Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.4 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4
(i) \(\sqrt{\mathbf{2 5 . 3}}\)
(ii) \(\sqrt{49.5}\)
(iii) \(\sqrt{0.6}\)
(iv) \((0.009)^{\frac{1}{3}}\)
(v) \((0.999)^{\frac{1}{10}}\)
(vi) \((15)^{\frac{1}{4}}\)
(vii) \((26)^{\frac{1}{3}}\)
(viii) \((255)^{\frac{1}{4}}\)
(ix) \((82)^{\frac{1}{4}}\)
(x) \((401)^{\frac{1}{2}}\)
(xi) \((0.0037)^{\frac{1}{2}}\)
(xii) \((\mathbf{2 6 . 5 7})^{\frac{1}{3}}\)
(xiii) \((81.5)^{\frac{1}{4}}\)
(xiv) \((3.968)^{\frac{3}{2}}\)
(xv) \((32.15)^{\frac{1}{5}}\)
Solution.
(i) \(\sqrt{25.3}\)
Consider y = √x. Let x = 25 and ∆x = 0.3.
Then ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}\)
= \(\sqrt{25.3}-\sqrt{25}=\sqrt{25.3}-5\)
⇒ √253 = ∆y + 5
Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right)\) ∆x
= \(\frac{1}{2 \sqrt{x}}\) (0.3) = 0.03
Hence, the approximate value of √25.3 is 0.03 + 5 = 5.03.
(ii) √49.5
Consider y = √x.
Let x = 49 and ∆x = 0.5
Then, ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}\)
= \(\sqrt{49.5}-\sqrt{49}=\sqrt{49.5}-7\)
⇒ √49.5 = 7 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(0.5)\) [as y = √x]
= \(\frac{1}{2 \sqrt{49}}\) (0.5)
= \(\frac{1}{14}\) (0.5) = 7.035.
(iii) √0.6
Consider y = √x.
Let x = 1 and ∆x = – 0.4.
Then, ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{0.6}-1\)
⇒ √0.6 = 1 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = (\(\frac{d y}{d x}\)) ∆x = \(\frac{1}{2 \sqrt{x}}\) (∆x) [as y = √x]
= \(\frac{1}{2}\) (- 0.4) = – 0.2
Hence, the approximate value of √0.6 is 1 + (- 0.2) = 1 – 0.2 = 0.8.
(iv) \((0.009)^{\frac{1}{3}}\)
Consider y = \((x)^{\frac{1}{3}}\).
Let x = 0.008 and ∆x = 0.001.
Then, ∆y = (x + ∆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)\(\frac{1}{3}\) = 0.2 + ∆Y
Now, dy is approximately equal ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{1}{3(x)^{\frac{2}{3}}}\) ∆x [as y = x\(\frac{1}{3}\)]
= \(\frac{1}{3 \times 0.04}(0.001)=\frac{0.001}{0.12}\) = 0.008
Hence, the approximate value of \((0.009)^{\frac{1}{3}}\) is 0.2 + 0.008 = 0.208.
(v) \((0.999)^{\frac{1}{10}}\)
Consider y = \((x^{\frac{1}{10}}\).
Let x = 1 and ∆x = – 0.001.
Then, ∆y = (x + ∆x)\(\frac{1}{10}\) – (x)\(\frac{1}{10}\)
= (0.999)\(\frac{1}{10}\) – 1
= (0.999)\(\frac{1}{10}\)
=1 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{10}\)]
= \(\frac{1}{10(x)^{\frac{9}{10}}}\) (∆x)
= \(\frac{1}{10}\) (- 0.001)
= – 0.0001
Hence, the approximate value of (0.999)\(\frac{1}{10}\) is 1 + (- 0.0001) = 0.0999.
(vi) \((15)^{\frac{1}{4}}\)
Consider y = (x)\(\frac{1}{4}\).
Let x = 16 and ∆x = – 1.
