   = 2.9629 \\ $

So, the value of \[{\left( {26} \right)^{\dfrac{1}{3}}}\] is \[2.9629\] .

(viii) \[\mathbf{{\left( {255} \right)^{\dfrac{1}{4}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{4}}}\] . Let \[x = 265\] and \[\Delta x =  - 1\] .

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{4}}} - {\left( x \right)^{\dfrac{1}{4}}} \\ $

$  \Delta y = {\left( {255} \right)^{\dfrac{1}{4}}} - {\left( {256} \right)^{\dfrac{1}{4}}} \\ $

$  \Delta y = {\left( {255} \right)^{\dfrac{1}{4}}} - 4 \\ $

$  {\left( {255} \right)^{\dfrac{1}{4}}} = \Delta y + 4 \\ $

Now, finding the value of \[\Delta y\].

As, \[y = {x^{\dfrac{1}{4}}}\]

On differentiating both side w.r.t. \[x\] .

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\]

As \[dy\] and \[dx\]are approximately equal to \[\Delta y\]and \[\Delta x\] , it is given as,

$   \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\Delta x \\ $

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{4{{\left( {256} \right)}^{\dfrac{3}{4}}}}}\left( { - 1} \right) \\ $

$  \Delta y = \dfrac{1}{{4 \times 64}} \times \left( { - 1} \right) \\ $

$  \Delta y =  - \dfrac{1}{{256}} \\ $

$  \Delta y =  - 0.0039 \\ $

Therefore,

$  {\left( {255} \right)^{\dfrac{1}{4}}} = \Delta y + 4 \\ $

$  {\left( {255} \right)^{\dfrac{1}{4}}} = 4 + \left( { - 0.0039} \right) \\ $

  $ = 3.9961 \\ $

So, the value of \[{\left( {255} \right)^{\dfrac{1}{4}}}\]is \[3.9961\] .

(ix) \[\mathbf{{\left( {82} \right)^{\dfrac{1}{4}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{4}}}\]. Let \[x = 81\] and \[\Delta x = 1\].

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{4}}} - {\left( x \right)^{\dfrac{1}{4}}} \\ $

$  \Delta y = {\left( {82} \right)^{\dfrac{1}{4}}} - {\left( {81} \right)^{\dfrac{1}{4}}} \\  $

$  \Delta y = {\left( {82} \right)^{\dfrac{1}{4}}} - 3 \\  $

$  {\left( {82} \right)^{\dfrac{1}{4}}} = \Delta y + 3 \\  $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{4}}}\], on differentiating both side w.r.t. \[x\] we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\Delta x \\ $

Now, substituting the value of \[x\]and \[\Delta x\].

$  \Delta y = \dfrac{1}{{4{{\left( {81} \right)}^{\dfrac{3}{4}}}}}\left( 1 \right) \\ $

$  \Delta y = \dfrac{1}{{4 \times 27}} \times \left( 1 \right) \\ $

$  \Delta y = \dfrac{1}{{108}} \\ $

$  \Delta y = 0.009 \\ $

 Therefore,

$  {\left( {82} \right)^{\dfrac{1}{4}}} = \Delta y + 3 \\ $

$  {\left( {82} \right)^{\dfrac{1}{4}}} = 3 + \left( {0.009} \right) \\ $

$   = 3.009 \\ $

 So, the value of \[{\left( {82} \right)^{\dfrac{1}{4}}}\] is \[3.009\].

(x)\[\mathbf{{\left( {401} \right)^{\dfrac{1}{2}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{2}}}\]. Let \[x = 400\] and \[\Delta x = 1\].

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{2}}} - {\left( x \right)^{\dfrac{1}{2}}} \\ $

$  \Delta y = {\left( {401} \right)^{\dfrac{1}{2}}} - {\left( {400} \right)^{\dfrac{1}{2}}} \\ $

$  \Delta y = {\left( {401} \right)^{\dfrac{1}{2}}} - 20 \\ $

$  {\left( {401} \right)^{\dfrac{1}{2}}} = \Delta y + 20 \\ $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{2}}}\], on differentiating both side w.r.t. \[x\] we get

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{2{{\left( x \right)}^{\dfrac{1}{2}}}}}\Delta x \\ $

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{2{{\left( {400} \right)}^{\dfrac{1}{2}}}}}\left( 1 \right) \\ $

$  \Delta y = \dfrac{1}{{2 \times 20}} \times \left( 1 \right) \\ $

$  \Delta y = \dfrac{1}{{40}} \\ $

$  \Delta y = 0.025 \\ $

 Therefore,

$  {\left( {401} \right)^{\dfrac{1}{2}}} = \Delta y + 20 \\ $

$  {\left( {401} \right)^{\dfrac{1}{2}}} = 20 + \left( {0.025} \right) \\ $

 $  = 20.025 \\ $

So, the value of \[{\left( {401} \right)^{\dfrac{1}{2}}}\] is \[20.025\].

