Let f(x) = (1-x+x^2) + (1+x+x^2)/x - x for x ≠ 0 . Then the value of f(0) so that f is continuous at x = 0 is: - Mathematics MCQ Mathematics Mock Test - 11 2026 | ExamTest.live

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Let f (x) = \frac{l o g ( 1 - x + x ^{2} ) + l o g ( 1 + x + x ^{2} )}{s e c x - c o s x} for x \neq = 0 . Then the value of f (0) so that f is continuous at x = 0 is:

Solution & Step-by-step Explanation

For f (x) to be continuous at x = 0, f (0) = lim_{x \to 0} f (x) .Using log properties: lo g A + lo g B = lo g (A B) . (1 - x + x^{2}) (1 + x + x^{2}) = ((1 + x^{2}) - x) ((1 + x^{2}) + x) = (1 + x^{2})^{2} - x^{2} = 1 + 2 x^{2} + x^{4} - x^{2} = 1 + x^{2} + x^{4} Denominator: sec x - cos x = \frac{1}{c o s x} - cos x = \frac{1 - c o s ^{2} x}{c o s x} = \frac{s i n ^{2} x}{c o s x} . x \to 0 lim f (x) = x \to 0 lim \frac{lo g ( 1 + x ^{2} + x ^{4} )}{\frac{s i n ^{2} x}{c o s x}} = x \to 0 lim \frac{lo g ( 1 + x ^{2} + x ^{4} )}{x ^{2} + x ^{4}} \cdot \frac{x ^{2} + x ^{4}}{x ^{2}} \cdot \frac{x ^{2}}{sin ^{2} x} \cdot cos x Using standard limits: lim_{t \to 0} \frac{l o g ( 1 + t )}{t} = 1 and lim_{x \to 0} \frac{s i n x}{x} = 1 . = 1 \cdot x \to 0 lim (1 + x^{2}) \cdot 1 \cdot 1 = 1 \cdot 1 \cdot 1 = 1 Thus, f (0) = 1 .

Try it yourself before checking the explanation above.

Let f (x) = \frac{l o g ( 1 - x + x ^{2} ) + l o g ( 1 + x + x ^{2} )}{s e c x - c o s x} for x \neq = 0 . Then the value of f (0) so that f is continuous at x = 0 is:

0

2

1

\frac{1}{2}

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