# Class 12 Maths MCQ – Exponential and Logarithmic Functions

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

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

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

2. Differentiate 8ecos2x w.r.t x.
a) 16 sin⁡2x ecos2x
b) -16 sin⁡2x ecos2x
c) -16 sin⁡2x e-cos⁡2x
d) 16 sin⁡2x e-cos⁡2x

Explanation: Consider y=8ecos2x
Differentiating w.r.t x by using chain rule, we get
$$\frac{dy}{dx}=\frac{d}{dx}$$(8ecos2x)
=8ecos2x$$\frac{d}{dx}$$ (cos⁡2x)
=8ecos2x(-sin⁡2x)$$\frac{d}{dx}$$ (2x)
=8ecos2x(-sin⁡2x)(2)
∴$$\frac{dy}{dx}$$=-16 sin⁡2x ecos2x

3. Differentiate 3 sin-1⁡(e2x) 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) $$\frac{6e^{-2x}}{\sqrt{1-e^{4x}}}$$

Explanation: Consider y=3 sin-1⁡(e2x)
$$\frac{dy}{dx}=\frac{d}{dx}$$(3 sin-1⁡(e2x))
$$\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\frac{d}{dx}$$(e2x)
$$\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )$$2e2x
∴$$\frac{dy}{dx}$$=$$\frac{6e^{2x}}{\sqrt{1-e^{4x}}}$$

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) –$$\frac{1}{x log⁡x^5}$$

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 3e3x3 w.r.t x.
a) 27x-2 e3x3
b) 27x2 e3x3
c) -27x2 e3x3
d) -27x-2 e3x3

Explanation: Consider y=3e3x3
$$\frac{dy}{dx}$$=$$\frac{d}{dx}$$(3e3x3)
$$\frac{dy}{dx}$$=3e3x3 $$\frac{d}{dx}$$(3x3)
$$\frac{dy}{dx}$$=3e3x3 (3(3x2))
$$\frac{dy}{dx}$$=27x2 e3x3.
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6. Differentiate 5ex2 tan⁡x w.r.t x.
a) 5ex2 (1+tan⁡x)2
b) -5ex2 (1+tan⁡x)2
c) 5ex2 (1-tan⁡x)2
d) -5ex2 (1-tan⁡x)2

Explanation: Consider y=5ex2 tan⁡x
Differentiating w.r.t x by using chain rule, we get
$$\frac{dy}{dx}$$=tan⁡x $$\frac{d}{dx}$$ (5ex2)+5ex2 $$\frac{d}{dx}$$ (tan⁡x)
$$\frac{dy}{dx}$$=tan⁡x (5ex2.2x)+5ex2 (sec2⁡x)
$$\frac{dy}{dx}$$=5ex2 (2x tan⁡x+sec2⁡x)
$$\frac{dy}{dx}$$=5ex2 (1+tan2⁡x+2x tan⁡x)
$$\frac{dy}{dx}$$=5ex2 (1+tan⁡x)2

7. Differentiate log⁡(e5x3) w.r.t x.
a) $$\frac{-15x^2}{e^{5x^3}}$$
b) $$\frac{15x^2}{e^{5x^3}}$$
c) 15x2
d) -15x2

Explanation: Consider y=log⁡(e5x3)
y=5x3 (∴log⁡ex=x)
⇒$$\frac{dy}{dx}$$=$$\frac{d}{dx} (5x^3)$$
∴$$\frac{dy}{dx}=5(3x^2)=15x^2$$

8. Differentiate 7 log⁡(x4.5ex3) 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) $$\frac{7(4+3x^3)}{x}$$

Explanation: Consider y=7 log⁡(x4.5ex3)
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 2ex4 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) –$$\frac{2e^{x^4} (4x^4 log⁡x+1)}{x}$$

Explanation: Consider y=2ex4 log⁡x
$$\frac{dy}{dx}$$=$$\frac{d}{dx}$$ (2ex4 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) –$$x^2 e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})$$

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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