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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 2 · Q32
(a) Find the value of 'k' for which the quadratic equation ( k +1) x^2-2(3 k +1) x +(8 k +1)=0 has real and equal roots.
(a)
Find the value of 'k' for which the quadratic equation $\displaystyle (\mathrm{k}+1) \mathrm{x}^{2}-2(3 \mathrm{k}+1) \mathrm{x}+(8 \mathrm{k}+1)=0$ has real and equal roots.
OR
(b)
A $\displaystyle 2$-digit number is such that the product of its digits is 18. When $\displaystyle 63$ is subtracted from the number, the digits interchange their places. Find the number.
Marking-scheme solution
For real and equal roots.
\[[-2(3 k+1)]^{2}-4(k+1)(8 k+1)=0
\]
\[\begin{array}{ll}
\Rightarrow & k^{2}-3 \mathrm{k}=0 \\
\therefore & \mathrm{k}=0, \mathrm{k}=3
\end{array}
\]
Let the required no. be \(\displaystyle 10 \mathrm{x}+\mathrm{y}\)
Here \(\displaystyle \mathrm{xy}=18\)---- (i)
\[\begin{aligned}
& (10 x+y)-63=10 y+x \\
& \text { or } x-y=7
\end{aligned}
\]
Solving (i) and (ii) to get
\[x=9 \text { and } y=2
\]
Hence required number is 92.
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.