Q33 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
If $\lim _{x \rightarrow 0} \frac{\mathrm{e}^{(\mathrm{a}-1) x}+2 \cos \mathrm{~b} x+(\mathrm{c}-2) \mathrm{e}^{-x}}{x \cos x-\log _{\mathrm{e}}(1+x)}=2$, then $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}$ is equal to :
- A5
- B9
- C3
- Correct answerD7
Solution
$\lim _{x \rightarrow 0} \frac{\left(1+(a-1) x+\frac{(a-1)^2 x^2}{2!}\right)+2\left(1-\frac{b^2 x^2}{2!}\right)+(c-2)\left(1-x+\frac{x^2}{2!}\right)}{x\left(1-\frac{x^2}{2!}\right)-\left(x-\frac{x^2}{2} \ldots\right)}=2$
$\lim _{x \rightarrow 0} \frac{(1+2+c-2)+x(a-1-c+2)+x^2\left(\frac{(a-1)^2}{2}-b^2+\left(\frac{c-2}{2}\right)\right)}{\frac{x^2}{2}-\frac{x^3}{2!}+\ldots}=2$
For which
$\because c+1=0 \Rightarrow c=-1$
$\because a-c=-1 \Rightarrow a=-2$
$\because \frac{(a-1)^2}{2}-b^2+\left(\frac{c-2}{2}\right)=1$
$\frac{9}{2}-b^2-\frac{3}{2}=1 \Rightarrow b^2=2$
$a^2+b^2+c^2=4+2+1=7$
$\lim _{x \rightarrow 0} \frac{(1+2+c-2)+x(a-1-c+2)+x^2\left(\frac{(a-1)^2}{2}-b^2+\left(\frac{c-2}{2}\right)\right)}{\frac{x^2}{2}-\frac{x^3}{2!}+\ldots}=2$
For which
$\because c+1=0 \Rightarrow c=-1$
$\because a-c=-1 \Rightarrow a=-2$
$\because \frac{(a-1)^2}{2}-b^2+\left(\frac{c-2}{2}\right)=1$
$\frac{9}{2}-b^2-\frac{3}{2}=1 \Rightarrow b^2=2$
$a^2+b^2+c^2=4+2+1=7$
Hence, the correct answer is option: 'D'.