Class 12 Maths Chapter 6 Application of Derivatives – Exercise 6.5 NCERT Solutions

Introduction

Exercise 6.5 focuses on approximations using derivatives. Students learn how to use linear approximation and differentials to estimate values of functions near a given point. This exercise is important for practical applications where exact computation is difficult.

Formulas Used

1. Linear Approximation: If $f(x)$ is differentiable at $x=a$, then for small $h$:

$$f(a+h)\approx f(a)+h{f}^{\mathrm{\prime }}(a)$$

2. Differentials:

$$dy=f^{\mathrm{\prime }}(x)dx$$

Used to approximate change in $y$ for small change in $x$.

Students Frequently Make Mistakes

• Forgetting to apply derivative at the correct point.

• Using large $h$ values where approximation fails.

• Confusing exact value with approximate value.

• Errors in simplification of linear approximation.

• Skipping domain restrictions for functions.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Approximate $\sqrt{4.1}$ using derivatives. Let $f(x)=\sqrt{x},a=4$.

$$f^{\mathrm{\prime }}(x)=\frac{1}{2\sqrt{x}},f^{\mathrm{\prime }}(4)=\frac{1}{4}$$

$$f(4.1)\approx f(4)+0.1\cdot f^{\mathrm{\prime }}(4)=2+0.1\cdot \frac{1}{4}=2.025$$

Q2. Approximate $\sqrt{9.2}$. $f(x)=\sqrt{x},a=9$.

$$f^{\mathrm{\prime }}(9)=\frac{1}{6},f(9.2)\approx 3+0.2\cdot \frac{1}{6}=3.033$$

Q3. Approximate $\sqrt{15.9}$. $f(x)=\sqrt{x},a=16$.

$$f^{\mathrm{\prime }}(16)=\frac{1}{8},f(15.9)\approx 4-0.1\cdot \frac{1}{8}=3.9875$$

Q4. Approximate $(1.02{)}^5$. $f(x)=x^5,a=1$.

$$f^{\mathrm{\prime }}(x)=5x^4,f^{\mathrm{\prime }}(1)=5$$

$$f(1.02)\approx 1+0.02\cdot 5=1.1$$

Q5. Approximate $(0.99{)}^3$. $f(x)=x^3,a=1$.

$$f^{\mathrm{\prime }}(1)=3$$

$$f(0.99)\approx 1-0.01\cdot 3=0.97$$

Q6. Approximate $\ln (1.1)$. $f(x)=\ln x,a=1$.

$$f^{\mathrm{\prime }}(1)=1$$

$$f(1.1)\approx 0+0.1\cdot 1=0.1$$

Q7. Approximate $\ln (0.9)$. $f(x)=\ln x,a=1$.

$$f^{\mathrm{\prime }}(1)=1$$

$$f(0.9)\approx 0-0.1\cdot 1=-0.1$$

Q8. Approximate $e^{0.01}$. $f(x)=e^x,a=0$.

$$f^{\mathrm{\prime }}(0)=1$$

$$f(0.01)\approx 1+0.01=1.01$$

Q9. Approximate $e^{1.02}$. $f(x)=e^x,a=1$.

$$f^{\mathrm{\prime }}(1)=e$$

$$f(1.02)\approx e+0.02e=e(1.02)$$

Q10. Approximate $\cos (0.1)$. $f(x)=\cos x,a=0$.

$$f^{\mathrm{\prime }}(0)=-\sin 0=0$$

$$f(0.1)\approx f(0)+0.1\cdot 0=1$$

FAQs (10)

FAQ1. What is linear approximation? Method to estimate function value near a point using tangent line.

FAQ2. What is differential? Small change in function value approximated by derivative.

FAQ3. Why use approximation? Exact computation may be complex.

FAQ4. When is approximation valid? For small changes near known point.

FAQ5. What is derivative role in approximation? It gives slope of tangent line.

FAQ6. Can approximation be exact? Yes, if function is linear.

FAQ7. Why choose nearby point $a$? To simplify calculation.

FAQ8. Is approximation always accurate? No, depends on size of change.

FAQ9. What is practical use of approximation? Physics, engineering, quick estimates.

FAQ10. Why is Exercise 6.5 important? It builds foundation for real‑life applications of calculus.

Conclusion

Exercise 6.5 has 10 solved questions and 10 FAQs that strengthen your understanding of approximations using derivatives. This completes the Applications of Derivatives chapter in Class 12 Maths.

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