(xi) \[\mathbf{{\left( {0.0037} \right)^{\dfrac{1}{2}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{2}}}\] . Let \[x = 0.0036\] and \[\Delta x = 0.0001\] .

Therefore,

$  \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{2}}} - {\left( x \right)^{\dfrac{1}{2}}} \\ $

$  \Delta y = {\left( {0.0037} \right)^{\dfrac{1}{2}}} - {\left( {0.0036} \right)^{\dfrac{1}{2}}} \\ $

$  \Delta y = {\left( {0.0037} \right)^{\dfrac{1}{2}}} - 0.06 \\ $

 $ {\left( {0.0037} \right)^{\dfrac{1}{2}}} = \Delta y + 0.06 \\ $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{2}}}\], on differentiating both sides w.r.t. \[x\] .

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$ \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{2{{\left( x \right)}^{\dfrac{1}{2}}}}}\Delta x \\  $

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{2{{\left( {0.0036} \right)}^{\dfrac{1}{2}}}}}\left( {0.0001} \right) \\ $

$  \Delta y = \dfrac{1}{{2 \times 0.06}} \times \left( {0.0001} \right) \\ $

$  \Delta y = \dfrac{{0.0001}}{{0.12}} \\ $

$  \Delta y = 0.00083 \\ $

Therefore,

$  {\left( {0.0037} \right)^{\dfrac{1}{2}}} = \Delta y + 0.06 \\ $

$  {\left( {0.0037} \right)^{\dfrac{1}{2}}} = 0.06 + \left( {0.00083} \right) \\ $

$   = 0.06083 \\ $

So, the value of \[{\left( {0.0037} \right)^{\dfrac{1}{2}}}\] is \[0.06083\].

(xii) \[\mathbf{{\left( {26.57} \right)^{\dfrac{1}{3}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{3}}}\]. Let \[x = 27\] and \[\Delta x =  - 0.43\].

Therefore,

$ \Delta y = {\left( {x + \vartriangle x} \right)^{\dfrac{1}{3}}} - {\left( x \right)^{\dfrac{1}{3}}} \\ $

 $ \Delta y = {\left( {26.57} \right)^{\dfrac{1}{3}}} - {\left( {27} \right)^{\dfrac{1}{3}}} \\ $

$  \Delta y = {\left( {26.57} \right)^{\dfrac{1}{3}}} - 3 \\ $

$  {\left( {26.57} \right)^{\dfrac{1}{3}}} = \Delta y + 3 \\ $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{3}}}\], on differentiating both side w.r.t. \[x\].

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$ \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$ \Delta y = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}\Delta x \\ $

 Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{3{{\left( {27} \right)}^{\dfrac{2}{3}}}}}\left( { - 0.43} \right) \\ $

$  \Delta y = \dfrac{1}{{3 \times 9}} \times \left( { - 0.43} \right) \\ $

$  \Delta y =  - \dfrac{{0.43}}{{27}} \\ $

$  \Delta y =  - 0.015 \\ $

Therefore,

$ {\left( {26.57} \right)^{\dfrac{1}{3}}} = \Delta y + 3 \\ $

$  {\left( {26.57} \right)^{\dfrac{1}{3}}} = 3 + \left( { - 0.015} \right) \\ $

$ = 2.984 \\ $

So, the value of \[{\left( {26.57} \right)^{\dfrac{1}{3}}}\] is \[2.984\].

(xiii) \[\mathbf{{\left( {81.5} \right)^{\dfrac{1}{4}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{4}}}\]. Let \[x = 81\] and \[\Delta x = 0.5\].

Therefore,

$  \Delta y = {\left( {x + \Delta x} \right)^{\dfrac{1}{4}}} - {\left( x \right)^{\dfrac{1}{4}}} \\ $

$  \Delta y = {\left( {81.5} \right)^{\dfrac{1}{4}}} - {\left( {81} \right)^{\dfrac{1}{4}}} \\ $

$  \Delta y = {\left( {81.5} \right)^{\dfrac{1}{4}}} - 3 \\ $

$  {\left( {81.5} \right)^{\dfrac{1}{4}}} = \Delta y + 3 \\ $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{1}{4}}}\], on differentiating both side w.r.t. \[x\] .

