CALCULAS 1ST ORDER DERIVATIVE

CALCULAS 1ST ORDER DERIVATIVE Part II

DIFFERENTIATION SOLUTION OF SNDEY SEM-3

অবকলন SEMESTER-3 Unit-3 CALCULAS 1st order derivative

1ST ORDER DERIVATIVE Part I দেখতে এখানে CLICK করো।

Fill in the Blanks

1. ^d/_dx cos(logx) = _______
Ⓐ – sin(logx) Ⓑ ¹/_x cos(logx)
Ⓒ –¹/_x sin(logx) Ⓓ ¹/_x sin(logx)

Solution: ^d/_dx cos(logx)
= – sin(logx). ^d/_dx(logx)
= – sin(logx). ¹/_x = –¹/_x sin(logx)
Ans: Ⓒ –¹/_x sin(logx)

2. ^d/_dxlog(sin x) = _______
Ⓐ tan x Ⓑ cot x Ⓒ – tan x Ⓓ – cot x

Solution: ^d/_dxlog(sin x)
= ¹/_sinx. ^d/_dx(sin x)
= ¹/_sinx.cosx = cotx
Ans: Ⓑ cot x

3. ^d/_dx(cot x°) = _______
Ⓐ -cosec³ x° Ⓑ cosec³ x°
Ⓒ –¹⁸⁰/_πcosec² x°
Ⓓ –^π/₁₈₀cosec² x°

Solution: ^d/_dx(cot x°)
=^d/_dx(cot ^πx/₁₈₀)
= – cosec² ^πx/₁₈₀. ^π/₁₈₀
= –^π/₁₈₀cosec² x°
Ans: Ⓓ –^π/₁₈₀cosec² x°

4. y = 10^{10^x} হলে, ^dy/_dx = _______
Ⓐ 10^{10^x}(log_e10)² Ⓑ 10^{10^x}log_e10
Ⓒ 10^{10^x}.10^xlog_e10 Ⓓ 10^{10^x}.10^x(log_e10)²

Solution: y = 10^{10^x}
∴^dy/_dx = 10^{10^x}.log_e10.^d/_dx(10^x)
= 10^{10^x}.log_e10.10^x.log_e10
= 10^{10^x}.10^x(log_e10)²
Ans: Ⓓ 10^{10^x}.10^x(log_e10)²

\(5.\ f(x) = \sqrt{x^2 + 1}\) হলে \(f'(-1) =\) _______

Ⓐ ¹/_√2 Ⓑ –¹/_√2
Ⓒ 1 Ⓓ -1

\(Solution:\ f(x) = \sqrt{x^2 + 1}\\∴ f'(x) =\frac{1}{2\sqrt{x^2 + 1}}.\frac{d}{dx}(x^2 + 1)\\=\frac{1}{2\sqrt{x^2 + 1}}.2x\\=\frac{x}{\sqrt{x^2 + 1}}\\∴ f'(-1) =\frac{-1}{\sqrt{(-1)^2 + 1}}=\frac{-1}{√2}\\Ans:\ Ⓑ\ \frac{-1}{√2}\)

6. y = ¹/_{3x – 1} হলে, x = 0 বিন্দুতে ^dy/_dx = _______
Ⓐ -3 Ⓑ ¹/₃
Ⓒ –¹/₉Ⓓ 3

Solution: y = ¹/_{3x – 1} = (3x – 1)^-1
∴ ^dy/_dx =-1(3x – 1)^-2.^d/_dx(3x – 1)
=-1(3x – 1)^-2.3
x = 0 বিন্দুতে,
^dy/_dx = -1(3.0 – 1)^-2.3
= -1(-1)^-2.3 = -3
Ans: Ⓐ -3

7. f(x) = log(tan^x/₂) হলে f'(x) = _______
Ⓐ sin x Ⓑ ¹ / _{2sin^x/₂cos ^x/₂}
Ⓒ -cosecx Ⓓ cosecx

Solution: f(x) = log(tan^x/₂)
∴ f'(x) = ¹ / _tan^x/₂ .sec² ^x/₂ .¹/₂
= ¹ / _{2sin^x/₂cos ^x/₂}
= ¹/_sinx = cosecx
Ans: Ⓓ cosecx

8. logx-এর সাপেক্ষে x² -এর অবকল সহগ = _______
Ⓐ 2x²Ⓑ 2
Ⓒ 2x Ⓓ 2x³

Solution: ^d/_d(logx)(x²)
= ^d/_dx(x²).¹/_{^d/_dx(logx)}
= 2x.¹/_{¹/_x} = 2x²
Ans: Ⓐ 2x²

9. যদি y = cosec³x হয়, তবে ^dy/_dx = _______
Ⓐ – 3cosecxcot x
Ⓑ – 3cosec³xcot³ x
Ⓒ – cosec³xcot x
Ⓓ – 3cosec³xcot x

Solution: y = cosec³x
∴ ^dy/_dx = 3cosec²x.^d/_dx(cosecx)
= 3cosec²x.(- cosecxcotx)
= – 3cosec³xcot x
Ans: Ⓓ – 3cosec³xcot x

10. যদি y = m^{ax2 + bx + c} হয়, তবে ^dy/_dx = _______
Ⓐ (2ax + b)m^{ax2 + bx + c}.log_em
Ⓑ (2ax + b)m^{ax2 + bx + c}
Ⓒ m^{ax2 + bx + c}.log_em
Ⓓ (2ax + b).log_em

Solution: y = m^{ax2 + bx + c}
∴ ^dy/_dx =m^{ax2 + bx + c}.log_em.^d/_dx(ax² + bx + c)
= m^{ax2 + bx + c}.log_em.(2ax + b)
Ans: Ⓐ (2ax + b)m^{ax2 + bx + c}.log_em

11. y = log(ax + b)³ হলে ^dy/_dx = _______
Ⓐ ³/_{ax + b} Ⓑ ^3x/_{ax + b}
Ⓒ ^3a/_{ax + b}Ⓓ এদের কোনোটিই নয়

Solution: y = log(ax + b)³ = 3log(ax + b)
∴ ^dy/_dx
= 3.¹/_{ax + b}.(a.1) = ^3a/_{ax + b}
Ans: Ⓒ ^3a/_{ax + b}

\(12.\ y = \sqrt{tan^{-1}x}\) হলে \(\frac{dy}{dx} =\ Ⓐ\ \frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\\Ⓑ\ \frac{2}{(1 + x^2)\sqrt{tan^{-1}x}}\\Ⓒ\ -\frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\\Ⓓ\ \frac{1}{\sqrt{tan^{-1}x}}\)

\(Solution:\ y = \sqrt{tan^{-1}x}\\∴ \frac{dy}{dx} =\frac{1}{2\sqrt{tan^{-1}x}}.\frac{d}{dx}(tan^{-1}x)\\=\frac{1}{2\sqrt{tan^{-1}x}}.\frac{1}{1 + x^2}\\=\frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\\Ans:\ Ⓐ\ \frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\)

13. যদি y = f(x)ⁿ হয়, তবে ^dy/_dx = ______
Ⓐ n{f(x)}^{n – 1}
Ⓑ n{f(x)}^{n – 1}.f'(x)
Ⓒ n{f(x)}ⁿ .f'(x)
Ⓓ nf'(x)

Solution: y = f(x)ⁿ
∴ ^dy/_dx = nf(x)^{n – 1}.^d/_dx{f(x)}
= n{f(x)}^{n – 1}.f'(x)
Ans: Ⓑ n{f(x)}^{n – 1}.f'(x)

14. যদি y = e^φ(x) হয়, তবে ^dy/_dx = ______
Ⓐ e^φ(x). φ'(x)
Ⓑ e^φ‘^(x) Ⓒ e^φ(x)
Ⓓ এদের কোনোটিই নয়

Solution: y = e^φ(x)
∴ ^dy/_dx
= e^φ(x).^d/_dx{φ(x)} = e^φ(x). φ'(x)
Ans: Ⓐ e^φ(x). φ'(x)

15. যদি \(y = log{\sqrt{f(x)}}\) হয়, তবে \(\frac{dy}{dx}=\\Ⓐ\ \frac{1}{2}f'(x)\quad Ⓑ\ \frac{1}{2f'(x)}\\Ⓒ\ \frac{f'(x)}{2f(x)}\quad Ⓓ\ \frac{2f'(x)}{f(x)}\\Solution:\ y = log{\sqrt{f(x)}}\\∴ \frac{dy}{dx}= \frac{1}{\sqrt{f(x)}}.\frac{d}{dx}(f(x))\\= \frac{1}{\sqrt{f(x)}}.\frac{1}{2\sqrt{f(x)}}.f'(x)\\= \frac{f'(x)}{2f(x)}\\Ans:\ Ⓒ\ \frac{f'(x)}{2f(x)}\)

SEMESTER-3
সূচিপত্র

👉 UNIT-1 সম্বন্ধ ও অপেক্ষক

1. সম্বন্ধ
2. অপেক্ষক
3. বিপরীত ত্রিকোণমিতিক অপেক্ষকসমূহ

👉 UNIT-2 বীজগণিত

1. ম্যাট্রিক্সের প্রকারভেদ ও ম্যাট্রিক্স বীজগণিত
- ম্যাট্রিক্স PART 1
- ম্যাট্রিক্স PART 2
2. নির্ণায়ক
- মাইনর ও সহ-উৎপাদক
3. একটি ম্যাট্রিক্সের অ্যাডজয়েন্ট ও বিপরীত ম্যাট্রিক্স এবং সরল সহসমীকরণের সমাধান

👉 UNIT-3 কলনবিদ্যা

1. সন্ততা এবং অন্তরকলনযোগ্যতা
- সন্ততা
- অন্তরকলনযোগ্যতা
2. অবকলন বা অন্তরকলন
- অবকলন – I
- অবকলন PART – II
3. দ্বিতীয় ক্রমের অন্তরকলজ
4. অন্তরকলজের ব্যাখ্যা
5. স্পর্শক ও অভিলম্ব
6. বর্ধিষ্ণু ও ক্ষয়িষ্ণু অপেক্ষক
. চরম ও অবম মান