Then, ∆y = (x + ∆x)\(\frac{1}{4}\) – (x)\(\frac{1}{4}\)
= (15)\(\frac{1}{4}\) – (16)\(\frac{1}{4}\)
= (15)\(\frac{1}{4}\) – 2
⇒ (15)\(\frac{1}{4}\) = 2 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{4(16)^{\frac{3}{4}}}(-1)=\frac{-1}{4 \times 8}=\frac{-1}{32}\)
= – 0.03125
Hence, the approximate value of \((15)^{\frac{1}{4}}\) is 2 + (- 0.03125) = 1.96875.
(vii) \((26)^{\frac{1}{3}}\)
Consider y = (x)\(\frac{1}{3}\).
Let x = 27 and ∆x = – 1
Then, ∆y= (x + ∆x)\(\frac{1}{3}\) – (x)\(\frac{1}{3}\)
= (26)\(\frac{1}{3}\) – (27)\(\frac{1}{3}\)
= (26)\(\frac{1}{3}\) – 3
⇒ (26)\(\frac{1}{3}\) = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{3}\)]
= \(\frac{1}{3(x)^{\frac{2}{3}}}\) (∆x)
= \(\frac{1}{3(27)^{\frac{2}{3}}}(-1)=\frac{-1}{27}\)
= – 0.0370.
Hence, the approximate value of \((26)^{\frac{1}{3}}\) is 3 + (- 0.03125) = 2.9629.
(viii) \((255)^{\frac{1}{4}}\)
Consider y = (x)\(\frac{1}{4}\).
Let x = 256 and ∆x = – 1.
Then, ∆y = (x + ∆x)\(\frac{1}{4}\) – (x)\(\frac{1}{4}\)
= (255)\(\frac{1}{4}\) – (256)\(\frac{1}{4}\)
= (255)\(\frac{1}{4}\) – 4
⇒ (255)\(\frac{1}{4}\) = 4 + ∆y
Now, dy is approximately equal to zy and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)
= \(\frac{1}{4(256)^{\frac{3}{4}}}(-1)=\frac{-1}{4 \times 4^{3}}\)
= – 0.0039.
Hence, the approximate value of (255)\(\frac{1}{4}\) is 4 + (- 00039) = 3.9961.
(ix) \((82)^{\frac{1}{4}}\)
Consider y = (x)\(\frac{1}{4}\).
Let x = 81 and ∆x = 1.
Then, ∆y = (x + ∆x)\(\frac{1}{4}\) – (x)\(\frac{1}{4}\)
= (82)\(\frac{1}{4}\) – (81)\(\frac{1}{4}\)
= (82)\(\frac{1}{4}\) – 3
(82)\(\frac{1}{4}\) = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{4(x)^{\frac{3}{4}}}\) (∆x)
= \(\frac{1}{4(81)^{\frac{3}{4}}}(1)=\frac{1}{4(3)^{3}}=\frac{1}{108}\)
= 0.009
Hence, the approximate value of \((82)^{\frac{1}{4}}\) is 3 + 0.009 = 3.009.
(x) \((401)^{\frac{1}{2}}\)
Consider y = (x)\(\frac{1}{4}\).
Let x = 400 and ∆x = 1.
Then, ∆y = \(\sqrt{x+\Delta x}-\sqrt{x}\)
= \(\sqrt{401}-\sqrt{400}=\sqrt{401}-20\)
⇒ √401 = 20 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(\Delta x)\) [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{2 \times 20}(1)=\frac{1}{40}\)
= 0.025
Hence, the approximate value of √401 is 20 + 0.025 = 20.025.
(xi) \((0.0037)^{\frac{1}{2}}\)
Consider y = (x)\(\frac{1}{2}\).
Let x = 0.0036 and ∆x = 0.0001.
Then,
∆y = (x + ∆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
(0.0037)\(\frac{1}{2}\) = 0.06 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{2 \sqrt{x}}(\Delta x)\)
= \(\frac{1}{2 \times 0.06}(0.0001)\)
= \(\frac{0.0001}{0.12}\) = 0.00083
Hence, the approximate value of \((0.0037)^{\frac{1}{2}}\) is 0.06 + 0.00083 = 0.06083.
