Test: Monotonicity (Competition Level) - 1
22 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Monotonicity (Competition Level) - 1
The interval in which the function x3 increases less rapidly than 6x2 + 15x + 5 is
The function 
is monotonically decreasing in
is monotonically decreasing inf(x)= {−(x−1)/x2 x<1, x(x−1)/x2 x≥1}
f'(x)= {(x−2)/x3 x<1, −(x−2)/x3 x≥1}
x<1, if f′(x)<0 (for f(x) to be monotonically decreasing
⇒ (x−2)/x3<0
⇒x∈(0,2)
But x<1 ⇒ x∈(0,1)
For x≥1, if f′(x)<0
⇒ −(x−2)]/x3<0
⇒(x−2)/x3 > 0
⇒x∈(−∞,0)∪(2,∞)
But, x≥1 ⇒x∈(2,∞)
Hence, x∈(0,1)∪(2,∞)
If y = (a + 2) x3 – 3ax2 + 9ax – 1 decreases monotonically 
x ∈ R then `a' lies in the interval
x ∈ R then `a' lies in the intervalThe true set of real values of x for which the function, f(x) = x ln x – x + 1 is positive is
f(x)=xlnx−x+1
f(1)=0
On differentiating w.r.t x, we get
f′(x)=x*1/x+lnx−1
=lnx
Therefore,f′(x)>0 for allx∈(1,∞)
f′(x)<0 for allx∈(0,1)
lim x→0 f(x)=1
f(x)>0
for allx∈(0,1)∪(1,∞)
The set of all x for which ln (1 + x) ≤ x is equal to
f(x) = ln(1+x) - x ≤ 0
f(x) = ln(1+x) - x
f’(x) = 1/(1+x) - 1
= (1 - 1 - x)/(1+x)
= -x/(1+x)
f’(x) ≤ 0
-x/(1+x) ≤ 0
0 ≤ x /(1+x)
for(x = 2) (-2)/(1-2)
= 2 > 0, therefore x > - 1
The curve y = f(x) which satisfies the condition f'(x) > 0 and f"(x) < 0 for all real x, is
f’(x) > 0
=> f(x) is increasing
f’’(x) < 0 => f(x) is convex
If the point (1, 3) serves as the point of inflection of the curve y = ax3 + bx2 then the value of `a' and `b' are
The function f(x) = x3 – 6x2 + ax + b satisfy the conditions of Rolle's theorem in [1, 3]. The value of a and b are
If f(x) = 
; g(x) = 
for a > 1, a ¹ 1 and x Î R, where {*} & [*] denote the fractional part and integral part functions respectively, then which of the following statements holds good for the function h(x), where (ln a) h(x) = (ln f(x) +ln g(x)).
Let f(x) = (x – 4) (x – 5) (x – 6) (x – 7) then,
Given that f is a real valued differentiable function such that f(x) f'(x) < 0 for all real x, it follows that
If f(x) = 
; g(x) = 
where 0 < x < 1, then
f : R → R be a differentiable function x ∈ R. If tangent drawn to the curve at any point x ∈ (a, b) always lie below the curve, then
A value of C for which the conclusion of Mean Value Theorem holds for the function f(x) = loge x on the interval [1, 3] is
The function f(x) = x(x + 3) e–x/2 satisfies all the conditions of Rolle’s theorem in [–3, 0]. The value of c which verifies Rolle’s theorem, is
f(x) = x (x + 3)e–x/2 is continuous in [–3, 0]
and f′(x)=(x2 + 3x) (– 1/2)e-x/2 + (2x + 3)e–x/2
= – 1/2(x2 – x – 6)e–x/2
Therefore f′(x) exists (i.e., finite) for all x
Also f(–3) = 0, f (0) = 0
So that f(–3) = f(0)
Hence all the three conditions of the theorem are satisfied.
Now consider f′(c)=0
i.e., –1/2(c2 – c – 6)e–c/2 = 0
c2 – c – 6 = 0
(c + 2) (c – 3) = 0 c = 3 or –2
Hence there exists – 2 ∈ (–3, 0) such that
f′(–2) = 0
x3 – 3x2 – 9x + 20 is
f(x) = x2 - x sin x is
The number of values of `c' of Lagrange's mean value theorem for the function,
f(x) = (x – 1) (x – 2) (x – 3), x ∈ (0, 4) is
If f(x) and g(x) are differentiable in [0, 1] such that f(0) = 2, g(0) = 0, f(1) = 6, g(1) = 2, then Rolle's theorem is applicable for which of the following
Equation 3x2 + 4ax + b = 0 has at least one root in (0, 1) if
Function for which LMVT is applicable but Rolle's theorem is not
LMVT is not applicable for which of the following ?
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