# Mathematics Questions and Answers – First Order Derivative – 2

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This set of Mathematics Question Papers for Class 11 focuses on “First Order Derivative – 2”.

1. If A (x1, y1) and B (x2, y2) be two points on the curve y = ax2 + bx + c, then as perLagrange’s mean value theorem whichof the following is correct?
a) At least one point C(x3, y3) where the tangent will be intersecting the chord AB
b) At least one point C(x3, y3) where the tangent will be overlapping to the chord AB
c) At least two points where the tangent will be parallel to the chord AB
d) At least one point C(x3, y3) where the tangent will be parallel to the chord AB

Explanation: Here, y = f(x) = ax2 + bx + c
As f(x) is a polynomial function, it is continuous and differentiable for all x.
So, according to geometrical interpretation of mean value theorem there will be at least one point C (x3, y3) between A (x1, y1) and B (x2, y2) where tangent will be parallel chord AB.

2. If $$\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}$$ = -1 then, what is the value of a?
a) 1
b) 2
c) 3
d) 4

Explanation: Let, y = xx
Thus, log y = x log x
Differentiating both sides with respect to x, we get,
1/y dy/dx = (x*1/x) + log x
=>dy/dx = y(1 + log x)
Or, dxx/dx = xx(1 + log x)
Using, L’Hospital’s rule,
$$\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}$$ = $$\lim\limits_{x \rightarrow a}\frac{(a^x*log⁡a-x^{a-1})}{x^x(1+logx)}$$
= $$\lim\limits_{x \rightarrow a}\frac{(a^a*log⁡a-a^{a})}{a^a(1+loga)}$$
= (log a – 1)/(log a + 1)
As per the question,
(log a – 1)/(log a + 1) = -1
Or, (log a – 1) = -log a – 1
Or, 2 log a = 0
Or, log a = 0
So, a = 1

3. If functions f(x) and g(x) are continuous in [a, b] and differentiable in (a, b) then which of the following is correct if there exists at least one point c, a < c < b, such that $$\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}$$?
a) (b + a)$$\begin{vmatrix}f(a) & f”(c) \\g(a) & g”(c) \end {vmatrix}$$
b) (b – a)$$\begin{vmatrix}f(a) & f”(c) \\g(a) & g”(c) \end {vmatrix}$$
c) (b + a)$$\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}$$
d) (b – a)$$\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}$$

Explanation: Let, F(x) = $$\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}$$ = f(a)g(x) – f(x)g(a) …..(1)
=> F’(x) = f’(a)g’(x) – f’(x)g(a)
Since, f(x) and g(x) are continuous in [a, b] and differentiable in (a, b),
So, F(x) is continuous in [a, b] and differentiable in (a, b)
Also from (1), F(a) = f(a)g(a) – f(a)g(a) = 0
And F(b) = f(a)g(b) – f(b)g(a)
Now, by the mean value theorem, there exists at least one point c, a < c < b, such that,
F’(c) = (F’(b) – F’(a)) / (b – a)
=> f(a) g’(c) – g(a) f’(c) = (f(a)g(b) – f(b)g(a) – 0)/b – a
Or, f(a)g(b) – f(b)g(a) = (b – a)( f(a) g’(c) – g(a) f’(c))
=>$$\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}$$ = (b – a)$$\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}$$
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4. What is the number of critical points of f(x) = |x2 – 1| / x2?
a) 0
b) 1
c) 2
d) 3

Explanation:Clearly f (x) is not differentiable at x = 1 and x = -1
And x = 0 is not a critical point not in the domain.
Therefore 1 and -1 are critical points.
Thus, there are 2 critical points.

5. What will be the value of dy/dx if x = asec2θ and y = atan3θ at θ = π/4?
a) 1/2
b) 3/4
c) 3/2
d) 1/4

Explanation: Since, x = asec2θ,
Therefore, dx/dθ = a*d/dx(sec2θ)
= 2asecθ*secθ tanθ
Again, dy/dθ = a*d/dx(tan3θ)
= a * 3 tan2θ * d/dθ(tanθ)
= 3a tan2θ sec2θ
Therefore, dy/dx = (dy/dθ)/(dx/dθ)
= 3a tan2θ sec2θ/2asecθ*secθ tanθ
Thus, at θ = π/4 we have,
dy/dx = 3/2(tan π/4)
= 3/2

6. What is the number of critical points for f(x) = max(sinx, cosx) for x belonging to (0, 2π)?
a) 2
b) 5
c) 3
d) 4

Explanation: We know that in the range of (0, 2π) the graph of sinx and cosx intersects each other in three points.
And we know that these points of intersection are only the critical points
Thus, there are 3 critical points.

7. If y = (3x – 4)/(x+2) then what s the value of dy/dx?
a) dy/dx
b) y
c) 1/ (dy/dx)
d) A constant

Explanation: It is given that y = (3x – 4)/(x + 2) ……….(1)
Now differentiating both the sides, we get that,
dy/dx = (x + 2)*3 – (3x – 4)/(x + 2)2
= 10/(x + 2)2
Again from (1) we get,
xy + 2y = 3x – 4
or, x = – 2(y + 2)
Thus dx/dy = -2* ((y – 3) – (y + 2))/ (y – 3)2
Or, y – 3 = (3x – 4)/(x + 2) – 3
= -10/(x + 2)
Thus, dx/dy = 10/(-10/(x + 2))2
= (x + 2)2/10, where,x ≠ 0 i.e. dx/dy ≠ 0
Therefore, dy/dx*dx/dy = 10/(x + 2)2 * -10/(x + 2)
= 1
=> dy/dx = 1/(dy/dx)

8. What is the value of (dy/dx)2 + 1 if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?
a) Tan2θ
b) Cosec2θ
c) Cot2θ
d) Sec2θ

Explanation: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)
=> x = 2a cos2θ*sin2θ and y = 2a sin2θ*cos2θ
Now differentiating x and y with respect to θ we get,
dx/dθ = 2a[cos2θ*2cos2θ + sin2θ*2cos2θ cos2θ]
= 4a cosθ (cosθ cos2θ – sinθ sin2θ )
= 4a cosθ cos(θ + 2θ)
= 4a cosθ cos3θ
dy/dθ = 2a[cos2θ*2cosθ sinθ + sin2θ (-2sin2θ)]
= 4a sinθ (cosθ cos2θ – sinθ sin2θ )
= 4a sinθ cos(θ + 2θ)
= 4a sinθ cos3θ
Thus, dy/dx = (dy/dθ)/(dx/dθ)
= (= 4a cosθ cos3θ)/( 4a sinθ cos3θ)
= tanθ
So, (dy/dx)2 + 1 = 1 + tan2θ = sec2θ

9. If, y = 1/(1 + x + x2 + x3), then what is the value of y’ at x = 0?
a) 0
b) 1
c) -1
d) ½

Explanation: Given, y = 1/(1 + x + x2 + x3)
Assuming x ≠ 1
Or, y = (x – 1)/(x – 1)( x3 + x2 + x + 1)
Differentiating both the sides with respect to x, we get,
dy/dx = [(x4 – 1)*1 – (x – 1)*4x3]/ (x4 – 1)2
Thus, putting x = 0 in the above equation, we get,
(dy/dx) = -1/(-1)2
= -1

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

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