Ex 6.4 class 12 maths ncert solutions || class 12 maths exercise 6.4 || class 12 maths ncert solutions chapter 6 exercise 6.4 || exercise 6.4 class 12 maths ncert solutions || application of derivatives class 12 ncert solutions (English Medium)
NCERT Solutions for Class 12 Maths – Exercise 6.4 (Application of Derivatives)
Exercise 6.4 of Class 12 Maths NCERT Solutions is a vital part of the chapter Application of Derivatives, which plays a significant role in understanding calculus concepts. This exercise primarily deals with practical problems involving maxima and minima, helping students apply derivative techniques to find optimal solutions. The Class 12 Maths Exercise 6.4 NCERT Solutions offer detailed, step-by-step explanations to make each concept crystal clear. Designed especially for English Medium students, these solutions strictly adhere to the CBSE curriculum. By studying from the Class 12 Maths NCERT Solutions Chapter 6 Exercise 6.4, students can build a strong foundation in solving real-life application problems, which are frequently seen in board exams and competitive tests alike.
ex 6.4 class 12 maths ncert solutions || class 12 maths exercise 6.4 || application of derivatives class 12 ncert solutions || class 12 maths ncert solutions chapter 6 exercise 6.4 || exercise 6.4 class 12 maths ncert solutions
Exercise 6.4
1A. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\(\sqrt{25.3}\)
\(\sqrt{25.3}\)
Answer
Consider \( y=\sqrt{x} \)
Let \( x=25 \) and \( \Delta x=0.3 \). Then, we get
\(\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\)
\(\Delta y=\sqrt{25.3}-\sqrt{25}\)
\(\Delta y=\sqrt{25.3}-5\)
\(\Delta y=\sqrt{25.3}=\Delta y+5\)
Now, dy is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(0.3)\)
\(=\frac{1}{\sqrt[2]{25}}(0.3)\)
\(=0.03\)
Therefore, the approximate value of \( \sqrt{25.3} \) is 5.03 .
1B. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\(\sqrt{49.5}\)
\(\sqrt{49.5}\)
Answer
Consider \( y=\sqrt{x} \)
Let \( x=49 \) and \( \Delta x=0.5 \). Then, we get
\(\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\)
\(=\sqrt{49.5}-\sqrt{49}\)
\(=\sqrt{49.5}-7\)
\(=\sqrt{49.5}=\Delta y+7\)
Now, dy is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(0.5)\)
\(=\frac{1}{\sqrt[2]{49}}(0.5)\)
\(=0.035\)
Therefore, the approximate value of \( \sqrt{49.5} \) is 7.035 .
1C. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\(\sqrt{0.6}\)
\(\sqrt{0.6}\)
Answer
Consider \( y=\sqrt{x} \)
Let \( x=1 \) and \( \Delta x=-0.4 \). Then, we get
\(\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\)
\(=\sqrt{0.6-1}\)
\(=\sqrt{0.6}=\Delta y+1\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(\Delta x)\)
\(=\frac{1}{2}(-0.4)\)
\(=-0.2\)
Therefore, the approximate value of \( \sqrt{0.6} \) is \(0.8\) .
1D. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\((0.009)^{\frac{1}{3}}\)
\((0.009)^{\frac{1}{3}}\)
Answer
Consider \( y=(x)^{\frac{1}{3}} \)
Let \( x=0.008 \) and \( \Delta x=0.001 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{3}}-(x)^{\frac{1}{3}}\)
\(=(0.009)^{\frac{1}{3}}-(0.008)^{\frac{1}{3}}=(0.009)^{\frac{1}{3}}-0.2\)
\(=(0.009)^{\frac{1}{3}}=\Delta y+0.2\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x)\)
\(=\frac{1}{3 \times 0.04}(0.001)\)
\(=0.208\)
Therefore, the approximate value of \( (0.009)^{\frac{1}{3}} \) is \(0.208 \).
1E. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (0.999)^{\frac{1}{10}} \)
\( (0.999)^{\frac{1}{10}} \)
Answer
Consider \( y=(x)^{\frac{1}{10}} \)
Let \( x=1 \) and \( \Delta x=-0.001 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{10}}-(x)^{\frac{1}{10}}\)
\(\Delta y=(0.999)^{\frac{1}{10}}-1\)
\((0.999)^{\frac{1}{10}}=\Delta y+1\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{10(x)^{\frac{9}{10}}}(\Delta x)\)
\(=\frac{1}{10}(-0.001)\)
\(=-0.0001\)
Therefore, the approximate value of \( (0.999)^{\frac{1}{10}} \) is \(0.9999 \).
1F. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (15)^{\frac{1}{4}} \)
\( (15)^{\frac{1}{4}} \)
Answer
Consider \( y=(x)^{\frac{1}{4}} \)
Let \( x=16 \) and \( \Delta x=-1 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{4}}\)
\((x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}\)
\(=(15)^{\frac{1}{4}}-(16)^{\frac{1}{4}}\)
\(=(15)^{\frac{1}{4}}-2\)
\(\Rightarrow(15)^{\frac{1}{4}}=\Delta y+2\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)
\(=\frac{1}{4(16)^{\frac{3}{4}}}(-1)\)
\(=\frac{-1}{4 \times 8}\)
\(=\frac{-1}{32}\)
\(=-0.03125\)
Therefore, the approximate value of \( (15)^{\frac{1}{4}} \) is \(-0.03125 \).
ex 6.4 class 12 maths ncert solutions || class 12 maths exercise 6.4 || application of derivatives class 12 ncert solutions || class 12 maths ncert solutions chapter 6 exercise 6.4 || exercise 6.4 class 12 maths ncert solutions
1G. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (26)^{\frac{1}{3}} \)
\( (26)^{\frac{1}{3}} \)
Answer
Consider \( y=(x)^{\frac{1}{3}} \)
Let \( x=27 \) and \( \Delta x=-1 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{3}}-(x)^{\frac{1}{3}}\)
\(=(26)^{\frac{1}{3}}-(27)^{\frac{1}{3}}\)
\(=(26)^{\frac{1}{3}}-3\)
\(=(26)^{\frac{1}{3}}=\Delta y+3\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x)\)
\(=\frac{1}{3(27)^{\frac{2}{3}}}(-1)\)
\(=\frac{-1}{27}\)
\(=0.0 \overline{370}\)
Therefore, the approximate value of \( (26)^{\frac{1}{3}} \) is \(2.9629 \).
1H. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (255)^{\frac{1}{4}} \)
\( (255)^{\frac{1}{4}} \)
Answer
Consider \( y=(x)^{\frac{1}{4}} \)
Let \( x=256 \) and \( \Delta x=-1 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}\)
\(=(225)^{\frac{1}{4}}-(256)^{\frac{1}{4}}\)
\(=(225)^{\frac{1}{4}}-4\)
\(=(225)^{\frac{1}{4}}=\Delta y+4\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)
\(=\frac{1}{4(256)^{\frac{3}{4}}}(-1)\)
\(=-0.0039\)
Therefore, the approximate value of \( (255)^{\frac{1}{4}} \) is \(3.9961 \).
1I. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (82)^{\frac{1}{4}} \)
\( (82)^{\frac{1}{4}} \)
Answer
Consider \( y=(x)^{\frac{1}{4}} \)
Let \( x=81 \) and \( \Delta x=1 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}\)
\(=(82)^{\frac{1}{4}}-(81)^{\frac{1}{4}}\)
\(=(82)^{\frac{1}{4}}-3\)
\(\Rightarrow(81)^{\frac{1}{4}}=\Delta y+3\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)
\(=\frac{1}{4(81)^{\frac{3}{4}}}(1)\)
\(=0.009\)
Therefore, the approximate value of \( (82)^{\frac{1}{4}} \) is \(3.009 \).
1j. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (401)^{\frac{1}{2}} \)
\( (401)^{\frac{1}{2}} \)
Answer
Consider \( y=(x)^{\frac{1}{2}} \)
Let \( x=400 \) and \( \Delta x=1 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{2}}-(x)^{\frac{1}{2}}\)
\(=(401)^{\frac{1}{2}}-(400)^{\frac{1}{2}}\)
\(=(401)^{\frac{1}{2}}-20\)
\(\Rightarrow(401)^{\frac{1}{2}}=\Delta y+20\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{\sqrt[2]{x}}(\Delta x)\)
\(=\frac{1}{2 \times 20}(1)\)
\(=\frac{1}{40}\)
\(=0.025\)
Therefore, the approximate value of \( (400)^{\frac{1}{2}} \) is \(20.025 \).