👉 UNIT-4 সম্ভাবনা

1. সম্ভাবনা
- সম্ভাবনা
- Bayes’ উপপাদ্য
2. সমসম্ভব চলক ও তার বিভাজন
3. দ্বিপদ বিভাজন

👉 Semester III -এর প্রশ্নপত্রের সম্পূর্ণ সমাধান

Semester III 2025 এর প্রশ্নপত্র ও তার সম্পূর্ণ সমাধান

16. যদি y = 3^u(x) হয়, তবে ^dy/_dx = ______
Ⓐ3^u(x).log3
Ⓑ 3^u(x).u'(x)
Ⓒ log3.u'(x)
Ⓓ 3^u(x).log3.u'(x)

Solution: y = 3^u(x)
∴ ^dy/_dx
=3^u(x)log3.^d/_dx{u(x)}
= 3^u(x).log3.u'(x)
Ans: Ⓓ 3^u(x).log3.u'(x)

17. যদি y = sin{f(x)} হয়, তবে ^dy/_dx = ______
Ⓐ cos{f(x)}.f'(x)
Ⓑ sec{f(x)}.f'(x)
Ⓒ sin^-1{f(x)}f'(x)
Ⓓএদের কোনোটিই নয়

Solution: y = sin{f(x)}
∴ ^dy/_dx
= cos f(x).^d/_dx{f(x)}
= cos{f(x)}.f'(x)
Ans: Ⓐ cos{f(x)}.f'(x)

18. যদি y = sec{φ(x)} হয়, তবে ^dy/_dx = ______
Ⓐ secφ(x).tan^-1φ(x).φ'(x)
Ⓑ sec^-1φ(x).tanφ(x).φ'(x)
Ⓒ sec^-1φ(x).ctan^-1φ(x).φ'(x)
Ⓓ secφ(x)tanφ(x).φ'(x)

Solution: y = sec{φ(x)}
∴ ^dy/_dx
= secφ(x)tanφ(x).^d/_dx{φ(x)}
= secφ(x)tanφ(x).φ'(x)
Ans: Ⓓ secφ(x)tan ¹φ(x).φ'(x)

\(19.\ \frac{d}{dx}(\sqrt{logx})^3 = \\Ⓐ\ \frac{3\sqrt{logx}}{2x}\quad Ⓑ\ \frac{2\sqrt{logx}}{3x}\\Ⓒ\ \frac{\sqrt{logx}}{2x}\quad Ⓓ\ \frac{\sqrt{logx}}{3x}\\Solution:\ \frac{d}{dx}(\sqrt{logx})^3\\=3(\sqrt{logx})^2.\frac{1}{2\sqrt{logx}}.\frac{1}{x}\\= \frac{3\sqrt{logx}}{2x}\\Ans:\ Ⓐ\ \frac{3\sqrt{logx}}{2x}\)

20. ^d/_dx(e^x⁹) = ______
Ⓐ 9x⁸.e^x⁹ Ⓑ x⁸.e^x⁹
Ⓒ 9e^x⁸ Ⓓ 8e^x⁸

Solution: ^d/_dx(e^x⁹)
= e^x⁹.^d/_dx(x⁹)
= e^x⁹.9x⁸ = 9x⁸.e^x⁹
Ans: Ⓐ 9x⁸.e^x⁹

21. ^d/_dxlog(cot x) = ______
Ⓐ secx cosecx
Ⓑ – secx cosecx
Ⓒ sinxcosx
Ⓓ – sinxcosx

Solution: ^d/_dxlog(cot x)
= ¹/_{cot x}.(-cosec² x)
= – secx cosecx
Ans: Ⓑ – secx cosecx

22. ^d/_dxlog(logx) (x > 1) = ______
Ⓐ ¹/_logx Ⓑ ¹/_x²
Ⓒ xlogx Ⓓ ¹/_xlogx

Solution: ^d/_dxlog(logx)
= ¹/_logx.^d/_dx(logx)
= ¹/_logx.¹/_x = ¹/_xlogx
Ans: Ⓓ ¹/_xlogx

23. ^d/_dx(e^{e^x}) = ______
Ⓐ e^{e^x} Ⓑ (e^x)² Ⓒ e^x.e^{e^x}
Ⓓএদের কোনোটিই নয়

Solution: ^d/_dx(e^{e^x})
= e^{e^x}.^d/_dx(e^x)
= e^{e^x}.e^x = e^x.e^{e^x}
Ans: Ⓒ e^x.e^{e^x}

24. x-এর সাপেক্ষে log(cosecx + cot x) এর অন্তরকলজ kcosecx হলে k = ______
Ⓐ 1 Ⓑ -1 Ⓒ 2 Ⓓ –¹/₂

Solution: ^d/_dx[log(cosecx + cot x)]
=¹/_{cosecx + cot x}.(- cosecxcot x – cosec²x)
= ^{– cosecx(cot x + cosecx)}/_{cosecx + cot x}
= -cosecx
Ans: Ⓑ -1

25. ^d/_dx(cosec⁶x) = ______
Ⓐ – cosec⁶xcotx
Ⓑ – 6cosec⁶xcotx
Ⓒ – 6cosecxcotx
Ⓓ- 6cosecxcot⁶x

Solution: ^d/_dx(cosec⁶x)
= 6cosec⁵x.(- cosecx.cotx)
= – 6cosec⁶xcotx
Ans: Ⓑ – 6cosec⁶xcotx

26. ^d/_dx(cot^-1x)⁴ = ^k/_{1 + x²} (cot^-1x)³ হলে k = ______
Ⓐ 2 Ⓑ -2 Ⓒ 4 Ⓓ -4

Solution: ^d/_dx(cot^-1x)⁴ = ^k/_{1 + x²}(cot^-1x)³
⇒ 4(cot^-1x)³.(- ¹/_{1 + x²}) = ^k/_{1 + x²}(cot^-1x)³
⇒ ^-4/_{1 + x²} = ^k/_{1 + x²}
∴ k = -4
Ans: Ⓓ -4

\(27. log⁡\left[ e^x\left( \frac{x – 2}{x + 2} \right)^\frac{3}{4} \right]\) অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ ______\(Ⓐ\ \frac{x^2- 1}{x^2 – 2}\quad Ⓑ\ \frac{x^2 – 1}{x^4 – 2}\\Ⓒ\ \frac{x^2 – 1}{x^2 – 4}\quad Ⓓ\ \frac{x^2 + 1}{x^2 + 4}\)

\(Solution: \frac{d}{dx}\left( log⁡\left[ e^x\left( \frac{x – 2}{x + 2} \right)^\frac{3}{4} \right] \right)\\=\frac{d}{dx}\left[ loge^x+\frac{3}{4}\left\{ log(x-2)-log(x+2) \right\} \right]\\=\frac{d}{dx}\left[x+\frac{3}{4}\left\{ log(x-2)-log(x+2) \right\} \right]\\=1+\frac{3}{4}\left[ \frac{1}{x-2}.1-\frac{1}{x+2}.1 \right]\\=1+\frac{3}{4}.\frac{x+2-x+2}{(x-2)(x+2)}\\=1+\frac{3}{4}.\frac{4}{x^2 – 4}\\=1+\frac{3}{x^2 – 4}\\=\frac{x^2 – 1}{x^2 – 4}\\Ans:\ Ⓒ\ \frac{x^2 – 1}{x^2 – 4}\)

28. tan^-1(^{sin x}/_{1 + cos x}) অপেক্ষকটির x -এর সাপেক্ষে অন্তরকলজ ______
Ⓐ ¹/₂ Ⓑ 1 Ⓒ 0 Ⓓ 2

Solution: ^d/_dx[tan^-1(^{sin x}/_{1 + cos x})]
= ^d/_dx[tan^-1(^{2sin ^x/₂ cos ^x/₂}/_{2cos² ^x/₂})]
=^d/_dx[tan^-1(tan ^x/₂)]
= ^d/_dx(^x/₂) = ¹/₂
Ans: Ⓐ ¹/₂

29. নীচের ______ অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ ^{2xlogxcosec2x – logtan x}/_x(logx)²
Ⓐ log_x sinx
Ⓑ log_sinx x
Ⓒ log_x tan x
Ⓓ log_{sin x}(sec(x) + 10^x²

\(Solution: \frac{d}{dx}(log_x⁡tanx)\\=\frac{d}{dx}\left( \frac{logtanx}{logx} \right)\\=\frac{\frac{1}{tanx}.sec^2⁡x.logx-logtanx.\frac{1}{x}}{(logx)^2}\\=\frac{\frac{xlogx}{sinxcosx}– logtanx}{x(logx)^2}\\=\frac{\frac{2xlogx}{sin2x}- logtanx}{x(logx)^2}\\=\frac{2xlogxcosec2x – logtanx}{x(logx)^2}\)

Ans: Ⓒ log_x tan x

30. ^d/_dx(x^x)^x = ______
Ⓐ x²(x + 2xlogx)
Ⓑ x^{x² + 1}(x + 2logx)
Ⓒ x^2x(x + 2xlogx)
Ⓓ x^x²(x + 2xlogx)

Solution: ^d/_dx(x^x)^x
= ^d/_dx(e^{xlogx^x})
= ^d/_dx(e^{x² logx})
=e^{x² logx}.^d/_dx(x² logx)
= (x^x)^x(2x.logx + x².¹/_x)
= x^x²(x + 2xlogx)
Ans: Ⓓ x^x²(x + 2xlogx)

31. y = tanx^sinx হলে, [^dy/_dx]_x_{= ^π/₄} = ______
Ⓐ 2 Ⓑ √2 Ⓒ √3 Ⓓ 1

Solution: y = tanx^sinx
∴ logy = logtanx^sinx = sinxlogtanx
∴ 1/y. ^dy/_dx = cosxlogtanx + sinx.¹/_tanxsec²x
⇒ ^dy/_dx = tanx^sinx [cosxlogtanx + secx]
[^dy/_dx]_{x = ^π/₄}
= (tan ^π/₄)^{sin ^π/₄} [cos ^π/₄ logtan ^π/₄ + sec ^π/₄]
= (1)^√2[¹/_√2 log1 + √2]
=1[0 + √2] = √2
Ans: Ⓑ √2

32. যদি xy + 1 = cos(xy) হয়, তবে x = ^π/₂, y = 0 বিন্দুতে ^dy/_dx = ______
Ⓐ 0 Ⓑ 1 Ⓒ 2 Ⓓ 3

Solution: xy + 1 = cos(xy)
∴ 1.y + x.^dy/_dx + 0 = -sin(xy)( 1.y + x.^dy/_dx)
x = ^π/₂, y = 0 বিন্দুতে,
1.0 + ^π/₂.^dy/_dx = -sin(^π/₂.0)( 1.0 + ^π/₂.^dy/_dx)
⇒ ^π/₂.^dy/_dx = -sin 0.^π/₂.^dy/_dx
⇒ ^π/₂.^dy/_dx = 0
∴ ^dy/_dx = 0
Ans: Ⓐ 0

33. x = a(cos t + logtan^t/₂) , y = a sint হলে ^dy/_dx = ______
Ⓐ sint Ⓑ cost
Ⓒ tant Ⓓ cot t

Solution: x = a(cos t + logtan^t/₂)

\(∴ \frac{dx}{dt}= a\left( -sint+ \frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2} \right)\\= a\left( -sint+ \frac{1}{2 sin⁡\frac{t}{2}.cos\frac{t}{2}} \right)\)

= a(-sint + ¹/_sin⁡t)
= a.^{1 – sin^2⁡⁡t}/_sin⁡t = ^{acos^2⁡⁡t}/_sin⁡t
y = a sint
∴ ^dy/_dt = acost

\(∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\= \frac{acost}{\frac{acos^2⁡t}{sin⁡t}}= tant\)

Ans: Ⓒ tant

34. যদি tan y = ^2t/_{1 – t²}, sin x = ^2t/_{1 + t²} হয়, তবে ^dy/_dx = ______
Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

Solution: tan y = ^2t/_{1 – t²}
⇒ y = tan^-1 ^2t/_{1 – t²} = 2tan^-1 t
sin x = ^2t/_{1 + t²}
⇒ x = sin^-1 ^2t/_{1 + t²} = 2tan^-1 t
∴ y = x
∴ ^dy/_dx = 1
Ans: Ⓑ 1

35. যদি x = a(θ + sin θ) , y = a(1 + cos θ) হয়, তবে [^dy/_dx]_θ_{= ^π/₂} ______
Ⓐ 0 Ⓑ -1 Ⓒ -2 Ⓓ 6

Solution: x = a(θ + sin θ)
∴ ^dx/_dθ = a(1 + cos θ),
y = a(1 + cos θ)
∴ ^dy/_dθ = – asin θ
∴ ^dy/_dx = ^dy/dθ / _dx/dθ
= ^{-asin θ} / _{a(1 + cos θ)}
= – ^{2sin ^θ/₂cos ^θ/₂} / _{2cos ^θ/₂}
=- tan ^θ/₂
∴ [^dy/_dx]_{θ = ^π/₂}
= – tan ^π/₄ = – 1
Ans: Ⓑ -1

36. যদি \(y= sin\sqrt{x^2+a^2}\) হয়, তবে \(\frac{dy}{dx} =\) ______

\(Ⓐ\ \frac{xsin\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\quad Ⓑ\ \frac{cos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\\Ⓒ\ \frac{xcos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\quad Ⓓ\ \frac{cos\sqrt{x^2+a^2}}{x^2- a^2}\)

\(Solution:\ y = sin\sqrt{x^2+a^2}\\∴ \frac{dy}{dx}= cos\sqrt{x^2+a^2}.\frac{1}{2\sqrt{x^2+a^2}}.2x=\frac{xcos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\\Ans:\ Ⓒ\ \frac{xcos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\)

\(37.\ y = \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}}\) হলে, \(\left[ \frac{dy}{dx} \right]_{x=4}=\)______

Ⓐ 100 Ⓑ 110
Ⓒ 120 Ⓓ 140

\(Solution: y = \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}}\\∴logy = log \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}} = 3logx + \frac{1}{2}log(x^2 – 12) – \frac{1}{3}log(20 – 3x)\\∴ \frac{1}{y}.\frac{dy}{dx}= \frac{3}{x}+ \frac{1}{2}.\frac{1}{(x^2 – 12)}.2x – \frac{1}{3}.\frac{1}{20 -3x}.(-3)\\⇒ \frac{dy}{dx} = \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}}\left[ \frac{3}{x} + \frac{x}{x^2 – 12} + \frac{1}{20 -3x} \right]\\∴ \left[ \frac{dy}{dx} \right]_{x=4}\\= \frac{4^3\sqrt{4^2 – 12}}{\sqrt[3]{20 – 4}}\left[ \frac{3}{4} + \frac{4}{16 – 12} + \frac{1}{20 -3.4} \right]\\= \frac{64.2}{2}[\frac{3}{4} + 1 + \frac{1}{8}]\)

= 48 + 64 + 8 = 120
Ans: Ⓒ 120

38. tan² y = ^{1 + cos 2x}/_{1 – cos 2x} হলে, ^dy/_dx = ______
Ⓐ 0 Ⓑ ±1
Ⓒ ±2 Ⓓ ±4

Solution: tan² y = ^{1 + cos 2x}/_{1 – cos 2x}
⇒ tan² y = cot²x
⇒ tan y = ± cotx = tan(^π/₂ ± x)
∴ y = ^π/₂ ± x
∴ ^dy/_dx = ±1
Ans: Ⓑ ±1

39. y = ^{1 + sin} ^θ ^{+ cos} ^θ/_{1 + sin} _θ _{– cos} _θ হলে ^dy/_dθ + ¹/_{1 – cos} _θ = ______
Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

Solution: y = ^{1 + sin} ^θ ^{+ cos} ^θ/_{1 + sin} _θ _{– cos} _θ
⇒ y = ^{1 + cos} ^θ ^{+ sin} ^θ/_{1 – cos} _{θ +} _sin _θ

\(⇒\frac{2cos^2 \frac{θ}{2} + 2sin \frac{θ}{2}.cos\frac{θ}{2}}{2sin^2\frac{θ}{2} + 2sin\frac{θ}{2}.cos\frac{θ}{2}}\\=\frac{2cos\frac{θ}{2}(cos\frac{θ}{2}+ sin\frac{θ}{2})}{2sin\frac{θ}{2}(sin\frac{θ}{2} + cos\frac{θ}{2})}\\=cot\frac{θ}{2}\\∴ \frac{dy}{dθ}=-cosec^2\frac{θ}{2}.\frac{1}{2} \\=\frac{1}{2sin^2\frac{θ}{2}}\\=\frac{1}{1 -cos⁡2.\frac{θ}{2}}\\=\frac{1}{1 – cosθ}\)

∴ ^dy/_dθ – ¹/_{1 – cosθ} = 0
Ans: Ⓐ 0

40. sin y = xsin(a + y) হলে, যদি ^dy/_dx = ^sina/_{1 – kxcosa + x²} হয়, তবে k = ______
Ⓐ 1 Ⓑ 2 Ⓒ -2 Ⓓ -1

Solution: sin y = xsin(a + y)
⇒ sin y = xsina cosy + xcosa siny
⇒sin y(1 – xcosa) = xsinacosy
⇒ tan y = ^xsina/_{1 – xcosa}
⇒ y = tan^-1 ^xsina/_{1 – xcosa}

\(∴ \frac{dy}{dx}=\frac{1}{1 + \left( \frac{xsina}{1 – xcosa} \right)^2}.\frac{sina(1 – xcosa) – xsina(-1.cosa)}{(1 – xcosa)^2}\\⇒ \frac{dy}{dx}=\frac{(1 – xcosa)^2}{(1 – xcosa)^2 + (xsina)^2}.\frac{sina -xsinacosa + xsinacosa}{(1 – xcosa)^2}\\⇒ \frac{dy}{dx}=\frac{sina}{1- 2xcosa + x^2 cos^2 a + x^2 sin^2 a}\\⇒ \frac{dy}{dx}=\frac{sina}{1 – 2xcosa + x^2}\\∴ k = 2\\Ans:\ Ⓑ\ 2\)

\(41.\ \frac{d}{dx}\left[\frac{sin x}{sin(x – \frac{π}{4})} \right]= kcosec^2 \left( x – \frac{π}{4} \right)\)

হলে k-এর মান ______
Ⓐ ¹/₂ Ⓑ –¹/_√2
Ⓒ ¹/₃ Ⓓ –¹/_√3

\(Solution:\ \frac{d}{dx}\left[\frac{sin x}{sin(x – \frac{π}{4})} \right]\\=\frac{cos x.sin(x – \frac{π}{4}) -sinx.cos(x – \frac{π}{4}).1}{sin^2 (x – \frac{π}{4})}\\= \frac{sin(x – \frac{π}{4} – x)}{sin^2 (x – \frac{π}{4})}\\= \frac{-sin \frac{π}{4}}{sin^2 (x – \frac{π}{4})}\)

= –¹/_√2cosec² (x – ^π/₄)
∴ k = – ¹/_√2
Ans: Ⓑ –¹/_√2

42. নীচের ______ সম্পর্কটির ক্ষেত্রে ^dy/_dx = ^y/_x হবে।
Ⓐ x^y = y^x Ⓑ x^y.y^x = 1
Ⓒ (cos x)^y = (sin y)^x
Ⓓ x³y⁴ = (x + y)⁷

Solution: x³y⁴ = (x + y)⁷
∴ log x³y⁴ = (x + y)⁷
⇒ 3logx + 4logy = 7log(x + y)
∴ ³/_x + ⁴/_y.^dy/_dx = ⁷/_{x + y}(1 + ^dy/_dx)
⇒ ^dy/_dx(⁴/_y – ⁷/_{x + y}) = ⁷/_{x + y} – ³/_x
⇒^dy/_dx.^{4x + 4y – 7y}/_{y(x + y)} = ^{7x – 3x – 3y}/_{x(x + y)}
⇒ ^dy/_dx.^{4x – 3y}/_y = ^{4x – 3y}/_x
∴ ^dy/_dx = ^y/_x
Ans: Ⓓ x³y⁴ = (x + y)⁷