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$ \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{4{{\left( x \right)}^{\dfrac{3}{4}}}}}\Delta x \\ $

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{4{{\left( {81} \right)}^{\dfrac{3}{4}}}}}\left( {0.5} \right) \\ $

$ \Delta y = \dfrac{1}{{4 \times 27}} \times \left( {0.5} \right) \\ $

$ \Delta y = \dfrac{{0.5}}{{108}} \\ $

$  \Delta y = 0.0046 \\ $

Therefore,

$ {\left( {81.5} \right)^{\dfrac{1}{4}}} = \Delta y + 3 \\ $ 

$  {\left( {82} \right)^{\dfrac{1}{4}}} = 3 + \left( {0.0046} \right) \\  $

$ = 3.0046 \\ $

So, the value of \[{\left( {81.5} \right)^{\dfrac{1}{4}}}\] is \[3.0046\].

(xiv)  \[\mathbf{{\left( {3.968} \right)^{\dfrac{3}{2}}}}\]

Ans: Let \[y = {x^{\dfrac{3}{2}}}\]. Let \[x = 4\] and \[\Delta x =  - 0.032\] .

Therefore,

$  \Delta y = {\left( {x + \Delta x} \right)^{\dfrac{3}{2}}} - {\left( x \right)^{\dfrac{3}{2}}} \\ $

$  \Delta y = {\left( {3.968} \right)^{\dfrac{3}{2}}} - {\left( 4 \right)^{\dfrac{3}{2}}} \\ $

$  \Delta y = {\left( {3.968} \right)^{\dfrac{3}{2}}} - 8 \\ $

$  {\left( {3.968} \right)^{\dfrac{3}{2}}} = \Delta y + 8 \\  $

Now, finding the value of \[\Delta y\].

As \[y = {x^{\dfrac{3}{2}}}\] , on differentiating both side w.r.t. \[x\] .

\[\dfrac{{dy}}{{dx}} = \dfrac{3}{2}{\left( x \right)^{\dfrac{1}{2}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

$  \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$ \Delta y = \dfrac{3}{2}{\left( x \right)^{\dfrac{1}{2}}}\Delta x \\ $

 Now, substituting the values of \[x\]and\[\Delta x\].

$ \Delta y = \dfrac{3}{2}{\left( 4 \right)^{\dfrac{1}{2}}}\left( { - 0.032} \right) \\ $

 $ \Delta y = \dfrac{3}{2} \times 2 \times \left( { - 0.032} \right) \\ $

$  \Delta y =  - 0.096 \\ $

Therefore,

$  {\left( {3.968} \right)^{\dfrac{3}{2}}} = \Delta y + 8 \\ $

$  {\left( {3.968} \right)^{\dfrac{3}{2}}} = 8 + \left( { - 0.096} \right) \\ $

 $  = 7.904 \\ $

So, the value of \[{\left( {3.968} \right)^{\dfrac{3}{2}}}\] is \[7.904\] .

(xv) \[\mathbf{{\left( {32.15} \right)^{\dfrac{1}{5}}}}\]

Ans: Let \[y = {x^{\dfrac{1}{5}}}\]. Let \[x = 32\] and \[\Delta x = 0.15\] .

Therefore,

$  \Delta y = {\left( {x + \Delta x} \right)^{\dfrac{1}{5}}} - {\left( x \right)^{\dfrac{1}{5}}} \\ $

$  \Delta y = {\left( {32.15} \right)^{\dfrac{1}{5}}} - {\left( {32} \right)^{\dfrac{1}{5}}} \\ $

$  \Delta y = {\left( {32.15} \right)^{\dfrac{1}{5}}} - 2 \\ $

$  {\left( {32.15} \right)^{\dfrac{1}{5}}} = \Delta y + 2 \\ $

Now, finding the value of \[\Delta y\].

As, \[y = {x^{\dfrac{1}{5}}}\]

On differentiating both side w.r.t. \[x\] ,

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{5{{\left( x \right)}^{\dfrac{4}{5}}}}}\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\] , it is given as,

$ \Delta y = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x \\ $

$  \Delta y = \dfrac{1}{{5{{\left( x \right)}^{\dfrac{4}{5}}}}}\Delta x \\ $ 

Now, substituting the value of \[x\] and \[\Delta x\].

