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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 : 95 of 120

Marks: +1, -0

What is the value of

p

?

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

Solution:

Concept:

Use continuity and continuity of

f'(x)

at the joining points to form equations in

p

and

q

.

Explanation:

Since

f(x)

is continuous at

x=3

, left-hand limit equals right-hand limit:

\lim_{x\to 3^-} f(x)=3-1=2

\lim_{x\to 3^+} f(x)=p(3)^2+q(3)+2=9p+3q+2

So,

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

.

Now

f'(x)

is continuous at

x=3

.

For

1 < x<3

,

f'(x)=1

.

For

x>3

,

f'(x)=2px+q

.

Therefore,

\lim_{x\to 3^-} f'(x)=\lim_{x\to 3^+} f'(x)

1=2p(3)+q=6p+q

Substitute

q=-3p

:

1=6p-3p=3p

p=\frac{1}{3}

Answer:

\frac{1}{3}

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