Match the following. (a) f : [ 0 , 4 ] → R is differentiable and a , b ∈ ( 0 , 4 ) are some constants. Then f ′ ( a ) f ( b ) ( f ( 4 ) ) 2 − ( f ( 0 ) ) 2 = (p) 2 (b) Let F ( x ) = f ( x ) g ( x )

Question

Match the following. (a) f : [ 0 , 4 ] → R is differentiable and a , b ∈ ( 0 , 4 ) are some constants. Then f ′ ( a ) f ( b ) ( f ( 4 ) ) 2 − ( f ( 0 ) ) 2 ​ = (p) 2 (b) Let F ( x ) = f ( x ) g ( x ) h ( x ) for all real x , where f , g , h are differentiable functions. At some point x 0 ​ , F ′ ( x 0 ​ ) = 21 F ( x 0 ​ ) , f ′ ( x 0 ​ ) = 4 f ( x 0 ​ ) , g ′ ( x 0 ​ ) = − 7 g ( x 0 ​ ) and h ′ ( x 0 ​ ) = kh ( x 0 ​ ) . Then 6 k ​ = (q) 4 (c) f ( x ) is a function such that ∣ f ( x 1 ​ ) − f ( x 2 ​ ) ∣ ≤ ∣ x 1 ​ − x 2 ​ ∣ 2 ∀ x 1 ​ , x 2 ​ ∈ R , and the normal at some point on y = f ( x ) passes through ( 1 , 2 ) . The equation of the normal is l x + m y − n = 0 , where l , m , n are co-prime integers and l > 0 . Then l m − mn + n l = (r) 8 (d) The normal at ( a t 1 2 ​ , 2 a t 1 ​ ) on the curve y 2 = 4 a x meets the curve again at ( a t 2 2 ​ , 2 a t 2 ​ ) . Then the least value of t 2 2 ​ is (s) 1

TestHub · Accepted Answer

(a) By LMVT, f ( 4 ) − f ( 0 ) = 4 f ′ ( a ) ; by the intermediate value theorem, 2 f ( 4 ) + f ( 0 ) ​ = f ( b ) for some b . So the expression = f ′ ( a ) f ( b ) 4 f ′ ( a ) ⋅ 2 f ( b ) ​ = 8 ( r ) .

Mathematics - Application of Derivative Question with Solution | TestHub

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Doubts & Discussion

(a) $f : [0, 4] \to R$ is differentiable and $a, b \in (0, 4)$ are some constants. Then $\frac{( f ( 4 ) ) ^{2} - ( f ( 0 ) ) ^{2}}{f ^{'} ( a ) f ( b )} =$	(p) 2
(b) Let $F (x) = f (x) g (x) h (x)$ for all real $x$ , where $f, g, h$ are differentiable functions. At some point $x_{0}$ , $F^{'} (x_{0}) = 21 F (x_{0})$ , $f^{'} (x_{0}) = 4 f (x_{0})$ , $g^{'} (x_{0}) = - 7 g (x_{0})$ and $h^{'} (x_{0}) = kh (x_{0})$ . Then $\frac{k}{6} =$	(q) 4
(c) $f (x)$ is a function such that $∣ f (x_{1}) - f (x_{2}) ∣ \leq ∣ x_{1} - x_{2} ∣^{2} \forall x_{1}, x_{2} \in R$ , and the normal at some point on $y = f (x)$ passes through $(1, 2)$ . The equation of the normal is $l x + m y - n = 0$ , where $l, m, n$ are co-prime integers and $l > 0$ . Then $l m - mn + n l =$	(r) 8
(d) The normal at $(a t_{1}^{2}, 2 a t_{1})$ on the curve $y^{2} = 4 a x$ meets the curve again at $(a t_{2}^{2}, 2 a t_{2})$ . Then the least value of $t_{2}^{2}$ is	(s) 1