$  \Delta y = \dfrac{1}{{5{{\left( {32} \right)}^{\dfrac{4}{5}}}}}\left( {0.15} \right) \\ $

$  \Delta y = \dfrac{1}{{5 \times 16}} \times \left( {0.15} \right) \\ $

$  \Delta y = \dfrac{{0.15}}{{80}} \\ $

$  \Delta y = 0.00187 \\ $

Therefore,

$  {\left( {32.15} \right)^{\dfrac{1}{5}}} = \Delta y + 2 \\ $

$  {\left( {32.15} \right)^{\dfrac{1}{5}}} = 2 + \left( {0.00187} \right) \\ $

$  = 2.00187 \\ $

So, the value of \[{\left( {32.15} \right)^{\dfrac{1}{5}}}\] is \[2.00187\].

2. Find the approximate value of \[\mathbf{f(2.01)}\], where \[\mathbf{f(x) = 4{x^2} + 5x + 2}\].

Ans: Consider \[x = 2\] and \[\Delta x = 0.01\] .

Therefore,

$  f(2.01) = f(x + \Delta x) \\ $

$ = 4{(x + \Delta x)^2} + 5(x + \Delta x) + 2 \\ $

Therefore,

$  \vartriangle y = f(x + \Delta x) - f(x) \\ $

$  f(x + \Delta x) = \Delta y + f(x)......(1) \\ $

Since \[f(x) = 4{x^2} + 5x + 2\]

Differentiating both side w.r.t \[x\] .

\[\dfrac{{dy}}{{dx}} = 8x + 5\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

\[\Delta y = \left( {8x + 5} \right)\Delta x\]

Therefore equation 1 can be written as

$  f(2.01) = 4{x^2} + 5x + 2 + \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) \\ $

$ = 28 + 0.21 \\ $

$ = 28.21 \\  $

So, the approximate value of \[f(2.01)\] is \[28.21\] .

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

Ans: Consider \[x = 5\] and \[\Delta x = 0.001\].

Therefore,

$  f(2.01) = f(x + \Delta x) \\ $

$ = {(x + \Delta x)^3} + 7{(x + \Delta x)^2} + 15 \\ $ 

Therefore,

$  \Delta y = f(x + \Delta x) - f(x) \\ $

$  f(x + \Delta x) = \Delta y + f(x)......(1) \\ $

Since \[f(x) = {x^3} + 7{x^2} + 15\]

Differentiating both side w.r.t \[x\] ,

\[\dfrac{{dy}}{{dx}} = 3{x^2} + 14x\]

As, \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\], it is given as,

\[\Delta y = \left( {3{x^2} - 14x} \right)\Delta x\]

Therefore equation 1 can be written as

$  f(5.001) = {x^3} - 7{x^2} + 15 + \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] \times \left( {0.001} \right) \\ $

$  =  - 35 + 0.005 \\ $

$   =  - 35.001 \\ $

$   =  - 34.995 \\ $

So, the value of \[f(5.001)\] is \[ - 34.995\] .

4. Find the approximate change in volume \[V\]of a cube side \[x{\text{ meter}}\]caused by increasing side by \[1\% \] .

Ans: The volume of a cube of side \[x{\text{ meter}}\] is given as,

\[V = {x^3}\]

Differentiating both side w.r.t \[x\],

\[\dfrac{{dV}}{{dx}} = 3{x^2}\]

As \[dx\] is approximately equal to \[\Delta x\], it is given as,

$ dV = \left( {3{x^2}} \right)\left( {\Delta x} \right) \\ $

$ = \left( {3{x^2}} \right)\left( {0.01x} \right) \\ $

$ = 0.03{x^3}{{\text{m}}^3} \\ $

Therefore, the approximate change in the volume of cube is \[0.03{x^2}{m^3}\] .

5. Find the approximate change in the surface area of a cube of side \[x{\text{ meter}}\] caused by decreasing the side by \[1\% \].

Ans:  Suface area of square of side \[x{\text{ meter}}\] is given as,

\[S = 6{x^2}\]

Differentiating both side w.r.t. \[x\],

\[\dfrac{{dS}}{{dx}} = 12x\]

As \[dx\] is approximately equal to \[\Delta x\], it is given as,

\[dS = \left( {12x} \right)\left( {\Delta x} \right)\]

As \[1\% \] of  is \[0.01x\].

$ dS = \left( {12x} \right)\left( {0.01x} \right) \\ $

$ = 0.12{x^2}{m^2} \\ $

Therefore, the approximate change in surface area of cube is \[0.12{x^2}{m^2}\] .