43. ^d/_dx[(sin x)^{cos x} + (cos x)^{sin x}] = ______
Ⓐ (sin x)^{sin x}[sin x tan x – cos xlog cos x] + (sin x)^{cos x}[tan xlog tan x – sin x tan x ]
Ⓑ (cos x)^{sin x}[cos x cotx + sin xlog cos x] + (sin x)^{sin x}[cot xlog tan x + sin x cot x ]
Ⓒ (sin x)^{cos x}[cos x cotx – sin xlog sin x] + (cos x)^{sin x}[cos xlog cos x – sin x tan x]
Ⓓ এদের কোনোটিই নয়

Solution: ^d/_dx[(sin x)^{cos x} + (cos x)^{sin x}]
= ^d/_dx(sin x)^{cos x} + ^d/_dx(cos x)^{sin x}
= ^d/_dx[e^{cosxlog(sin x)}] + ^d/_dx[e^{sin xlog(cosx)}]
=e^{cosxlog(sin x)}.[ cosx.¹/_sinx.cosx + (-sinx).log(sinx)] + e^{sin xlog(cosx)}.[cosx.log(cosx) + sinx.¹/_cosx.(-sinx)]
= (sin x)^{cos x}[cos x cotx – sin xlog sin x] + (cos x)^{sin x}[cos xlog cos x – sin x tan x]
Ans: Ⓒ (sin x)^{cos x}[cos x cotx – sin xlog sin x] + (cos x)^{sin x}[cos xlog cos x – sin x tan x]

\(44.\ y = sin[\left[ 2tan^{-1}\sqrt{\frac{1-x}{1 + x}} \right]⁡\) হলে, \(\frac{dy}{dx}= ——\\Ⓐ\ \frac{x}{\sqrt{1 – x^2}}\quad Ⓑ\ -\frac{x}{\sqrt{1 – x^2}}\\Ⓒ\ -\frac{1}{\sqrt{1 – x^2}}\quad Ⓓ\ \frac{1}{\sqrt{1 – x^2}}\)

Solution: ধরি, x = cos θ

\(∴ y = sin\left[ 2tan^{-1}\sqrt{\frac{1-x}{1 + x}} \right]\\=sin\left[ 2tan^{-1}\sqrt{\frac{1-cos θ}{1 + cos θ}} \right]\\=sin\left[ 2tan^{-1}\sqrt{\frac{2sin^2 \frac{θ}{2}}{2cos^2 \frac{θ}{2}}} \right]\\=sin\left[ 2tan^{-1}\sqrt{tan \frac{θ}{2}} \right]\\=sin2.\frac{θ}{2}\\=sinθ= \sqrt{1-x^2}\\∴ \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}}.(-2x)= -\frac{x}{\sqrt{1-x^2}}\\Ans:\ Ⓑ\ -\frac{x}{\sqrt{1 – x^2}}\)

45. x = ^{t – e^-t²}/_2t² হলে t ^dx/_dt + 2x – ¹/_2t = ______
Ⓐ e^-t Ⓑ e^{^-t²}
Ⓒ e^-2t Ⓓ e^-2t²

Solution: x = ^{t – e^-t²}/_2t²
⇒ x.2t² = t – e^-t²
⇒ ^dx/_dt.2t² + 4tx = 1 – e^-t²(-2t)
⇒t.^dx/_dt + 2x = ¹/_2t + e^-t²
⇒ t.^dx/_dt + 2x – ¹/_2t = e^-t²
Ans: Ⓑ e^-t²

Column Matching

1. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।স্তম্ভ Aস্তম্ভ B

স্তম্ভ A	স্তম্ভ B
[i] ^d/_dx(sin^{-1 2x}/_{1 + x}²) =	[a] ¹_{/(1 + x)√x}
[ii] ^d/_dx(sin^-1(e^x2) =	[b] ²/_{1 + x2}
[iii] ^d/_dx(cos^-1 ^{1 – x}/_{1 + x}) =	[c] ^2x/_{1 + x⁴}
[iv] ^d/_dx(tan^-1(x²) =	[d] ^2xex2/_{sqt(1 – e(2×2))}

Ⓐ [i] – [b], [ii] – [d], [iii] – [a], [iv] – [c]
Ⓑ [i] – [b], [ii] – [a], [iii] – [d], [iv] – [c]
Ⓒ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]
Ⓓ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]

Solution: [i] ^d/_dx(sin^{-1 2x}/_{1 + x}²)
= ^d/_dx(tan^-12x)
= ²/_{1 + x2} → [b]

\([ii]\ \frac{d}{dx}\left( sin^{-1}⁡\left( {e^{x^2}} \right) \right)\\= \frac{1}{\sqrt{1-(e^{x^2})^2}}.e^{x^2}.2x\\=\frac{2xe^{x^2}} {\sqrt{1-e^{2x^2}}}\) → [d]

[iii] ^d/_dx(cos^-1 ^{1 – x}/_{1 + x})
= ^d/_dx(2tan^-1 √x)
= 2.¹/_{1 + (√x)²}.¹/_2√x = ¹_{/(1 + x)√x} → [a]
[iv] ^d/_dx(tan^-1(x²)
= ¹/_{1 + (x²)²}.2x
= ^2x/_{1 + x⁴} → [a]
Ans: Ⓐ [i] – [b], [ii] – [d], [iii] – [a], [iv] – [c]

2. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

স্তম্ভ A	স্তম্ভ B
[i] ^d/_dx(sin 4xcos x) =	[a] 8(2sin8x – sin4x)
[ii] ^d/_dx(4sin2xsin 6x) =	[b] – 2sin4x – sin2x
[iii] ^d/_dx(sinxsin2xsin3x) =	[c] ¹/₂(5cos5x + 3cos3x)
[iv] ^d/_dx(cosxcos 3x) =	[d] ¹/₂(cos2x + 2cos4x – 3cos6x)

Solution: [i] ^d/_dx(sin 4xcos x)
=^d/_dx(¹/₂.2sin 4xcos x)
= ¹/₂×^d/_dx(sin 5x + sin 3x)
= ¹/₂(5cos5x + 3cos3x) → [c]
[ii] ^d/_dx(4sin2xsin 6x)
= 2×^d/_dx(2sin2xsin6x)
= 2×^d/_dx(cos4x – cos8x)
=2(-4sin4x + 8sin8x)
= 8(2sin8x – sin4x)→ [a]

[iii] ^d/_dx(sinxsin2xsin3x)
= ^d/_dx[¹/₂.(2sinxsin2x)sin3x]
= ¹/₂×^d/_dx[(cosx – cos3x)sin3x]
=¹/₄[(2cosx sin3x) – (2cos3xsin3x)]
= ¹/₄[sin 4x + sin 2x – sin6x]
=¹/₄(4cos4x + 2cos2x – 6cos6x)
= ¹/₂(cos2x + 2cos4x – 3cos6x) → [d]
[iv] ^d/_dx(cosxcos 3x)
=^d/_dx(¹/₂.2cos3xcos x)
=¹/₂×^d/_dx(-4cos4x – 2cos2x)
= ¹/₂(-4sin4x-2sin2x)
= -2sin4x – sin2x→ [b]
Ans: Ⓒ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]

3. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

স্তম্ভ A	স্তম্ভ B
[i] ^d/_dx(e^ax cos bx) =	[a] sin^{p – 1}x cos^{q – 1}x×(pcos²x – qsin²x)
[ii] ^d/_dx (sin^px cos^q x) =	[b] e^ax(acos bx – bsin bx)
[iii] ^d/_dx(sin^px/cos^q x) =	[c] ¹/₂ e^xcos x(sin 2x + 3cos 2x – 1)
[iv] ^d/_dx(e^xsinxcos²x) =	[d] sin^{p – 1}x/cos^{q + 1}x (pcos²x + qsin²x)

Ⓐ [i] – [b], [ii] – [], [iii] – [a], [iv] – [c]
Ⓑ [i] – [b], [ii] – [a], [iii] – [d], [iv] – [c]
Ⓒ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]
Ⓓ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]

Solution: [i] ^d/_dx(e^ax cos bx)
= e^ax.a.cos bx + e^ax (-sinbx.b)
= e^ax(acos bx – bsin bx) → [b]
[ii] ^d/_dx(sin^px cos^q x)
= (psin^{p – 1}x.cosx)cos^qx + sin^px.(qcos^{q – 1}x.-sinx)
=sin^{p – 1}x.cos^{q – 1}x.pcos²x + sin^{p – 1}x.cos^{q – 1}x.qsin²x
= sin^{p – 1}xcos^{q – 1}x(pcos²x – qsin²x) → [a]

\([iii]\ \frac{d}{dx}\left( \frac{sin^p⁡x}{cos^q⁡x} \right)\\= \frac{(psin^{p-1}x.cosx)cos^qx – sin^p⁡x(q cos^{q-1}⁡x.(-sinx))}{(cos^q⁡x)^2}\\= \frac{psin^{p-1}x cos^{q+1}x + q sin^{p+1}x cos^{q-1}⁡x}{cos^{2q}⁡x}\\ = \frac{sin^{p-1}x cos^{q-1}x (pcos^2 x + q sin^2x)}{cos^2q⁡x}\\= \frac{sin^{p-1}⁡x}{cos^{q+1})x}(pcos^2 x+qsin^2⁡x)\) → [d]

[iv] ^d/_dx(e^xsinxcos²x)
= e^xsinx. ^d/_dx(cos²x) + e^x.^d/_dx(sinx)cos²x + ^d/_dx(e^x)sinxcos²x
=e^xsinx.2cosx.(-sinx) + e^x.cosx.cos²x + e^xsinxcos²x
= ¹/₂e^xcos x(-4sin²x + 2cos²x + 2sinxcosx)
= ¹/₂e^xcos x[-2(1 – cos2x) + (1 + cos2x) + sin2x]
== ¹/₂e^xcos x[-2 + 2cos2x + 1 + cos2x + sin2x]
= ¹/₂e^xcos x(sin 2x + 3cos 2x – 1) → [c]
Ans: Ⓑ [i] – [b], [ii] – [a], [iii] – [d], [iv] – [c]

4. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।স্তম্ভ Aস্তম্ভ B

স্তম্ভ A	স্তম্ভ B
[i] logx-এর সাপেক্ষে x² -এর অবকল সহগ	[a] xsec²x
[ii] logx-এর সাপেক্ষে tanx-এর অবকল সহগ	[b] secx
[iii] cosecx-এর সাপেক্ষে cotx -এর অবকল সহগ	[c] 2x²
[iv] tan^-1x-এর সাপেক্ষে log₁₀x-এর অবকল সহগ	[d] ^{1 + x²}/_x log₁₀e

Ⓐ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]
Ⓑ [i] – [c], [ii] – [b], [iii] – [a], [iv] – [d]
Ⓒ [i] – [a], [ii] – [c], [iii] – [b], [iv] – [d]
Ⓓ [i] – [c], [ii] – [a], [iii] – [b], [iv] – [d]

Solution: [i] ^d/_d(logx)(x²)
= ^d/_dx(x²).¹/_{^d/_dx(logx)}
= 2x.¹/_x = 2x² → [c]
[ii] ^d/_d(logx)(tanx
= ^d/_dx(tanx).¹/_{^d/_dx(logx)}
= sec²x.¹/_{¹/_x} = xsec²x → [a]
[ii] ^d/_d(cosecx)(cotx)
= ^d/_dx(cotx).¹/_{^d/_dx(cosecx)}
= -cosec²x.¹/_cosecxcotx = secx → [b]

[ii] ^d/_d(tan^-1x)(log₁₀x)
= ^d/_dx(log_x10).¹/_{^d/_dx(tan^-1x)}
=^d/_dx(log_ex. log₁₀e).¹/_{¹/_{1 + x²}}
= log₁₀e.¹/_x(1 + x²)
= ^{1 + x²}/_x log₁₀e → [d]
Ans: Ⓓ [i] – [c], [ii] – [a], [iii] – [b], [iv] – [d]

5. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

স্তম্ভ A	স্তম্ভ B
[i] x² + y² = 4 হলে, ^dy/_dx =	=[a] y³
[ii] y = ^x/_{√(1 + x²)} হলে x^{3 dy}/_dx =	[b] ^y²/_{x(y – x)}
[iii] y= e^{^y/_x} হলে ^dy/_dx =	[c] ^y/_{x(1 + y)}
[iv] ye^y হলে ^dy/_dx =	[d] – ^x/_y

Ⓐ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]
Ⓑ [i] – [d], [ii] – [a], [iii] – [c], [iv] – [b]
Ⓒ [i] – [a], [ii] – [d], [iii] – [b], [iv] – [c]
Ⓓ [i] – [d], [ii] – [b], [iii] – [c], [iv] – [a]

Solution: [i] x² + y² = 4
∴ 2x + 2y^dy/_dx = 0
⇒ 2y^dy/_dx = – ^x/_y → [d]

\([ii]\ y\ = \frac{x}{\sqrt{1 + x^2}}\\∴ \frac{dy}{dx} = \frac{1.\sqrt{1 + x^2}-x.\frac{1}{2\sqrt{1 + x^2}}.2x}{(\sqrt{1 + x^2})^2}\\=\frac{1 + x^2- x^2}{(1 + x^2)\sqrt{1 + x^2}}=\frac{1}{(\sqrt{1 + x^2})^3}\\∴ x^3\frac{dy}{dx}=\frac{x^3}{(\sqrt{1 + x^2})^3}\\⇒ x^3\frac{dy}{dx} = \left( \frac{x}{\sqrt{1 + x^2}} \right)^3=y^3 → [a]\)

[iii] y= e^{^y/_x}

\(∴ \frac{dy}{dx} = e^\frac{y}{x}.\frac{\frac{dy}{dx}.x – y.1}{x^2}\\⇒ \frac{dy}{dx} = y.\frac{x\frac{dy}{dx} – y}{x^2}\)

⇒ x^2dy/_dx = xy ^dy/_dx – y²
⇒ x(x – y)^dy/_dx = – y²
⇒^dy/_dx = ^{– y²}/_{x(x – y)}
⇒ ^dy/_dx = ^y²/_{x(y – x)}→ [b]

[iv] ye^y = x
∴ ^dy/_dx. e^y + y. e^y. ^dy/_dx = 1
⇒ e^y .^dy/_dx(1 + y) = 1
⇒ ^x/_y.^dy/_dx(1 + y) = 1
⇒^dy/_dx = → [c]
Ans: Ⓐ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]

Rearrangement of Sentences/Events

1. cos y = xcos(a + y) থেকে ^dy/_dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

[i] x = ^{cos y}/_{cos(a + y)}
[ii] ^dx/_dy = ^{sin(a + y – y)}/_{cos²(a + y)
[}iii] ^dx/_dy = ^{– sin ycos(a + y) + cos ysin(a + y)}/_{cos²(a + y)}[iv] ^dy/_dx = ^{cos²(a + y)}/_{sin a} ধাপগুলির সঠিক ক্রমটি হবে-
Ⓐ [i] – [ii] – [iii] – [iv]
Ⓑ [i] – [iv] – [iii] – [ii]
Ⓒ [i] – [ii] – [iv] – [iii]
Ⓓ [i] – [iii] – [ii] – [iv]

Solution: cos y = xcos(a + y)
⇒ x = ^{cos y}/_{cos(a + y)} → [i]
∴ ^dx/_dy = ^{– sin ycos(a + y) + cos ysin(a + y)}/_{cos²(a + y)}→ [iii]
⇒ ^dx/_dy = ^{sin(a + y – y)}/_{cos²(a + y)} → [iii]
⇒ ^dx/_dy = ^{sin a}/_{cos²(a + y)}
∴ ^dy/_dx = ^{cos²(a + y)}/_{sin a}→ [iv]
Ans: Ⓓ [i] – [iii] – [ii] – [iv]

2. y^z = e^{y – x} থেকে ^dy/_dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

    [i] xlogy = y – x
  [ii] ^dy/_dx = ^ylogey/_{y – x}
    [iii] ^d/_dx(xlog y) = ^d/_dx(y – x)
   [iv] (1 – ^x/_y) ^dy/_dx = 1 + logy
    [v] ^dy/_dx = ^{log ey}/_{log y/_logey} = ^(logey)²/_logy
    [vi] ^dy/_dx = ^logey/_{1 – x/_y}
    [vii] logy + ^x/_y ^dy/_dx = ^dy/_dx – 1
      ধাপগুলির সঠিক ক্রমটি হবে –
Ⓐ [i] – [ii] – [iii] – [iv] – [v] – [vi] [vii]
Ⓑ [i] – [iii] – [vii] – [iv] – [ii] – [vi] – [v]
Ⓒ [i] – [iii] – [iv] – [vii] – [vi] – [ii] – [v]
Ⓓ [i] – [iii] – [vi] – [ii] – [iv] – [vii] – [v]

Solution: y^x = e^{y – x}
বা, logy^x = loge^{y – x}
বা, xlogy = (y – x)loge = (y – x) → [i]
    ^d/_dx(xlog y) = ^d/_dx(y – x) → [iii]
∴ 1.logy + x. ¹/_y ^dy/_dx = ^dy/_dx – 1
⇒ logy + ^x/_y ^dy/_dx = ^dy/_dx – 1 → [vii]
⇒(1 – ^x/_y)^dy/_dx= 1 + logy → [iv]
⇒ ^{y – x}/_y^dy/_dx = log_ee + logy = logey
⇒ ^dy/_dx = ^ylogey/_{y – x} → [ii]
⇒^dy/_dx = ^logey/_{1 – x/_y} → [vi]    Ans: Ⓑ [i] – [iii] – [vii] – [iv] – [ii] – [vi] – [v]

3. f(x) = (^{a + x}/_{b + x})^x থেকে f'(0) -এর মান নির্ণয় করার ধাপগুলি নীচে দেওয়া হল –
[i] f'(x) = ^du/_dx + ^dv/_dx নির্ণয় করতে হবে।
[ii] u = (^{a + x}/_{b + x})^x এবং v = cos x ধরতে হবে।
[iii] f'(x) -এ x = 0 বসাতে হবে।
[iv] ^du/_dx এবং ^dv/_dx নির্ণয় করতে হবে।

ধাপগুলির সঠিক ক্রমটি হবে-
Ⓐ [i] – [ii] – [iv] – [iii]
Ⓑ [ii] – [iv] – [i] – [iii]
Ⓒ [ii] – [i] – [iv] – [iii]
Ⓓ [ii] – [i] – [iii] – [iv]

Solution: f(x) =(^{a + x}/_{b + x})^x + cosx
[ii] u = (^{a + x}/_{b + x})^x এবং v = cos x
[i] f'(x) = ^du/_dx + ^dv/_dxনির্ণয় করতে হবে।
[iv] ^du/_dx এবং ^dv/_dx নির্ণয় করতে হবে।
u = (^{a + x}/_{b + x})^x
⇒ logu = log(^{a + x}/_{b + x})^x
           = x[log(a + x) – log(b + x)]
∴ ¹/_u ^du/_dx = 1.[ log(a + x) – log(b + x)] + x.[¹/_{a + x} – ¹/_{b + x}]
⇒ ^du/_dx = (^{a + x}/_{b + x})^x[log(^{a + x}/_{b + x}) + ^{x(b – a)}/_{(a + x)(b + x)}]
     v = cos x
∴ ^dv/_dx = -sinx
[iii] f'(x) -এ x = 0 বসাতে হবে।
∴ f'(x) = (^{a + x}/_{b + x})^x[log(^{a + x}/_{b + x}) + ^{x(b – a)}/_{(a + x)(b + x)}] – sinx
∴ f'(0) = (^{a + 0}/_{b + 0})⁰ [log(^{a + 0}/_{b + 0}) + ^{0(b – a)}/_{(a + 0(b + 0)}] – sin 0
            = 1.log(^a/_b) + 0 – 0 = log(^a/_b)
Ans: Ⓒ [ii] – [i] – [iv] – [iii]

