Exponential and Logarithmic Functions - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Exponential and Logarithmic Functions”.

1. Differentiate 8e^-x+2e^x w.r.t x.
a) 2e^-x+8e^x
b) 2e^x+8e^-x
c) 2e^-x-8e^x
d) 2e^x-8e^-x
View Answer

Answer: d
Explanation: To solve:y=(8e^-x+2e^x)
Differentiating w.r.t x we get,
\(\frac{dy}{dx}\)=8(-e^-x+2e^x)
∴\(\frac{dy}{dx}\)=2e^x-8e^-x.

2. Differentiate 8e^cos2x w.r.t x.
a) 16 sin⁡2x e^cos2x
b) -16 sin⁡2x e^cos2x
c) -16 sin⁡2x e^-cos⁡2x
d) 16 sin⁡2x e^-cos⁡2x
View Answer

Answer: b
Explanation: Consider y=8e^cos2x
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}=\frac{d}{dx}\)(8e^cos2x)
=8e^cos2x\(\frac{d}{dx}\) (cos⁡2x)
=8e^cos2x(-sin⁡2x)\(\frac{d}{dx}\) (2x)
=8e^cos2x(-sin⁡2x)(2)
∴\(\frac{dy}{dx}\)=-16 sin⁡2x e^cos2x

3. Differentiate 3 sin^-1⁡(e^2x) w.r.t x.
a) \(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)
b) \(\frac{2e^2x}{\sqrt{1-e^{4x}}}\)
c) –\(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)
d) \(\frac{6e^{-2x}}{\sqrt{1-e^{4x}}}\)
View Answer

Answer: a
Explanation: Consider y=3 sin^-1⁡(e^2x)
\(\frac{dy}{dx}=\frac{d}{dx}\)(3 sin^-1⁡(e^2x))
\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\frac{d}{dx}\)(e^2x)
\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\)2e^2x
∴\(\frac{dy}{dx}\)=\(\frac{6e^{2x}}{\sqrt{1-e^{4x}}}\)

4. Differentiate log⁡(log⁡x⁵) w.r.t x.
a) –\(\frac{5}{x log⁡x^5}\)
b) \(\frac{1}{log⁡x^5}\)
c) \(\frac{5}{x log⁡x^5}\)
d) –\(\frac{1}{x log⁡x^5}\)
View Answer

Answer: c
Explanation: Consider y=(log⁡(log⁡(x⁵)))
\(\frac{dy}{dx}=\frac{1}{log⁡x^5} \frac{d}{dx} (log⁡x^5)\)
\(\frac{dy}{dx}=\frac{1}{log⁡x^5}.\frac{1}{x^5}.\frac{d}{dx} (x^5)\)
\(\frac{dy}{dx}=\frac{1}{log⁡x^5}.\frac{1}{x^5}.5x^4\)
∴\(\frac{dy}{dx}=\frac{5}{x log⁡x^5}\)

5. Differentiate 3e^3x³ w.r.t x.
a) 27x^-2 e^3x³
b) 27x² e^3x³
c) -27x² e^3x³
d) -27x^-2 e^3x³
View Answer

Answer: b
Explanation: Consider y=3e^3x³
\(\frac{dy}{dx}\)=\(\frac{d}{dx}\)(3e^3x³)
\(\frac{dy}{dx}\)=3e^3x³ \(\frac{d}{dx}\)(3x³)
\(\frac{dy}{dx}\)=3e^3x³ (3(3x²))
\(\frac{dy}{dx}\)=27x² e^3x³.

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6. Differentiate 5e^x² tan⁡x w.r.t x.
a) 5e^x² (1+tan⁡x)²
b) -5e^x² (1+tan⁡x)²
c) 5e^x² (1-tan⁡x)²
d) -5e^x² (1-tan⁡x)²
View Answer

Answer: a
Explanation: Consider y=5e^x² tan⁡x
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=tan⁡x \(\frac{d}{dx}\) (5e^x²)+5e^x² \(\frac{d}{dx}\) (tan⁡x)
\(\frac{dy}{dx}\)=tan⁡x (5e^x².2x)+5e^x² (sec^2⁡x)
\(\frac{dy}{dx}\)=5e^x² (2x tan⁡x+sec^2⁡x)
\(\frac{dy}{dx}\)=5e^x² (1+tan^2⁡x+2x tan⁡x)
\(\frac{dy}{dx}\)=5e^x² (1+tan⁡x)²

