Mathematics Questions and Answers – Calculus Application – Maxima and Minima – 2

«
»

This set of Mathematics Aptitude Test for Engineering Entrance Exams focuses on “Calculus Application – Maxima and Minima – 2”.

1. What will be the nature of the equation sin(x + α)/sin(x + β)?
a) Possess only minimum value
b) Possess only maximum value
c) Does not possess a maximum or minimum value
d) Data inadequate
View Answer

Answer: c
Explanation: Let, y = sin(x + α)/sin(x + β)
Then,
dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin2(x + β)
= sin(x+β – x-α)/sin2(x + β)
Or sin(β – α)/sin2(x + β)
So, for minimum or maximum value of x we have,
dy/dx = 0
Or sin(β – α)/sin2(x + β) = 0
Or sin(β – α) = 0 ……….(1)
Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).
Therefore, y has neither a maximum or minimum value.
advertisement

2. Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its maximum?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: c
Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.

3. Given, f(x) = x3 – 12x2 + 45x + 8. What is the maximum value of f(x)?
a) 61
b) 62
c) 63
d) 54
View Answer

Answer: b
Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.
Putting, x = 3 in (1)
Thus, its maximum value is,
f(3) = 33 – 12*32 + 45*3 + 8 = 62.
advertisement
advertisement

4. Given, f(x) = x3 – 12x2 + 45x + 8. At which point does f(x) has its minimum?
a) 1
b) 7
c) 3
d) 5
View Answer

Answer: d
Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.

5. Given, f(x) = x3 – 12x2 + 45x + 8. What is the minimum value of f(x)?
a) -1
b) 0
c) 1
d) Value does not exist
View Answer

Answer: c
Explanation: We have, f(x) = x3 – 12x2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x2 – 24x + 45
3x2 – 24x + 45 = 0
Or x2 – 8x + 15 = 0
Or(x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
So, f(x) has minimum at 5.
Putting, x = 5 in (1)
Thus, its maximum value is,
f(3) = 53 – 12*52 + 45*5 + 8 = 58.
advertisement

6. At which point does f(x) = |x – 1| has itslocal minimum?
a) They are unequal
b) They are equal
c) Depend on the numbers
d) Can’t be predicted
View Answer

Answer: b
Explanation: The given function is f(x) = ∣x − 1∣, x ∈ R.
It is known that a function f is differentiable at point x = c in its domain if both
\(\lim\limits_{h \rightarrow 0^-}\) hf(c + h) – f(c)
And
\(\lim\limits_{h \rightarrow 0^+}\) hf(c + h) – f(c) are finite and equal.
To check the differentiability of the function at x = 1,
LHS,
Consider the left hand limit of f at x=1
\(\lim\limits_{h \rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h}\)
= \(\lim\limits_{h \rightarrow 0^-}\frac{|h|}{h}\)
= \(\lim\limits_{h \rightarrow 0^-}\frac{-h}{h}\)
= −1
RHS,
Consider the right hand limit of f at x − 1
\(\lim\limits_{h \rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h}\)
= \(\lim\limits_{h \rightarrow 0^+}\frac{|h|}{h}\)
= 1
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.
As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.
Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.
Therefore, f(x) has a local minima at x = 1.

7. A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: c
Explanation: Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
advertisement

8. A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity?
a) 1 cm/sec
b) 2 cm/sec
c) 3 cm/sec
d) 4 cm/sec
View Answer

Answer: a
Explanation: Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Putting t = 3 in (1) we get,
33/3 – 2(3)2/ + 3(3) + 1
= 1 cm/sec.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

advertisement

To practice Mathematics Aptitude Test for Engineering Entrance Exams, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter