# Mathematics Questions and Answers – First Order Derivative – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “First Order Derivative – 1”.

1. At which point will f(x) attain a minima where, $$\mbox{f}(x)=\left\{ \begin{array} (x^3+x^2+10x & \mbox{x \gt 0} \\ 3 sinx, & \mbox{x \leq 0} \end{array} \right.$$?
a) f(x) has a local minima at x = 0
b) f(x) has a local minima at x > 0
c) f(x) has a local minima at x < 0
d) f(x) has a local minima at x ≥ 0

Explanation: We can see that f(x) is continuous at x = 0 ……(1)
Differentiating (1) with respect to x at -0 and +0, we get
$$\mbox{f}(x)=\left\{ \begin{array} (3x^2+2x+10 & \mbox{x \gt 0} \\ 3 cosxx \leq 0 \end{array} \right.$$
So, f’(0) = $$\lim\limits_{h \rightarrow 0} \frac{f(0 – h) – f(0)}{-h}$$

=> f’ (0) = -10
And, f’ +(0) = $$\lim\limits_{h \rightarrow 0} \frac{f(0 + h) – f(0)}{-h}$$

=> f’ +(0) = 3
Thus f(x) has a local minima at x = 0.

2. At which point will f(x) attain a maxima where, $$\mbox{f}(x)=\left\{ \begin{array} (sin(\frac{\pi x}{2}), & \mbox{x \gt 1} \\ 3 – 2x, & \mbox{x \geq 1} \end{array} \right.$$?
a) x = -1
b) x = 0
c) x = 1
d) x = ½

Explanation: Clearly, f(x) is continuous as x = 1as,
f(1 – 0) = f(1 + 0) = f(1) = 0
Now, f’-(1) = $$\lim\limits_{h \rightarrow 0} \frac{f(1 – h) – f(0)}{-h}$$
= $$\lim\limits_{h \rightarrow 0} \frac{sin(\frac{\pi}{2})(1-h)-1}{-h}$$
So, it is clear that this can be written as,
= $$\lim\limits_{h \rightarrow 0} \frac{cos(\frac{\pi}{2})-1}{-h}$$
= $$\lim\limits_{h \rightarrow 0} \frac{2 sin^2(\frac{\pi h}{4})}{h}$$
= 0
Now, f’ +(1) = $$\lim\limits_{h \rightarrow 0} \frac{f(1 + h) – f(1)}{h}$$
= $$\lim\limits_{h \rightarrow 0} \frac{3-2(1+h)-1}{h}$$ = -2
Now, from here we can say that f’ (1) = 0 whereas, f’ +(1) = -2
Thus, it is clear that there is a local maxima at x = 1.

3. f(x) is a polynomial of degree 4, vanishes at x = -1 and has local minimum/maximum at x = 1, x = 2, and x = 3. If, -22 f(x) dx = -1348/15. Then what is the value of f(x)?
a) x4 – 8x3 + 22x2 – 24x – 55
b) x4 – 8x3 + 22x2 – 24x + 55
c) x4 – 8x3 – 22x2 – 24x – 55
d) Data not sufficient

Explanation: Here it is given that f(x) is a polynomial of degree 4
So, it has critical points at x=1, x=2 and x=3
So, f’(x) = A(x – 1)(x – 2)(x – 3)
= A(x3 – 6x2 + 11x -6)
On integrating this, gives,
f(x) = A/4(x4 – 8x3 + 22x2 – 24x) + B
Also, f(-1) = 0
So, we get,
B = -554A/4
Also, it is given that -22 f(x) dx = -1348/15
So, A/4 -22(x4 – 8x3 + 22x2 – 24x – 55)dx = -1348/15
By changing the range of the integrals, we get,
=> A/2 02(x4 + 22x2 – 55) dx = -1348/15
=> -337A/15 = -1348/15
=> A = 4
Thus, f(x) = x4 – 8x3 + 22x2 – 24x – 55.

4. What will be the maximum area of an isosceles triangle inscribed in the ellipse x2/a2 + y2/b2 = 1 if its vertex at one end of the major axis?
a) 3√3/4 ab sq units
b) 3√3/2 ab sq units
c) √3/2 ab sq units
d) 3/4 ab sq units

Explanation: Let A, B, C be the vertices of the isosceles triangle.
Let B = (a cosθ, b sinθ)
=> C – (a cosθ. -b sinθ)  (by symmetry).
Now, area of a triangle is given by
= ½(2b sinθ)(1 + cosθ)
=> dA/dθ = ab[cosθ(1 + cosθ) – sin2θ]
= ab(1 + cosθ)(2 cosθ – 1)
Now, putting dA/dθ = 0,
We get, θ = π/3, π
Now, for θ = π, triangle ABC is not possible.
And, dA/dθ]θ = π/3 < 0,
Thus, for θ = π/3, the area of isosceles triangle will be maximum.
So, A = 3√3/4 ab sq units.