1K. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (0.0037)^{\frac{1}{2}} \)
\( (0.0037)^{\frac{1}{2}} \)
Answer
Consider \( y=(x)^{\frac{1}{2}} \)
Let \( x=0.0036 \) and \( \Delta x=0.0001 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{2}}-(x)^{\frac{1}{2}}\)
\(=(0.0037)^{\frac{1}{2}}-(0.0036)^{\frac{1}{2}}\)
\(=(0.0037)^{\frac{1}{2}}-0.06\)
\(\Rightarrow(0.0037)^{\frac{1}{2}}=\Delta y+0.06\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{\sqrt[2]{x}}(\Delta x)\)
\(=\frac{1}{2 \times 0.06}(0.0001)\)
\(=\frac{0.0001}{0.12}\)
\(=0.00083\)
Therefore, the approximate value of \( (0.0037)^{\frac{1}{2}} \) is \(0.06083 \).
ex 6.4 class 12 maths ncert solutions || class 12 maths exercise 6.4 || application of derivatives class 12 ncert solutions || class 12 maths ncert solutions chapter 6 exercise 6.4 || exercise 6.4 class 12 maths ncert solutions
1L. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (26.57)^{\frac{1}{3}} \)
\( (26.57)^{\frac{1}{3}} \)
Answer
Consider \( y=(x)^{\frac{1}{3}} \)
Let \( x=27 \) and \( \Delta x=-0.43 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{3}}-(x)^{\frac{1}{3}}\)
\(=(26.57)^{\frac{1}{3}}-(27)^{\frac{1}{3}}\)
\(=(26.57)^{\frac{1}{3}}-3\)
\(\Rightarrow(26.57)^{\frac{1}{3}}=\Delta y+3\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x)\)
\(=\frac{1}{3(27)^{\frac{2}{3}}}(-0.43)\)
\(=\frac{1}{3(3)^{2}}(-0.43)\)
\(=\frac{1}{3(9)}(-0.43)\)
\(=\frac{-0.43}{27}\)
\(=-0.015\)
Therefore, the approximate value of \( (26.57)^{\frac{1}{3}} \) is \(2.984 \).
1M. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (81.5)^{\frac{1}{4}} \)
\( (81.5)^{\frac{1}{4}} \)
Answer
Consider \( y=(x)^{\frac{1}{4}} \)
Let \( x=81 \) and \( \Delta x=0.5 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}\)
\(=(81.05)^{\frac{1}{4}}-(81)^{\frac{1}{4}}\)
\(=(81.05)^{\frac{1}{4}}-3\)
\(\Rightarrow(81.05)^{\frac{1}{4}}=\Delta y+3\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)
\(=\frac{1}{4(81)^{\frac{3}{4}}}(0.5)\)
\(=\frac{1}{4(3)^{3}}(0.5)\)
\(=\frac{1}{4(27)}(0.5)\)
\(=0.0046\)
Therefore, the approximate value of \( (81.05)^{\frac{1}{4}} \) is \(3.0046 \).
1N. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (3.962)^{\frac{3}{2}} \)
\( (3.962)^{\frac{3}{2}} \)
Answer
Consider \( y=(x)^{\frac{3}{2}} \)
Let \( x=4 \) and \( \Delta x=-0.032 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{3}{2}}-(x)^{\frac{3}{2}}\)
\(=(3.968)^{\frac{3}{2}}-(4)^{\frac{3}{2}}\)
\(=(3.968)^{\frac{3}{2}}-8\)
\(\Rightarrow(3.968)^{\frac{3}{2}}=\Delta y+8\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{3}{2}(x)^{\frac{1}{2}}(\Delta x)\)
\(=\frac{3}{2}(4)^{\frac{1}{2}}(-0.032)\)
\(=\frac{3}{2}(2)(-0.032)\)
\(=3(-0.032)\)
\(=-0.096\)
Therefore, the approximate value of \( (3.968)^{\frac{3}{2}} \) is \(7.904 \).
1O. Using differentials, find the approximate value of each of the following up to 3 places of decimal.