(xii) (26.57)\(\frac{1}{3}\)
Consider y = (x)\(\frac{1}{3}\).
Let x = 27 and ∆x = – 0.43.
Then, ∆y = (x + ∆x)\(\frac{1}{3}\) – (x)\(\frac{1}{3}\)
= (26.57)\(\frac{1}{3}\) – (27)\(\frac{1}{3}\)
= (26.57)\(\frac{1}{3}\) – 3
⇒ (26.57)\(\frac{1}{3}\) = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x)\)
= \(\frac{1}{3(9)}(-0.43)=\frac{-0.43}{27}\) = – 0.016
Hence, the approximate value of (26.57)\(\frac{1}{3}\) is 3 + (- 0.0 16) = 2.984.
(xiii) \((81.5)^{\frac{1}{4}}\)
Consider y = (x)\(\frac{1}{4}\).
Let x = 81 and Ax 0.5.
Then, Ay = (x + &)\(\frac{1}{4}\) – (x)\(\frac{1}{4}\)
= (81.5)\(\frac{1}{4}\) – (81)\(\frac{1}{4}\)
= (81.5)\(\frac{1}{4}\) – 3
⇒ \((81.5)^{\frac{1}{4}}\) = 3 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{1}{4(x)^{4}}(\Delta x)\)
= \(\frac{1}{4(3)^{3}}(0.5)=\frac{0.5}{108}\)
= 0.0046
Hence, the approximate value of (81.5)\(\frac{1}{4}\) is 3 + (0.0046) = 3.0046.
(xiv) \((3.968)^{\frac{3}{2}}\)
Consider y = (x)\(\frac{3}{2}\).
Let x = 4 and ∆x = – 0.032.
Then, ∆y = (x + ∆x)\(\frac{3}{2}\) – (x)\(\frac{3}{2}\)
= (3.968)\(\frac{3}{2}\) – (4)\(\frac{3}{2}\)
= (3.968)\(\frac{3}{2}\) – 8
⇒ (3.968)\(\frac{3}{2}\) = 8 + ∆y
Now, dy is approximately equal ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{4}\)]
= \(\frac{3}{2}\) (2) (- 0.032)
= – 0.096
Hence, the approximate value of \((3.968)^{\frac{3}{2}}\) is 8 + (- 0.096) = 7.904.
(xv) \((32.15)^{\frac{1}{5}}\)
Consider y = (x)\(\frac{1}{5}\).
Let x = 32 and ∆x = 0.15
Then, ∆y = \((x+\Delta x)^{\frac{1}{5}}-(x)^{\frac{1}{5}}\)
= \((32.15)^{\frac{1}{5}}-(32)^{\frac{1}{5}}=(32.15)^{\frac{1}{5}}-2\)
⇒ (32.15)\(\frac{1}{5}\) = 2 + ∆y
Now, dy is approximately equal to ∆y and is given by
dy = \(\frac{d y}{d x}\) ∆x [as y = x\(\frac{1}{5}\)]
= \(\frac{1}{5(x)^{\frac{4}{5}}}(\Delta x)\)
= \(\frac{1}{5 \times(2)^{4}}(0.15)=\frac{0.15}{80}\)
= 0.00187
Hence, the approximate value of \((32.15)^{\frac{1}{5}}\) is 2 + 0.00187 = 2.00187.
Question 2.
Find the approximate value of f(2.01), where f(x) = 4x2 + 5x + 2.
Solution.
Let x = 2 and ∆x = 0.01
Then, we have f(2.01) = f(x + ∆x)
= 4(x + ∆x)2 + 5(x + ∆x) + 2
Now, ∆y = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
~ f(x) + f'(x) . ∆x [as dx = ∆x]
⇒ f(2.01) = (4x2 + 5x + 2) + (8x + 5)∆x
= [4(2)2 + 5(2) + 2] + [8(2) + 5] (0.01) [as x = 2, ∆x = 0.01]
= (16 + 10 + 2) + (16 + 5) (0.01)
= 28 + (21) (0.01)
= 28 + 0.21 = 28.21
Hence, the approximate value of f(2.01) is 28.21.