5. What is the value of (∞)0 and 1?
a) (0, 1)
b) (1, 0)
c) They are indeterminate form
d) (1, 1)

Explanation: If there are forms like ∞-∞, 0*∞, 00, (∞)0 and 1 then they are said to be indeterminate form.
Then to evaluate the limiting value of the function at the point, L’Hospital’s rule is used.

6. If f(x) = |4x – x2 – 3| when x € [0, 4], then, which of the following is correct?
a) x = 1 is the global maximum
b) x = 2 is the global maximum
c) x = 3 is the global maximum
d) x = 0 is the global maximum

Explanation: Clearly, x = 1, 3 are the points of global minimum (as the values being equal).
And, x = 0, 4 are the points of global maximum (as the values being equal).
And, x = 2, is the point of local maximum (as the values being equal).
Thus, x = 3 is the global maximum.

7. At which point does f(x) will attain local minima if f(x) = 0x (t+1)(et – 1)(t – 2)(t + 4) dt?
a) 0
b) -1
c) 1
d) -4

Explanation: Here, f(x) = 0x (t + 1)(et – 1)(t – 2)(t + 4) dt
So, f’(x) = (x + 1)(ex – 1)(x – 2)(x + 4)
So, ────+───-4|───────────-1|──────+──────0|─────────────2|────+────
Clearly, x = -1 and x = 2 are points of local minima.
As, x = 2 is not present so it is -1.

8. What will be the value of $$\lim\limits_{x \rightarrow 1} x^{\frac{1}{1-x}}$$?
a) 1/e
b) e
c) 0
d) 1

Explanation: We have, y = x(1/(1 – x))
So, log y = (log x)/(1 – x)
So, $$\lim\limits_{x \rightarrow 1}$$log y = $$\lim\limits_{x \rightarrow 1}$$(log x)/(1 – x)
Now, using L’Hospital’s rule,
$$\lim\limits_{x \rightarrow 1}$$log y = $$\lim\limits_{x \rightarrow 1}$$(($$\frac{1}{x})$$ / – 1)
=>, $$\lim\limits_{x \rightarrow 1}$$log y = -1
So, $$\lim\limits_{x \rightarrow 1} x^{\frac{1}{1-x}}$$ = 1/e.

9. Is Rolle’s theorem valid for f(x) = x2 – 3x + 4 in the interval [1, 2]?
a) Yes
b) No
c) Depends on x
d) Data not sufficient

Explanation: Obviously, f(x) is continuous at [1, 2]
And, f(x) differentiable at [1, 2]
Also, f(1) = f(2) = 2
Now, f’(x) = 0
=> 2x – 3 = 0
=> x = 3/2
Thus, x belongs to [1, 2]
Hence, it is verified.

10. Let f (x) = (x – a) (x – b) (x – c), a < b < c. Then f'(x) = 0 has two roots. At which interval does these roots belongs?
a) Both the roots in (a, b)
b) At least one root in (a, b) and at least one root in (b, c)
c) Both the roots in (b, c)
d) Neither in (a, b) nor in (b, c)

Explanation: Here, f (x) being a polynomial is continuous and differentiable for all real values of x.
We also have f(a) = f(b) = f(c). If we apply Rolle’s theorem to f (x) in [a, b] and [b, c] we will observe that f'(x) = 0 will have at least one root in (a, b) and at least one root in (b, c).
But f'(x) is a polynomial of degree two, so that f'(x) = 0 can’t have more than two roots. It implies that exactly one root of f'(x) = 0 will lie in (a, b) and exactly one root off'(x) = 0 will lie in (b, c).
Let y = f(x) be a polynomial function of degree n. If f (x) = 0 has real roots only, then f'(x) = 0, f”(x) = 0, … , fn-1(x) = 0 will have real roots. It is in fact the general version of above mentioned application, because if f (x) = 0 have all real roots, then between two consecutive roots of f(x) = 0, exactly one root of f'(x) = 0 will lie.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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