Relationship between Statements ____

প্রতিটি প্রশ্নে বিবৃতি A এবং বিবৃতি B দেওয়া আছে। নীচের কোন বিকল্পটি বিবৃতি A এবং বিবৃতি B-এর মধ্যে সঠিক সম্পর্ক নির্দেশ করে?
Ⓐ বিবৃতি A এবং বিবৃতি B পরস্পরবিরোধী
Ⓑ বিবৃতি B হল বিবৃতি A এর কারণ
Ⓒ বিবৃতি A হল সত্য কিন্তু বিবৃতি B মিথ্যা
Ⓓ বিবৃতি A এবং বিবৃত্তি B পরস্পর নির্ভরশীল নয়

1. বিবৃতি-A: \(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\) হলে \(\frac{dy}{dx} = \begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\) (যেখানে l, m, n, a, b, c ধ্রুবক)

বিবৃতি-B: \(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}\) হলে \(\frac{dy}{dx} = \begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}\)

Solution: বিবৃতি-A:

\(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\\∴\frac{dy}{dx}=\begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\0\quad 0\quad 0\\a\quad b\quad c\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\l\quad m\quad n\\0\quad 0\quad 0\end{vmatrix}\\=\begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}+0+0\\=\begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\)→ বিবৃতিটি সত্য

বিবৃতি-B:

\(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}\\\frac{dy}{dx} = \begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\F^{‘}(x)\ G^{‘}(x)\ H^{‘}(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\F(x)\ G(x)\ H(x)\\ϕ^{‘}(x)\ ψ^{‘}(x)\ τ^{‘}(x)\end{vmatrix}\)→ বিবৃতিটি মিথ্যা

Ans: Ⓒ বিবৃতি A হল সত্য কিন্তু বিবৃতি B মিথ্যা

2. বিবৃতি-A: tan y = ^{tan x + secx – 1}/_{tan x – secx + 1} হলে ^dy/_dx = ¹/₂
বিবৃতি-B: y = tan^-1(^{a cos x – b sin x}/_{b cos x + a sin x}) হলে ^dy/_dx = – 1

Solution: বিবৃতি-A: ^{tan x + secx – 1}/_{tan x – secx + 1}
= ^{tan⁡x+ secx – (sec²x – tan²⁡x}/_{tan x – secx + 1}
= ^{(tan⁡x+ secx)(1 –tan⁡x+ secx)}/_{tan x – secx + 1}
⇒ tanx + secx
⇒ ^{1 + sin x}/_cosx
= ^{(cos ^x/₂ + sin ^{^x/₂})²}/_{cos^²_^{^x}/₂} _{– sin^²⁡ _^{^x}/₂}
= ^{(cos ^x/₂ + sin ^{^x/₂})²}/_{(cos_^{^x}/₂} _{– sin⁡ _^{^x}/₂)(cos_^{^x}/₂} _{+ sin⁡ _^{^x}/₂)}
=^{cos ^x/₂ + sin ^{^x/₂}}/_{cos_^{^x}/₂} _{– sin⁡ _^{^x}/₂}
= ^{1 + tan ^{^x/₂}}/_{1 – tan_^{^x}/₂}
⇒ ^{tan ^π/₄ + tan ^{^x/₂}}/_{1 – tan^π_/₄}._{tan⁡ _^{^x}/₂}
⇒ tan(^π/₄ + ^x/₂)
∴ tan y = tan(^π/₄ + ^x/₂)
⇒ y = ^π/₄ + ^x/₂
∴ ^dy/_dx = ¹/₂ → বিবৃতিটি সত্য

বিবৃতি-B: y = tan^-1(^{a cos x – b sin x}/_{b cos x + a sin x})
= tan^-1(^{^a/_b – tan x}/_{1 + ^a/_b tan x})
=tan^-1(^{^a/_b}) – tan^-1(tan x)
= ^{^a/_b} – x
∴ ^dy/_dx = 0 – 1 = -1 → বিবৃতিটি সত্য
Ans: Ⓓ বিবৃতি A এবং বিবৃত্তি B পরস্পর নির্ভরশীল নয়

Click here to visit our Facebook

True and False ___

1. বিবৃতি-I: যদি \(y = cos^{-1}{f(x)}\) হয়, তবে \(\frac{dy}{dx}=-\frac{f^{‘}(x)}{\sqrt{1 – [f(x)]^2}}\)

বিবৃতি-II: যদি \(y = tan^{-1}{ϕ(x)}\) হয়, তবে \(\frac{dy}{dx}=\frac{ϕ^{‘}(x)}{\sqrt{1 + [ϕ(x)]^2}}\)

Solution: বিবৃতি-I: y = cos^-1{f(x)}

\(∴\frac{dy}{dx}=-\frac{f^{‘}(x)}{\sqrt{1 – [f(x)]^2}}\)

→ বিবৃতিটি সত্য
বিবৃতি-II: y = tan^-1{ϕ(x)}

\(∴\frac{dy}{dx}=\frac{ϕ^{‘}(x)}{1 + [ϕ(x)]^2}\)

→ বিবৃতিটি মিথ্যা
Ans: Ⓑ বিবৃতি I সত্য

2. বিবৃতি-I: ^d/_dx(2x² + 3)⁴ = 16x(2x² + 3)³

বিবৃতি-II: \(\frac{d}{dx}\sqrt{x^2+a^2}= \frac{x}{\sqrt{x^2+a^2}}\)

Ⓐ বিবৃতি I ও II সত্য
Ⓑ বিবৃতি । সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓒ বিবৃতি I মিথ্যা কিন্তু বিবৃতি II সত্য
Ⓓ বিবৃতি I ও II মিথ্যা

Solution: বিবৃতি-I: ^d/_dx(2x² + 3)⁴
= 4(2x² + 3)³.2.2x
= 16x(2x² + 3)³ → বিবৃতিটি সত্য
বিবৃতি-II:

\(\frac{d}{dx}\sqrt{x^2+a^2}\\=\frac{1}{2\sqrt{x^2+a^2}}.2x\\= \frac{x}{\sqrt{x^2+a^2}}\) → বিবৃতিটি সত্য

Ans: Ⓐ বিবৃতি I ও II সত্য

3. বিবৃতি-I: x-এর সাপেক্ষে log_ax এর অন্তরকলজ ¹/_x log_ae
বিবৃতি-II: x-এর সাপেক্ষে log_xe এর অন্তরকলজ – ¹/_{x(log x)²}
Ⓐ উভয় বিবৃতি সত্য
Ⓑ উভয় বিবৃতি মিথ্যা
Ⓒ বিবৃতি I সত্য এবং II মিথ্যা
Ⓓ বিবৃতি I মিথ্যা এবং II সত্য

Solution: বিবৃতি-I: ^d/_dx(log_ax)
= ^d/_dx(log_ex. log_ae)
= ¹/_x log_ae→ বিবৃতিটি সত্য
বিবৃতি-II: ^d/_dx(log_xe)
= ^d/_dx(¹/_{log x})
= ¹/_{(log x)²}.¹/_x=¹/_{x(log x)²}→ বিবৃতিটি সত্য
Ans: Ⓐ উভয় বিবৃতি সত্য

4. বিবৃতি-I: ^d/_dx (x^x) = x^x(1 + logx)
বিবৃতি-II: ^d/_dx(2e^x – 3^x + 10^x ln(10) = (2e^x – 3m^xlog_e^m + 10^xlog_e¹⁰)
বিবৃতি-III: ^d/_dx (2^{x + 2} – e^{x + 1} + log₁₀x = 2^{x + 2}log_e² – e^x + ^log10e/_x

Ⓐ বিবৃতি I, II, III সত্য
Ⓑ বিবৃতি I, II, III মিথ্যা
Ⓒ বিবৃতি I, II সত্য কিন্তু III সত্য নয়
Ⓓ বিবৃতি II, III সত্য কিন্তু । সত্য নয়

Solution: বিবৃতি-I: ^d/_dx(x^x)
= ^d/_dx(e^xlogx)
=e^xlogx.( x.¹/_x +1.logx)
= x^x(1 + logx + 1) → বিবৃতিটি সত্য
বিবৃতি-II: ^d/_dx(2e^x – 3m^x + 10^x)
= 2e^x – 3m^x.log_em+ 10^x. log_e10 → বিবৃতিটি সত্য
বিবৃতি-III: ^d/_dx(2^{x + 2} – e^{x + 1} + log₁₀x)
= 2^{x + 2}log_e2 – e^{x + 1} + ¹/_x .log₁₀e → বিবৃতিটি মিথ্যা
Ans: Ⓒ বিবৃতি I, II সত্য কিন্তু III সত্য নয়

5. বিবৃতি-I: ^d/_dx(10^x .x¹⁰) = 10^x.x⁹ (10 + xlog10)

বিবৃতি-II:\(\frac{d}{dx}\left(x\frac{e^x + e^{3x}}{e^x + e^{-x}} \right) = e^{3x}(1 + 2x)\)

Solution: বিবৃতি-I: ^d/_dx(10^x .x¹⁰)
= 10^x.log_e10. x¹⁰ + 10^x .10x⁹
= 10^x.x⁹ (xlog10 +10) → বিবৃতিটি সত্য

বিবৃতি-II:

\(\frac{d}{dx}\left(x\frac{e^x + e^{3x}}{e^x + e^{-x}} \right)\\=\frac{d}{dx}\left(x\frac{e^{2x}(e^{-x} + e^x)}{e^x + e^{-x}} \right)\)

= ^d/_dx(xe^2x)
=1. e^2x + x.e^2x.2
= e^2x(1 + 2x) → বিবৃতিটি মিথ্যা
Ans: Ⓑ বিবৃতি । সত্য এবং II মিথ্যা

