begin array -1- 1- - Whine-Column-I- -Column-IINhhlineP-The shortest distance between -1- 6 - -origin - and - the curve - is X - 2- y - 2- xy -60 is- - 11 hlineR-On -0-2- the maximum - value of -3 - 3 - f x - max left x - x -1-3 x-right - is - ff x -left matrix -x-x- - when -x- - is - 0 d d - x - x -1 - when - x - is weven - 8lendmatrix -right- - - - -Then the value of int -2-4- - f x dx -3 - is - -lend array A- P-2- Q-1- R-3- S-4B- P-2- Q-4- R-1- S-4C- P-4- Q-2- R-1- S-1D- P-4- Q-3- R-1- S-2

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Byju's Answer

Standard XII

Mathematics

Sign of Quadratic Expression

begin array |...

Question

$\begin{matrix} C o l u m n - I & C o l u m n - I I (P) & T h e s h o r t e s t d i s t a n c e b e t w e e n & (1) & 6 o r i g i n a n d t h e c u r v e i s x^{2} + y^{2} + x y = 60 i s (Q) & T h e v a l u e o f \int_{- 2011}^{2011} \frac{d x}{1 + x^{9} + \sqrt{1 + x^{18}}} - 2011 i s & (2) & 0 (R) & O n [0, 2] t h e m a x i m u m v a l u e o f & (3) & 3 f (x) = m a x {x^{,} x - 1, 3 x} i s (S) & L e t f : R \to R b e g i v e n b y & (4) & \sqrt{40} f (x) = {\begin{matrix} | x - [x] | & w h e n [x] i s o d d | x - [x] - 1 & w h e n [x] i s e v e n \end{matrix} T h e n t h e v a l u e o f \int_{- 2}^{4} f (x) d x - 3 i s \end{matrix}$

P-2, Q-1, R-3, S-4

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P-2, Q-4, R-1, S-4

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P-4, Q-2, R-1, S-1

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P-4, Q-3, R-1, S-2

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Solution

The correct option is C
P-4, Q-2, R-1, S-1

Conceptual

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Similar questions

Q. Match the following:

\begin{matrix} C o l u m n - I & C o l u m n - I I (P) & M i n i m u m o f \frac{x^{4}}{y^{4}} + \frac{y^{4}}{x^{4}}; & (1) & \frac{1}{4} (x, y \neq 0; x, y > 0) i s (Q) & x ϵ (0, 1), t h e f u n c t i o n x^{25} (1 - x)^{75} & (2) & 2 t a k e s t h e m a x i m u m v a l u e a t x (R) & T h e m i n i m u m v a l u e o f & (3) & 8^{2} \times 16^{4} 2^{(x^{2} - 3)^{3} + 27} i s (S) & I f x + y = 24, (x > 0, y > 0), t h e n & (4) & 1 t h e m a x i m u m v a l u e o f x^{2} y^{4} i s \end{matrix}

Matching type questions:

$\begin{matrix} C o l u m n - I & C o l u m n - I I (P) & N o . o f p o i n t s o f i n t e r s e c t i o n o f c u r v e s | y | = l n | x | a n d (x - 1)^{2} + y^{2} - 4 = 0 & (1) & 1 (Q) & S_{n} = \sum_{r = 1}^{n} (n > 6) t h e n S_{n} - 7 [\frac{S_{n}}{7}] i s (w h e r e [x] d e n o t e s i n t e g e r l e s s t h a n o r e q u a l t o x) & (2) & 2 (R) & E q u a t i o n o f t a n g e n t a t z_{0} t o t h e c i r c l e | z | = r i s R e (\frac{z}{z_{0}}) = λ, t h e n l i s & (3) & 3 (S) & N o r m a l d r a w n a t a n y p o i n t o f a n e l l i p s e \frac{x^{2}}{25} + \frac{y^{2}}{9}, i s t a n g e n t t o t h e & (4) & 5 c i r c l e x^{2} + y^{2} = r^{2} t h e n m a x i m u m v a l u e o f r i s (5) & 5 \end{matrix}$

\begin{matrix} L i s t I & L i s t I I P . & A r e a b o u n d e d b y y = x | x |, x - a x i s, & 1. & \frac{10}{3} x = 1 a n d x = - 1 Q . & A r e a b o u n d e d b y x - y + 2, & 2. & \frac{64}{3} x = \sqrt{y} a n d x = 0 R . & A r e a e n c l o s e d b y y^{2} = x & 3. & \frac{2}{3} a n d y = | x | S . & A r e a b o u n d e d b y y^{2} = x & 4. & \frac{1}{6} y = 4 a n d x = 0 \end{matrix}

Q. Match the following from column 1 and column 2.

\begin{matrix} Column 1 & Column 2 P. Crista and macula & 1. Spinal cord Q. Knee-jerk reflex & 2. Cochlea R. Scala media & 3. Photoreceptor cells S. Rods and cons & 4. Vestibular apparatus \end{matrix}

Given $lim x \to a f (x) = 2$ and $lim x \to a g (x) = 3$

Match th columns

$\begin{matrix} Column 1 & Column 2 1. {lim}_{x \to a} (f (x) \times g (x)) & P.8 2. {lim}_{x \to a} (f (x) - g (x)) & Q .10 3. {lim}_{x \to a} (\frac{f (x)}{g (x)}) \times 15 & R . - 1 4. {lim}_{x \to a} (f (x)^{g (x)}) & S .6 \end{matrix}$

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The correct option is C P-4, Q-2, R-1, S-1 Conceptual

The correct option is C
P-4, Q-2, R-1, S-1

Conceptual