\( (32.15)^{\frac{1}{5}} \)
\( (32.15)^{\frac{1}{5}} \)
Answer
Consider \( y=(x)^{\frac{1}{5}} \)
Let \( x=32 \) and \( \Delta x=0.15 \). Then, we get
\(\Delta y=(x+\Delta x)^{\frac{1}{5}}-(x)^{\frac{1}{5}}\)
\(=(32.15)^{\frac{1}{5}}-(32)^{\frac{1}{5}}\)
\(=(32.15)^{\frac{1}{5}}-2\)
\(\Rightarrow(32.15)^{\frac{1}{5}}=\Delta y+2\)
Now, \(dy\) is approximately equal to \( \Delta y \) and is given by:
\(\mathrm{dy}=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{5(x)^{\frac{4}{5}}}(\Delta x)\)
\(=\frac{1}{5(32)^{\frac{4}{5}}}(0.15)\)
\(=\frac{1}{5(2)^{4}}(0.15)\)
\(=\frac{1}{5(16)}(0.15)\)
\(=0.00187\)
Therefore, the approximate value of \( (32.15)^{\frac{1}{5}} \) is \(2.00187 \).
2. Find the approximate value of \( \mathrm{f}(2.01) \), where \( \mathrm{f}(x)=4 x^{2}+5 x+2 \).
Answer
Let \( x=2 \) and \( \Delta x=0.01 \). Then, we get,
\( \mathrm{f}(2.01)=\mathrm{f}(x+\Delta x)=4(x+\Delta x)^{2}+5(x+\Delta x)+2 \)
Now, \( \Delta y=\mathrm{f}(x+\Delta x)-\mathrm{f}(x) \)
\(\Rightarrow \mathrm{f}(x+\Delta x)=\mathrm{f}(x)+\Delta y\)
\(\approx \mathrm{f}(x)+\mathrm{f}^{\prime}(x) \cdot \Delta x(\text { as }dx =\Delta x)\)
\(\Rightarrow \mathrm{f}(2.01) \approx\left(4 x^{2}+5 x+2\right)+(8 x+5) \Delta x\)
\(=\left[4(2)^{2}+5(2)+2\right]+[8(2)+5](0.01)\)
\(=(16+10+2)+(16+5)(0.01)\)
\(=28+0.21\)
\(=28.21\)
Therefore, the approximate value of \( f(2.01) \) is \(28.21 \).
3. Find the approximate value of \( f(5.001) \), where \( f(x)=x^{3}-7 x^{2}+15 \).
Answer
Let \( x=5 \) and \( \Delta x=0.001 \). Then, we get,
\(f(5.001)=f(x+\Delta x)=(x+\Delta x)^{3}-7(x+\Delta x)^{2}+15\)
Now, \( \Delta y=f(x+\Delta x)-f(x) \)
\(\Rightarrow \mathrm{f}(x+\Delta x)=\mathrm{f}(x)+\Delta y\)
\(\approx \mathrm{f}(x)+\mathrm{f}^{\prime}(x) \cdot \Delta x(\text { as } \mathrm{dx}=\Delta x)\)
\(\Rightarrow \mathrm{f}(5.001) \approx\left(x^{3}-7 x^{2}+15\right)+\left(3 x^{2}-14 x\right) \Delta x\)
\(=\left[(5)^{2}-7(5)^{2}+15\right]+\left[3(5)^{2}-14(5)\right](0.001)\)
\(=(125-175+15)+(75-70)(0.001)\)
\(=-35+0.005\)
\(=-34.995\)
Therefore, the approximate value of \( \mathrm{f}(5.001) \) is \(-34.995 \).
4. Find the approximate change in the volume V of a cube of side \(x\) metres caused by increasing the side by \( 1 \% \).
Answer
It is given that the volume of a cube \( (V) \) of a side \(x\)
\(\Rightarrow \mathrm{V}=x^{3}\)
\(\Rightarrow \mathrm{dV}=\left(\frac{d V}{d x}\right) \Delta x\)
\(=\left(3 x^{2}\right) \Delta x\)
\(=\left(3 x^{2}\right)(0.01 x)\)
\(=0.03 x^{3}\)
Therefore, the approximate change in the volume \(V\) of a cube of side \(x\) metres caused by increasing the side by \( 1 \% \) is \( 0.03 x^{3} \).
5. Find the approximate change in the surface area of a cube of side \( x \) metres caused by decreasing the side by \( 1 \% \).
Answer
It is given that the surface area of cube (S) of a side \( x \)
\(\Rightarrow S=6 x^{2}\)
\(\Rightarrow \mathrm{dS}=\left(\frac{d S}{d x}\right) \Delta x\)
\(=(12 x) \Delta x\)
\(=(12 x)(0.01 x)\)
\(=0.12 x^{2}\)
Therefore, the approximate change in the surface area of a cube of side \(x\) metres caused by decreasing the side by \( 1 \% \) is \( 0.12 x^{2} \).
