2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.5

Students can Download Basic Maths Exercise 18.5 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.5

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.5 Two or Three Marks Questions and Answers

Question 1.
\(x^{\sqrt{x}}\)
Answer:
Let y = \(x^{\sqrt{x}}\) taking log^m both sides
log y = \(\sqrt{x} \cdot \log x\)

Question 2.
x^{sin x}
Answer:
Let y = x^{sin x}, taking log^m both sudes
log y = sinx log x, Diff w.r.t x
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = sin x . \(\frac{1}{x}\) + log x . cos x
\(\frac{d y}{d x}\) = y[\(\frac{\sin x}{x}\) + log x . cos x]

Question 3.
(sin x)^x.
Answer:
Let y = (sin x)^x, Taking log^m
log y = x log sin x
\(\frac{1}{y} \frac{d y}{d x}\) = x . \(\frac{1}{\sin x}\) . cos x + logsin x.1
\(\frac{d y}{d x}\) = y[x cot x + logsin x]

Question 4.
x^{5+log x}
Answer :
Let y = x^{5+log x}, Taking log^m both sides
log y = (5 + log x). log x,
Differentiate W.r.t x
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = (5 + log x) . \(\frac{1}{x}\) + log(x) . \(\frac{1}{x}\)
\(\frac{d y}{d x}\) = y \(\left[\frac{5+\log x+\log x}{x}\right]\)
= y\(\left[\frac{5+2 \log x}{x}\right]\)

Question 5.
x^{(sinx – cosx)}.
Answer:
Let y = x^{(sinx – cosx)} ; log y = (sin x – cos x) log x diff w.r.t x.
\(\frac{1}{y} \frac{d y}{d x}\) = (sinx – cosx) . \(\frac{1}{x}\) + log x (cos x + sin x);
\(\frac{d y}{d x}\) = y [\(\frac{\sin x-\cos x}{x}\) + log x.(cos x + sin x)]

Part – B

2nd PUC Basic Maths Differential Calculus Ex 18.5 Five Marks Questions and Answers

Question 1.
x^logx + (log x)^x
Answer:
Let y = x^logx + (log x)^x
y = u + v
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) …(1)
Where u = x and v = (log x)^x
Taking log^m both sides
log u = log x . log x log v = x log (log x)
Differentiate both w.r.t. x

Question 2.
x² . e^x2 . log x.
Answer:
Let y = x² . e^x2 . log x. taking log^m both sides
log y = log(x² . e^x2 . log x) log y = log (x² . e^x2 . log x) differentiate w.r.t x
log y = log x² + log e^x2 + log(log x)

Question 3.
(x + 1)² (x + 2)³ (x + 3)⁴.
Answer:
Let y = (x + 1)² (x + 2)³ (x + 3)⁴, taking log^m
log y = log(x + 1)² + (x + 2)³ + (x + 3)⁴
log y = 2 log (x + 1) + 3 log (x + 2) + 4 log (x + 3)
Differentiate w.r.t x

Question 4.
x^2x + x^x2.
Answer:
Let y = x^2x + x^x2
y = u + v
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ….(1)
Where u = x^2x and v = x^x2
Taking log^m both sides
log u = 2x logx logv = x² log x
Differentiate both w.r.t x

Question 5.
\(=x^{\left(1+\frac{1}{x}\right)}+\left(1+\frac{1}{x}\right)^{x}\).
Answer:
Let y = \(=x^{\left(1+\frac{1}{x}\right)}+\left(1+\frac{1}{x}\right)^{x}\) y = u + v \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Where u = \(x^{1+\frac{1}{x}}\) v = \(\left(1+\frac{1}{x}\right)^{x}\)
Taking log^m both sides

Question 6.
\(\sqrt{\frac{\sqrt{(x-1)(x-2)}}{(x-3)(x-4)(x-5)}}\)
Answer:
Let y = \(\sqrt{\frac{\sqrt{(x-1)(x-2)}}{(x-3)(x-4)(x-5)}}\)
Taking log^m both sides
log y = \(\frac { 1 }{ 2 }\) [log (x – 1) + log (x – 2) – log (x – 3) – log (x – 4) – log (x – 5)]

Question 7.
x³ . e^2x . sec²x.
Answer:
Let y = x³ . e^2x . sec²x.
Taking log^m boyh sides
log y = log x³ + log ^2x + log sec²x
Differentiate w.r.t x → log y = 3 log x + 2x log e + 2 log sec x

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