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UPSC NDA I 2026 Mathematics Paper

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Question Numbers: 95-96

For the next two (02) items that follow:

Let

f(x)=\left\{\begin{array}{lr}a x(x-1), & x < 1 \\ x-1, & 1 \leqslant x \leqslant 3 \\ p x^{2}+q x+2, & x > 3\end{array}\right.

Given that

f(x)

is continuous for all

x

but not differentiable at

x=1

. Further

f^{\prime}(x)

is continuous at

x=3

.

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Question : 96 of 120

Marks: +1, -0

What is the value of

q

?

-1
$-\frac{1}{3}$
$\frac{1}{3}$
1

Solution:

Concept:

Use continuity of

f(x)

and continuity of

f'(x)

at

x=3

to solve for

q

.

Explanation:

Since

f(x)

is continuous at

x=3

,

f(3)=2=\lim_{x\to 3^+}(px^2+qx+2)

9p+3q+2=2 \Rightarrow 3p+q=0 \Rightarrow q=-3p

Since

f'(x)

is continuous at

x=3

,

Left derivative of

f(x)

at

x=3

is

1

.

Right derivative is

6p+q

.

So,

6p+q=1

Substitute

q=-3p

:

6p-3p=1 \Rightarrow 3p=1 \Rightarrow p=\frac13

Thus,

q=-3\left(\frac13\right)=-1

Answer:

q=-1

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