Question:
Consider the following two curves in the x-y plane:\[ y = x^3 + x^2 + 5 \]
\[ y = x^2 + x + 5 \]
Which of following statements is true for \( -2 \leq x \leq 2 \) ?
Consider the following two curves in the x-y plane:
\[ y = x^3 + x^2 + 5 \]
\[ y = x^2 + x + 5 \]
Which of following statements is true for \( -2 \leq x \leq 2 \) ?Show Hint
Equate the two expressions for y and factorise what is left. Then check each root against the allowed range of x.
Updated On: Nov 7, 2026
- The two curves intersect once.
- The two curves intersect twice.
- The two curves do not intersect
- The two curves intersect thrice.
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Two curves meet where they have the same y value for the same x. So we set the two expressions for y equal and solve for x. Only the solutions that lie between -2 and 2 count.
Step 2: Key Formula or Approach:
Equate the two right-hand sides, move everything to one side and factorise.
Step 3: Set up and solve.
\[ x^3 + x^2 + 5 = x^2 + x + 5 \] The \( x^2 \) terms and the 5 cancel on both sides, leaving \[ x^3 = x \Rightarrow x^3 - x = 0 \Rightarrow x(x - 1)(x + 1) = 0 \] So \( x = 0, 1 \text{ or } -1 \).
Step 4: Check the interval.
All three values, -1, 0 and 1, lie inside \( -2 \leq x \leq 2 \). The y values at the meeting points are 5, 5 and 7 for x = -1, 0 and 1 respectively, for example at x = 1 both curves give \( 1 + 1 + 5 = 7 \). So the curves meet at three distinct points.
Step 5: Check the options.
Once (option 1), twice (option 2) and never (option 3) all undercount the three solutions of the cubic. The statement 'intersect thrice' is the one that matches, which is option 4.
Final Answer:
The curves intersect at three points, option 4. \[ \boxed{\text{Option 4: intersect thrice}} \]
Two curves meet where they have the same y value for the same x. So we set the two expressions for y equal and solve for x. Only the solutions that lie between -2 and 2 count.
Step 2: Key Formula or Approach:
Equate the two right-hand sides, move everything to one side and factorise.
Step 3: Set up and solve.
\[ x^3 + x^2 + 5 = x^2 + x + 5 \] The \( x^2 \) terms and the 5 cancel on both sides, leaving \[ x^3 = x \Rightarrow x^3 - x = 0 \Rightarrow x(x - 1)(x + 1) = 0 \] So \( x = 0, 1 \text{ or } -1 \).
Step 4: Check the interval.
All three values, -1, 0 and 1, lie inside \( -2 \leq x \leq 2 \). The y values at the meeting points are 5, 5 and 7 for x = -1, 0 and 1 respectively, for example at x = 1 both curves give \( 1 + 1 + 5 = 7 \). So the curves meet at three distinct points.
Step 5: Check the options.
Once (option 1), twice (option 2) and never (option 3) all undercount the three solutions of the cubic. The statement 'intersect thrice' is the one that matches, which is option 4.
Final Answer:
The curves intersect at three points, option 4. \[ \boxed{\text{Option 4: intersect thrice}} \]
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