The number of solutions of equations 3x^2 + xsinx + cosx = 0 *
Answers
(a) x+ y = e
(b) e (x + y) = 1
(c) y + ex = 1
(d) None of these
2. The least natural number a for which x + ax–2 > 2 for all x Î R+
(a) 1
(b) 2
(c) 5
(d) 9
3. The number of solutions of equations 3x2 + xsinx + cosx = 0
(a) 3
(b) 2
(c) 1
(d) 0
4. The equation of the tangent to the curve y = 4 + sin2 x at x =0 is
(a) y = 2
(b) y = 3
(c) y = 4
(d) y = 6
5. Let f (x) = MCQs on Applications of Derivatives if f (x) has local minimum at x = 2, then
(a) a £ –2
(b) 1 £ a £ 10
(c) 0 £ a £ 9
(d) None of these
6. If y = f (x2 – 1) and f¢ (x) = x2 + 1, then at x = –1 is
(a) –2
(b) –4
(c) –6
(d) –8
7. A particle moves along a straight line according to the law s = – 3t2 + 9t + 17, where s is in meters and t in seconds. Velocity decreases when
(a) 0 < t < 5
(b) 0 < t < 3
(c) t > 5
(d) t > 3
8. The maximum and minimum value of ab sin x +b. cos x + c lie in the interval where < 1, b > 0
(a) [b –c, b + c]
(b) (b – c , b + c)
(c) [c –b , b + c]
(d) None
9.The function MCQs on Applications of Derivatives has no maximum or minimum if (where k Î I)
(a) b – a = k p
(b) b – a ¹ kp
(c) b – a = 2k p
(d) None of these
10. The angle of intersection of the curves y = x2, 6y = 7 –x3 at (1, 1) is
(a)
(b)
(c)
(d) None of these
11. Let f (x) = x3 –6x2 + 9x + 18, then f (x) is strictly decreasing in
(a) (–¥, 1]
(b) [3, ¥)
(c) (–¥, 1] È [3, ¥)
(d) (1, 3)
12. On the interval the function log sin x is
(a) Increasing
(b) Decreasing
(c) Neither increasing nor decreasing
(d) None of these
13. Let f (x) = 1 + 2x2 + 22x4 + …… + 210x20. Then f (x) has
We can conclude that the equation has exactly one solution in the interval [0, π/2], and hence, it has exactly two solutions in the real line (since we know that there exists at least one solution by the IVT, and there are no other solutions outside the interval [0, π/2])
Unfortunately, there is no general formula for finding the exact number of solutions of a polynomial equation of degree higher than 2.
However, we can use some mathematical techniques to analyze the given equation and determine the number of solutions.
One way to approach this problem is to use the Intermediate Value Theorem (IVT) and the fact that the function is continuous.
First, notice that and = Therefore, by the IVT, there exists at least one solution of the equation f(x) = 0 in the interval (0, π/2).
Next, observe that f(x) is an even function, i.e., f(-x) = f(x) for all x. Therefore, any solution x of the equation f(x) = 0 implies the existence of a solution -x. Hence, we can focus on the interval [0, π/2].
Taking the derivative of f(x), we obtain:
Notice that f'(0) = 0 and f'(x) > 0 for all x in the interval (0, π/2]. Therefore, f(x) is increasing on the interval (0, π/2], and it can have at most one root in that interval.
Combining these observations, we can conclude that the equation has exactly one solution in the interval [0, π/2], and hence,
It has exactly two solutions in the real line (since we know that there exists at least one solution by the IVT, and there are no other solutions outside the interval [0, π/2])
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