Q: The number of critical points of f (x) = max ( sin x , cos x) for x ∈ (0 , 2 π)
(A) 2
(B) 5
(C) 3
(D) none of these
Click to See Answer :
Hence (C) is correct.
Science and Education
Math Solutions
Q: The number of critical points of f (x) = max ( sin x , cos x) for x ∈ (0 , 2 π)
(A) 2
(B) 5
(C) 3
(D) none of these
Hence (C) is correct.
Q: If 3(a + 2c) = 4(b + 3d) ≠ 0 then the equation ax3 + bx2 + cx + d = 0 will have
(A) no real solution
(B) at least one real root in (–1, 0)
(C) at least one real root in (0, 1)
(D) none of these
Sol: Let $\large f(x) = \frac{a x^4}{4} + \frac{b x^3}{3}+ \frac{c x^2}{2}+ dx $ ; which is continuous and differentiable.
f(0) = 0 ,
$\large f(-1) = \frac{a}{4} – \frac{b}{3} + \frac{c}{2} – d$
$\large = \frac{1}{4}(a+2c)-\frac{1}{3}(b+3d) = 0 $
so, according to Rolle’s theorem, there exist atleast one root of f'(x)= 0 in (–1, 0).
Q: f : [1 , ∞) → R : f(x) is a monotonic and differentiable function and f(1) = 1, then number of solutions of the equation $\large f(f(x)) = \frac{1}{x^2 – 2x + 2} $ is /are
(A) 2
(B) 1
(C) infinite
(D) zero
Ans: (B)
Sol: g (x) = f (f (x))
g'(x) = f'(f (x)) f’ (x) > 0
g (1) = f (f (1)) = f (1) = 1
g (x) ≥ 1
$\large \frac{1}{(x-1)^2 + 1} \le 1 $
Hence (B) is correct
Q: If $\large f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t} , x > 0 , \ne 1$ then
(A) f(x) is an increasing function
(B) f(x) has a minima at x = 1
(C) f(x) is a decreasing function
(D) f(x) has a maxima at x = 1
Ans: (A)
Sol: $\large f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t}$
For increasing or decreasing function,
$\large f'(x) = \frac{1}{ln x^3} . 3x^2 – \frac{1}{ln x^2} . 2x $ (using Leibnitz formula)
$\large f'(x) = \frac{1}{3 ln x} . 3x^2 – \frac{1}{2 ln x} . 2x $
$\large = \frac{1}{ln x}(x^2 – x) $
Since f'(x) > 0 for x > 0 , x ≠ 1 hence f(x) is increasing function.
Hence (A) is correct.
Q : If tanA, tanB, tanC are the solutions of the equation x3 – k2 x2 – px + 2k + 1 = 0, then ΔABC is
(A) an isosceles triangle
(B) an equilateral triangle
(C) a right angled triangle
(D) none of these
Ans: (D)
Sol. tanA + tanB + tanC = k2 and tanA tanB tanC = -2k – 1
In a ΔABC
tanA + tanB + tanC = tanA tanB tanC
⇒ k2 = -2k-1 ⇒ k = -1
⇒ tanA + tanB + tanC = tanA tanB tanC = 1
which is not possible.
Q : If ax2 + bx + 6 = 0 does not have two distinct real roots where a, b ∈ R, then the least value of 3a + b is
(A) 4
(B) -1
(C) 1
(D) -2
Ans: (D)
Sol. D ≤ 0 ⇒ f(x) ≥ 0 or, f(x) ≤ 0 for all real x, but f(0) = 6 > 0
⇒ f(x) ≥ 0 ∀ x ∈ R. In particular f(3) ≥ 0 ⇒ 9a + 3b + 6 ≥ 0
⇒ 3a + b ≥ – 2.