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

1. If A (x_{1}, y_{1}) and B (x_{2}, y_{2}) be two points on the curve y = ax^{2} + bx + c, then as perLagrange’s mean value theorem whichof the following is correct?

a) At least one point C(x_{3}, y_{3}) where the tangent will be intersecting the chord AB

b) At least one point C(x_{3}, y_{3}) 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(x_{3}, y_{3}) where the tangent will be parallel to the chord AB

View Answer

Explanation: Here, y = f(x) = ax

^{2}+ 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 (x

_{3}, y

_{3}) between A (x

_{1}, y

_{1}) and B (x

_{2}, y

_{2}) 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

View Answer

Explanation: Let, y = x

^{x}

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, dx

^{x}/dx = x

^{x}(1 + log x)

Using, L’Hospital’s rule,

\(\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}\) = \(\lim\limits_{x \rightarrow a}\frac{(a^x*loga-x^{a-1})}{x^x(1+logx)}\)

= \(\lim\limits_{x \rightarrow a}\frac{(a^a*loga-a^{a})}{a^a(1+loga)}\)

= (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)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)

View Answer

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)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)

4. What is the number of critical points of f(x) = |x^{2} – 1| / x^{2}?

a) 0

b) 1

c) 2

d) 3

View Answer

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 = asec^{2}θ and y = atan^{3}θ at θ = π/4?

a) 1/2

b) 3/4

c) 3/2

d) 1/4

View Answer

Explanation: Since, x = asec

^{2}θ,

Therefore, dx/dθ = a*d/dx(sec

^{2}θ)

= 2asecθ*secθ tanθ

Again, dy/dθ = a*d/dx(tan

^{3}θ)

= a * 3 tan

^{2}θ * d/dθ(tanθ)

= 3a tan

^{2}θ sec

^{2}θ

Therefore, dy/dx = (dy/dθ)/(dx/dθ)

= 3a tan

^{2}θ sec

^{2}θ/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

View Answer

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

View Answer

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) Tan^{2}θ

b) Cosec^{2}θ

c) Cot^{2}θ

d) Sec^{2}θ

View Answer

Explanation: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)

=> x = 2a cos

^{2}θ*sin2θ and y = 2a sin

^{2}θ*cos2θ

Now differentiating x and y with respect to θ we get,

dx/dθ = 2a[cos

^{2}θ*2cos2θ + sin2θ*2cos2θ cos2θ]

= 4a cosθ (cosθ cos2θ – sinθ sin2θ )

= 4a cosθ cos(θ + 2θ)

= 4a cosθ cos3θ

dy/dθ = 2a[cos2θ*2cosθ sinθ + sin

^{2}θ (-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 + tan

^{2}θ = sec

^{2}θ

9. If, y = 1/(1 + x + x^{2} + x^{3}), then what is the value of y’ at x = 0?

a) 0

b) 1

c) -1

d) ½

View Answer

Explanation: Given, y = 1/(1 + x + x

^{2}+ x

^{3})

Assuming x ≠ 1

Or, y = (x – 1)/(x – 1)( x

^{3}+ x

^{2}+ x + 1)

Differentiating both the sides with respect to x, we get,

dy/dx = [(x

^{4}– 1)*1 – (x – 1)*4x

^{3}]/ (x

^{4}– 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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