6. বিবৃতি-I: ^d/_dxlog(ax)^x = log(ax) + 1

বিবৃতি-II: \(\frac{d}{dx}logsin(\sqrt{x^2+1})= \frac{x}{\sqrt{x^2+1}}cot\sqrt{x^2+1}\)

Ⓐ বিবৃতি । সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓑ বিবৃতি I, II উভয়ই সত্য
Ⓒ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য
Ⓓ বিবৃতি I, II উভয়ই মিথ্যা

Solution: বিবৃতি-I: ^d/_dxlog(ax)^x
= ^d/_dx[x(loga + logx)]
= 1.(loga + logx) + x(0 + ¹/_x)
= log(ax) + 1→ বিবৃতিটি সত্য
বিবৃতি-II:

\(\frac{d}{dx}logsin(\sqrt{x^2+1})\\= \frac{1}{sin\sqrt{x^2+1}}.cos\sqrt{x^2+1}.\frac{1}{2\sqrt{x^2+1}}.2x\\=\frac{x}{\sqrt{x^2+1}}cot\sqrt{x^2+1}\) → বিবৃতিটি সত্য

Ans: Ⓑ বিবৃতি I, II উভয়ই সত্য

7. বিবৃতি-I: \(\frac{d}{dx}\left( 2xtan^{-1}x – log(1 + x2) \right) = \frac{2tan^{-1}⁡x}{\sqrt{1 + x^2}}\)

বিবৃতি-II:\(\frac{d}{dx}\left( \frac{2x+1}{x^3}-logtan x + x sin^{-1}(x^2 )\right) = -\frac{4}{x^3}-\frac{3}{x^4}-secxcosecx + sin^{-1}⁡(x^2 ) +\frac{2x^2}{\sqrt{1 – x^4}}\)

Solution: বিবৃতি-I:

\(\frac{d}{dx}\left( 2xtan^{-1}x – log(1 + x2) \right)\\=2.1.tan^{-1}x + 2x.\frac{1}{1 + x^2}- \frac{1}{1 + x^2}.2x = 2.tan^{-1}x\)→ বিবৃতিটি মিথ্যা

বিবৃতি-II:

\(\frac{d}{dx}\left( \frac{2x+1}{x^3}-logtan x + x sin^{-1}(x^2 )\right)\\=\frac{d}{dx}\left(\frac{2}{x^2} +\frac{1}{x^3} -logtan x + x sin^{-1}⁡(x^2 )\right)\\=-\frac{4}{x^3}-\frac{3}{x^4} -\frac{1}{tanx}.sec^2⁡x+1.sin^{-1}(x^2)+x.\frac{1}{\sqrt{1 – x^4}}).2x\\=-\frac{4}{x^3}-\frac{3}{x^4}-secxcosecx +sin^{-1}⁡(x^2 ) +\frac{2x^2}{\sqrt{1 – x^4}}\)→ বিবৃতিটি সত্য

8. বিবৃতি-I: \(\sqrt{1 + sin x}+\sqrt{\frac{1 – x^2}{1 + x^2}}\) অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ \(\frac{1}{2}\left( cos\frac{x}{2} – sin\frac{x}{2} \right))- \frac{2x}{(1 + x^2)\sqrt{1 – x^4}}\)

বিবৃতি-II: \(\frac{1}{\sqrt{x + a}-\sqrt{x + b}}+ x^2sinx^3\) অপেক্ষকটির x3 -এর সাপেক্ষে অন্তরকলজ \(\frac{1}{2(a – b)}\left[ \frac{1}{\sqrt{x + a}}+\frac{1}{\sqrt{x + b}} \right]+ 2xsinx^3 + 3x^4cos x^3\)

Solution: বিবৃতি- I:

\(\frac{d}{dx}\left[ \sqrt{1 + sin x}+\sqrt{\frac{1 – x^2}{1 + x^2}} \right]\\\frac{d}{dx}\left[cos⁡\frac{x}{2}+sin⁡\frac{x}{2} \right]+\frac{d}{dx}\left[ \left( \frac{1 – x^2}{1 + x^2} \right)^{\frac{1}{2}}\right]\\=-sin\frac{x}{2}.\frac{1}{2}+cos\frac{x}{2}.\frac{1}{2}+\frac{1}{2}.\left( \frac{1 – x^2}{1 + x^2} \right)^{-\frac{1}{2}}.\frac{-2x(1 + x^2 )-(1- x^2 ).2x}{(1 + x^2)^2}\\=-\frac{1}{2}\left(sin⁡\frac{x}{2}-cos\frac{x}{2}\right)+\frac{1}{2}\left( \frac{1 + x^2}{1 – x^2} \right)^{\frac{1}{2}}.\frac{-4x}{(1 + x^2 )^2}\\=\frac{1}{2}\left(cos⁡\frac{x}{2}-sin\frac{x}{2}\right)-\frac{1}{2}.\frac{\sqrt{1 + x^2}}{\sqrt{1 – x^2}}.\frac{4x}{(1 + x^2)^2}\\=\frac{1}{2}\left(cos⁡\frac{x}{2}-sin\frac{x}{2}\right)-\frac{1}{\sqrt{1 – x^2}}.\frac{2x}{\sqrt{1 + x^2}(1 + x^2)}\\=\frac{1}{2}\left(cos⁡\frac{x}{2}-sin\frac{x}{2}\right)-\frac{2x}{(1 + x^2)\sqrt{1 – x^4}}\) → বিবৃতিটি সত্য

বিবৃতি-II:

\(\frac{d}{dx}\left[ \frac{1}{\sqrt{x + a} -\sqrt{x + b}}+x^2 sinx^3 \right]\\=\frac{d}{dx}\left[ \frac{\sqrt{x + a})+\sqrt{x + b}}{(\sqrt{x + a} -\sqrt{x + b})(\sqrt{x + a}+\sqrt{x + b})} \right]+\frac{d}{dx}[x^2 sinx^3]\\=\frac{d}{dx}\left[ \frac{\sqrt{x + a}+\sqrt{x + b}}{a-b} \right]+2xsinx^3+x^2.cosx^3.3x^2\\=\frac{1}{(a – b)}\left[\frac{1}{2\sqrt{x + a}}.1+\frac{1}{2\sqrt{x + b}}.1 \right]+ 2xsinx^3 + 3x^4cos x^3\\=\frac{1}{2(a – b)}\left[\frac{1}{\sqrt{x + a}}+\frac{1}{\sqrt{x + b}}\right]+ 2xsinx^3 + 3x^4cos x^3\)→ বিবৃতিটি সত্য

Ans: Ⓑ বিবৃতি I, II উভয়ই সত্য

9. বিবৃতি-Ⅰ: e^xsecx অপেক্ষকটির x -এর সাপেক্ষে অন্তরকলজ e^xsecx(1 + tan x)।
বিবৃতি-II: (cos x)^logx অপেক্ষকটির x এর সাপেক্ষে অন্তরকলজ (log x)^cosx[^cosx/_xlogx – sin x – log(logx)]|

Ⓐ বিবৃতি I, II উভয়ই সত্য
Ⓑ বিবৃতি I, II উভয়ই মিথ্যা
Ⓒ বিবৃতি। সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

Solution: বিবৃতি- I: ^d/_dx(e^xsecx)
= e^xsecx + e^xsecx.tanx
= e^xsecx(1 + tan x) → বিবৃতিটি সত্য
বিবৃতি- II: ^d/_dx [(cos x)^logx]
= ^d/_dx[e^{log⁡{(cos x)^logx}}]
= ^d/_dx[e^{logx⁡{log⁡(cosx)}}]
== e^{logx⁡{log⁡(cosx)}} .^d/_dx[logx⁡{log⁡(cosx) }]
= (cos x)^logx.(¹/_x.log⁡(cosx) + logx.¹/_cosx.(-sinx))
= (cos x)^logx(¹/_x log⁡(cosx)- logxtanx)→ বিবৃতিটি মিথ্যা
Ans: Ⓒ বিবৃতি। সত্য কিন্তু বিবৃতি II মিথ্যা

10. বিবৃতি-I: tan^-1x এর সাপেক্ষে ^{tan^-1x}/_{1 + tan^-1x} এর অবকল সহগ ¹/_{(1 – tan^-1x)²}

বিবৃতি-II: \(tan^{-1}x\) এর সাপেক্ষে \(tan^{-1}⁡\frac{\sqrt{1 + x^2}- 1}{x}\) এর অবকল সহগ 1

Ⓐ বিবৃতি। সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓑ বিবৃতি I, II উভয়ই সত্য
Ⓒ বিবৃতি I, II উভয়ই মিথ্যা
Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

Solution: বিবৃতি-I: ধরি, y = tan^-1x এবং

\(z = \frac{tan^{-1}⁡x}{1+tan^{-1}⁡x}=\frac{y}{1+y}\\∴ \frac{dz}{dy}=\frac{1(1 + y)- y.(0 + 1)}{(1 + y)^2} \\= \frac{1}{(1 + y)^2} =\frac{1}{(1 + tan^{-1}⁡x)^2}\)→ বিবৃতিটি মিথ্যা

বিবৃতি-II: ধরি, y = tan^-1x
∴ x = tany

\(z = tan^{-1}\frac{\sqrt{1 + x^2}– 1}{x}\\=tan^{-1}\frac{secy – 1}{tany}\\= tan^{-1}⁡\frac{1 – cosy}{siny}\\=tan^{-1}⁡\frac{2sin^2\frac{y}{2}}{2sin \frac{y}{2}.cos \frac{y}{2}}\)

11. বিবৃতি-I: sin^-1 ^2x/_{1 + x²} -এর সাপেক্ষে cos^-1 ^{1 – x²}/_{1 + x²} এর অবকল সহগ ¹/₂

বিবৃতি-II: \(sin^{-1⁡}x\) এর সাপেক্ষে \(x^{sin^{-1⁡}⁡x}\) এর অবকল সহগ \(x^{sin^{-1⁡}⁡x}\left[ \frac{\sqrt{1- x^2}}{x}sin^{-1⁡}x+ logx \right]\)

