The maximum value $M$ of $3^{x} + 5^{x} - 9^{x} + 15^{x} - 25^{x}$ , as $x$ varies over reals, satisfies

3 < M < 5

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0 < M < 2

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9 < M < 25

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5 < M < 9

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Solution

The correct option is C $0 < M < 2$
${M = 3}^{x} + 5^{x} - 9^{x} - 15^{x} - 25^{x} a = 3^{x}, b = 5^{x} {M = 3}^{x} + 5^{x} - {(3 \times 3)}^{x} - {(3 \times 5)}^{x} - {(5 \times 5)}^{x} M = 3^{x} + 5^{x} - 3^{2 x} - {({3)}^{x} (5)}^{x} - 5^{2 x} M = a + b - a^{2} - a b - b^{2}$
$M = \frac{1}{2} [2 - {(a - 1)}^{2} - {(b - 1)}^{2} - {(a - b)}^{2}]$
All the terms inside the bracket are prefect squares which can be minimum $0$ so that our whole term is maximum so
$M \leq \frac{1}{2} \times 2 M \leq 1$
So according to given options option $B$ is correct

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Similar questions

3^{x} + 5^{x} - 9^{x} + 15^{x} - 25^{x},

x =

5^{x + 1} = 25^{x - 2}

3^{x - 3} \times 2^{3 - x}

2 \frac{1}{m}

m

Complete the last column of the table.

S. No.	Equation	Value	Say, whether the equation is satisfied. (Yes/No)
(i)	x + 3 = 0	x = 3	-
(ii)	x + 3 = 0	x = 0	-
(iii)	x + 3 = 0	x = − 3	-
(iv)	x − 7 = 1	x = 7	-
(v)	x − 7 = 1	x = 8	-
(vi)	5x = 25	x = 0	-
(vii)	5x = 25	x = 5	-
(viii)	5x = 25	x = − 5	-
(ix)		m = − 6	-
(x)		m = 0	-
(xi)		m = 6	-

x^{2} + y^{2} = 25

x^{3} - 2 x^{2} - 9; 5 x^{3} + 2 x + 9

- 7 m^{4} + 5 m^{3} + \sqrt{2}; 5 m^{4} - 3 m^{3} + 2 m^{2} + 3 m - 6

2 y^{2} + 7 y + 5; 3 y + 9; 3 y^{2} - 4 y - 3

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