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

Ans: In this question it is given that ,

radius\[\left( r \right) = 7m\]

\[\Delta r = 0.02m\]

So, the volume of sphere is given as,

\[V = \dfrac{4}{3}\pi {r^3}\]

Differentiating both side w.r.t. \[r\],

\[\dfrac{{dV}}{{dr}} = 4\pi {r^2}\]

As \[dr\] is approximately equal to \[\Delta r\], it is given as,

$ dV = \left( {4\pi {r^2}} \right)\left( {\Delta r} \right) \\ $

$ = 4\pi {\left( 7 \right)^2}\left( {0.02} \right) \\ $

$ = 3.92\pi {m^2} \\ $

Therefore, the approximate error in calculated volume of sphere is \[3.92\pi {m^2}\].

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

Ans: In this question, it is given that,

radius\[\left( r \right) = 9m\]

\[\Delta r = 0.03m\]

So, surface area of sphere is given as,

\[S = 4\pi {r^2}\]

Differentiating both side w.r.t. \[r\],

\[\dfrac{{dS}}{{dr}} = 8\pi r\]

As \[dr\] is approximately equal to \[\Delta r\], it is given as,

$ dS = \left( {8\pi r} \right)\left( {\Delta r} \right) \\ $

$ = 8\pi \left( 9 \right)\left( {0.03} \right) \\ $

$ = 2.16\pi {m^2} \\ $

Therefore, the approximate error in calculated surface area of sphere is \[2.16\pi {m^2}\].

8. If \[\mathbf{f(x) = 3{x^2} + 15x + 5}\], then the approximate value of \[f\left( {3.02} \right)\] is \[\mathbf{{\text{(A) 47}}{\text{.66      (B) 57}}{\text{.66       (C) 67}}{\text{.66        (D) 77}}{\text{.66}}}\]

Ans: Consider \[x = 3\] and \[\Delta x = 0.02\].

Therefore,

$ f(3.02) = f(x + \Delta x) \\ $

$ = 3{(x + \Delta x)^2} + 15(x + \Delta x) + 5 \\ $

 Therefore,

$  \Delta y = f(x + \Delta x) - f(x) \\ $

$  f(x + \Delta x) = \Delta y + f(x)......(1) \\ $

Since \[f(x) = 3{x^2} + 15x + 5\], differentiating both side w.r.t \[x\] .

\[\dfrac{{dy}}{{dx}} = 6x + 15\]

As \[dy\] and \[dx\] are approximately equal to \[\Delta y\] and \[\Delta x\] , it is given as,

\[\Delta y = \left( {6x + 15} \right)\Delta x\]

Therefore, equation 1 can be written as

$  f(3.02) = 3{x^2} + 15x + 5 + \left( {6x + 15} \right)\vartriangle x \\ $

$   = \left[ {3{{\left( 3 \right)}^2} + 15\left( 3 \right) + 5} \right] + \left[ {6\left( 3 \right) + 15} \right]\left( {0.02} \right) \\ $

$   = 77 +  + 0.66 \\ $

$   = 77.66 \\ $

So, the approximate value of \[f(3.02)\] is \[77.66\].

Therefore, the correct option is \[{\text{(D)}}\] .

9. Find theapproximate change in volume \[V\]of a cube side \[x{\text{ meter}}\]caused by increasing side by \[3\% \] is

\[\mathbf{{\text{(A) 0}}{\text{.06}}{x^3}{{\text{m}}^3}{\text{      (B) 0}}{\text{.6}}{x^3}{{\text{m}}^3}{\text{       (C) 0}}{\text{.09}}{x^3}{{\text{m}}^3}{\text{        (D) 0}}{\text{.9}}{x^3}{{\text{m}}^3}}\]

Ans: The volume of a cube of side \[x{\text{ meter}}\] is given as,

\[V = {x^3}\]

Differentiating both side w.r.t \[x\],

\[\dfrac{{dV}}{{dx}} = 3{x^2}\]

As \[dx\]is approximately equal to \[\Delta x\] , it is given as,

\[dV = \left( {3{x^2}} \right)\left( {\Delta x} \right)\]

\[3\% \]of\[x\] is \[0.03x\]

$ dV = \left( {3{x^2}} \right)\left( {0.03x} \right) \\ $

 $ = 0.09{x^3}{{\text{m}}^3} \\ $

Therefore, the approximate change in the volume of cube is \[0.09{x^3}{{\text{m}}^3}\] .

Therefore, the correct option is \[(C)\] .

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