250+ TOP MCQs on Exponential and Logarithmic Functions | Class 12 Maths

Mathematics Multiple Choice Questions & Answers (MCQs) 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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}}})
Answer: a
Clarification: 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})
Answer: c
Clarification: 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
Answer: b
Clarification: 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.

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
Answer: a
Clarification: 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
Answer: c
Clarification: 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})
Answer: d
Clarification: 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})
Answer: c
Clarification: 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}))
Answer: a
Clarification: 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}))