MCQMediumJEE 2024Differentiability

JEE Mathematics 2024 Question with Solution

Let aa and bb be real constants such that the function ff defined by f(x)={x2+3x+a,x1bx+2,x>1f(x)=\begin{cases}x^2+3x+a, & x\leq 1\\ bx+2, & x>1\end{cases} is differentiable on R\mathbb{R}. Then, the value of 22f(x)dx\int_{-2}^{2} f(x) \, dx equals:

  • A

    1515

  • B

    196\frac{19}{6}

  • C

    2121

  • D

    1717

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: f(x)={x2+3x+a,x1bx+2,x>1f(x)=\begin{cases}x^2+3x+a, & x\leq 1\\ bx+2, & x>1\end{cases} is differentiable on R\mathbb{R}.

Find: The value of 22f(x)dx\int_{-2}^{2} f(x) \, dx.

For differentiability at x=1x=1, the function must be continuous and have equal left and right derivatives.

From continuity at x=1x=1:

4+a=b+24+a=b+2

So,

ba=2b-a=2

From differentiability at x=1x=1:

ddx(x2+3x+a)=2x+3\frac{d}{dx}(x^2+3x+a)=2x+3

Hence the left derivative at x=1x=1 is

f(1)=2(1)+3=5f'_-(1)=2(1)+3=5

For x>1x>1,

ddx(bx+2)=b\frac{d}{dx}(bx+2)=b

So the right derivative at x=1x=1 is

f+(1)=bf'_+(1)=b

Therefore,

b=5b=5

Using ba=2b-a=2,

5a=25-a=2 a=3a=3

Thus,

f(x)={x2+3x+3,x15x+2,x>1f(x)=\begin{cases}x^2+3x+3, & x\leq 1\\ 5x+2, & x>1\end{cases}

Now split the integral at x=1x=1:

22f(x)dx=21(x2+3x+3)dx+12(5x+2)dx\int_{-2}^{2} f(x) \, dx = \int_{-2}^{1} (x^2+3x+3) \, dx + \int_{1}^{2} (5x+2) \, dx

For the first part,

21(x2+3x+3)dx=[x33+3x22+3x]21=152\int_{-2}^{1} (x^2+3x+3) \, dx = \left[\frac{x^3}{3}+\frac{3x^2}{2}+3x\right]_{-2}^{1} = \frac{15}{2}

For the second part,

12(5x+2)dx=[5x22+2x]12=192\int_{1}^{2} (5x+2) \, dx = \left[\frac{5x^2}{2}+2x\right]_{1}^{2} = \frac{19}{2}

Adding,

22f(x)dx=152+192=17\int_{-2}^{2} f(x) \, dx = \frac{15}{2}+\frac{19}{2}=17

Therefore, the value of the integral is 1717, so the correct option is D.

Continuity and Differentiability Conditions

Given: The function changes definition at x=1x=1.

Find: Use continuity and differentiability to determine aa and bb, then evaluate the integral.

  1. Continuity at x=1x=1 gives
limx1f(x)=limx1+f(x)\lim_{x\to 1^-} f(x)=\lim_{x\to 1^+} f(x)

That is,

12+3(1)+a=b(1)+21^2+3(1)+a=b(1)+2 4+a=b+24+a=b+2
  1. Differentiability at x=1x=1 gives equality of one-sided derivatives:
f(1)=f+(1)f'_-(1)=f'_+(1)

Since

f(x)=2x+3forx<1f'(x)=2x+3 \quad \text{for} \quad x<1

we get

f(1)=5f'_-(1)=5

And for the second part,

f(x)=bforx>1f'(x)=b \quad \text{for} \quad x>1

so

f+(1)=bf'_+(1)=b

Hence,

b=5b=5

Substituting into the continuity equation,

4+a=74+a=7 a=3a=3
  1. Evaluate the integral:
22f(x)dx=21(x2+3x+3)dx+12(5x+2)dx\int_{-2}^{2} f(x) \, dx = \int_{-2}^{1} (x^2+3x+3) \, dx + \int_{1}^{2} (5x+2) \, dx

The first integral equals 152\frac{15}{2} and the second equals 192\frac{19}{2}. Thus,

152+192=17\frac{15}{2}+\frac{19}{2}=17

Therefore, the correct option is D.

Common mistakes

  • Using only continuity at x=1x=1 and forgetting the differentiability condition. Continuity gives one equation, but differentiability requires equality of derivatives as well. Always apply both conditions for a differentiable piecewise function.

  • Not splitting the integral at x=1x=1. The definition of f(x)f(x) changes there, so integrating with a single expression over [2,2][-2,2] is incorrect. Break the integral into [2,1][-2,1] and [1,2][1,2].

  • Computing the derivative of x2+3x+ax^2+3x+a incorrectly by treating aa as variable-dependent. Since aa is a constant, its derivative is 00. Therefore, the derivative is 2x+32x+3.

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