Question 3.
Find the approximate value of f(5.001), where f(x) = x3 – 7x2 +15. ,
Solution.
Let x = 5 and ∆x = 0.001
Then, we have f(5.001) = f(x + ∆x)
= (x + ∆x)3 – 7(x + ∆x)2 + 15
Now, ∆y = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
~ f(x) + f'(x) . ∆x [as dx = ∆x]
⇒ f(5.001) = (x3 – 7x2 + 15) + (3x2 – 14x)∆x
= [(5)3 – 7(5)2 + 15] + [3(5)2 – 14(5)] (0.001) [as x = 5, ∆x = 0.001]
= (125 – 175 + 15) + (75 – 70) (0.001)
= – 35 + (5) (0.001)
= – 35 + 0.005
= – 34.995
Hence, the approximate value of f(5.001) is – 34.995.
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 = x3.
dV = (\(\frac{d V}{d x}\)) ∆x
= (3x2) ∆x
= (3x2) (0.01 x) [as 1% of x = 0.01 x]
= 0.03 x3
Hence, the approximate change in the volume of the cube is 0.03 x3 m3.
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 = 6x2.
∴ \(\frac{d S}{d x}=\left(\frac{d S}{d x}\right) \Delta x\)
= (12 x) ∆x
= (12 x) (- 0.01 x)[as 1% of x = 0.01 x]
= – 0.12 x2
Hence, the approximate change in the surface area of the cube is 0.12x2 m2.
Question 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.
Solution.
Let r be the radius of the sphere and ∆r be the error in measuring the radius.
Then, r = 7 m and ∆r = 0.02 m
Now, the volume V of the sphere is given by
V = \(\frac{4}{3}\) πr3
⇒ \(\frac{d V}{d r}\) = 4πr2
⇒ dV = (\(\frac{d V}{d r}\)) ∆r
= (4πr2) ∆r
= 4π(7)2 (0.02) m3
= 3.92 π m3
Hence, the approximate error in calculating the volume is 3.92 π m3.
Question 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.
Solution.
Let r be the radius of the sphere and ∆r be the error in measuring the radius.
Then, r = 9 m and ∆r = 0.03 m
Now, the surface area of the sphere (S) is given by
S = 4πr2
⇒ \(\frac{d S}{d r}\) = 8πr
⇒ dS = (\(\frac{d S}{d r}\)) ∆r
= (8πr) ∆r
= 8π (9) (0.03) m2
= 2.16K m2
Hence, the approximate error in calculating the surface area is 2.16 π m2.
Question 8.
If fix) = 3x2 + 15x + 5, then the approximate value of f(3.02) is
(A) 47.66
(B) 57.66
(C) 67.66
(D) 77.66
Solution.
Let x = 3 and ∆x = 0.02. Then, we have
f(3.02) = f(x + ∆x)
= 3(x + ∆x)2 + 15(x + ∆x) + 5
Now, ∆y = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
≈ f(x) + f'(x)∆x [as dx = ∆x]
⇒ f(3.02) = (3x2 + 15x + 5) + (6x + 15) ∆x
= [3(3)2 +15(3) + 5] + [6(3) +15] (0.02) [as x = 3, ∆x = 0.02]
= (27 + 45 + 5) + (18 +15) (0.02)
= 77 + (33) (0.02) = 77 + 0.66 = 77.66
Hence, the approximate value of f(3.02) is 77.66.
The correct answer is (D).
Question 9.
The approximate change in the volume of a cube of side x metres caused by increasing the side of 3% is
(A) 0.06 x3 m3
(B) 0.6 x3 m3
(C) 0.09 x3 m3
(D) 0.9 x3 m3
Solution.
The volume of a cube (V) is side x is given by V = x3.
∴ dV = (\(\frac{d V}{d x}\)) ∆x
= (3x2) ∆x
= (3x2) (0.03 x) [as 3% of x = 0.03x]
= 0.09 x3 m3.
Hence, the approximate change in the volume of the cube is 0.09 x3 m33.
The correct answer is (C).