6. If the radius of a sphere is measured as 7 m with an error of 0.02 m , then find the approximate error in calculating its volume.
Answer
Let \( r \) be the radius of the sphere and \( \Delta r \) be the error in measuring the radius.
Now, it is given that \( \mathrm{r}=7 \mathrm{~m} \) and \( \Delta \mathrm{r}=0.02 \mathrm{~m} \)
We know that volume of sphere \( (\mathrm{V})=\frac{4}{3} \pi r^{3} \)
Now, \( \frac{d V}{d r}=4 \pi r^{2} \)
\(\Rightarrow d V=\left(\frac{d V}{d r}\right) \Delta r\)
\(=\left(4 \pi r^{2}\right) \Delta r\)
\(=4 \pi\left(7^{2}\right)(0.02) \mathrm{m}^{3}\)
\(=3.92 \pi \mathrm{m}^{3}\)
Therefore, the approximate error in calculating its volume is \( 3.92 \pi \mathrm{m}^{3} \).
7. If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Answer
Let \( r \) be the radius of the sphere and \( \Delta r \) be the error in measuring the radius.
Now, it is given that \( \mathrm{r}=9 \mathrm{~m} \) and \( \Delta \mathrm{r}=0.03 \mathrm{~m} \)
We know that surface area of sphere \( (S)=4 \pi r^{2} \)
Now, \( \frac{d s}{d r}=8 \pi r \)
\(\Rightarrow \mathrm{dS}=\left(\frac{d S}{d r}\right) \Delta r\)
\(=(8 \pi \mathrm{r}) \Delta \mathrm{r}\)
\(=8 \pi(9)(0.03) \mathrm{m}^{2}\)
\(=2.16 \pi \mathrm{m}^{3}\)
Therefore, the approximate error in calculating its surface area is \( 2.16 \pi \) \( \mathrm{m}^{3} \).
8. If \( f(x)=3 x^{2}+15 x+5 \), then the approximate value of \( f(3.02) \) is
A. 47.66 B. 57.66 C. 67.66 D. 77.66
A. 47.66 B. 57.66 C. 67.66 D. 77.66
Answer
Let \( x=3 \) and \( \Delta x=0.02 \). Then, we get,
\(f(3.02)=f(x+\Delta x)=3(x+\Delta x)^{2}+15(x+\Delta x)+5\)
Now, \( \Delta y=\mathrm{f}(x+\Delta x)-\mathrm{f}(x) \)
\(\Rightarrow \mathrm{f}(x+\Delta x)=\mathrm{f}(x)+\Delta y\)
\(\approx \mathrm{f}(x)+\mathrm{f}^{\prime}(x) \cdot \Delta x(\text { as } \mathrm{dx}=\Delta x)\)
\(\Rightarrow \mathrm{f}(3.02) \approx\left(3 x^{2}+15 x+5\right)+(6 x+15) \Delta x\)
\(=\left[3(3)^{2}+15(3)+5\right]+[6(3)+15](0.02)\)
\(=(27+45+5)+(18+15)(0.02)\)
\(=77+0.66\)
\(=77.66\)
Therefore, the approximate value of \( \mathrm{f}(3.02) \) is \(77.66 \).
9. The approximate change in the volume of a cube of side \(x\) metres caused by increasing the side by \( 3 \% \) is
A. \( 0.06 x^{3} \mathrm{~m}^{3} \) B. \( 0.6 x^{3} \mathrm{~m}^{3} \) C. \( 0.09 x^{3} \mathrm{~m}^{3} \) D. \( 0.9 x^{3} \mathrm{~m}^{3} \)
A. \( 0.06 x^{3} \mathrm{~m}^{3} \) B. \( 0.6 x^{3} \mathrm{~m}^{3} \) C. \( 0.09 x^{3} \mathrm{~m}^{3} \) D. \( 0.9 x^{3} \mathrm{~m}^{3} \)
Answer
It is given that the volume of a cube \( (V) \) of a side \(x\)
\(\Rightarrow \mathrm{V}=x^{3}\)
\(\Rightarrow \mathrm{dV}=\left(\frac{d V}{d x}\right) \Delta x\)
\(=\left(3 x^{2}\right) \Delta x\)
\(=\left(3 x^{2}\right)(0.03 x)\)
\(=0.09 x^{3}\)
Therefore, the approximate change in the volume of a cube of side \(x\) metres caused by increasing the side by \( 3 \% \) is \( 0.09 x^{3} \).
