GATE CSE 2026 | Set 2 | Question: 17

Question

For a real number $a$, let $I(a)=\int_{-1}^{1}\left(3 x^{2}-a x+1\right) d x$. Which of the following statements is/are true?

• A. The value of $I(a)$ is independent of the value of $a$
• B. The value of $I(a)$ can vary with the value of $a$
• C. There exists $a \in(-\infty,+\infty)$ such that $I(a)$ is a positive real number
• D. There exists $a \in(-\infty,+\infty)$ such that $I(a)$ is a negative real number

Comments

∫-1 to 1 (3x^2-ax+1)dx.

∫-1 to 1 (3x^2) dx - ∫-1 to1 (ax) dx + ∫-1 to 1 (1) dx 

3 ∫-1 to 1 (x^2) dx - a ∫-1 to1 (x) dx + ∫-1 to 1 (1) dx

integral formulae for above equation :

1) ∫ x^n dx = x^n+1/n+1 + c  , c is constant .

2)  ∫ 1 dx = x + c  , c is constant .

3* [(x^3)/3] -1 to 1 + a*[(x^2)/2] -1 to 1 + [x] -1 to 1.

for  3*[(x^3)/3] =  3*(2)/3 = 2 is the result for the first indifinite integration .  

for a*[(x^2)/2] =   a*(1-1)/2 = a*0 = 0   is the result for the second indifinite integration .

for  [x] = 1-(-1) = 2 is the result for the third indifinite integration .

I(a) = 2+0+2 = 4 . 

I(a) is independent of a . 

a ∈ (-∞ , ∞) doesn,t matter . I(a) is a positive real number.

A and C are  the answers 

 

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It's an MSQ, beware !

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What does "independent of a" mean?

It means:

The value does NOT change when a changes.

For example,

If I(a) = 4

for every value of a, then I(a) is independent of a.

So whether a=1,10,−5,100 the answer remains the same.

Answer 1

Answer : (A,C)

Answer 2

We are given $I(a) = \int_{-1}^{1} (3x^2 - ax + 1) dx$.

Step-by-Step Integration

$$I(a) = \left[ \frac{3x^3}{3} - \frac{ax^2}{2} + x \right]_{-1}^{1}$$

$$I(a) = \left[ x^3 - \frac{ax^2}{2} + x \right]_{-1}^{1}$$

Now, substitute the limits:

• Upper limit (1): $(1)^3 - \frac{a(1)^2}{2} + 1 = 1 - \frac{a}{2} + 1 = 2 - \frac{a}{2}$

• Lower limit (-1): $(-1)^3 - \frac{a(-1)^2}{2} + (-1) = -1 - \frac{a}{2} - 1 = -2 - \frac{a}{2}$

Subtracting them:

$$I(a) = (2 - \frac{a}{2}) - (-2 - \frac{a}{2})$$

$$I(a) = 2 - \frac{a}{2} + 2 + \frac{a}{2}$$

$$I(a) = 4$$

Evaluating Statements

• A. The value of $I(a)$ is independent of the value of $a$: True. The result is always 4.

• B. The value of $I(a)$ can vary with the value of $a$: False. The $a$ terms cancel out.

• C. There exists $a \in (-\infty, +\infty)$ such that $I(a)$ is a positive real number: True. Since $I(a) = 4$ for all $a$, it is always positive.

• D. There exists $a \in (-\infty, +\infty)$ such that $I(a)$ is a negative real number: False. The value is fixed at 4.

Correct Options: A and C.

Answer 3

 

There is a Trick to easily solve these kinds of Problems.

     Integration extends from $(-\text{something})$ to $(+\text{something})$
    
     Now it has $-ax$ :
    
    $f(x) = -ax$
    
    If we replace $x$ with $-x$, then :
    
    $f(-x) = -a(-x) = ax$
    
    Look closely at our result ($ax$). It is the exact opposite (the negative version) of our starting function ($-ax$). Now, $x$ is to the power of 1 (an odd number), this entire term is an odd function. $a$ vanishes from existence. HENCE CANCEL THIS. Final answer is independent of $a$. (Option A is True)

Now the remaining function :

\[
\int_{-1}^{1} (3x^2 + 1)\,dx
\]

Looking at the above function we can directly say : function is strictly hovering above the $x$-axis. This means, Final Area is a positive real number. (Option C is True)

A + C = CORRECT
 

Answer 4

Answer Below: