Mathematics Question Papers for Class 11 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
Answer: d
Clarification: 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 (limlimits_{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
Answer: a
Clarification: 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,
(limlimits_{x rightarrow a}frac{(a^x-x^a)}{x^x-a^a}) = (limlimits_{x rightarrow a}frac{(a^x*loga-x^{a-1})}{x^x(1+logx)})
= (limlimits_{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})
Answer: d
Clarification: 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) = |x2 – 1| / x2?
a) 0
b) 1
c) 2
d) 3
Answer: c
Clarification: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
Answer: c
Clarification: 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