Ⓐ বিবৃতি । সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓑ বিবৃতি I, II উভয়ই সত্য
Ⓒ বিবৃতি I, II উভয়ই মিথ্যা
Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

Solution: বিবৃতি-I: ধরি, y = cos^-1 ^{1 – x²}/_{1 + x²} = 2tan^-1x
এবং z = sin^-1 ^2x/_{1 + x²} = 2tan^-1x
∴ y = z
∴ ^dy/_dz = 1 → বিবৃতিটি মিথ্যা
বিবৃতি-II: ধরি, y = x^{sin^-1x} = e^{logx^{sin^-1x}}= e^{^{sin^-1x}logx}

\(∴ \frac{dy}{dx}= e^{sin^{-1}⁡x logx}\left( \frac{1}{\sqrt{1 -x^2}}logx+sin^{-1⁡}x.\frac{1}{x} \right)\\=x^{sin^{-1}⁡x}\left( \frac{logx}{\sqrt{1 -x^2}}+\frac{sin^{-1⁡}x}{x} \right)\)এবং \(z = sin^{-1}⁡x \\∴ \frac{dz}{dx}= \frac{1}{\sqrt{1 – x^2}}\\∴ \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}} \\=\frac{x^{sin^{-1}⁡x}\left( \frac{logx}{\sqrt{1 -x^2}}+\frac{sin^{-1⁡}x}{x} \right)}{\frac{1}{\sqrt{1 – x^2}}}\\=x^{sin^{-1}⁡x}\left[logx+ \sqrt{1- x^2}.\frac{sin^{-1}⁡x}{x}\right]\\=x^{sin^{-1}⁡x}\left[ \frac{\sqrt{1- x^2}}{x}sin^{-1}x +logx \right]\)→বিবৃতিটি সত্য

Ans: Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

12 বিবৃতি-I: যদি \(y = e^{\sqrt{cos x}}\) হয়, তবে \(\frac{dy}{dx} = \frac{sin xe^{\sqrt{cos x}}}{2\sqrt{cos x}}\\\)বিবৃতি-II: যদি \(y = \frac{1}{\sqrt{log(secx)}}\) হয়, তবে \(\frac{dy}{dx} = – \frac{1}{2}tan x[log(secx)]^{-\frac{3}{2}}\)

Solution: বিবৃতি-I: y = e^{√cos x}
∴ ^dy/_dx = e^{√cos x}.¹/_{2√cos x}.(-sin x)
= – ^{sinxe^{√cos x}}/_2√cosx → বিবৃতিটি মিথ্যা

বিবৃতি-II: \(y = \frac{1}{\sqrt{log(secx)}}=log(secx)^{-\frac{1}{2}}\)

∴ ^dy/_dx = –¹/₂[log(secx)] ^–³/₂.¹/_secx.secx.tanx
= –¹/₂tanx[log(secx)] ^–³/₂ → বিবৃতিটি সত্য
Ans: Ⓑ বিবৃতি II সত্য

Case Based ___

1. y = f(t) এবং x = Φ(t) হলে ^dy/_dx = ^f’(t)/_Φ’(t) যেখানে Φ’(t) ≠ 0

Solution: x = acos θ
∴ ^dx/_dθ = – asin θ
y = bsin θ
∴ ^dy/_dθ = bcos θ
∴ ^dy/_dx = ^dy/dθ/_dx/dθ = ^bcosθ/_-asinθ = – ^b/_acot θ
Ans: Ⓒ – ^b/_acot θ

[ii] x = acos ^θ/₂; y = bsin ^θ/₂ হলে ^dy/_dx =
Ⓐ ^b/_atan ^θ/₂ Ⓑ –^b/_atan ^θ/₂
Ⓒ – ^b/_acot ^θ/₂ Ⓓ এদের কোনোটিই নয়

Solution: x = acos ^θ/₂
∴ ^dx/_dθ = – asin ^θ/₂. ¹/₂
y = bsin ^θ/₂
∴ ^dy/_dθ = bcos ^θ/₂. ¹/₂
∴ ^dy/_dx = ^dy/dθ/_dx/dθ = ^bcos ^{^θ/2}/_{-asin ^θ/2} = – ^b/_acot θ
Ans: Ⓒ – ^b/_acot ^θ/₂

Solution: x = acos³ θ
∴ ^dx/_dθ = a.3cos² θ.(- sin θ) = – 3acos² θ.sin θ
y = asin³ θ
∴ ^dy/_dθ = a.3sin² θ cos θ = 3asin² θ cos θ
∴ ^dy/_dx = ^dy/dθ/_dx/dθ = ^3asin² ^θ ^cos ^θ/_– _3acos² _θ.sin _θ = -tan θ
Ans: Ⓒ -tan θ

2. মনে করো, y = f(x) অপেক্ষকের বিপরীত অপেক্ষক x = g(y) এবং f'(x) = ¹/_{1 + x²}

Solution: y = f(x)
এবং x = g(y)
⇒ x = g{f(x)}
∴ 1 = g'{f(x)}. f'(x)
⇒ 1 = g'{f(x)}. ¹/_{1 + x²}
⇒g'{f(x)} = 1 + x²
⇒ g'(y) = 1 + [g(y)]²
∴ g'(x) = 1 + [g(x)]²
Ans: Ⓐ 1 + [g(x)]²

Solution: f(x) = tanx হলে f'(x) = ¹/_{1 + x2} হয়।
Ans: Ⓐ tanx

3. মনে করো, f(x) = (^{a + x}/_{1 + x})^{a + 1 + 2x}
[i] f'(x) = kf(x) হলে k =
Ⓐ ^{(1 – a)(1 + a + x)}/_{(1 + x)(a + x)} + 2log(^{a + x}/_{1 + x})
Ⓑ ^{(1 – a)(1 + a + 2x)}/_{(1 + x)(a + x)} + log(^{a + x}/_{1 + x})
Ⓒ ^{(1 – a)(1 + a + 2x)}/_{(1 + x)(a + x)} + 2log(^{a + x}/_{1 + x})
Ⓓ এদের কোনোটিই নয়

Solution: f(x) = (^{a + x}/_{1 + x})^{a + 1 + 2x}
⇒ logf(x) = log(^{a + x}/_{1 + x})^{a + 1 + 2x}
⇒logf(x) = (a + 1 + 2x)[log(a + x) – log(1 + x)]
⇒ ¹/_f(x)f'(x) = 2[log(a + x) – log(1 + x)] + (a + 1 + 2x)[¹/_{a + x} – ¹/_{1 + x}]
⇒ ¹/_f(x)f'(x) = 2log(^{a + x}/_{1 + x}) + (a + 1 + 2x)[^{1 + x – a – x}/_{(a + x)(1 + x)}]
⇒¹/_f(x)f'(x) = 2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}
⇒ f'(x) = f(x)[2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}]
∴ k = 2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}
Ans: Ⓒ ^{(1 – a)(1 + a + 2x)}/_{(1 + x)(a + x)} + 2log(^{a + x}/_{1 + x})

[ii] f'(0) =
Ⓐ a^{a + 1} [2log a + ^{1 – a2}/_a]
Ⓑ a^{a + 1} [log(a) + ^{1 – a2}/_a]
Ⓒ a^{a + 1} [2 loga – ^{1 – a2}/_a]
Ⓓ এদের কোনোটিই নয়

Solution: f'(x) = f(x)[2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}]
∴ f'(0) = (^{a + 0}/_{1 + 0})^{a + 1 + 2.0}[2log(^{a + 0}/_{1 + 0}) + ^{(a + 1 + 2.0)(1 – a)}/_{(a + 0)(1 + 0)}]
= a^{a + 1}[2log a + ^{(a + 1)(1 – a)}/_a]
= a^{a + 1}[2log a + ^{1 – a2}/_a]
Ans: Ⓐ a^{a + 1} [2log a + ^{1 – a2}/_a]

4. মনে করো, (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . +C_nxⁿ
[i] C₀ + 2C₁ + 3C₂ + . . . + (n +1)C_n =
Ⓐ (n + 2)2^{n + 1}
Ⓑ (n + 2)2ⁿ
Ⓒ (n + 1)2^{n + 1}
Ⓓ (n + 2)2^{n – 1}

Solution: (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . +C_nxⁿ
∴ x(1 + x)ⁿ = xC₀ + C₁x² + C₂x³ + . . . + C_nx^{n + 1} x -এর সাপেক্ষে অবকলন করে পাই,
1.(1 + x)ⁿ + x.n(1 + x)^{n – 1} = 1.C₀ + 2xC₁x + 3xC₂x² + . . . + (n + 1)C_nxⁿ x = 1 বসিয়ে পাই,
(1 + 1)ⁿ + 1.n(1+1)^{n – 1} = C₀ + 2.1.C₁ + 3.1.C₂.1² + . . . + (n + 1)C_n.1ⁿ⇒ 2ⁿ + n.2^{n – 1} = C₀ + 2C₁ + 3C₂ + . . . + (n + 1)C_n
⇒ 2.2^{n – 1} + n.2^{n – 1} = C₀ + 2C₁ + 3C₂ + . . . + (n + 1)C_n
⇒ (n + 2)2^{n – 1} = C₀ + 2C₁ + 3C₂ + . . . + (n + 1)C_nAns: Ⓓ (n + 2)2^{n – 1}

Solution: (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . + C_nxⁿ x -এর সাপেক্ষে অবকলন করে পাই,
n(1 + x)^{n – 1} = C₁ + 2C₂x + 3C₃x² + . . . + nC_nx^{n -1}x = -1 বসিয়ে পাই,
n(1 – 1)^{n – 1} = C₁ + 2C₂(-1) + 3C₃(-1)² + . . . + nC_n(-1)^{n -1}
⇒ 0 = C₁ – 2C₂ + 3C₃ + . . . + (-1)^{n -1}nC_nAns: Ⓑ 0

2025 SEMESTER III MATHEMATICS QUESTION PAPER WITH SOLUTION →

← অবকলন DIFFERENTIATION-PART I

Tag: CALCULAS 1ST ORDER DERIVATIVE