7. Differentiate log⁡(e^5x³) w.r.t x.
a) \(\frac{-15x^2}{e^{5x^3}}\)
b) \(\frac{15x^2}{e^{5x^3}}\)
c) 15x²
d) -15x²
View Answer

Answer: c
Explanation: Consider y=log⁡(e^5x³)
y=5x³ (∴log⁡e^x=x)
⇒\(\frac{dy}{dx}\)=\(\frac{d}{dx} (5x^3)\)
∴\(\frac{dy}{dx}=5(3x^2)=15x^2\)

8. Differentiate 7 log⁡(x⁴.5e^x³) w.r.t x.
a) \(\frac{7(4+3x^3)}{x^2}\)
b) \(\frac{7(4-3x^3)}{x}\)
c) –\(\frac{7(4+3x^3)}{x}\)
d) \(\frac{7(4+3x^3)}{x}\)
View Answer

Answer: d
Explanation: Consider y=7 log⁡(x⁴.5e^x³)
y=\(7(log⁡x^4 +log⁡5e^{x^3})\)
y=\(7(4 log⁡x+log⁡5e^{x^3})\)
\(\frac{dy}{dx}=7(4 \frac{d}{dx} (log⁡x)+\frac{d}{dx} (log⁡5e^{x^3}))\)
\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}} \frac{d}{dx} (5e^{x^3}))\)
\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}}.5e^{x^3}.\frac{d}{dx} {x^3})\)
\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}}.5e^{x^3}.3x^2)\)
\(\frac{dy}{dx}=7(\frac{4}{x}+3x^2)\)
∴\(\frac{dy}{dx}=\frac{7(4+3x^3)}{x}\)

9. Differentiate 2e^x⁴ log⁡x w.r.t x.
a) \(\frac{2e^{x^4} (4x^4 log⁡x+1)}{x^2}\)
b) \(\frac{e^{x^4} (4x^4 log⁡x+1)}{x}\)
c) \(\frac{2e^{x^4} (4x^4 log⁡x+1)}{x}\)
d) –\(\frac{2e^{x^4} (4x^4 log⁡x+1)}{x}\)
View Answer

Answer: c
Explanation: Consider y=2e^x⁴ log⁡x
\(\frac{dy}{dx}\)=\(\frac{d}{dx}\) (2e^x⁴ log⁡x)
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=2(log⁡x \(\frac{d}{dx} (e^{x^4})+e^{x^4}\frac{d}{dx}\) (log⁡x))
\(\frac{dy}{dx}\)=\(2(log⁡x.e^{x^4}\frac{d}{dx} {x^4}+e^{x^4}.\frac{1}{x})\)
\(\frac{dy}{dx}\)=\(2(log⁡x.e^{x^4}.4x^3+e^{x^4}.\frac{1}{x})\)
\(\frac{dy}{dx}\)=\(2e^{x^4} (4x^3 log⁡x+\frac{1}{x})\)
∴\(\frac{dy}{dx}=\frac{(2e^{x^4} (4x^4 log⁡x+1))}{x}\)

10. Differentiate \(log⁡(cos⁡(sin⁡(e^{x^3})))\) w.r.t x.
a) –\(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)
b) \(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)
c) –\(3e^{x^3} \,cos⁡e^{x^3} \,cos⁡(sin⁡e^{x^3})\)
d) –\(x^2 e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)
View Answer

Answer: a
Explanation: Consider y=\(log⁡(cos⁡(sin⁡(e^{x^3})))\)
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=\(\frac{d}{dx} (log⁡(cos⁡(sin⁡(e^{x^3}))))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} \frac{d}{dx} (cos⁡(sin⁡e^{x^3})))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}) \frac{d}{dx}(sin⁡e^{x^3})))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3}) \frac{d}{dx} (e^{x^3})))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3})(e^{x^3}) \frac{d}{dx} {x^3}))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}).cos⁡e^{x^3} .e^{x^3}.3x^2)\)
\(\frac{dy}{dx}\)=-\((\frac{3x^2 e^{x^3} cos⁡e^{x^3} sin⁡(sin⁡e^{x^3})}{cos⁡(sin⁡e^{x^3})})\)
\(\frac{dy}{dx}\)=-\(3x^2 e^{x^3} cos⁡e^{x^3} tan⁡(sin⁡e^{x^3})\)

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