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Limits and Derivatives

Multiple Choice Questions

41.

$\lim_{x \to α} {(\tan (x) \cot (x))}^{\frac{1}{x - α}}$ is equal to

$2 \csc (2 α)$
$\frac{1}{2} \sin (2 α)$
$- 2 \csc (2 α)$
None of these

None of these

$We have, \lim_{x \to α} {(\tan (x) \cot (α))}^{\frac{1}{x - α}} = \lim_{x \to α} {\{1 + (\tan (x) \cot (α) - 1)\}}^{\frac{1}{x - α}} = e^{\lim_{x \to α} \frac{(\tan (x) \cot (α) - 1)}{x - α} \times \frac{1}{\tan (α)}} = e^{\lim_{x \to α} \frac{(\tan (x) - \tan (α))}{x - α} \times \frac{1}{\tan (α)}} = e^{\sec^{2} (α) / \tan (α)} = e^{2 \csc (2 α)}$

42.

For each t ∈R, let [t] be the greatest integer less than or equal to t. Then

$\underset{x \to 0^{+}}{l i m} x ([\frac{1}{x}] + [\frac{2}{x}] + . . . . . . + [\frac{15}{x}])$

does not exist (in R)
is equal to 0
is equal to 15
is equal to 120

is equal to 120

$\underset{x \to 0^{+}}{l i m} x ([\frac{1}{x}] + [\frac{2}{x}] + . . . . . . . + [\frac{15}{x}]) \Rightarrow N o w \frac{1}{x} - 1 < [\frac{1}{x}] \leq \frac{1}{x} \frac{2}{x} - 1 < [\frac{2}{x}] \leq \frac{2}{x} \frac{15}{x} - 1 < [\frac{15}{x}] \leq \frac{15}{x} \underset{x \to 0^{+}}{l i m} x (\frac{\sum_{}^{} 15}{x} - 15) < \underset{x \to 0^{+}}{l i m} x ([\frac{1}{x}] + [\frac{2}{x}] + . . . . . + [\frac{15}{x}]) \leq \underset{x \to 0^{+}}{l i m} x (\frac{\sum_{} 15}{x}) \underset{x \to 0^{+}}{l i m} x \sum_{} 15 - 5 x < L \leq \underset{x \to 0^{+}}{l i m} \sum_{} 15 L 120$

43.

stack lim space with straight x space rightwards arrow straight pi over 2 below space fraction numerator cot space straight x space minus cos space straight x over denominator left parenthesis straight pi minus 2 straight x right parenthesis cubed end fraction space equals

1/4
1/24
1/16
1/16

1/16

limit as straight x rightwards arrow straight pi over 2 of space fraction numerator cot space straight x space left parenthesis 1 minus sin space straight x right parenthesis over denominator negative 8 space open parentheses straight x space minus begin display style straight pi over 2 end style close parentheses cubed end fraction space equals space limit as straight x rightwards arrow fraction numerator begin display style straight pi end style over denominator begin display style 2 end style end fraction of space fraction numerator tan begin display style space end style begin display style open parentheses straight pi over 2 minus straight x close parentheses end style over denominator 8 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction space fraction numerator open parentheses 1 minus space cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses close parentheses over denominator open parentheses begin display style straight pi over 2 minus straight x end style close parentheses end fraction space equals space 1 over 8.1.1 half space equals space 1 over 16

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44.

If 5(tan²x – cos²x) = 2cos ²x + 9, then the value of cos⁴x is

-7/9
-3/5
1/3
1/3

-7/9

5 space open square brackets fraction numerator 1 minus straight t over denominator straight t end fraction minus straight t close square brackets space equals space 2 left parenthesis 2 straight t minus 1 right parenthesis space plus space 9 left curly bracket space Let space cos squared space straight x space equals space straight t right curly bracket

⇒5(1 – t – t²) = t(4t + 7)
⇒ 9t² + 12t – 5 = 0
⇒ 9t² + 15t – 3t – 5 = 0
⇒ (3t – 1) (3t + 5) = 0
⇒ t = t/3 as t≠-5/3.
cos2x = 2(1/3)-1 = -1/3
$cos space 4 straight x space equals space 2 space open parentheses negative 1 third close parentheses squared minus 1 space equals space minus 7 over 9$

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